url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://scicomp.stackexchange.com/questions/11220/why-do-qm-programs-use-redundant-internal-coordinates-for-geometry-optimization | 1,632,017,882,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056656.6/warc/CC-MAIN-20210919005057-20210919035057-00026.warc.gz | 556,142,168 | 38,168 | # Why do QM programs use redundant internal coordinates for geometry optimization?
Brief explanation of QM geometry optimization
Quantum mechanics packages are often tasked with optimizing a chemical structure. The problem is essentially this: Given a set of points in 3D space and potentials that are strongly correlated with chemical structure (bond lengths, bond angles, torsion angles), minimize the total potential with the points as degrees of freedom. Let's just assume you know who's bonded to who already. The potentials relate pairs of points that are bonded and nonbonded but the bonded ones are especially strong.
The question
QM programs often use redundant internal coordinates for geometry optimization. For a single molecule, conformations can be parameterized nonredundantly in internal coordinates in a tree structure. The degrees of freedom are bond lengths, bond angles, and torsion angles. (Some chemical bonds may not have an explicit bond length parameter in this scheme.) Redundant internal coordinates have additional parameters that are not strictly necessary to describe the chemical topology, ex: nonbonded distances.
Redundant internal coordinates have apparently proven more efficient empirically, especially for systems that have bonded cycles.
The optimizations are typically done with BFGS.
Qualitatively, how can this be? Why would adding unnecessary information improve optimization efficiency?
If you imagine the equations of just two bodies that are bound to each other along a fixed connection and float through space without external forces, and if you denote their centers as $x_1(t),x_2(t)$ both of which are 3-dimensional coordinates, then you can write the equations of motion as $$\ddot x_1(t) = \lambda(x_2-x_1), \\ \ddot x_2(t) = -\lambda(x_2-x_1), \\ \|x_1-x_2\|^2 - d^2 = 0.$$ This is known as a Differential-Algebraic Equation (DAE). Note its pleasant simplicity -- the worst part of it is the quadratic constraint. The force on the right hand side is in the direction $x_2-x_1$, i.e., in the direction of the fixed connection. We don't know the Lagrange multiplier $\lambda$ yet, but that comes out of the constraint.
Now imagine how to write the same equations with non-redundant variables. These would be, for example the position $x_1(t)$ and, because the distance between the two bodies is fixed, the two angles by which to describe their relative position. An alternative would be the center of mass $X(t)$ and the relative position of one of the bodies to the center of mass, again in terms of angles (the other body's position then is opposite). In the latter coordinates, we have the equation $$\ddot X(t) = 0$$ for the center of mass. We also need to write down the equations for the two angles, which I will not do here because they have a rather complex form. The take-away message if one did, however, would be that in these non-redundant coordinates, the equations will turn out to be (i) highly nonlinear, (ii) have singularities in the coordinate system such as the north pole when describing relative positions with angles, and (iii) are correspondingly difficult to solve. This is opposed to the rather simple form of the DAE for the redundant coordinates which only have to work with a quadratic constraints and computing the force (the Lagrange multiplier) that results from it. | 720 | 3,355 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-39 | latest | en | 0.882125 |
https://math.stackexchange.com/questions/3324402/connection-on-a-manifold-or-a-principal-bundle-on-the-manifold-from-notion-of/3328529 | 1,576,465,830,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541315293.87/warc/CC-MAIN-20191216013805-20191216041805-00154.warc.gz | 451,673,522 | 32,556 | # Connection on a manifold (or a principal bundle on the manifold) from notion of “parallel transport”
Let $$M$$ be a manifold.
A connection on a manifold $$M$$ (connection on vector bundle $$TM\rightarrow M$$) gives, among other things, an isomorphism $$T_aM\rightarrow T_bM$$ for each path $$\gamma$$ in $$M$$ with $$\gamma(0)=a$$ and $$\gamma(1)=b$$.
For some of physics students I know, a connection is precisely some isomorphism $$T_aM\rightarrow T_bM$$ for each path $$\gamma$$ in $$M$$ with $$\gamma(0)=a$$ and $$\gamma(1)=b$$. I am not able to convince/motivate them that this is not the whole data.
One justification they almost got convinced is that, with this data, I can not "pullback" connections even along surjective submersions.
Can some one suggest some way to motivate/convince about the full data that comes with connection on a manifold?
One can look for same in case of principal $$G$$-bundles.
Let $$\omega:P\rightarrow \Lambda^1_{\mathfrak{g}}T^*P$$ be a connection on the Principal $$G$$-bundle. Given a path $$\gamma$$ in $$M$$, we have an isomorphism $$\pi^{-1}_{\gamma(0)}\rightarrow \pi^{-1}_{\gamma(1)}$$.
Suppose we are given a collection of isomorphisms $$\mathcal{C}=\{\pi^{-1}_{\gamma(0)}\rightarrow \pi^{-1}_{\gamma(1)}\}$$ in $$P$$, indexed by paths $$\gamma$$ in $$M$$. Can we then think of some connection whose "parallel transport" is given by $$\mathcal{C}$$? Some obvious restrictions are, these $$\pi^{-1}_{\gamma(0)}\rightarrow \pi^{-1}_{\gamma(1)}$$ are actually $$G$$-equivariant diffeomorphisms. What other restrictions are reasonable (from view of some one seeing for first time) to impose on this collection $$\mathcal{C}$$ to hope for a connection on $$P(M,G)$$.
• Can a connection not be uniquely recovered from this set of isomorphisms? – user7530 Aug 15 at 18:58
• @user7530 May be I am missing something.. Can you clarify how does one uniquely recover a connection? – user537667 Aug 15 at 19:00
• Suppose $\pi:N\rightarrow M$ be a smooth map which is a surjective submersion... Let $\{T_{\gamma(a)}\rightarrow T_{\gamma(b)}|\gamma:[0,1]\rightarrow M\}$... For a curve $\gamma$ in $N$, how does one associate an isomorphism $T_{\gamma(0)}N\rightarrow T_{\gamma(1)}N$? One can take the image $F(\alpha)$ in $M$, then take the isomorphism $T_{F(\alpha)(0)}M\rightarrow T_{F(\alpha)(1)}M$.. But this would not give any obvious choice of isomorphism $T_{\alpha(0)}N\rightarrow T_{\alpha(1)}N$ as $T_{\alpha(0)}N\rightarrow T_{F(\alpha(0))}M$ is only surjective and not an isomorphism.. – user537667 Aug 15 at 20:21
• Also you recover a connection from its parallel transport via $$D_v X = \left.\frac{d}{dt}\right|_{t=0} P_{-\gamma} (X \circ \gamma (t) )$$ where $X$ is a vector field and $\gamma$ a curve with $\gamma'(0) = v$. I guess it boils down to checking how perverse you can be in choosing parallel transport such that this still gives you a connection. – Carlos Esparza Aug 16 at 22:16
• @Piquito what should I do with it? – user537667 Aug 18 at 4:23
I find the wording of your question sufficiently ambiguous that it could lead to confusion already among mathematicians, without even going so far as to interfacing with physicists!
It is clear to me that any connection is determined by its parallel transport: in other words, for every function which assigns to each (smooth) path a (smoothly varying) isomorphism from the initial to the terminal tangent spaces, there exists at most one connection having this function as its parallel transport. (Indeed, the connection - if it exists - can be determined by differentiating this smoothly varying isomorphism, in some sense.)
It is also clear to me that not every such function is equal to the parallel transport of a connection - at the very least, one would need to start imposing compositional conditions (the isomorphism from $$a$$ to $$b$$ by following $$\gamma$$, composed with the isomorphism from $$b$$ to $$c$$ following $$\gamma'$$, should coincide with the isomorphism from $$a$$ to $$c$$ following the concatenation of $$\gamma$$ and $$\gamma'$$).
When you say the word "precisely", I think you mean what I wrote in the last paragraph: that just having a function from paths to isomorphisms does not necessarily mean that it can be written as the parallel transport of some connection. But then later when you say "this is not the whole data", it is misleading - since (as explained in my second paragraph) there is sufficient data to recover the connection if it exists - and if it does not exist, there is sufficient data to prove that it does not exist.
Moreover, I believe that once things are explained in this way (existence vs uniqueness), it does not matter if the audience is mathematicians or physicists - the idea is clear and intuitive either way. It is essentially the same as saying: a function from a group $$G$$ to a group $$H$$ is enough to determine a group homomorphism, but not every such function is equal to a group homomorphism.
• Thanks for your answer... My question was how to motivate to fill the structure that is not present in an arbitrary collection of isomorphisms $\{\pi^{-1}_{\gamma(0)}\rightarrow \pi^{-1}_{\gamma(1)}\}$ indexed by paths $\gamma$ in $M$ to give a connection on the principal/vector bundle over $M$... – user537667 Aug 20 at 6:07
• I'm not sure what motivation is needed. Obviously if I parallel transport along a curve, it cannot be any different than if I stop the transport partway then allow it to continue. But that is precisely the same thing as the concatenation rule I wrote. And it is obvious that if you just have arbitrary isomorphisms, they won't satisfy this condition - for the same reason if I write down three arbitrary matrices, the product of two of them will not equal the third in general. The group homom ex. - think of $G$ via path concatenation, $H$ via composition of isomorphisms. – pre-kidney Aug 20 at 6:16 | 1,567 | 5,920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 40, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-51 | latest | en | 0.831037 |
http://math.stackexchange.com/questions/102327/examples-of-continuous-growth-rates-greater-than-exponential/102330 | 1,436,057,807,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375097038.44/warc/CC-MAIN-20150627031817-00270-ip-10-179-60-89.ec2.internal.warc.gz | 156,479,197 | 19,814 | # Examples of continuous growth rates greater than exponential
I read on Wikipedia that growth rate of a function can sometimes be greater than exponential. Can you give me some examples of such functions (preferably continuous ones)?
Obviously $x^x$ grows faster than normal exponential, and $x^{x^x}$ even more so - does this concept have a name, and can an arbitrary/infinite amount of such "exponentiality" be expressed with a mathematical expression?
Any other interesting functions to be aware of?
-
For the name of the concept: look for "powertower" in mathworld or wikipedia, and for attempts to the extension to continuous "heights" (the number of x in your question as the variable parameter) look for the term "tetration" in wikipedia (note, that there is not yet an accepted extension for general powertowers to interpolated "heights") – Gottfried Helms Jan 25 '12 at 20:40
The Gamma function defined by $\Gamma(x) = \int_0^\infty e^{-t}t^{x-1}dt$ is a continuous function (and further, an analytic function) for which $\Gamma(n) = (n-1)!$ for all $n \in \mathbb{N}$. In particular, it grows faster than any exponent.
Asymptotically,
$\Gamma(z) \cong \sqrt{z} \cdot (\frac{z}{e})^z$
-
You can compose the exponential function any number of times with itself, say $f_0(x)=x$ and $f_{n+1}(x)=\exp(f_n(x))$ for all $n\in\mathbf N$, to get ever faster growing functions. Every $f_n$ is an analytic function, so everywhere (in $\mathbf C$) indefinitely differentiable. For increasing arguments $x\in\mathbf R$, these functions still grow much slower than even $2\uparrow\uparrow m$ does as a function of $m\in\mathbf N$, because the latter composes $x\mapsto 2^x$ un unbounded number of times (namely $m$ times) with itself (and then applies to $1$), whereas each $f_n$ only has a fixed number $n$ compositions of $\exp$. Lacking a continuous version of function composition (composing a function $x$ times with itself), I'm not sure one can match the growth of $2\uparrow\uparrow m$ with an analytic function.
-
$@$Marc: There is a standard result in complex analysis (in the neighborhood of Weierstrass Factorization and Mittag-Leffler) that given any closed discrete subset $\Omega \subset \mathbb{C}$ and any function $f: \Omega \rightarrow \mathbb{C}$, $f$ can be extended to an entire function. Taking $\Omega = \mathbb{N}$, we see that we can interpolate the up-arrow stuff by an analytic function. – Pete L. Clark Jan 25 '12 at 19:41
There is a standardized system of writing very large numbers, to be found here. The example from the page looks like this:
$$2 \uparrow\uparrow\uparrow 4 = \begin{matrix} \underbrace{2_{}^{2^{{}^{.\,^{.\,^{.\,^2}}}}}}\\ \qquad\quad\ \ \ 65,536\mbox{ copies of }2 \end{matrix} \approx (10\uparrow)^{65,531}(6.0 \times 10^{19,728}) \approx (10\uparrow)^{65,533} 4.3 ,$$ where $(10\uparrow)^n$ denotes a functional power of the function $f(n) = 10n$.
The $\uparrow$ is Knuth's up-arrow notation. With it $x^x$ translates to $x\uparrow\uparrow2$ and $x^{x^x}$ to $x\uparrow\uparrow3$.
-
Aside from the fact that that it seems like you're trolling because I already provided examples of continuous functions that grow faster, you're also wrong to imply descrete operators can't produce continous functions. x+x^2+x^3... may require transcendental operations to be described in closed form as a descrete summation, but it can still be applied to any infinitesimally small number to produce a continuous curve. In fact, every taylor series is a descrete summation of polynomials, yet, they all produce continuous curves that approximate continuous functions. Also you can apply a single summation operation to each value as I mentioned before. You could use the sum of n=1 of k/n^2 to n=infinity, where k has a domain of all real numbers, so you could replace k one by one with all possible valus and get a continuous curve from that summation operator. Such an operation is commonly used as and reffered to by anyone who has taken first semester calculus as an "integral."
- | 1,084 | 4,035 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2015-27 | longest | en | 0.848518 |
http://plottablejs.org/components/plots/rectangle/ | 1,713,045,262,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00362.warc.gz | 26,102,744 | 3,140 | # Rectangle Plot
``````var xScale = new Plottable.Scales.Category();
var yScale = new Plottable.Scales.Category();
var colorScale = new Plottable.Scales.InterpolatedColor();
colorScale.range(["#BDCEF0", "#5279C7"]);
var data = [
{ x: 1, y: 1, val: 2 }, { x: 1, y: 2, val: 1 }, { x: 1, y: 3, val: 3 },
{ x: 2, y: 1, val: 3 }, { x: 2, y: 2, val: 2 }, { x: 2, y: 3, val: 1 },
{ x: 3, y: 1, val: 1 }, { x: 3, y: 2, val: 3 }, { x: 3, y: 3, val: 2 }
]
var plot = new Plottable.Plots.Rectangle()
.x(function(d) { return d.x; }, xScale)
.y(function(d) { return d.y; }, yScale)
.attr("fill", function(d) { return d.val; }, colorScale)
.renderTo("svg#example");
plot.redraw();
});
``````
### Public Methods
• `.x2()`/`.x2(number|accessor)`/`.x2(number|accessor, scale)`
Gets/sets the value of x2 for the current plot. The first parameter can be a number or an `Accessor` function that returns a number. If the parameter `scale` is provided, results will be scaled.
If `x2` is set, the rectangle plot will assume that x values are treated as ranges from `x` to `x2`, as opposed as distinct items as described by `x`.
• `.y2()`/`.y2(number|accessor)`/`.y2(number|accessor, scale)`
Gets/sets the value of y2 for the current plot. The first parameter can be a number or an `Accessor` function that returns a number. If the parameter `scale` is provided, results will be scaled.
If `y2` is set, the rectangle plot will assume that y values are treated as ranges from `y` to `y2`, as opposed as distinct items as described by `y`. | 506 | 1,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.501019 |
https://ipsc.ksp.sk/2014/real/problems/h.html | 1,726,148,182,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00115.warc.gz | 290,081,908 | 3,450 | # Internet Problem Solving Contest
## Problem H – Hashsets
There are not many data structures that are used in practice more frequently than hashsets and hashmaps (also known as associative arrays). They get a lot of praise, and deserve most of it. However, people often overestimate their capabilities. The Internet is full of bad advice such as “just use a hashset, all operations are O(1)” and “don’t worry, it always works in practice”. We hope that you know better. And if you don’t, now you’ll learn.
Let’s start by looking at a sample program in two of the most common modern programming languages: C++11 and Java 7.
``````// C++11
#include <iostream>
#include <unordered_set>
int main() {
long long tmp;
std::unordered_set<long long> hashset;
while (std::cin >> tmp) hashset.insert(tmp);
}
// Java
import java.io.*;
import java.util.*;
public class HashSetDemo {
public static void main(String[] args) throws IOException {
HashSet<Long> hashset = new HashSet<Long>();
for (String x = in.readLine(); x != null ; x = in.readLine())
}
}``````
Both programs do the same thing: they read a sequence of signed 64-bit values from the standard input and they insert them into a hashset.
We compiled and ran both programs on our server. If we used the sequence 1, 2, …, 50 000 as the input, the C++ program needed about 0.05 seconds and the Java program needed about 0.25 seconds. After we increased the input to 1, 2, …, 1 000 000, the C++ program ran in 0.6 seconds and the Java program took 0.8 seconds. That’s fast, right?
Based on this evidence and their limited understanding, many people would claim that the above programs will process any sequence of n integers in O(n) time. Are you one of those people?
### Problem specification
Submit a sequence of 64-bit integers. The length of your sequence can be at most 50 000.
To solve the easy subproblem H1, your sequence must force at least one of our sample programs to run for at least 2 seconds on our server.
To solve the hard subproblem H2, your sequence must force both of our sample programs to run for at least 10 seconds on our server.
(The hard subproblem would certainly be solvable even if the limits were three times as large. We don’t require you to find the worst possible input – any input that’s bad enough will do.)
### Technical details
For the purpose of this problem, there is no “the C++” or “the Java”. As the internals of hashsets are implementation-defined, we have to go with specific compiler versions. We did our best to choose the most common ones.
The officially supported versions are gcc 4.8.1, gcc 4.7.2, gcc 4.6.4, OpenJDK 1.7.0_40, and OpenJDK 1.6.0_24. Solutions that work with any combination of these compilers should be accepted.
Up to our best knowledge, any version of gcc in the 4.6, 4.7, and 4.8 branches should be OK. Any OpenJDK version of Java 6 or 7 should also be OK. However, the only officially supported versions are the ones explicitly given above. For reference, below are sources of two of the officially supported versions.
• gcc 4.8.1
• OpenJDK 7u40
• Should one of the above links fail for some reason, try our local mirror instead.
• You can also find sources of other versions and even browsable repositories on-line if you look hard enough.
### Output specification
The file you submit should contain a whitespace-separated sequence of at most 50,000 integers, each between − 263 and 263 − 1, inclusive. (Don’t worry about the type of whitespace, we will format it properly before feeding it into our Java program.) | 877 | 3,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-38 | latest | en | 0.897211 |
http://blogs.scientificamerican.com/the-curious-wavefunction/2013/01/22/an-eternity-of-infinities-the-power-and-beauty-of-mathematics/?tab=read-posts | 1,429,598,285,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246640550.27/warc/CC-MAIN-20150417045720-00254-ip-10-235-10-82.ec2.internal.warc.gz | 35,714,212 | 30,166 | Musings on chemistry and the history and philosophy of science
The Curious Wavefunction Home
An eternity of infinities: the power and beauty of mathematics
The views expressed are those of the author and are not necessarily those of Scientific American.
The biggest intellectual shock I ever received was in high school. Someone gifted me a copy of the physicist George Gamow’s classic book “One two three…infinity”. Gamow was not only a brilliant scientist but also one of the best science popularizers of the late twentieth century. In his book I encountered the deepest and most utterly fascinating pure intellectual fact I have ever known; the fact that mathematics allows us to compare ‘different infinities’. This idea will forever strike awe and wonder in me and I think is the ultimate tribute to the singularly bizarre and completely counter-intuitive worlds that science and especially mathematics can uncover.
Gamow starts by alerting us to the Hottentot tribe in Africa. Members of this tribe cannot formally count beyond three. How then do they compare commodities such as animals whose numbers are greater than three? By employing one of the most logical and primitive methods of counting- the method of counting by one-to-one correspondences or put more simply, by pairing objects with each other. So if a Hottentot has ten animals and she wishes to compare these with animals from a rival tribe, she will pair off each animal with its counterpart. If animals are left over in her own collection, she wins. If they are left over in her rival’s collection, she has to admit the rival tribe’s superiority in sheep.What is remarkable is that this simplest of counting methods allowed the great German mathematician Georg Cantor to discover one of the most stunning and counter-intuitive facts ever divined by pure thinking. Consider the set of natural numbers 1, 2, 3… Now consider the set of even numbers 2, 4, 6…If asked which set is greater, commonsense would quickly point to the former. After all the set of natural numbers contains both even and odd numbers and this would of course be greater than just the set of even numbers, wouldn’t it? But if modern science and mathematics have revealed one thing about the universe, it’s that the universe often makes commonsense stand on its head. And so it is the case here. Let’s use the Hottentot method. Line up the natural numbers and the even numbers next to each other and pair them up.
1 2 3 4 5…
2 4 6 8 10…
So 1 pairs up with 2, 2 pairs up with 4, 3 pairs up with 6 and so on. It’s now obvious that every natural number n will always pair up with an even number 2n. Thus the set of natural numbers is equal to the set of even numbers, a conclusion that seems to fly in the face of commonsense and shatters its visage. We can extend this conclusion even further. For instance consider the set of squares of natural numbers, a set that would seem even ‘smaller’ than the set of even numbers. By similar pairings we can show that every natural number n can be paired with its square n2, again demonstrating the equality of the two sets. Now you can play around with this method and establish all kinds of equalities, for instance that of whole numbers (all positive and negative numbers) with squares.
But what Cantor did with this technique was much deeper than amusing pairings. The set of natural numbers is infinite. The set of even numbers is also infinite. Yet they can be compared. Cantor showed that two infinities can actually be compared and can be shown to be equal to each other. Before Cantor infinity was just a place card for ‘unlimited’, a vague notion that exceeded man’s imagination to visualize. But Cantor showed that infinity can be mathematically precisely quantified, captured in simple notation and expressed more or less like a finite number. In fact he found a precise mapping technique with which a certain kind of infinity can be defined. By Cantor’s definition, any infinite set of objects which has a one-to-one mapping or correspondence with the natural numbers is called a ‘countably’ infinite set of objects. The correspondence needs to be strictly one-to-one and it needs to be exhaustive, that is, for every object in the first set there must be a corresponding object in the second one. The set of natural numbers is thus a ruler with which to measure the ‘size’ of other infinite sets. This countable infinity was quantified by a measure called the ‘cardinality’ of the set. The cardinality of the set of natural numbers and all others which are equivalent to it through one-to-one mappings is called ‘aleph-naught’, denoted by the symbol ℵ0.
The set of natural numbers and the set of odd and even numbers constitute the ‘smallest’ infinity and they all have a cardinality of ℵ0. Sets which seemed disparately different in size could all now be declared equivalent to each other and pared down to a single classification. This was a towering achievement.
The perplexities of Cantor’s infinities led the great mathematician David Hilbert to propose an amusing situation called ‘Hilbert’s Hotel’. Let’s say you are on a long journey and, weary and hungry, you come to a fine-looking hotel. The hotel looks like any other but there’s a catch: much to your delight, it contains a countably infinite number of rooms. So now when the manager at the front desk says “Sorry, but we are full”, you have a response ready for him. You simply tell him to move the first guest into the second room, the second guest into the third room and so on, with the nth guest moving into the (n+1)th room. Easy! But now what if you are accompanied by your friends? In fact, what if you are so popular that you are accompanied by a countably infinite number of friends? No problem! You simply ask the manager to move the first guest into the second room, the second guest into the fourth room, the third guest into the sixth room…and the nth guest into the 2nth room. Now all the odd-numbered rooms are empty, and since we already know that the set of odd numbers is countably infinite, these rooms will easily accommodate all your countably infinite guests, making you even more popular. Mathematics can bend the laws of the material world like nothing else.
But the previous discussion leaves a nagging question. Since all our infinities are countably infinite, is there something like an ‘uncountably’ infinite set? In fact, what would such an infinity even look like? The ensuing discussion probably constitutes the gem in the crown of infinities and it struck infinite wonder in my heart when I read it.
Let’s consider the set of real numbers, numbers defined with a decimal point as a.bcdefg… The real numbers consist of the rational and the irrational numbers. Is this set countably infinite? By Cantor’s definition, to demonstrate this we would have to prove that there is a one-to-one mapping between the set of real numbers and the set of natural numbers. Is this possible? Well, let’s say we have an endless list of rational numbers, for instance 2.823, 7.298, 4.001 etc. Now pair up each one of these with the natural numbers 1, 2, 3…, in this case simply by counting them. For instance:
S1 = 2.823
S2 = 7.298
S3 = 4.001
S4 = …
Have we proved that the rational numbers are countably infinite? Not really. This is because I can construct a new real number not on the list using the following prescription: construct a new real number such that it differs from the first real number in the first decimal place, the second real number in the second decimal place, the third real number in the third decimal place…and the nth real number in the nth decimal place. So for the example of three numbers above the new number can be:
S0 = 3.942
(9 is different from 8 in S1, 4 is different from 9 in S2 and 2 is different from 1 in S3)
Thus, given an endless list of real numbers counted from 1, 2, 3…onwards, one can always construct a number which is not on the list since it will differ from the 1st number in the first decimal place, 2nd number in the second decimal place…and from the nth number in the nth decimal place.
Cantor called this argument the ‘diagonal argument’ since it really constructs a new real number from a line that’s diagonally drawn across all the relevant numbers after the decimal points in each of the listed numbers. The image from the Wikipedia page makes the picture clearer:
In this picture, the new number is constructed from the red numbers on the diagonal. It’s obvious that the new number Eu will be different from every single number E1…En on the list. The diagonal argument is an astonishingly simple and elegant technique that can be used to prove a deep truth.
With this comparison Cantor achieved something awe-inspiring. He showed that one infinity can be greater than another, and in fact it can be infinitely greater than another. This really drives the nail in the coffin of commonsense, since a ‘comparison of two infinities’ appears absurd to the uninformed mind. But our intuitive ideas about sets break down in the face of infinity. A similar argument can demonstrate that while the rational numbers are countably infinite, the irrational numbers are uncountably so. This leads to another shattering comparison; it tells us that the tiny line segment between 0 and 1 on the number line containing real numbers (denoted by [0, 1]) is ‘larger’ than the entire set of natural numbers. A more spectacular case of David obliterating Goliath I have never seen.
The uncountably infinite set of reals comprises a separate cardinality from the cardinality of countably infinite objects like the naturals which was denoted by ℵ0. Thus one might logically expect the cardinality of the reals to be denoted by ℵ1. But as usual reality thwarts logic. This cardinality is actually denoted by ‘c’ and not by the expected ℵ1. Why this is so is beyond my capability to understand, but it is fascinating. While it can be proven that 2^ℵ0 = c, the hypothesis that c = ℵ1 is actually just a hypothesis, not a proven and obvious fact of mathematics. This hypothesis is called the ‘continuum hypothesis’ and happens to be one of the biggest unsolved problems in pure mathematics. The problem was in fact the first of the 23 famous problems for the new century proposed by David Hilbert in 1900 during the International Mathematical Congress in France (among others on the list were the notorious Riemann hypothesis and the fond belief that the axioms of arithmetic are consistent, later demolished by Kurt Gödel). The brilliant English mathematician G H Hardy put the continuum at the top of his list of things to do before he died (he did not succeed). A corollary of the hypothesis is that there are no sets with cardinality between ℵ0 and c. Unfortunately the continuum hypothesis may be forever beyond our reach. The same Gödel and the Princeton mathematician Paul Cohen damned the hypothesis by proving that, assuming the consistency of the basic foundation of set theory, the continuum hypothesis is undecidable and therefore it cannot be proved nor disproved. This is assuming that there are no contradictions in the basic foundation of set theory, something that itself is ‘widely believed’ but not proven. Of course all this is meat and drink for mathematicians wandering around in the most abstract reaches of thought and it will undoubtedly keep them busy for years.
But it all starts with the Hottentots, Cantor and the most primitive methods of counting and comparison. I happened to chance upon Gamow’s little gem yesterday, and all this came back to me in a rush. The comparison of infinities is simple to understand and is a fantastic device for introducing children to the wonders of mathematics. It drives home the essential weirdness of the mathematical universe and raises penetrating questions not only about the nature of this universe but about the nature of the human mind that can comprehend it. One of the biggest questions concerns the nature of reality itself. Physics has also revealed counter-intuitive truths about the universe like the curvature of space-time, the duality of waves and particles and the spooky phenomenon of entanglement, but these truths undoubtedly have a real existence as observed through exhaustive experimentation. But what do the bizarre truths revealed by mathematics actually mean? Unlike the truths of physics they can’t exactly be touched and seen. Can some of these such as the perceived differences between two kinds of infinities simply be a function of human perception, or do these truths point to an objective reality ‘out there’? If they are only a function of human perception, what is it exactly in the structure of the brain that makes such wondrous creations possible? In the twenty-first century when neuroscience promises to reveal more of the brain than was ever possible, the investigation of mathematical understanding could prove to be profoundly significant.
Blake was probably not thinking about the continuum hypothesis when he wrote the following lines:
To see a world in a grain of sand,
And a heaven in a wild flower,
Hold infinity in the palm of your hand,
And eternity in an hour.
But mathematics would have validated his thoughts. It is through mathematics that we can hold not one but an infinity of infinities in the palm of our hand, for all of eternity.
About the Author: Ashutosh (Ash) Jogalekar is a chemist interested in the history and philosophy of science. He considers science to be a seamless and all-encompassing part of the human experience. Follow on Twitter @curiouswavefn.
The views expressed are those of the author and are not necessarily those of Scientific American.
Previous: Monday morning levity: Louisiana senator asks if E. coli evolve into persons MoreThe Curious Wavefunction Next: Study indicates that scientific fraud may have a male bias
Rights & Permissions
1. 1. Shecky R. 7:58 pm 01/22/2013
nice summary of Cantorian thought… further mind-blowing is the “Cantor Set” or “Cantor’s Dust” which simultaneously ‘seems’ to contain ‘nothing’ or an infinity of elements:
2. 2. rloldershaw 9:42 pm 01/22/2013
“Blake was probably not thinking about the continuum hypothesis when he wrote the following lines:
To see a world in a grain of sand,
And a heaven in a wild flower,
Hold infinity in the palm of your hand,
And eternity in an hour. ”
————————————-
No, probably he was thinking about nature.
Which brings up an interesting application of Canor’s method to the physical world.
If the cosmos is a discrete self-similar (i.e., fractal)hierarchy, an important question would be whether or not the hierarchy was infinite in scale, or whether the hierarchy had cutoffs.
It might seem like one could never scientifically test the finite vs infinite issue, but there is a way to do so in the specific case of exact self-similarity. In this special case, the fractal hierarchy must be infinite and this can be tested using a limited region of the hierarchy. Only for an infinite hierarchy can a specific higher-level system and its specific self-similar lower-scale analogue have exact self-similarity.
The proof involves Cantor’s matching technique applied to the number of subsystem levels contained within the higher-level system and its lower-scale analogue. Only for an infinite hierarchy can analogue systems at different levels have the same number of levels of subsystems.
Robert L. Oldershaw
http://www3.amherst.edu/~rloldershaw
Discrete Scale Relativity/Fractal Cosmology
3. 3. KaiRoesner 6:50 am 01/23/2013
Let’s try out this famous Hilbert’s Hotel and ask the manager to move all his guests by one room. How is he going to do that? He can’t possibly walk down the corridor, knock on each door and ask the residents to move to the next room because that would take an infinite time and I would never be able to move in. Ok, so maybe the manager has previously installed a PA system in his hotel by which he now orders all his guests at once to move to the next room. But wait, the signal that travels from his mike to the speakers in all the rooms will take longer and longer the further down the corridor it goes. It will again take an infinite time until it reaches the guests at the (non-existing) end of the corridor. Hm… last try: The manager knocks at the first door, asks the resident to (please) step out, go to the next room and ask the guest there to do the same. While the first guest does as she was told I move into the first room – mission accomplished! Uh… but now there is still always (i.e. for an infinite time) at least one person who does not have a room. So I can’t really think of a practical way to fit one additional guest into Hilbert’s Hotel without kicking one out. But I guess “practical” is not a requirement here…
4. 4. marclevesque 11:46 am 01/23/2013
If I’m following:
((1,2,3…) + (1,2,3…) + (1,2,3…) …) > (1,2,3…)
and assuming:
(1,2,3…) = (2,4,6…)
then I’m not sure what > means in the first case, the cardinality is greater in what way ?
5. 5. marclevesque 11:46 am 01/23/2013
Correction
((1,2,3…), (1,2,3…), (1,2,3…)…) > (1,2,3…)
6. 6. Allen Hazen 5:31 pm 01/23/2013
Re:
“The problem was in fact the first of the 23 famous problems for the new century proposed by David Hilbert in 1900 during the International Mathematical Congress in France (among others on the list were the notorious Riemann hypothesis and the fond belief that the axioms of arithmetic are consistent, later demolished by Kurt Gödel)”
Slightly misleading way of putting it. Gödel didn’t demolish the belief that the axioms of arithmetic are consistent in his famous 1931 paper (his later work contains two proofs that the axioms of (first order Peano) Arithmetic are consistent): he demolished Hilbert’s hope that this consistency itself could be proved by certain restricted means.
Not to distract people from the main point though. Is Gamow’s book still in print, preferably as a cheap paperback? The leading mathematician of MY high school class had it, shared ideas with me: we both learned from it. It’s a book bright high school students should be able to come across!
7. 7. curiouswavefunction 9:48 pm 01/23/2013
Thanks. Yes, Gamow’s book is still available as a cheap, nice Dover paperback.
8. 8. dilefante 1:06 am 01/24/2013
It’s a pity that Gamow wrote his book long before Archimedes’ palimpsest was re-discovered. Because, somewhere between our Hottentot-like past and Cantor, there is an interesting tidbit from Siracuse. We didn’t know it in the ’90s, but now we know that Archimedes anticipated Cantor in at least two things: First, he dealt with “actual infinities” (instead of the “potential” infinity of unbounded things, usually supposed to be the ceiling over Greek thought and still the only notion of infinity allowed by many of Cantor’s contemporaries). Second, when he happened to extend a proportion by means of equating two infinite sets, he felt the need to first show… that those sets could be put in a one-to-one relationship! Alas, his intuition was way too ahead of his time, and two millennia had to pass before the ideas reappeared.
The story of the re-discovery and the analysis of the palimpsest, around ~2000, is wonderfully told in “The Archimedes Codex”.
9. 9. S. N. Tiwary 3:04 am 01/24/2013
Power and beauty of mathematics is well known to all.
Mathematics is the queen of all subjects.
God is mathematician, hence mathematics is beautiful and powerful.
S. N. Tiwary
Director
10. 10. kbkop 11:31 pm 04/26/2013
While many have attempted to prove that real numbers are countable by putting forth “proofs” that
Georg Cantor’s Diagonal proof is flawed, it also seems that nearly every attempt to do so has
involved complicated explanations that tend to cloud the clarity of the assertion. After much
analysis and contemplation about the problem, I will now make my attempt at showing clearly and
concisely why Cantor’s proof is flawed, and why real numbers are indeed infinitely countable.
11. 11. kbkop 11:51 pm 04/26/2013
Actually, the entire notion that an infinity has any “size” at all or that one is smaller or larger than another is just erroneous. By its very definition, SIZE is a limit, while infinity has no limits. It is like talking about the density of air in a total vacuum…there is no such thing! To say there are “less” even or odd integers than the number of total integers is just as wrong. There is NO number of even integers and NO number of integers. When it comes to infinity, the concept of NO-THING and EVERY-THING is equivalent, and that is precisely what makes infinity…NO count, NO quantity.
12. 12. kbkop 1:54 am 04/27/2013
And of course there is a one-to-one matching between the members of different infinities, and while that may imply they are countable, they are not quantifiable. And for fractions, we are actually talking about a single infinite sequence of finite sets, where each set is made up of a finite number of progressively smaller parts, as follows: 1/2, 1/3, 2/3, 1/4, 2/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/6, 3/6, 4/6, 5/6, etc., so while the size of each piece gets infinitely smaller, the number of pieces increases infinitely. So, in reality, the infinite sequence is a linear set of values representing how many times the whole is divided into equal parts. Then for each of those values, there is a finite number representing how many of those equal parts there are.
You must sign in or register as a ScientificAmerican.com member to submit a comment. | 4,902 | 21,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2015-18 | latest | en | 0.948095 |
https://oeis.org/A320485 | 1,620,398,211,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988793.99/warc/CC-MAIN-20210507120655-20210507150655-00024.warc.gz | 454,692,626 | 5,179 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A320485 Keep just the digits of n that appear exactly once; write -1 if all digits disappear. 12
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, -1, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, -1, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, -1, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, -1, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, -1, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, -1, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, -1, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, -1, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, -1, 1, 0, 102, 103, 104, 105, 106, 107, 108, 109, 0, -1, 2, 3, 4, 5, 6, 7, 8, 9, 120, 2, 1, 123, 124, 125, 126, 127, 128, 129, 130, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Digits that appear more than once in n are erased. Leading zeros are erased unless the result is 0. If all digits are erased, we write -1 for the result. The map n -> a(n) was invented by Eric Angelini and described in a posting to the Sequence Fans Mailing List on Oct 24 2018. More than the usual number of terms are shown in order to reach some interesting examples. a(n) = -1 mostly. - David A. Corneth, Oct 24 2018 REFERENCES Eric Angelini, Posting to Sequence Fans Mailing List, Oct 24 2018 LINKS Robert Israel, Table of n, a(n) for n = 0..10000 N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence) FORMULA From Rémy Sigrist, Oct 24 2018: (Start) a(n) = n iff n belong to A010784. a(n) <= 9876543210 with equality iff n = 9876543210. (End) If n > 9876543210, then a(n) < n. If a(n) < n, then a(n) <= 99n/1000. - Chai Wah Wu, Oct 24 2018 EXAMPLE 1231 becomes 23, 1123 becomes 23, 11231 becomes 23, and 11023 becomes 23 (as we don't accept leading zeros). Note that 112323 disappears immediately and we get -1. 101 and 110 become 0 while 11000 and 10001 become -1. MAPLE f:= proc(n) local F, S; F:= convert(n, base, 10); S:= select(t -> numboccur(t, F)>1, [\$0..9]); if S = {} then return n fi; F:= subs(seq(s=NULL, s=S), F); if F = [] then -1 else add(F[i]*10^(i-1), i=1..nops(F)) fi end proc: map(f, [\$0..200]); # Robert Israel, Oct 24 2018 MATHEMATICA Array[If[# == {}, -1, FromDigits@ #] &@ Map[If[#[[-1]] > 1, -1, #[[1]] ] /. -1 -> Nothing &, Tally@ IntegerDigits[#]] &, 131] (* Michael De Vlieger, Oct 24 2018 *) PROG (PARI) a(n) = {my(d=digits(n), v = vector(10), res = 0, t = 0); for(i=1, #d, v[d[i]+1]++); for(i=1, #d, if(v[d[i]+1]==1, t = 1; res=10 * res + d[i])); res - !t + !n} \\ David A. Corneth, Oct 24 2018 (Python) def A320485(n): return (lambda x: int(x) if x != '' else -1)(''.join(d if str(n).count(d) == 1 else '' for d in str(n))) # Chai Wah Wu, Nov 19 2018 CROSSREFS See A320486 for another version. Cf. A010784, A115853, A321008-A321012. Sequence in context: A118541 A084905 A180409 * A321802 A337864 A106612 Adjacent sequences: A320482 A320483 A320484 * A320486 A320487 A320488 KEYWORD sign,base,look AUTHOR N. J. A. Sloane, Oct 24 2018 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified May 7 10:32 EDT 2021. Contains 343650 sequences. (Running on oeis4.) | 1,402 | 3,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.812567 |
http://hackage.haskell.org/package/Agda-2.3.2.2/docs/Agda-Termination-SparseMatrix.html | 1,474,795,311,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660158.72/warc/CC-MAIN-20160924173740-00133-ip-10-143-35-109.ec2.internal.warc.gz | 126,225,529 | 4,372 | Agda-2.3.2.2: A dependently typed functional programming language and proof assistant
Agda.Termination.SparseMatrix
Description
Sparse matrices.
We assume the matrices to be very sparse, so we just implement them as sorted association lists.
Synopsis
# Basic data types
data Matrix i b Source
Type of matrices, parameterised on the type of values.
Instances
Functor (Matrix i) (Eq i, Eq b) => Eq (Matrix i b) (Ord i, Ord b) => Ord (Matrix i b) (Ord i, Integral i, Enum i, Ix i, Show i, Show b, HasZero b) => Show (Matrix i b) (Arbitrary i, Num i, Integral i, Arbitrary b, HasZero b) => Arbitrary (Matrix i b) (Show i, Ord i, Integral i, Enum i, Ix i, CoArbitrary b, HasZero b) => CoArbitrary (Matrix i b) (Show i, Integral i, Ix i, HasZero b, Pretty b) => Pretty (Matrix i b)
matrixInvariant :: (Num i, Ix i) => Matrix i b -> BoolSource
data Size i Source
Size of a matrix.
Constructors
Size Fieldsrows :: i cols :: i
Instances
Eq i => Eq (Size i) Ord i => Ord (Size i) Show i => Show (Size i) (Arbitrary i, Integral i) => Arbitrary (Size i) CoArbitrary i => CoArbitrary (Size i)
sizeInvariant :: (Ord i, Num i) => Size i -> BoolSource
data MIx i Source
Type of matrix indices (row, column).
Constructors
MIx Fieldsrow :: i col :: i
Instances
Eq i => Eq (MIx i) Ord i => Ord (MIx i) Show i => Show (MIx i) Ix i => Ix (MIx i) (Arbitrary i, Integral i) => Arbitrary (MIx i) CoArbitrary i => CoArbitrary (MIx i)
mIxInvariant :: (Ord i, Num i) => MIx i -> BoolSource
No nonpositive indices are allowed.
# Generating and creating matrices
fromLists :: (Ord i, Num i, Enum i, HasZero b) => Size i -> [[b]] -> Matrix i bSource
`fromLists sz rs` constructs a matrix from a list of lists of values (a list of rows).
Precondition: `length rs == rows sz && all ((== cols sz) . length) rs`.
fromIndexList :: (Ord i, HasZero b) => Size i -> [(MIx i, b)] -> Matrix i bSource
Constructs a matrix from a list of (index, value)-pairs.
toLists :: (Show i, Ord i, Integral i, Enum i, Ix i, HasZero b) => Matrix i b -> [[b]]Source
Converts a matrix to a list of row lists.
matrix :: (Arbitrary i, Integral i, Arbitrary b, HasZero b) => Size i -> Gen (Matrix i b)Source
Generates a matrix of the given size.
Arguments
:: (Arbitrary i, Integral i, Arbitrary b, HasZero b) => Size i -> (i -> Gen [b]) The generator is parameterised on the size of the row. -> Gen (Matrix i b)
Generates a matrix of the given size, using the given generator to generate the rows.
# Combining and querying matrices
size :: Matrix i b -> Size iSource
square :: Ix i => Matrix i b -> BoolSource
`True` iff the matrix is square.
isEmpty :: (Num i, Ix i) => Matrix i b -> BoolSource
Returns `True` iff the matrix is empty.
isSingleton :: (Num i, Ix i) => Matrix i b -> Maybe bSource
Returns 'Just b' iff it is a 1x1 matrix with just one entry `b`.
add :: Ord i => (a -> a -> a) -> Matrix i a -> Matrix i a -> Matrix i aSource
`add (+) m1 m2` adds `m1` and `m2`. Uses `(+)` to add values.
No longer precondition: `size m1 == size m2`.
intersectWith :: Ord i => (a -> a -> a) -> Matrix i a -> Matrix i a -> Matrix i aSource
`intersectWith f m1 m2` build the pointwise conjunction `m1` and `m2`. Uses `f` to combine non-zero values.
No longer precondition: `size m1 == size m2`.
mul :: (Enum i, Num i, Ix i, Eq a) => Semiring a -> Matrix i a -> Matrix i a -> Matrix i aSource
`mul semiring m1 m2` multiplies `m1` and `m2`. Uses the operations of the semiring `semiring` to perform the multiplication.
Precondition: `cols (size m1) == rows (size m2)`.
transpose :: Ord i => Matrix i b -> Matrix i bSource
diagonal :: (Show i, Enum i, Num i, Ix i, HasZero b) => Matrix i b -> Array i bSource
`diagonal m` extracts the diagonal of `m`.
No longer precondition: `square m`.
# Modifying matrices
addRow :: (Num i, HasZero b) => b -> Matrix i b -> Matrix i bSource
`addRow x m` adds a new row to `m`, after the rows already existing in the matrix. All elements in the new row get set to `x`.
addColumn :: (Num i, HasZero b) => b -> Matrix i b -> Matrix i bSource
`addColumn x m` adds a new column to `m`, after the columns already existing in the matrix. All elements in the new column get set to `x`. | 1,257 | 4,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-40 | longest | en | 0.637317 |
https://learnindex.com/counting/ | 1,708,924,423,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474650.85/warc/CC-MAIN-20240226030734-20240226060734-00570.warc.gz | 349,628,308 | 23,048 | # Counting
Counting is the foundation of all mathematical operations. It is an essential skill that we use in our daily lives, from counting money to telling time. Counting is the process of determining the number of objects in a set or group.
## Objectives:
• Understand the concept of counting
• Count up to 100
• Practice counting by twos, fives, and tens
• Use counting in real-life situations
Counting Numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100.
Counting by Twos: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100.
Counting by Fives: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.
Counting by Tens: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.
## Real-Life Examples:
• Counting the number of apples in a basket
• Counting money
• Telling time on a clock
• Counting the number of people in a room
• Counting the number of days in a week
## Practice Exercise:
1. Count from 1 to 50.
2. Count by twos from 2 to 20.
3. Count by fives from 5 to 30.
4. Count by tens from 10 to 50.
5. Count the number of pencils in a box.
## Types of Counting There are different types of counting, including:
1. Counting by Ones: This is the most basic form of counting, where we simply add one to the previous number to get the next number in the sequence. For example, 1, 2, 3, 4, 5, and so on.
2. Counting by Twos: This involves adding two to the previous number to get the next number in the sequence. For example, 2, 4, 6, 8, 10, and so on.
3. Counting by Fives: This involves adding five to the previous number to get the next number in the sequence. For example, 5, 10, 15, 20, 25, and so on.
4. Counting by Tens: This involves adding ten to the previous number to get the next number in the sequence. For example, 10, 20, 30, 40, 50, and so on.
## Counting Strategies There are different strategies that can be used for counting, including:
1. Finger Counting: This involves using fingers to represent numbers. For example, to count up to 5, we can hold up five fingers on one hand.
2. Group Counting: This involves grouping objects into sets and counting the number of sets. For example, if we have 20 pencils and we group them into sets of 5, we can count that we have 4 sets of 5 pencils, which gives us a total of 20 pencils.
3. Skip Counting: This involves counting by a specific number, such as twos, fives, or tens, to get to the desired number quickly.
## Practice Exercise Let’s try a practice exercise to test your counting skills:
Count the number of apples in the following picture:
Answer: There are 6 apples in the picture.
## Conclusion:
Counting is an essential skill in mathematics that is used in many areas of study. By understanding the different types of counting and the strategies for counting, you can improve your ability to solve mathematical problems and tackle everyday tasks that involve counting. | 1,210 | 3,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-10 | latest | en | 0.895706 |
https://forum.usaco.guide/t/usaco-bronze-january-2020-problem-1/540 | 1,660,772,833,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573118.26/warc/CC-MAIN-20220817213446-20220818003446-00034.warc.gz | 274,079,928 | 6,314 | # USACO Bronze January 2020 Problem 1
I am working on this problem:
My solution to this problem
My code does not seem to give the correct output.
pls help
We aren’t just going to debug your code for you. What does your code attempt to do? What’s a case on where it fails? What happens? These are all pieces of information we expect you to provide when you ask a question.
(idk how to repost that asking for help post)
1 Like
I am not asking to debug it…For me the code seems correct but doesnt work…i tried checking how the input is processed but still no clue what is gone wrong. If you could what is wrong I could just check and change the code accordingly
EDIT: I took a quick look at your code, and it seems that you aren’t printing out any newlines (the very essence of this problem). Is this intentional?
1 Like
If you misunderstood me, I am not telling to edit and rewrite my code. I am telling to tell what is wrong with my algorithm. Later I can change the code accordingly.
1 Like
@Saicharan08
This is my code at least:
// Created by Jesse Choe - Bronze Template
#include <bits/stdc++.h>
using namespace std;
// Type aliases
using ll = long long;
using str = string;
using vi = vector<int>;
using pi = pair<int,int>;
using vpi = vector<pi>;
using si = set<int>;
// Vector Operations
#define sz (int)(x).size()
#define pb push_back
#define all(x) begin(x), end(x)
#define sor(x) sort(all(x))
#define rev(x) reverse(all(x))
#define del(x, i) erase(begin(x)+i)
#define rem(x,i) erase(begin(x), begin(x)+i)
// Pairs
#define mp make_pair
#define sec second
#define fir first
// Sets and Maps
#define ins insert
#define ez erase
#define cnt count
// Math
#define MOD 1e9+7
#define MAX_INT 1e18*9
// Permutation
#define possibilities(x) while(next_permutation(all(x))
// Loops
#define Trav(a,x) for (auto& a: x)
#define For(i,a,b) for (int i = (a); i < (b); ++i)
// Input/Output - s is basically the file name without the ".in" and ".out"
void setIO(string s) {
ios_base::sync_with_stdio(0); cin.tie(0);
freopen((s+".in").c_str(),"r",stdin);
freopen((s+".out").c_str(),"w",stdout);
}
int main(){
setIO("word");
int n, k; cin >> n >> k;
int curLength = 0;
int words = 0;
For(i, 0, n){
string s; cin >> s;
curLength+=s.length();
if(curLength <= k){
words++;
if(words != 1){
cout << " ";
}
cout << s;
} else {
cout << endl << s;
words = 1;
curLength = s.length();
}
}
}
Hope this helps!
Whether the issue is with the idea or the implementation, it’s still debugging.
2 Likes
oh ok sorry…i am new to this forum so perhaps I did not tell all the details about how much I have done in this question. I solved the question on my own too | 731 | 2,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-33 | latest | en | 0.771897 |
https://www.numwords.com/words-to-number/en/255000 | 1,606,814,718,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00186.warc.gz | 807,276,673 | 2,556 | NumWords.com
# How to write Two hundred fifty-five thousand in numbers in English?
We can write Two hundred fifty-five thousand equal to 255000 in numbers in English
< Two hundred fifty-four thousand nine hundred ninety-nine :||: Two hundred fifty-five thousand one >
Five hundred ten thousand = 510000 = 255000 × 2
Seven hundred sixty-five thousand = 765000 = 255000 × 3
One million twenty thousand = 1020000 = 255000 × 4
One million two hundred seventy-five thousand = 1275000 = 255000 × 5
One million five hundred thirty thousand = 1530000 = 255000 × 6
One million seven hundred eighty-five thousand = 1785000 = 255000 × 7
Two million forty thousand = 2040000 = 255000 × 8
Two million two hundred ninety-five thousand = 2295000 = 255000 × 9
Two million five hundred fifty thousand = 2550000 = 255000 × 10
Two million eight hundred five thousand = 2805000 = 255000 × 11
Three million sixty thousand = 3060000 = 255000 × 12
Three million three hundred fifteen thousand = 3315000 = 255000 × 13
Three million five hundred seventy thousand = 3570000 = 255000 × 14
Three million eight hundred twenty-five thousand = 3825000 = 255000 × 15
Four million eighty thousand = 4080000 = 255000 × 16
Four million three hundred thirty-five thousand = 4335000 = 255000 × 17
Four million five hundred ninety thousand = 4590000 = 255000 × 18
Four million eight hundred forty-five thousand = 4845000 = 255000 × 19
Five million one hundred thousand = 5100000 = 255000 × 20
Sitemap | 412 | 1,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-50 | latest | en | 0.805299 |
https://coderanch.com/t/269556/java-programmer-OCPJP/certification/explain | 1,475,120,700,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661778.22/warc/CC-MAIN-20160924173741-00007-ip-10-143-35-109.ec2.internal.warc.gz | 880,605,466 | 10,204 | Win a copy of Functional Reactive Programming this week in the Other Languages forum!
# who can explain me this please ??
Ranch Hand
Posts: 189
I was trying by mysel the following :
int i =10;
private byte k = i ;
this won't compile, but when declare i final, it compiles fine
final int i =10;
private byte k = i ;
This will compile
Could you explain me why ??
Deepak Bala
Bartender
Posts: 6663
5
A final int is a compile time constant. What that means is that you cannot change the value of that int at runtime. So the compiler now knows what the value of int is and whether it can fit into the byte. So it compiles.
However if the int were not final its value may not be in the range of a byte. So the compiler cannot know for sure if the assignment is legal. Try these different code snippets
int i = 10;
byte b = i;
final int i = 10;
byte b = i;
final int i = 256;
byte b = i;
Ralph Jaus
Ranch Hand
Posts: 342
A final int is a compile time constant
Note that John is using a different definition of a compile time constant than K&B (see chap. 5 under "Legal expressions for switch and case", page 324): There a compile time constant is final and the value is assigned in the declaration.
So it compiles
This is true only for compile time constants in the sense of K&B. For example
final int i;
i = 10;
byte b = i;
won't compile, but
final int i = 10;
byte b = i;
will.
Ranch Hand
Posts: 189
Thanks my friends, I got it.
Regards
Faber Siagian
Ranch Hand
Posts: 52
Hi John,
I think compile-time constant is a prerequisite for "case" legal argument, and nothing to do with this case.
the value of i is still in the range of byte, and yes, it won't be compiled successfully, because you need an explicit cast on it.
It should be fine.
kaushik vira
Ranch Hand
Posts: 102
public class Test {
final int i =10;
private byte k = i ;
}
This is very simple.. compiler doing optimizations on code. it`s final so it`s value not got to be change. onces you compile this code then latter when you decompile this code. you will get code like following.
public class Test
{
public Test()
{
k = 10;
}
final int i = 10;
private byte k;
}
now there is no relation between int and byte like k=i; like. now it`s totally independent and not affect each other. and 10 is in range of byte so again no need to cast it.
every compiler doing compile time optimization on code.
for future detail go Java performance tunning.
[ July 21, 2008: Message edited by: kaushik vira ]
[ July 21, 2008: Message edited by: kaushik vira ]
[ July 21, 2008: Message edited by: kaushik vira ] | 682 | 2,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-40 | latest | en | 0.869615 |
https://science.slashdot.org/story/13/01/24/0228204/the-mathematics-of-the-lifespan-of-species?sdsrc=nextbtmprev | 1,527,205,110,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866894.26/warc/CC-MAIN-20180524224941-20180525004941-00320.warc.gz | 630,174,739 | 53,160 | typodupeerror
## The Mathematics of the Lifespan of Species158
skade88 writes "NPR is reporting on a study in which the author claims to have found the formula to predict the average life span of members of a species. It does not apply to specific individuals of that species, only to the average life span of members of the species as a whole. From the article: 'It's hard to believe that creatures as different as jellyfish and cheetahs, daisies and bats, are governed by the same mathematical logic, but size seems to predict lifespan. The formula seems to be nature's way to preserve larger creatures who need time to grow and prosper, and it not only operates in all living things, but even in the cells of living things. It tells animals for example, that there's a universal limit to life, that though they come in different sizes, they have roughly a billion and a half heart beats; elephant hearts beat slowly, hummingbird hearts beat fast, but when your count is up, you are over.'"
This discussion has been archived. No new comments can be posted.
## The Mathematics of the Lifespan of Species
• #### That's why I don't exercise (Score:5, Funny)
on Thursday January 24, 2013 @12:23AM (#42677457) Journal
Keep my heart rate to a minimum...
• #### Re: (Score:3)
Ditto.
But in all seriousness, even the summary says that this only applies to a species as a whole. Even if there was a hard quota on how many heartbeats you had, there's no point saving up your heartbeats not exercising just to die early from a heart attack.
• #### Re: (Score:3)
But I am surprised the article does not mention even one of the obvious -- one might even say glaring -- exceptions.
Take various species of tortoise for example. They can easily live to be 150 years old, yet weigh (many of them anyway) far less than a human. Same with many species of parrot.
Then there's the hydra [wikipedia.org] ... 100 million years old or so, in its current form (just a wild guess... it could be a billion but I don't think that's likely), but every hydra is budded from its "parent"... so each indiv
• #### Re:That's why I don't exercise (Score:4, Interesting)
on Thursday January 24, 2013 @07:33AM (#42678927)
Even if there was a hard quota on how many heartbeats you had, there's no point saving up your heartbeats not exercising just to die early from a heart attack.
Actually, I don't exercise for crap, I'm overweight, and my resting heart rate is riduclously high. Sure, exercise would get my heart rate up in the short term, but if I had a stronger, more athletic heart, built through exercising, I would conserve heartbeats over time. Mom was very athletic, and her resting heart rate was something scary slow.
Of course Mom died when she was 55. Oops.
• #### short term pleasure versus long term benefits (Score:2)
I've always exercised for short term well-being feeling and the fun of sport. As long as it doesnt seem to damage you in the long term (there are some open questions about this) or even lengthens you life, then all the better. I feel sorry for those slog through unpleasant exercise thinking they'll live a few years longer.
• #### Re: (Score:2)
You don't need to be sprint running or mountain climbed to keep fit. You just need to walk 30 minutes every day. Walking up and down staircases is even better. Avoid high-fructose corn syrup drinks and processed meat. Then the weight just melts off.
• #### Re:That's why I don't exercise (Score:5, Insightful)
on Thursday January 24, 2013 @01:08AM (#42677709)
Funny but I don' t think the math adds up. Let's look at a 50 year period and assume a constant heart rate and skip pesky leap
So for Joe Average, that's 26,280,000 minutes at 70 bpm or 1,839,600,000 beats.
Now in that time Frankie Fitness works out 5 hrs per week for 50 yrs at a heart rate of 150 beats per minute so 780,000 min or 117,000,000 beats during exercise.
Assume that drops his average heart rate to 60 bpm so over 50 years, the number of heartbeats outside when not exercising would be 60 * 25,500,000 = 1,530,000,000.
So Frankie's total heartbeats over 50 years would be 1,647,000,000 so he saves close to 200 million beats over Joe Average.
• #### apropos Redd Foxx quote (Score:5, Insightful)
on Thursday January 24, 2013 @01:50AM (#42677909) Journal
"Health nuts are going to feel stupid someday, lying in hospitals dying of nothing."
• #### Re:That's why I don't exercise (Score:5, Insightful)
on Thursday January 24, 2013 @02:58AM (#42678117) Homepage
Wrong. By exercising one can lower their heart rate. Around 72 is generally considered the number of beats per minute. From daily exercise I lowered mine to low 50s bpm. What is the difference of the number of beats a year for example between both resting heart rates? Around 11 037 600 beats, looks quite staggering.
• #### Re: (Score:2)
These formulas tend not to work for humans. We have a much longer lifespan than mammals of similar size.
• #### Re: (Score:2)
These formulas tend not to work for humans. We have a much longer lifespan than mammals of similar size.
Humans are an exception because unlike most other species on earth, we use science to prolong lifespans. Modern medical science and other things have basically doubled our expected lifespans over 200 years (from around 40-ish in the 1800s to 80+ today). Even the lifespan in the 1900s generally haven't been all that much better over 1800. Though, kids born in the 21st century have a shorter expected lifesp
• #### Re: (Score:2)
You're resting HR is probably higher if you don't exercies. I would guess somewhere between 70 and 90 BPM depending on your body composition. I'll give you the benefit of the doubt and say that your resting HR is 70BPM. Which is about 100,800 beats per day.
I exercise and because of that, my resting HR is between 45 and 55 BPM. If I exercise for 1 hour per day and my HR during that hour is about 140BPM, my total daily beats are 69,000 (23 hours) + 8,400 (1 hour) = 77,400 beats.
Exercise and you're going to li
• #### exercise drastically cuts heart rate (Score:2)
A very fit adult human will have a pulse rate of 50 or under, compared to the so-called medical average of 70. Say you triple your heart rate for one hour a day during vigorous exercise for the 25% reduction the other 23 hours. That is still a 13% pulse rate reduction over all.
• #### This is not new (Score:5, Informative)
on Thursday January 24, 2013 @12:24AM (#42677465)
Isaac Asimov wrote an essay about this a long time ago (in the 1960's IIRC), and I doubt the idea originated with him.
I believe Asimov was talking about 3 billion heartbeats or so as the limit; 1.5 billion heartbeats is only about 60 years for a human, and we tend to live longer than that under good conditions.
• #### Re:This is not new (Score:4, Interesting)
on Thursday January 24, 2013 @12:41AM (#42677563)
Aye... I remember reading that article. Perhaps in an Analog.. Perhaps in an IASFM.
Amazing that these scientists are now "discovering" this "new" fact.
Wonder how much knowledge we lose and have to rediscover.
• #### Re:This is not new (Score:5, Informative)
by Anonymous Coward on Thursday January 24, 2013 @01:02AM (#42677673)
True, the essay was published in The Magazine of Fantasy and Science Fiction, and the title was "The Slowly Moving Finger".
• #### Re:This is not new (Score:4, Insightful)
on Thursday January 24, 2013 @12:46AM (#42677579)
Or correct.
1.5 billion heartbeats for someone who has a constant heart rate of 72bpm would, according to this theory, only have them living for 39.6 years. So color me skeptical.
And frankly, if my heart rate never deviated from 72bpm, I can't say I'd call that living. I'm still going out for a run tomorrow morning.
• #### Re:This is not new (Score:5, Interesting)
on Thursday January 24, 2013 @01:09AM (#42677719) Journal
the summary says that the result is valid for species, not individuals. even that is wrong; it's not exactly valid for every species; the result is actually that there is a significant power-law trend across species which is that the mortality rate and birth rate both scale approximately as -0.25*(dry mass) on a log-log scale. however there is also significant variation from the log-log line-of-best-fit; the r^2 is around 0.8, though i don't care enough to read exactly how they designed the study. http://www.pnas.org/content/104/40/15777.full [pnas.org]
humans have, of course, cheated death to some extent, so we're outliers, though it is worth noting that prehistoric humans had a max. lifespan of around 40 years...
this is an old result for animal species; the `result' here is that they checked the extrapolated fit for ~700 plant species and validated it in that domain. scientists generally make small extensions or validate previous conjectures; since the public doesn't understand what they're building from, the media has to present the history as the novelty. it's kind of funny, really.
i remember reading a paper (from sante fe institute, of course) ~20 years ago or so which tried to define a `generalized heartbeat' for cities and nation-states to see if the scaling law would extrapolate. of course, the problem is you can define such a thing however you want.
• #### Re:This is not new (Score:4, Insightful)
<calum@callingthetune.co.uk> on Thursday January 24, 2013 @05:09AM (#42678513) Homepage
• #### Re: (Score:2)
humans have, of course, cheated death to some extent, so we're outliers, though it is worth noting that prehistoric humans had a max. lifespan of around 40 years...
No. prehistoric humans had a life expectancy of 25-40 years [wikipedia.org]. Life expectancy is the mean age at death not the maximum lifespan. Given that we are genetically identical to prehistoric man, I think it's fair to say that they're maximum lifespan was somewhere between 100 and 120 years just like us.
• #### Re: (Score:2)
But then you can't really compare animals to humans at all. Medicine and nutrition has gotten so good that we have basically doubled our lifespan over our ancestors. You could probably compare humans to some domesticated animals, but even that's not a good comparison. If a dog get's cancer, most people wouldn't pay for chemo, and would just put the dog down, whereas for certain cancers, the survival rate is getting pretty high. Think about all the other medical procedures that would prolong your life af
• #### Re: (Score:2)
pedantry. life span can refer to life expectancy, and that is what I meant. the point was just that animals don't have hospitals. if you take away the hospitals, humans fall in line with other animals.
• #### Re: (Score:2)
My DNA is different to that of a homosapien 50,000 years ago.
I'm of European decent, so I've got somewhere between 2 and 5% Neanderthals DNA.
• #### Re: (Score:2)
however there is also significant variation from the log-log line-of-best-fit; the r^2 is around 0.8
An R^2 value of 0.8 is actually pretty low. And looking at the graph, it's really only three points, even though it looks like a hundred points. They have one big blob for phytoplankton, one for trees, and a third blob in the middle for everything else. This is really not that impressive. If you throw three baseballs in a microwave and observe the resulting random positions, they will often come pretty close to lying on a single line (which is what the R^2 measures).
Within each blob, the correlation looks l
• #### Re: (Score:2)
yeah that's what I meant by significant variation.
you're right about the effective sample size... I feel really stupid for missing it.
• #### Re: (Score:2)
Take away healthcare, and everything humans have learned to extend their lives, then come back and tell us if you're still skeptical.
• #### Re:This is not new (Score:4, Interesting)
on Thursday January 24, 2013 @12:49AM (#42677601)
The 1960's was "a long time ago"? We have a much more accurate value than Asimov's in Genesis 6:3, applying to all the billions of human lives since, and verifiably correct to the two significant digits of precision indicated.
In terms of specific methodology to arrive at that figure, though, I cannot say beyond the obvious. ;)
• #### 120 years till the flood (Score:4, Informative)
<tepples@gmail.BOHRcom minus physicist> on Thursday January 24, 2013 @11:19AM (#42680279) Homepage Journal
We have a much more accurate value than Asimov's in Genesis 6:3
"After that Jehovah said: 'My spirit shall not act toward man indefinitely in that he is also flesh. Accordingly his days shall amount to a hundred and twenty years.'" (Genesis 6:3, NWT) That verse could be referring to the fact that God was about to flood the inhabited parts of Earth 120 years later [answersingenesis.org] to wipe away the interference of the Nephilim, right after Lamech and Methuselah were about to die. Noah was born when Lamech was 182, Methuselah died when Lamech was 782, and Lamech died at 777. (Genesis 5:25-31) So both Methuselah and Lamech died fairly shortly before Noah turned 600 and the flood came. (Genesis 7:6) The parallel view of Genesis 6:3 [bible.cc] suggests that the authors of some paraphrase translations, such as the New Living Translation, didn't consider this possibility, even despite Abraham's over 170-year life.--Genesis 25:7.
• #### Re: (Score:2)
Understood that there are alternate interpretations. Depending on one's worldview, it does, however, pose something of a argumentative dilemma to invoke the bible as giving factual data points as one's only option in the process of denying it has a factual projection. For all the -verifiable- data points (such is googling "oldest living human") the prediction is spot-on.
If we switch the context of discussion to a specifically theological context of the bible and the premise of accuracy, it is not difficul
• #### Re: (Score:2)
One suspects a word meaning "moon months" got mistranslated as "years" somewhere along the line, since that would bring these elder folks down to a more typical human lifespan.
• #### Re: (Score:2)
This is not new indeed.
And humans are a known exeption to this rule, living about twice as long as this formula would predict: at about 60 heartbeats per minute there are over 2.5 bln in the average life span of about 80 years for humans.
• #### Re: (Score:2)
I think the new thing he has there is that he added plants to the equation. I don't know how he counts heartbeats for plants, but apparently they love us all.
• #### Re: (Score:2)
The precise limit is 2147483647, because some lazy engineer decided that a 32 bit int counter was good enough. I plan on upgrading to the 64 bit heart. That will give me 9*10^18 or so heart beats before overflowing. Plenty enough for me.
• #### So if I want to increase my lifespan (Score:1, Funny)
...I should gain a couple hundred pounds?
• #### Re: (Score:2)
And avoid doing "healthy" things that make your heart beat more and faster, like walking.
• #### Dunbar in Catch-22 (Score:5, Funny)
on Thursday January 24, 2013 @12:29AM (#42677499) Homepage Journal
Time passes faster when you're having fun. If we have a limited number of heartbeats, the trick is to stay as miserable as possible, so that the time will pass more slowly.
• #### Re: (Score:2)
Should probably get high all of the time too, as apparently it makes time seem to pass more slowly. [youtube.com] Then again I don't know how miserable you can be when you're always stoned, so it probably offsets any gains.
• #### Re: (Score:1)
If we have a limited number of heartbeats, the trick is to stay as miserable as possible, so that the time will pass more slowly.
I knew pr0n would be the death of me!
• #### Re: (Score:2)
True - I can imagine nothing more boring that to live on the Arctic ocean floor for 500+ [wikipedia.org] years.
• #### Re: (Score:1)
That's not a viable strategy because your heart beat frequency also goes up if you are angry on something. Indeed, about every emotion increases it. What you therefore would need is an emotionless life.
• #### Re: (Score:2)
Time passes faster when you're having fun. If we have a limited number of heartbeats, the trick is to stay as miserable as possible, so that the time will pass more slowly.
I think the old joke was if the doctor tells you that you have a year to live move to Montana and marry a Jewish girl. It'll seem like forever and you'll be glad be glad when you finally die.
• #### "Formula" = Log-Log Regression (Score:5, Insightful)
on Thursday January 24, 2013 @12:33AM (#42677513)
He plots lifetime in days to entity mass in grams on a log-log plot and slaps a line on it. Note that some of the scatter in the vertical axis is up to 3 *orders of magnitude*. Had this been plotted on linear scale it would have looked like Jackson Pollock sneezed on the page. All that can be extracted is that big critters tend to live longer than small critters. So what is new here?
• #### Re: (Score:1)
by Anonymous Coward
Bingo! I kept thinking about the bristlecone pine & tortoises while reading this. Both huge exceptions to this sort of thinking.
• #### Daisies? (Score:2)
How does one quantify the heartbeat of a daisy?
• #### Re: (Score:2)
Yeah, yeah. I know. RTA. :(
• #### Re: (Score:3)
The period is directly proportional to the loveliness of a summer's day, perchance?
• #### Just proves (Score:2, Funny)
All life was designed by God.
• #### Re: (Score:2)
All life was designed by God.
Yes... it's a miracle, but how the fuck do magnets work?
• #### Re: (Score:1)
Mod parent +1 funny. Classic.
• #### Re: (Score:2)
God holds them together (or pushes them apart, appropriately). Duh. ;-)
• #### This is why I went back to school (Score:2)
I don't accept that we should see this as inevitable. We are learning a lot very rapidly about nanotech and biotech and some of those advances are in the fields of things like regeneration, cures, and life extension.
I fully intend to work on developing this technology and trying to fix this problem. Just because our DNA is built this way doesn't mean that we can change it.
Bioengineering and Nanoengineering are going to be some of the coolest things to do for a long time to come.
• #### Re: (Score:2)
Just because our DNA is built this way doesn't mean that we can change it.
Bioengineering and Nanoengineering are going to be some of the coolest things to do for a long time to come.
True, brother, but not for you... not for you.
It almost can feel some slightly trembling of your hands as you fingers miss some key while typing, a certain lack of attention and all that... signs of age catching up with you; you can no more change that DNA of yours, you simply don't have enough time to do significant discoveries... and posting on /. won't give you more of that precious time.
• #### Re: (Score:2)
Just because GPP has a 5 digit ID number does not mean you have to be an asshole. FOAD.
• #### Re: (Score:2)
You are guaranteed to lose if you don't try.
Our cybernetics and regeneration are improving at insane rates.
I am easily young enough to see these advancements extend my life in order to work on further advancements.
Overall I would prefer to replace body parts with cybernetic parts than organic upgrades.
• #### Re: (Score:2)
But, if humans develop immortality, won't your people come along and blow up the ship we're carrying the secret (and the person who developed it) on? Then you'll just tell us: "You are not ready for immortality." Then I'll just have to say "Whatever. Anyway, we're going to order some pizza, but we all want different toppings. What do you want?" and you'll get all huffy and maybe violent and say "NEVER ASK THAT QUESTION!"
• #### Relax (Score:4, Funny)
on Thursday January 24, 2013 @01:10AM (#42677725)
That thumping sound you hear in your chest?
If that sounds worrying you shouldn't worry, the worrying only just makes your heart beat faster and brings your inevitable demise that much closer.
That worry is very dangerous, even if you stop now you've already shortened your lifespan, and for every second you worry longer you're losing more and more of your life. This worry and stress is literally killing you and it won't stop unless you stop getting stressed out.
• #### Re: (Score:3)
That thumping sound you hear in your chest?
If that sounds worrying you shouldn't worry, the worrying only just makes your heart beat faster and brings your inevitable demise that much closer....
Is your Heart attacking you? Well tell it to Beat It!
Don't Despair, and Don't Delay!
Get your AbioCor Pulse-less blood pump today!
Our patented double helix flow system monitors and regulates your blood pressure smoothly, for all your oxygenation and cooling needs.
Now your blood can course through your veins without the noisy and annoying pounding in your ears!
Other implants have embarrassing charging wires, but you won't have a mess hanging out of your chest with our new wireless Transcutaneous Ene
• #### Re: (Score:2)
http://www.nytimes.com/2010/07/20/health/20docs.html?pagewanted=all&_r=0 [nytimes.com]
and dick cheney who had a HeartMate II device fitted in 2010 and lived without a pulse for 15 months , he now has had a heart transplant and might get another 10 years out of that.
http://en.wikipedia.org/wiki/Dick_Cheney#Health_problems [wikipedia.org]
He had his first heart attack in 1978 at the age of 37 followed by others in 84, 88, 2000 and 2010 he will have his 72nd birthday in about 6 days time.
• #### Re: (Score:2)
dick cheney who had a HeartMate II device fitted in 2010 and lived without a pulse for 15 months
So what you are saying is that Dick Cheney is a zombie. Not that I'm surprised.
• #### Re: (Score:2)
You'll die if you worry but you'll die if you don't - so why worry? enjoy yourself while you can
• #### Exercise (Score:5, Informative)
on Thursday January 24, 2013 @01:43AM (#42677871)
I came up with a similar theory years ago as an excuse not to exercise, for exercising increases one's heart rate. I concluded that exercise would therefore shorten my life. My girlfriend at the time didn't buy my logic. As a step aerobics instructor and science graduate student, she assured me that exercising only temporarily increases one's heart rate and that people who exercise regularly have slower heart rates during the non-exercising parts of their lives. I hate it when people use my own logic against me.
• #### Re: (Score:2)
She is wrong.
It's true that if you train your body to sustain strong effort, the every day life will require a lot less effort and your heart will require less effort.
But aerobics is very bad for the heart, because if you have some hidden health problem, it will expose it.
Yogis also believe that the human body has a limited number of heart-beats, so they try to slow down its beating.
Their theory is that you can become immortal when you stop your heart (this is done by control of the breathing).
I also heard
• #### Re: (Score:2)
Yogis also believe that the human body has a limited number of heart-beats
I don't think Yogis _also_ believe that. I think Yogis believe that and the idea has gotten lodged into our cultural meme-trap and non-scientific "studies" like this one pop up every few years to try to "prove" it and get splashed around as pop-science. It's annoying because it's not really substantial enough to disprove. Our tissues wear out, the heart is made of tissues and it wears out. There's an approximate amount of time this takes, as well as an approximate amount of time we live and an approximate n
• #### Re: (Score:2)
Yeah my resting heart rate is 45 beats per minute thanks to my love of running and some very good genes! At the computer right now it's reporting 50 on my iPhone, so it more likely hovers around that during the day.
• #### Re:Exercise (Score:4, Funny)
on Thursday January 24, 2013 @06:02AM (#42678659)
There is that Reality Distortion Field again.
On my Android phone, it says your heart rate is 56 beats a minute.
• #### Re: (Score:2)
haha, I've had it professional measured and can even count it myself. Also it's called exercise, you only have yourself to blame!
• #### Wrong wrong wrong (Score:5, Interesting)
on Thursday January 24, 2013 @01:43AM (#42677875)
There are so many exceptions to this "rule" that it is at best an interesting pattern. There are big turtles that live great long lives and big turtles that don't with little turtles being all over the place as well. There are birds of all kinds of sizes with small birds that live 80 years and big birds that live under 20. Some bacteria seem to be nearly immortal and others live days. Within dogs the big ones hardly outlast green bananas while the little ratty ones go on for decades. Poplar trees grow huge and die fast, oaks go on and on but some smaller trees are thousands of years old.
Even humming birds live a few years at crazy heartbeats as high as 1200 bpm (look it up if you don't buy that mind blowing number) yet other bigger birds with much slower heart beats live for the same length of time. So it isn't size or heartbeats.
If I had to suspect anything lifespan will be an evolutionary advantage like anything else. If you are surrounded by ever changing dangers a short fast life-cycle is probably best. But if you are fairly safe in steady environment a long life is probably safer. Turtles have slow metabolisms which allow them to survive long periods without food and are fairly safe from predictors so they don't have to worry about adapting too much. Rabbits are basically the forest's McNuggets so they need to continuously adapt in numbers and probably other things such as coloring; hence a fast short life cycle. We have created civilization where we are nearly 100% safe from predators and with things like food storage are not so buffeted by a changing nature; so we are getting longer an longer lived.
• #### Re: (Score:2)
Koi fish are another counter-argument, even if they never freeze into suspended animation.
• #### Stats for average member of a species (Score:2)
So, for your particular species, your best bet would be to select your parents for longevity. If you are into that sort of thing.
• #### Regression Formula (Score:1)
So pedometers should count backwards....
• #### 2038 (Score:5, Funny)
on Thursday January 24, 2013 @02:44AM (#42678073) Journal
"roughly a billion and a half heart beats...when your count is up, you are over.'"
But I have a Unix heart; the counter flips over to zero in 2038.
• #### Lets be accurate, it is 19 days later (Score:2)
The furthest time that can be represented this way is 03:14:07 UTC on Tuesday, 19 January 2038.
• #### Re: (Score:2)
the counter flips over to zero in 2038
I do believe that 03:14:07 UTC on Tuesday, 19 January 2038 will indeed be in 2038...
• #### Re: (Score:2)
It's official: 2038 is the year of the singularity.
• #### Outliers (Score:2)
Parrots
Galapagos Tortoises
Dogs (small ones tend to outlive larger ones by a factor sometimes approaching 2)
(and those are just the 1st 3 examples which spring to mind in 30 s)
• #### The real news. (Score:2)
"creatures as different as jellyfish and cheetahs, daisies and bats"
Daisies have heartbeats!
Bob.
• #### Heartless? (Score:2)
What about all these politicians? Are we stuck with them forever?
Before anyone goes off on left vs. right being more heartless...it's a joke, get over yourself.
• #### Interesting but I suspect its a symptom (Score:2)
It is an interesting concept however it does appear to have some easily named exceptions as others have pointed out. It also though has a lot of overlap with reproduction selectivity (r/K theory). Namely animals with shorter lifespans breed quicker and have more offspring than animals that live longer and have longer gestation periods.
Think about it... you have an elephant which takes nearly 2 years to come to term, it would make no sense to have a 10 year life cycle. Instead, they can live 60-70 years. Thi
• #### Okay (Score:2)
Explain to me the similar lifespans of gray parrots and elephants?
What about snakes, and other reptiles that continue growing with age. Do they die younger when young, and live longer when old?
• #### cause-effect has been assumed. (Score:2)
So I RTFA, which is mostly fluffy but has a graph and some links to the real research....BUT at the center of it is an assumption that goes unquestioned:
So Geoffrey West and his colleagues found that nature gives larger creatures a gift: more efficient cells. Literally.
Why not ask if the first step is when an organism hits upon a mutation to improve its efficiency and the consequence is larger/longer lasting individuals?
• #### reverse causation (Score:2)
We dont live longer because we are larger, but are larger because we live longer. Both could be tied to a third factor such as inherent genetic metabolic rate.
• #### Away with euphemism! (Score:2)
who need time to grow and breed, successfully
FTFY
#### Related LinksTop of the: day, week, month.
Marvelous! The super-user's going to boot me! What a finely tuned response to the situation!
Working... | 7,098 | 29,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-22 | latest | en | 0.954858 |
https://www.magischvierkant.com/two-dimensional-eng/6x6/medjig-method/ | 1,714,006,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00143.warc.gz | 779,640,423 | 32,578 | ### Medjig method
The first grid is a 2x2 'blown up' pure 3x3 magic square. Construct the second grid using 9 Medjig tiles.
In a (2x2) medjig tile are all the numbers from 0 up to 3, but each time in a different order. Take care that the sum of the numbers in each row/column/diagonal is (6 x 1,5 =) 9.
Take 1x number from first grid and add 9x number from the same cell of the second grid.
1x number + 9x number = pure magic 6x6 square
2 2 9 9 4 4 2 3 0 2 0 2 20 29 9 27 4 22 2 2 9 9 4 4 1 0 3 1 3 1 11 2 36 18 31 13 7 7 5 5 3 3 3 1 1 2 2 0 34 16 14 23 21 3 7 7 5 5 3 3 0 2 0 3 3 1 7 25 5 32 30 12 6 6 1 1 8 8 3 2 2 0 0 2 33 24 19 1 8 26 6 6 1 1 8 8 0 1 3 1 1 3 6 15 28 10 17 35
Use this method to construct even magic squares.
See 6x68x810x1012x1214x1416x1618x1820x2022x2224x2426x2628x2830x30 en 32x32
6x6, Medjig method.xls | 436 | 865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-18 | longest | en | 0.520923 |
https://www.coursehero.com/file/198604/ACOUSTICS-11newest1-1/ | 1,493,416,320,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123097.48/warc/CC-MAIN-20170423031203-00351-ip-10-145-167-34.ec2.internal.warc.gz | 877,466,955 | 95,951 | ACOUSTICS-11newest1-1
# ACOUSTICS-11newest1-1 - ACOUSTICS Sound requires a source...
This preview shows pages 1–5. Sign up to view the full content.
CSD 101 1 ACOUSTICS Sound requires a source and a medium To vibrate, a source must have two properties A) mass (m) B) elasticity (E) To transmit sound, a medium must be capable of being set into vibration To vibrate the medium must have the same two properties A) mass (m) B) elasticity (E)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
CSD 101 2 ACOUSTICS Mass (m) – 1)The amount of mass present applies 2) Mass contrasted with weight – weight is an attractive gravitational pull; where mass is the quantity of matter present (for example 160 pounds on earth = 27 pounds on the moon because the force of gravity is 6:1 Weight is similar to mass BUT they are different concepts Weight is a force Mass is the quantity of matter present Air has mass and weight
CSD 101 3 ACOUSTICS Elasticity (E) 1) property that enables RECOVERY from distortion of shape and volume 2) The ability to resist changes in shape or volume 3) Concept of “Elastic Limit” if applied force exceeds elastic limit- the distortion is permanent 4) With air, elasticity refers to the tendency of air volume to return to its former volume AFTER compression.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
CSD 101 4 ACOUSTICS 1) VIBRATORY MOTION OF TUNING FORK Strike the fork – vibrations occur The tuning fork tines are moved (displaced) from equilibrium Amplitude of displacement is proportional to the force applied 2) DISPLACEMENT FROM EQUILIBRIUM Newton’s First Law of Motion: INERTIA All bodies remain at rest or in a state of uniform motion unless another force acts in opposition Inertia is the tendency of a body at rest to stay at rest and the of a body in motion to stay in motion Magnitude of inertia is directly proportional to the mass More mass = more inertia Mass is a measure of inertia
This is the end of the preview. Sign up to access the rest of the document.
## This note was uploaded on 06/09/2008 for the course CSD 101 taught by Professor Blood,ingridmari during the Summer '07 term at Penn State.
### Page1 / 20
ACOUSTICS-11newest1-1 - ACOUSTICS Sound requires a source...
This preview shows document pages 1 - 5. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 615 | 2,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-17 | longest | en | 0.855122 |
https://www.smore.com/fxnh3 | 1,521,403,087,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645943.23/warc/CC-MAIN-20180318184945-20180318204945-00300.warc.gz | 850,692,772 | 13,228 | Question
Hypothesis
I think, that the element that needs less time to freeze of this 4 elements with the water it's the distilled water. Because the distilled water only have a element (water), I think less as elements have the water more faster freezing.
Background Research
Internet research:
I researched in wikipedia some elements to know the freeze temperature of this elements.
Alcohol
Distilled water
Salt with water
In my home there are some ancient physics books and enciclophedies to know charactheristics of the elements.
Experiment
The first thing that we know in this experiment it's the temperature.
Ambient temperature: 21 Celsius degrees
Freezer's temperature: - 20 Celsius degrees
First we take 4 equals glasses and we put the following items.
In the first glass we put 50 mililitres of distilled water.
In the second glass we put 4o mililitres of water and 10 grams of salt.
In the third glass deposit 40 mililitres of water and 10 mililitres of alcohol.
In the fourth glass we put 10 mililitres of alcohol.
And we put the 4 glasses in the freezer.
Results
The first glass to be frozen has been the glass of distilled water. At 2 hours after put the glass in the freezer.
The second glass (water with salt) 1:15 hours later it turned in a pasty substance. And freeze completly 12 hours later.
The third glass (water with alcohol) 3:00 hours later it turned in a pasty substance and at 4:15 hours later have the frozen surface. The third glass didn't freeze completly,. Because contains 20% of alcohol.
The fourth glass (alcohol) didn't freeze. It remained liquid, because the alcohol freeze at -114 Celsius degrees. And the limit freezer's temperature it's -20 degrees.
.
Conclusions
I do the experiment many times, and in all the times the first glass to freeze is the glass of distilled water. The conclusion of this experiment is that dissolving substances or chemicals in the water can change its freezing temperature , which in theory is at 0 degrees Celsius. These elements dissolved in water cause this freeze at a lower temperature of 0 degrees . This is useful for example to prevent the formation of ice on the surface of the road , throwing salt. In automobiles or other vehicles prevent freezing of cooling system, because it contains an antifreeze liquid composed of water and alcohol. | 515 | 2,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-13 | longest | en | 0.930915 |
http://murzim.net/QM/QM24.html | 1,550,694,754,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247496080.57/warc/CC-MAIN-20190220190527-20190220212527-00067.warc.gz | 181,452,435 | 2,401 | Mauro Murzi's pages on Philosophy of Science - Quantum mechanics
Applications of the indeterminacy principle
# [5. Heisenberg indeterminacy principle.]
## Applications of the indeterminacy principle.
We can use Heisenberg indeterminacy principle to understand why a quantum particle is able to escape from a potential well. The position q of the particle inside a potential well is strongly localized, thus the indetermination of its momentum p is high. Since the kinetic energy T of the particle is a function of its momentum p (T = p²/2m), the indetermination of the energy is high too. As a consequence of this indetermination, the particle can have enough energy to escape.
Another application of the indeterminacy principle concerns the zero-point energy. A quantum particle cannot be completely at rest, because in such case the indetermination of its position and momentum will be zero. This is true also for a particle at absolute zero. In classical physics, a particle at absolute zero has no energy at all; in quantum mechanics, a particle at absolute zero is not at rest, so it has a residual energy.
The indeterminacy principle is applicable also to energy E and time t:
When applied to the electromagnetic field, formula (5.8) states that also in empty space the electromagnetic field has an energy, called the vacuum energy. Given the equivalence between energy and mass, the vacuum energy can manifests itself creating couples of short-lived particles and anti-particles, which disappear in a time interval t determined by equation (5.8). | 317 | 1,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-09 | longest | en | 0.912411 |
https://www.unitconverters.net/fuel-consumption/gigameter-liter-to-meter-liter.htm | 1,519,601,634,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817523.0/warc/CC-MAIN-20180225225657-20180226005657-00371.warc.gz | 969,711,850 | 3,413 | Home / Fuel Consumption Conversion / Convert Gigameter/liter to Meter/liter
# Convert Gigameter/liter to Meter/liter
Please provide values below to convert gigameter/liter [Gm/L] to meter/liter [m/L], or vice versa.
From: gigameter/liter To: meter/liter
### Gigameter/liter to Meter/liter Conversion Table
Gigameter/liter [Gm/L]Meter/liter [m/L]
0.01 Gm/L10000000 m/L
0.1 Gm/L100000000 m/L
1 Gm/L1000000000 m/L
2 Gm/L2000000000 m/L
3 Gm/L3000000000 m/L
5 Gm/L5000000000 m/L
10 Gm/L10000000000 m/L
20 Gm/L20000000000 m/L
50 Gm/L50000000000 m/L
100 Gm/L100000000000 m/L
1000 Gm/L1000000000000 m/L
### How to Convert Gigameter/liter to Meter/liter
1 Gm/L = 1000000000 m/L
1 m/L = 1.0E-9 Gm/L
Example: convert 15 Gm/L to m/L:
15 Gm/L = 15 × 1000000000 m/L = 15000000000 m/L | 304 | 778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-09 | longest | en | 0.700533 |
https://www.physicsforums.com/threads/velocity-of-satellite.284265/ | 1,542,534,400,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744348.50/warc/CC-MAIN-20181118093845-20181118115845-00387.warc.gz | 972,146,545 | 13,068 | # Homework Help: Velocity of satellite
1. Jan 11, 2009
### Maiia
1. The problem statement, all variables and given/known data
A satellite is in a circular orbit about the Earth at a distance of one Earth radius above the surface. What is the velocity of the satellite?
3. The attempt at a solution
I assumed the orbital radius was twice the radius of the earth so
GMearth/r= velocity
but that leaves me with a very small velocity...
2. Jan 11, 2009
### G01
Can you show your answer and your work? If you do have a mistake, I can't find it without seeing both.
3. Jan 11, 2009
### Maiia
(GMearth/r)^.5= v
I plugged in (6.67*10^-11(5.98*10^24)/6.37*10^6)6.5= 7913.048m/s
4. Jan 11, 2009
### G01
Why are you multiplying by 6.5? Also, you never doubled the radius when you plugged in the numbers. According the the formula you give, which is correct, the velocity should be:
$$v=\sqrt{\frac{GM}{2r_e}}$$
Plugging in the numbers you give, I get an answer around 5000m/s.
5. Jan 11, 2009
### Maiia
why do you double the radius?
6. Jan 11, 2009
### G01
You said it yourself before. The orbital radius is twice the radius of Earth. In your formula, you used the radius of Earth, not double the radius of Earth. | 364 | 1,223 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-47 | latest | en | 0.879666 |
https://www.enotes.com/homework-help/people-laugh-mathematician-who-says-that-he-can-452981 | 1,627,749,210,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00218.warc.gz | 765,226,053 | 18,403 | People laugh at a mathematician who says that he can give money to an infinte number of people if the first person takes \$100 and each subsequent person takes one half of the earlier person. Is it really possible for this to be done without going bankrupt?
The mathematian would be able to give out money to infinite number of people, as described above, if the unit of money was infinitely divisible. Then, in fact, he could give out \$200 because it is a sum of a geometric sequence.
In real world, however, once he gets to 1/2 of \$6.25, he would have to give out \$3.125, which is 3 dollars and 12 and a half cents. Further division would result in a quarter of a cent, and so on. Such units of money do not exist.
Thus, while it is mathematically possible to write 200 as a sum of infinite series, it is physically impossible to break down 20000 descrete units (cents) into infinite number of groups because one cannot produce half or other fraction of a cent.
Approved by eNotes Editorial Team
The mathematician claims to be able to give money to an infinite number of people if the first person takes \$100 and the subsequent person is given half the amount.
The amount that is being given out is 100, 50, 25, 12.5, 6.25 ...
This is a geometric series with first term a = 100 and common ratio r = (1/2). The sum of an infinite terms of a geometric series with common ratio less than one is `S_oo = a/(1 - r)` . The total amount that the mathematician has to give out is only equal to `100/(1 - 1/2)` = 2*100 = 200.
The mathematician can do what he claims if he has \$200 with himself.
Approved by eNotes Editorial Team | 399 | 1,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-31 | latest | en | 0.966163 |
https://numberworld.info/14971 | 1,638,283,792,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359037.96/warc/CC-MAIN-20211130141247-20211130171247-00164.warc.gz | 497,966,803 | 3,802 | # Number 14971
### Properties of number 14971
Cross Sum:
Factorization:
Divisors:
1, 11, 1361, 14971
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
3a7b
Base 32:
ejr
sin(14971)
-0.96644376850935
cos(14971)
-0.25687826359864
tan(14971)
3.7622637079927
ln(14971)
9.6138702754498
lg(14971)
4.1752508103616
sqrt(14971)
122.35603785674
Square(14971)
### Number Look Up
Look Up
14971 which is pronounced (fourteen thousand nine hundred seventy-one) is a very impressive number. The cross sum of 14971 is 22. If you factorisate the number 14971 you will get these result 11 * 1361. The figure 14971 has 4 divisors ( 1, 11, 1361, 14971 ) whith a sum of 16344. 14971 is not a prime number. 14971 is not a fibonacci number. The figure 14971 is not a Bell Number. 14971 is not a Catalan Number. The convertion of 14971 to base 2 (Binary) is 11101001111011. The convertion of 14971 to base 3 (Ternary) is 202112111. The convertion of 14971 to base 4 (Quaternary) is 3221323. The convertion of 14971 to base 5 (Quintal) is 434341. The convertion of 14971 to base 8 (Octal) is 35173. The convertion of 14971 to base 16 (Hexadecimal) is 3a7b. The convertion of 14971 to base 32 is ejr. The sine of the number 14971 is -0.96644376850935. The cosine of the number 14971 is -0.25687826359864. The tangent of the figure 14971 is 3.7622637079927. The square root of 14971 is 122.35603785674.
If you square 14971 you will get the following result 224130841. The natural logarithm of 14971 is 9.6138702754498 and the decimal logarithm is 4.1752508103616. You should now know that 14971 is very great number! | 613 | 1,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-49 | latest | en | 0.76696 |
en.coinpick.com | 1,716,488,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058653.47/warc/CC-MAIN-20240523173456-20240523203456-00588.warc.gz | 198,622,715 | 16,185 | 메뉴 건너뛰기
## 5. Fibonacci Retracement
Fibonacci retracement?
The Fibonacci retracement was derived from a series of numbers discovered by the 13th century mathematician Leonardo Fibonacci. This mathematical relationship shows a sequence of numbers: 0%, 23.6%, 38.2%, 61.8%, 78.6%, 100%.
What is an array?
An array is the sum of the leading and trailing numbers.
1,1,2,3,5,8,13,21,34,55,89,144 It goes like this.
So what about chart indicators? What is the Fibonacci retracement used as a concept?
Arrangement in Charts
Even on a rising chart, corrections always occur and take a break. Think of this as a retracement concept, and it is a term with a number by grafting the adjustment ratio to the Fibonacci ratio. 'Fibonacci retracement'
13/21,21/34,34/55 The leading number divided by the trailing number is 0.618.
21/55,34/89,55/144 The leading number divided by the trailing number is 0.382.
When expressed in %, the values of 38.2% and 61.8% are obtained. The theory is that adding 50% to these two percentages will result in a retracement rate of 38.2%, 61,8%, and 50% when adjusted after an increase.
This is used to predict the timing of a rebound after correction. If the retracement is short, the next rise is large, and if the retracement is large, the rise is generally low.
Use of Fibonacci Trading
Fibonacci retracements are usually used in bull markets. I use the lows and highs in succession. Personally, I do not use the Fibonacci retracement as a lagging auxiliary indicator for trading. By using the Fibonacci retracement indicator, you can find a section with support and a section with resistance. In this way, you can set the timing of buying and selling.
The Fibonacci retracement indicator could be used for trading, but it is not necessarily the answer. If the Fibonacci retracement indicator could accurately predict it, all traders would have used it. Therefore, in order to know only what the Fibonacci retracement indicator is and use it for trading, it would be much more accurate to use other indicators together.
However, 0.5 and 0.618 are important values when using the Fibonacci retracement. This is because this section (see image above) is most likely a strong support or resistance section. In other words, if you understand easily, it can be said that the points of 1/3 and 2/3 of the chart are 0.382 and 0.618 points.
Typically, these studies are used to predict levels of support and resistance through the relationship between Fibonacci numbers. The most used relationships are:
0.618 (61.8%): The Fibonacci number divided by the next number is about 0.618.
0.382 (38.2%): The Fibonacci number divided by the number occupying two postfixes is about 0.3820.
0.2360 (23.60%): The Fibonacci number divided by the number occupying three postfixes is approximately 0.2360.
0.764 (76.4%): The difference between 38.2% and 23.6% is obtained by adding 61.8%, i.e. 76.4% = 61.8% + (38.2% – 23.6%).
0.5 (50%): About half of the main flow, the average is 38.2% to 61.8%.
0.0 (0%): The start of the market movement.
1.9 (100%): End of market movement.
G
M
T
언어를 감지갈리시아어구자라트어그루지야어그리스어네덜란드어네팔어노르웨이어덴마크어독일어라오어라트비아어라틴어러시아어루마니아어리투아니아어마라티어마오리어마케도니아어말레이어말라가시어말라얄람어몰타어몽골어몽어바스크어버마어베트남어벨로루시어벵골의보스니아어불가리아어세르비아어세부아노세소토어소말리아어순다어스와힐리어스웨덴어스페인어슬로바키아어슬로베니아어신할라어아랍어아르메니아어아이슬란드어아이티프랑스말아일랜드어아제르바이잔어아프리칸스어알바니아어에스토니아어에스페란토말영어요루바어우르두어우즈베크어우크라이나어웨일즈어이그보어이디시어이탈리아어인도네시아어일본어자바어줄루어중국어(간체)중국어(번체)체와어체코어카자흐어카탈로니아어칸나다어칸나다어크로아티아어타갈로그어타밀어타직어태국어터키어텔루구어펀자브어페르시아어포르투갈어폴란드어프랑스어핀란드어하우사어한국어헝가리어히브리어힌디어 갈리시아어구자라트어그루지야어그리스어네덜란드어네팔어노르웨이어덴마크어독일어라오어라트비아어라틴어러시아어루마니아어리투아니아어마라티어마오리어마케도니아어말레이어말라가시어말라얄람어몰타어몽골어몽어바스크어버마어베트남어벨로루시어벵골의보스니아어불가리아어세르비아어세부아노세소토어소말리아어순다어스와힐리어스웨덴어스페인어슬로바키아어슬로베니아어신할라어아랍어아르메니아어아이슬란드어아이티프랑스말아일랜드어아제르바이잔어아프리칸스어알바니아어에스토니아어에스페란토말영어요루바어우르두어우즈베크어우크라이나어웨일즈어이그보어이디시어이탈리아어인도네시아어일본어자바어줄루어중국어(간체)중국어(번체)체와어체코어카자흐어카탈로니아어칸나다어칸나다어크로아티아어타갈로그어타밀어타직어태국어터키어텔루구어펀자브어페르시아어포르투갈어폴란드어프랑스어핀란드어하우사어한국어헝가리어히브리어힌디어
음성 기능은 200자로 제한됨
옵션 : 역사 : 피드백 : Donate 닫기
2021년 해외코인거래소 순위 및 추천 거래량 11조(1위) 3.5조 9.5조 레버리지 최대100배 최대100배 최대125배 수수료 - 지정가 : 0.02% - 시장가 : 0.04%(1위) - 지정가 : -0.025% - 시장가 : 0.075% - 지정가 : -0.025% - 시장가 : 0.075% 거래방법 현물+선물+마진 선물+주식+FOREX 선물 회원가입 회원가입 회원가입 회원가입 할인코드 20% 할인 5% 할인 20% 할인 | 1,753 | 4,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-22 | latest | en | 0.932162 |
https://oeis.org/A273634 | 1,670,227,541,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711013.11/warc/CC-MAIN-20221205064509-20221205094509-00417.warc.gz | 467,030,169 | 4,299 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 59th year, we have over 358,000 sequences, and we’ve crossed 10,300 citations (which often say “discovered thanks to the OEIS”). Other ways to Give
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A273634 Decimal expansion of (Pi/6)^(1/3), the sphericity of the cube. 4
8, 0, 5, 9, 9, 5, 9, 7, 7, 0, 0, 8, 2, 3, 4, 8, 2, 0, 3, 5, 8, 4, 8, 3, 4, 2, 3, 3, 1, 9, 6, 4, 2, 4, 6, 9, 4, 7, 2, 3, 0, 7, 0, 3, 6, 1, 6, 1, 9, 3, 0, 7, 7, 7, 8, 4, 6, 1, 4, 6, 0, 3, 7, 6, 8, 9, 4, 7, 5, 4, 8, 2, 5, 2, 8, 5, 7, 2, 6, 3, 7, 1, 2, 3, 0, 7 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS Wikipedia, Sphericity. FORMULA Equals cube root of A019673. - Michel Marcus, May 27 2016 EXAMPLE 0.80599597700823482035848342331964246947230703616193077784614603... PROG (PARI) default(realprecision, 50080); my(x=(Pi/6)^(1/3)); for(k=1, 100, my(d=floor(x)); x=(x-d)*10; print1(d, ", ")) CROSSREFS Cf. A273633, A273635, A273636, A273637. Sequence in context: A190281 A107950 A336798 * A121839 A010517 A021851 Adjacent sequences: A273631 A273632 A273633 * A273635 A273636 A273637 KEYWORD nonn,cons AUTHOR Felix Fröhlich, May 27 2016 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified December 5 02:30 EST 2022. Contains 358572 sequences. (Running on oeis4.) | 693 | 1,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-49 | latest | en | 0.722004 |
http://reference.wolfram.com/mathematica/ref/GoodmanKruskalGammaTest.html | 1,386,748,593,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164033438/warc/CC-MAIN-20131204133353-00059-ip-10-33-133-15.ec2.internal.warc.gz | 152,377,838 | 8,580 | BUILT-IN MATHEMATICA SYMBOL
GoodmanKruskalGammaTest
GoodmanKruskalGammaTest[v1, v2]
tests whether the vectors and are independent.
GoodmanKruskalGammaTest[..., "property"]
returns the value of .
Details and OptionsDetails and Options
• GoodmanKruskalGammaTest performs a hypothesis test on and with null hypothesis that the vectors are independent, and alternative hypothesis that they are not.
• By default, a probability value or -value is returned.
• A small -value suggests that it is unlikely that is true.
• The arguments and can be any real-valued vectors of equal length.
• GoodmanKruskalGammaTest is based on the Goodman-Kruskal gamma coefficient which is computed by GoodmanKruskalGamma[v1, v2].
• GoodmanKruskalGammaTest[v1, v2, "HypothesisTestData"] returns a HypothesisTestData object htd that can be used to extract additional test results and properties using the form htd["property"].
• GoodmanKruskalGammaTest[v1, v2, "property"] can be used to directly give the value of .
• Properties related to the reporting of test results include:
• "PValue" the -value of the test "PValueTable" formatted table containing the -value "ShortTestConclusion" a short description of the conclusion of the test "TestConclusion" a description of the conclusion of the test "TestData" a list containing the test statistic and -value "TestDataTable" formatted table of the -value and test statistic "TestStatistic" the test statistic "TestStatisticTable" formatted table containing the test statistic
• The following options can be used:
• AlternativeHypothesis "Unequal" the inequality for the alternative hypothesis Method Automatic the method to use for computing -values SignificanceLevel 0.05 cutoff for diagnostics and reporting
• For tests of independence, a cutoff is chosen such that is rejected only if . The value of used for the and properties is controlled by the SignificanceLevel option. By default, is set to .
ExamplesExamplesopen allclose all
Basic Examples (1)Basic Examples (1)
Test whether two vectors are independent:
Out[2]=
Out[3]= | 467 | 2,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2013-48 | longest | en | 0.763863 |
https://functions.wolfram.com/ElementaryFunctions/Cosh/21/01/14/01/01/22/0002/ | 1,723,163,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00605.warc.gz | 209,251,122 | 10,211 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
Cosh
http://functions.wolfram.com/01.20.21.4489.01
Input Form
Integrate[z^n Sinh[b z^2 + e] Cosh[c z^2 + g]^v, z] == 2^(-2 - v) z^(1 + n) (Binomial[v, v/2] (E^e ((-b) z^2)^((1/2) (-1 - n)) Gamma[(1 + n)/2, (-b) z^2] - ((b z^2)^((1/2) (-1 - n)) Gamma[(1 + n)/2, b z^2])/E^e) (-1 + Mod[v, 2]) - Sum[E^(-e + 2 g s - g v) Binomial[v, s] (E^(2 e - 4 g s + 2 g v) ((-b + 2 c s - c v) z^2)^((1/2) (-1 - n)) Gamma[(1 + n)/2, (-b + 2 c s - c v) z^2] - E^(2 g (-2 s + v)) ((b + 2 c s - c v) z^2)^((1/2) (-1 - n)) Gamma[(1 + n)/2, (b + 2 c s - c v) z^2] + E^(2 e) ((-b - 2 c s + c v) z^2)^ ((1/2) (-1 - n)) Gamma[(1 + n)/2, (-b - 2 c s + c v) z^2] - ((b - 2 c s + c v) z^2)^((1/2) (-1 - n)) Gamma[(1 + n)/2, (b - 2 c s + c v) z^2]), {s, 0, Floor[(1/2) (-1 + v)]}]) /; Element[n, Integers] && n >= 0 && Element[v, Integers] && v > 0
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["\[Integral]", RowBox[List[SuperscriptBox["z", "n"], RowBox[List["Sinh", "[", RowBox[List[RowBox[List["b", " ", SuperscriptBox["z", "2"]]], "+", "e"]], "]"]], SuperscriptBox[RowBox[List["Cosh", "[", RowBox[List[RowBox[List["c", " ", SuperscriptBox["z", "2"]]], "+", "g"]], "]"]], "v"], RowBox[List["\[DifferentialD]", "z"]]]]]], "\[Equal]", RowBox[List[SuperscriptBox["2", RowBox[List[RowBox[List["-", "2"]], "-", "v"]]], " ", SuperscriptBox["z", RowBox[List["1", "+", "n"]]], " ", RowBox[List["(", RowBox[List[RowBox[List[RowBox[List["Binomial", "[", RowBox[List["v", ",", FractionBox["v", "2"]]], "]"]], " ", RowBox[List["(", RowBox[List[RowBox[List[SuperscriptBox["\[ExponentialE]", "e"], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["-", "b"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]], "-", RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List["-", "e"]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List["b", " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List["b", " ", SuperscriptBox["z", "2"]]]]], "]"]]]]]], ")"]], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "+", RowBox[List["Mod", "[", RowBox[List["v", ",", "2"]], "]"]]]], ")"]]]], "-", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["s", "=", "0"]], RowBox[List["Floor", "[", RowBox[List[FractionBox["1", "2"], RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "+", "v"]], ")"]]]], "]"]]], RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List["-", "e"]], "+", RowBox[List["2", " ", "g", " ", "s"]], "-", RowBox[List["g", " ", "v"]]]]], " ", RowBox[List["Binomial", "[", RowBox[List["v", ",", "s"]], "]"]], " ", RowBox[List["(", RowBox[List[RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List["2", " ", "e"]], "-", RowBox[List["4", " ", "g", " ", "s"]], "+", RowBox[List["2", " ", "g", " ", "v"]]]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["2", " ", "c", " ", "s"]], "-", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["2", " ", "c", " ", "s"]], "-", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]], "-", RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "g", " ", RowBox[List["(", RowBox[List[RowBox[List[RowBox[List["-", "2"]], " ", "s"]], "+", "v"]], ")"]]]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List["b", "+", RowBox[List["2", " ", "c", " ", "s"]], "-", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["(", RowBox[List["b", "+", RowBox[List["2", " ", "c", " ", "s"]], "-", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]], "+", RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "e"]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "-", RowBox[List["2", " ", "c", " ", "s"]], "+", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "-", RowBox[List["2", " ", "c", " ", "s"]], "+", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]], "-", RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List["b", "-", RowBox[List["2", " ", "c", " ", "s"]], "+", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["(", RowBox[List["b", "-", RowBox[List["2", " ", "c", " ", "s"]], "+", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]]]], ")"]]]]]]]], ")"]]]]]], "/;", RowBox[List[RowBox[List["n", "\[Element]", "Integers"]], "\[And]", RowBox[List["n", "\[GreaterEqual]", "0"]], "\[And]", RowBox[List["v", "\[Element]", "Integers"]], "\[And]", RowBox[List["v", ">", "0"]]]]]]]]
MathML Form
z n sinh ( b z 2 + e ) cosh v ( c z 2 + g ) z 2 - v - 2 z n + 1 ( ( v v 2 ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["v", Identity]], List[TagBox[FractionBox["v", "2"], Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( e ( - b z 2 ) 1 2 ( - n - 1 ) Γ ( n + 1 2 , - b z 2 ) - - e ( b z 2 ) 1 2 ( - n - 1 ) Γ ( n + 1 2 , b z 2 ) ) ( v mod 2 \$CellContext`v 2 - 1 ) - s = 0 v - 1 2 - e + 2 g s - g v ( v s ) TagBox[RowBox[List["(", GridBox[List[List[TagBox["v", Identity]], List[TagBox["s", Identity]]]], ")"]], InterpretTemplate[Function[Binomial[Slot[1], Slot[2]]]]] ( 2 e Γ ( n + 1 2 , ( - b - 2 c s + c v ) z 2 ) ( ( - b - 2 c s + c v ) z 2 ) 1 2 ( - n - 1 ) - ( ( b - 2 c s + c v ) z 2 ) 1 2 ( - n - 1 ) Γ ( n + 1 2 , ( b - 2 c s + c v ) z 2 ) + 2 e - 4 g s + 2 g v ( ( - b + 2 c s - c v ) z 2 ) 1 2 ( - n - 1 ) Γ ( n + 1 2 , ( - b + 2 c s - c v ) z 2 ) - 2 g ( v - 2 s ) ( ( b + 2 c s - c v ) z 2 ) 1 2 ( - n - 1 ) Γ ( n + 1 2 , ( b + 2 c s - c v ) z 2 ) ) ) /; n v + Condition z z n b z 2 e c z 2 g v 2 -1 v -2 z n 1 Binomial v v 2 -1 e -1 b z 2 1 2 -1 n -1 Gamma n 1 2 -1 -1 b z 2 -1 -1 e b z 2 1 2 -1 n -1 Gamma n 1 2 -1 b z 2 \$CellContext`v 2 -1 -1 s 0 v -1 2 -1 -1 e 2 g s -1 g v Binomial v s 2 e Gamma n 1 2 -1 -1 b -1 2 c s c v z 2 -1 b -1 2 c s c v z 2 1 2 -1 n -1 -1 b -1 2 c s c v z 2 1 2 -1 n -1 Gamma n 1 2 -1 b -1 2 c s c v z 2 2 e -1 4 g s 2 g v -1 b 2 c s -1 c v z 2 1 2 -1 n -1 Gamma n 1 2 -1 -1 b 2 c s -1 c v z 2 -1 2 g v -1 2 s b 2 c s -1 c v z 2 1 2 -1 n -1 Gamma n 1 2 -1 b 2 c s -1 c v z 2 n v SuperPlus [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["\[Integral]", RowBox[List[RowBox[List[SuperscriptBox["z_", "n_"], " ", RowBox[List["Sinh", "[", RowBox[List[RowBox[List["b_", " ", SuperscriptBox["z_", "2"]]], "+", "e_"]], "]"]], " ", SuperscriptBox[RowBox[List["Cosh", "[", RowBox[List[RowBox[List["c_", " ", SuperscriptBox["z_", "2"]]], "+", "g_"]], "]"]], "v_"]]], RowBox[List["\[DifferentialD]", "z_"]]]]]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List[SuperscriptBox["2", RowBox[List[RowBox[List["-", "2"]], "-", "v"]]], " ", SuperscriptBox["z", RowBox[List["1", "+", "n"]]], " ", RowBox[List["(", RowBox[List[RowBox[List[RowBox[List["Binomial", "[", RowBox[List["v", ",", FractionBox["v", "2"]]], "]"]], " ", RowBox[List["(", RowBox[List[RowBox[List[SuperscriptBox["\[ExponentialE]", "e"], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["-", "b"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]], "-", RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List["-", "e"]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List["b", " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List["b", " ", SuperscriptBox["z", "2"]]]]], "]"]]]]]], ")"]], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "+", RowBox[List["Mod", "[", RowBox[List["v", ",", "2"]], "]"]]]], ")"]]]], "-", RowBox[List[UnderoverscriptBox["\[Sum]", RowBox[List["s", "=", "0"]], RowBox[List["Floor", "[", RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "+", "v"]], ")"]]]], "]"]]], RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List["-", "e"]], "+", RowBox[List["2", " ", "g", " ", "s"]], "-", RowBox[List["g", " ", "v"]]]]], " ", RowBox[List["Binomial", "[", RowBox[List["v", ",", "s"]], "]"]], " ", RowBox[List["(", RowBox[List[RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List[RowBox[List["2", " ", "e"]], "-", RowBox[List["4", " ", "g", " ", "s"]], "+", RowBox[List["2", " ", "g", " ", "v"]]]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["2", " ", "c", " ", "s"]], "-", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "+", RowBox[List["2", " ", "c", " ", "s"]], "-", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]], "-", RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "g", " ", RowBox[List["(", RowBox[List[RowBox[List[RowBox[List["-", "2"]], " ", "s"]], "+", "v"]], ")"]]]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List["b", "+", RowBox[List["2", " ", "c", " ", "s"]], "-", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["(", RowBox[List["b", "+", RowBox[List["2", " ", "c", " ", "s"]], "-", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]], "+", RowBox[List[SuperscriptBox["\[ExponentialE]", RowBox[List["2", " ", "e"]]], " ", SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "-", RowBox[List["2", " ", "c", " ", "s"]], "+", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["(", RowBox[List[RowBox[List["-", "b"]], "-", RowBox[List["2", " ", "c", " ", "s"]], "+", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]], "-", RowBox[List[SuperscriptBox[RowBox[List["(", RowBox[List[RowBox[List["(", RowBox[List["b", "-", RowBox[List["2", " ", "c", " ", "s"]], "+", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]], ")"]], RowBox[List[FractionBox["1", "2"], " ", RowBox[List["(", RowBox[List[RowBox[List["-", "1"]], "-", "n"]], ")"]]]]], " ", RowBox[List["Gamma", "[", RowBox[List[FractionBox[RowBox[List["1", "+", "n"]], "2"], ",", RowBox[List[RowBox[List["(", RowBox[List["b", "-", RowBox[List["2", " ", "c", " ", "s"]], "+", RowBox[List["c", " ", "v"]]]], ")"]], " ", SuperscriptBox["z", "2"]]]]], "]"]]]]]], ")"]]]]]]]], ")"]]]], "/;", RowBox[List[RowBox[List["n", "\[Element]", "Integers"]], "&&", RowBox[List["n", "\[GreaterEqual]", "0"]], "&&", RowBox[List["v", "\[Element]", "Integers"]], "&&", RowBox[List["v", ">", "0"]]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2002-12-18 | 5,192 | 13,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-33 | latest | en | 0.171758 |
https://mattermodeling.stackexchange.com/questions/9875/is-an-exact-double-hybrid-density-the-same-as-the-exact-dft-density | 1,716,318,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00495.warc.gz | 323,540,046 | 27,526 | # Is an "exact" double hybrid density the same as the "exact" DFT density?
Double hybrid approximate functionals have "unoccupied" Kohn-Sham orbitals in their formulations due to their MP2 component; however, the "exact functional" depends only on the "occupied" K-S orbitals.
Thus the only way double hybrid approximations could approximate the "exact" functional is that the "exact" DFT density be theoretically invariant under MP2, or at least MPn where n tends to infinity.
This would mean that- should one apply MPn to a set of K-S orbitals, let n tend to infinity, and sum the squares of the resulting "MPn occupied orbitals"(like one theoretically does when trying to compute the density from an "unperturbed" K-S set), one should get the same density as the one computed without MPn being applied.
Is it mathematically true?
*I'm assuming exact functional here, not accounting for the fact that its exact form is unknown.
• +1. You might find this paper relevant, but it only talks about MPn with n=155 at most (not n=infinity). Perhaps you'd like to ask the question "Does MPn converge to FCI as n -> infinity" in a separate question. But I think this question should probably just stick to MP2 (or MPn with finite 'n') since otherwise it seems that you're asking two different questions: whether it's true for finite n, and whether it's true for infinitely large n. Oct 29, 2022 at 21:58
• As for a more direct answer to the question: wouldn't the exact double-hybrid functional need to be different from the exact ordinary-DFT functional, because the former starts with MP2? You may also be interested in density-corrected DFT, in which a more accurate density is used: mattermodeling.stackexchange.com/a/1260/5 Oct 29, 2022 at 22:04
• I think that the more I think about this question, the more confused I get. What do you mean by "exact DFT density"? Do you mean the "exact physical density"? Why would the "exact double hybrid density" be different? Oct 30, 2022 at 0:12
• @NikeDattani the density one gets if one sums the squares of the occupied K-S orbitals Oct 30, 2022 at 5:36
• I got confused by the word "exact". Oct 30, 2022 at 12:34 | 536 | 2,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.942419 |
https://www.mathhomeworkanswers.org/252444/find-the-limit | 1,542,493,183,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743854.48/warc/CC-MAIN-20181117205946-20181117231946-00146.warc.gz | 901,468,994 | 11,686 | find lim -> -1 6x+5/5x-6
When x=-1 the value of the expression is (-1)/(-11)=1/11. This is also the limit.
answered Jun 5 by Top Rated User (588,900 points) | 57 | 158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-47 | longest | en | 0.922717 |
https://uk.mathworks.com/matlabcentral/cody/problems/41-cell-joiner/solutions/1283442 | 1,603,968,613,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00594.warc.gz | 572,729,382 | 17,065 | Cody
# Problem 41. Cell joiner
Solution 1283442
Submitted on 9 Oct 2017 by Ratchet_Hamster
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = {'hello', 'basic', 'test', 'case'}; y_correct = 'hello basic test case'; assert(isequal(cellstr_joiner(x, ' '),y_correct))
2 Pass
x = {'this', 'one', '', 'has', ' ', 'some tricky', 'stuff'}; y_correct = 'this one has some tricky stuff'; assert(isequal(cellstr_joiner(x, ' '),y_correct))
3 Pass
x = {'delimiters', 'are', 'not', 'always', 'spaces'}; y_correct = 'delimiters?are?not?always?spaces'; assert(isequal(cellstr_joiner(x, '?'),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 239 | 845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-45 | latest | en | 0.516219 |
https://algebraworksheets.co/algebra-2-factoring-practice-worksheet-2/ | 1,657,061,707,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104628307.87/warc/CC-MAIN-20220705205356-20220705235356-00034.warc.gz | 133,069,815 | 14,385 | # Algebra 2 Factoring Practice Worksheet
Algebra 2 Factoring Practice Worksheet – Algebra Worksheets are designed to help students understand math. The subject of study is the study of mathematical symbols and the rules to manipulate them. It is the central thread that unites the disciplines of math, geometry and physics. It is a vital element of your education. This section will teach you everything you need to know about algebra, while focusing on the most significant subjects. These worksheets can help you enhance your abilities as student.
The worksheets teach students how to tackle algebraic word problems. They can employ fractions, integers and decimals in order to solve problems, modify equations, and analyze formulas. These worksheets can be used in both metric and traditional units. The worksheets also cover the various forms of linear equations, as well as the ability to find the x-intercept and solve word problems involving parallellines.
The most effective algebra worksheets will assist you in mastering the basics of solving quadratic equations. Many of them are printed, which means you don’t require additional software. These worksheets are designed for students to grasp the basics of algebra without worrying over using a calculator or spending long hours searching for the correct answer. It is possible to use these algebra worksheets to reinforce concepts and increase your knowledge. The worksheets are designed to strengthen concepts and enhance your math abilities.
These worksheets will give students an understanding of mathematical fundamentals. They show students how to make use of fractions in order to represent numbers. They also teach students how to solve equations with several unknown variables. These printables can help you learn math. These worksheets can help you comprehend equations more clearly. If you require help in improving your math skills Make sure you are using a top-quality set of worksheets.
You can improve your math skills by making use of algebra worksheets free. The worksheets are available for free and are a great way to practice basic math principles. These worksheets will help you learn the fundamentals of algebra and how to use them in your everyday activities. Consider them as a great tool for your education, in the event that you are a student. They can help you become an even better and more successful student. These printables will delight your children! | 439 | 2,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | latest | en | 0.934039 |
https://www.transtutors.com/questions/1-tco-2-royal-company-uses-budgeted-overhead-rates-to-apply-overhead-to-individual-j-4650233.htm | 1,611,678,153,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704800238.80/warc/CC-MAIN-20210126135838-20210126165838-00042.warc.gz | 1,038,733,815 | 13,989 | # 1.(TCO 2) Royal Company uses budgeted overhead rates to apply overhead to individual jobs. It uses a
1.(TCO 2) Royal Company uses budgeted overhead rates to apply overhead to individual jobs. It uses a system based on machine hours. Last year, the company made the following estimates for this year. Direct labor costs $9,900,000Factory overhead costs$11,748,000Direct Labor Hours $550,000Machine Hours$660,000(a) What is the budgeted overhead rate for the company? (Show your work)(b) If Job #31255 had the costs listed below, what would be the total cost of Job #31255? (Show your work)Material costs were $300,000; Direct labor costs were$216,000; Direct labor hours were 12,000; and Machine hours were 15,000. (Points : 30) Question 2.2.(TCO 3) James Company uses process costing to track its costs in two sequential production departments: Forming and Finishing. The following information is provided regarding the Forming Department:Forming DepartmentMonth Ended July 31Unit informationBeginning work in process, July 1 — 6,000Started into production during July — 18,000Completed and transferred to Finished Department during July — 16,000Ending work in process, July 31 (30% complete as to direct materials and 40% complete as to conversion costs) — 8,000 Cost information:Beginning work in process as of July 1 (consists of $5,000 of direct materials costs and$9,500 of conversion costs) — $14,500Direct materials used in July —$27,936Conversion costs incurred in July — $34,350 Required:(a) Calculate the equivalent units for direct materials. (Show your work)(b) Calculate the cost per equivalent unit for direct materials. (Show your work) (Points : 30) Question 3.3.(TCO 6) Classify each of the costs below within the ABC costing hierarchy as either unit-level, batch-level, product-level, or facility-level. (a) Factory insurance(b) Direct labor(c) Costs to develop new products(d) Painting each product(e) Receiving raw materials from suppliers (f) Training costs on sexual harassment in the workplace (Points : 30) Question 4.4.(TCO 2) Drake Company incurred costs of$77,000 for direct materials (raw) purchased. Direct labor was $45,000 and factory overhead was$22,000 for March.Inventories were as follows:Raw materials beginning $10,000; raw materials ending$14,000Work-in-process beginning $36,000; work-in-process ending$24,000Finished goods beginning $5,000; finished goods ending$10,000Required: Calculate the cost of goods manufactured for March for Drake Company. Show your work. (Points : 30) | 616 | 2,521 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-04 | latest | en | 0.945702 |
https://edulissy.com/product/homework-assignment-1-solution-17/ | 1,685,374,041,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644867.89/warc/CC-MAIN-20230529141542-20230529171542-00506.warc.gz | 271,132,858 | 18,605 | # Homework Assignment 1 Solution
\$30.00 \$24.00
Quantity
## Description
1. Consider program P, which runs on a 4 GHz machine M in 500 seconds. Consider an optimization to P that replaces all instances of multiplying a value by 4 (mult X,X,4) with two instructions that set x to x+x twice (add X,X; add X,X). Assume that every multiply instruction takes 4 cycles to execute, and every add instruction takes just 1 cycle. After recompiling, the program now runs in 400 seconds on machine M. Determine how many multiplies were replaced by this optimization. In your answer, show all of your calculations and analysis. [20%]
1. Your company could speed up a Java program on their new computer by adding hardware support for garbage collection. Assume that garbage collection currently comprises 50% of the cycles of the program. You have two possible changes to consider. The first one would be to automatically handle garbage collection in hardware, causing the increase in cycle time by a factor of 1.2. The second option would be to add the new instructions to the ISA that could be used during garbage collection. This would halve the number of cycles needed for garbage collections, but would increase the cycle time by a factor of 1.1. Which of these two options, if any, should you choose? [20%]
1. Consider two possible improvements that can be used to enhance a machine. You can either make multiply instructions run four times faster than before, or make memory access instructions run two times faster than before. You repeatedly run a program that takes 100 seconds to execute. Of this time, 25% is used for multiplication, 40% for memory access instructions, and 35% for other tasks.
1. What will the speedup be if you improve only multiplication? [5%]
2. What will the speedup be if you improve only memory access? [5%]
3. What will the speedup be if both improvements are made? [10%]
1. Assume that multiply instructions take 4 cycles to execute and account for 20% of the instructions in a typical program and that the other 80% of the instructions require an average of 2 cycles for each instruction.
1. What is the percentage of time that the CPU spends doing multiplication? [5%]
2. Estimate the CPI value for a typical program executing on this machine [5%].
3. Assume that it is possible to reduce the number of cycles required for multiplication from 4 to 2, but this will require a 20% increase in the cycle time. Nothing else will be affected. Should we proceed with this modification? Explain why or why not [10%].
1. Suppose that we designed a new floating point unit that accelerates the floating point instructions by a factor of 5. We are now looking for a benchmark to show off the new floating-point unit and we want the overall benchmark to show a speedup of 3. One benchmark we are considering runs for 100 seconds with the old floating-point hardware. How much of the execution time would floating-point instructions have to account for in this program in order to yield our desired speedup on this benchmark? [20%]
0 | 675 | 3,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.919388 |
http://faculty.kfupm.edu.sa/MATH/afarahat/Math%20Courses/Kfupm/Math002/HW-One/Chapter%208/hw-1-ch8.htm | 1,516,214,516,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886952.14/warc/CC-MAIN-20180117173312-20180117193312-00576.warc.gz | 136,314,977 | 5,522 | Note that to complete the square for x we added not
Here we have an ellipse with major axis parallel to the y axis since the bigger denominator is under the y term.
==>
==>
==>
Center:
vertices:
and
foci:
and
39+. Find the equation of the ellipse with center (3, 4), major axis of length 4, foci (3, 3) and (3, 5)
Solution:
Solution:
The bigger number (25) is under the . This tell us that the major axis of the ellipse is parallel to the y-axis (a vertical ellipse).
==>
==>
==>
Therefore,
vertices:
and
foci:
and
23. Find the vertices and foci of the ellipse
Solution:
We need to complete the square for both variable.
Regroup and move the constant to the other side.
Pull out the coefficient of and leave room for completing the square.
Square both sides
Multiply both sides by
Divide by 81
This is the equation of an ellipse with major axis parallel to the y axis.
center;
vertices:
and
foci:
and
We now know that the given equation is the equation of half an ellipse. Which half? From the equation we can see that all the x values will be greater than -5. Thus, the equation represents the right side of the ellipse with the above properties.
major axis of length 4 ==>
The equation is:
49. Find the equation of the ellipse with eccentricity 2/5, foci at (-1, 3) and (3, 3).
Solution:
==>
==>
(1)
The center of the ellipse will be at the middle point between the two foci.
c is half the distance between the two foci. ==>
Substitute c in equation (1), we get
Thus, the equation of the ellipse is
E1. Sketch the graph of
Solution:
==>
Rearrange the equation to make the quadratic variable on one side and the other variable on the other side.
Take 2 as a common factor and leave room for completing the square.
Factor out the coefficient of y
Divide by 2
Thus,
This is a equation of a parabola that opens upward, with:
and
vertex
focus;
directrix:
axis of symmetry:
Solution of Homework Problems Minus One
Chapter 8
8.1
10+. Find the vertex, focus and directrix of the parabola
Solution:
The equation is not in the standard form . Thus, we must first right it in the standard form.
Take 2 as a common factor
Use the property
Divide the equation by 4
This is the standard form of a parabola that opens to the left.
vertex:
focus:
directrix:
21+. Find the vertex, focus and directrix of the parabola
Solution:
Multiply the equation by -1 to make the squared term positive and rearrange the equation
==>
36+. The LNB (in instrument to collect satellite signals) is to be placed at the focus of a paraboloid dish. If the dish has diameter 120 cm and a depth of 40 cm, how far from the vertex of the paraboloid should the LNB be place.
Solution:
This problem was discussed in class.
E1. Sketch the graph of
Solution:
==>
This is the equation of a parabola that opens to the left.
==>
vertex:
focus:
directrix:
Axis of symm:
Thus, the given equation represents half a parabola. The question is: "Is it the upper half or the lower half?" A closer look at the given equation tells us that the y values will always be greater than 2. This helps us choose the upper half of the parabola. The graph below illustrates the two halves of the parabola.
8.2
1. Find the vertices and foci of the ellipse
31. Find the equation in standard form of the parabola with focus (3, -3) and directrix y = -5 .
Solution:
The equation of the directrix y = constant ==> the parabola opens either upward or downward. Moreover, the directrix is below the focus (by comparing the values -5 and -3) tells us that the parabola opens upward.
Therefore, the equation of the parabola will have the form
Now, the point on the directrix directly below the focus will have coordinates (3, -5). We can now use the midpoint formula to find the vertex, which is in the middle between the point (3, -5) and the focus (3, -3)
To find p we have to choices:
1) Compute the distance between the vertex and the focus
2) Compute half the distance between the focus and the point (3, -5) on the directrix.
Thus, the equation of the parabola is given by:
33. Find the equation in standard form of the parabola that has vertex (-4 , 1), has its axis of symmetry parallel to the y-axis, and passes through the point (-2, 2)
Solution:
A parabola with axis of symmetry parallel to the y-axis has the form:
Substitute the vertex
The parabola passes through the point (-2, 2) ==> the point satisfies the equation of the parabola.
Substitute the point in the equation to compute p
==>
The equation is: | 1,134 | 4,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-05 | latest | en | 0.910623 |
https://rd.springer.com/book/10.1007%2F978-3-642-46410-2 | 1,550,796,713,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247511573.67/warc/CC-MAIN-20190221233437-20190222015437-00252.warc.gz | 684,392,746 | 13,513 | # Approximate Behavior of Tandem Queues
• Gordon F. Newell
Book
Part of the Lecture Notes in Economics and Mathematical Systems book series (LNE, volume 171)
1. Front Matter
Pages N2-XI
2. Gordon F. Newell
Pages 1-42
3. Gordon F. Newell
Pages 43-89
4. Gordon F. Newell
Pages 90-148
5. Gordon F. Newell
Pages 149-217
6. Gordon F. Newell
Pages 218-230
7. Gordon F. Newell
Pages 231-312
8. Gordon F. Newell
Pages 313-380
9. Gordon F. Newell
Pages 381-402
10. Back Matter
Pages 403-413
### Introduction
The following monograph deals with the approximate stochastic behavior of a system consisting of a sequence of servers in series with finite storage between consecutive servers. The methods employ deterministic queueing and diffusion approximations which are valid under conditions in which the storages and the queue lengths are typically large compared with 1. One can disregard the fact that the customer counts must be integer valued and treat the queue as if it were a (stochastic) continuous fluid. In these approximations, it is not necessary to describe the detailed probability distribution of service times; it suffices simply to specify the rate of service and the variance rate (the variance of the number served per unit time). Specifically, customers are considered to originate from an infinite reservoir. They first pass through a server with service rate ~O' vari ance rate ~O' into a storage of finite capacity c . They then pass l through a server with service rate ~l' variance rate ~l' into a storage of capacity c ' etc., until finally, after passing through an nth server, 2 they go into an infinite reservoir (disappear). If any jth storage become , n , the service at the j-lth server is interrupted full j = 1, 2, and, of course, if a jth storage becomes empty the jth server is inter rupted; otherwise, services work at their maximum rate.
### Keywords
Distribution Image Queues Warteschlange equilibrium evaluation merger probability probability distribution science and technology service value-at-risk
#### Authors and affiliations
• Gordon F. Newell
• 1
1. 1.Institute of Transportation StudiesUniversity of CaliforniaBerkeleyUSA
### Bibliographic information
• DOI https://doi.org/10.1007/978-3-642-46410-2
• Copyright Information Springer-Verlag Berlin Heidelberg 1979
• Publisher Name Springer, Berlin, Heidelberg
• eBook Packages
• Print ISBN 978-3-540-09552-1
• Online ISBN 978-3-642-46410-2
• Series Print ISSN 0075-8442
• Buy this book on publisher's site
Industry Sectors
Pharma
Materials & Steel
Automotive
Biotechnology | 641 | 2,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-09 | latest | en | 0.853513 |
https://learnzillion.com/lesson_plans/2874-9-slice-three-dimensional-figures-creates-two-dimensional-faces-c | 1,529,700,560,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864795.68/warc/CC-MAIN-20180622201448-20180622221448-00204.warc.gz | 636,547,031 | 39,023 | Lesson plan
# 9. Slice three-dimensional figures creates two-dimensional faces (C)
teaches Common Core State Standards CCSS.Math.Content.7.G.A.3 http://corestandards.org/Math/Content/7/G/A/3
teaches Common Core State Standards CCSS.Math.Practice.MP4 http://corestandards.org/Math/Practice/MP4
teaches Common Core State Standards CCSS.Math.Practice.MP5 http://corestandards.org/Math/Practice/MP5
teaches Common Core State Standards CCSS.Math.Practice.MP7 http://corestandards.org/Math/Practice/MP7
## You have saved this lesson plan!
Here's where you can access your saved items.
Dismiss
Lesson objective: Understand that slicing a three-dimensional figure creates a new two-dimensional face.
Students bring prior knowledge of using nets to represent three-dimensional figures from CCSS 6.G.A.4. This prior knowledge is extended to three-dimensional figures that are sliced into two smaller figures as students identify the two-dimensional shape of the new face. A conceptual challenge students may encounter is visualizing a three dimensional figure from a two-dimensional image.
The concept is developed through work with a model pyramid, which shows that the shape of the new face of a sliced figure, is dependent on the shape of the original figure as well as where it is sliced.
This work helps students deepen their understanding of equivalence because the shape of a three-dimensional figure determines the shape of the new faces created when the figure is sliced.
Students engage in Mathematical Practice 4 (model with mathematics) as they apply their undersatnding of three-dimensional figures to identify the shape of the new face.
Key vocabulary:
• face
• three-dimensional figure
• two-dimensional figure
• slice | 364 | 1,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-26 | latest | en | 0.883513 |
https://math.stackexchange.com/questions/1632171/how-can-define-analytic-function-on-simply-connected-domain | 1,621,168,205,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00125.warc.gz | 387,003,138 | 38,603 | # How can define analytic function on simply connected domain?
Let $U ⊂ \mathbb{C}$ be a simply-connected region, and suppose $\exp(f(z))=z ∈ U$. Then there is a unique analytic function $f(z) : U → C$ such that $f(z) = \operatorname{Log} (z)= \ln|z|+i \operatorname{Arg}(z)$ with $f(-1)=+i\pi$ for all $z ∈ U$ with $0<\operatorname{Arg}(z) \leq 2\pi$.
If we take a closed set $L_{0},$ such that, $L_{0}= \{1+iy : -\infty <y\leq 0\}.~$ There is a unique analytic function in $~ R=U \setminus L_{0},~$ such that $f(z)= \ln|z-1|+i \operatorname{Arg}(z-1),~$ and satisfy that $exp(f(z))= z-1,~$ with $f(-2)=\ln|-2|+i\pi,~$ for all $z ∈ R,~$ with $\frac{- \pi}{2}<\operatorname{Arg}(z-1) \leq \frac{3 \pi}{2}$.
My question, is it tru to write that $f(z)= \operatorname{Log}(z-1)+ 2 \pi i k ~$ for all $z$ in the domain $D_{k}$ and $k=0,1$ such that $~D_0=\{z \in \mathbb{C} : \operatorname{Re}(z)>1 , \operatorname{Im}(z)\leq 0 \}~$ and $~D_1= R_0 \setminus D_0$
Or $f(z)= \operatorname{Log}(z-1)+ 2 \pi i k +\pi i$ for all $z$ in the domain $D_{k}$ and $k=-1,0$ such that $D_{-1}=\{z \in \mathbb{C} : \operatorname{Re}(z)>1 , \operatorname{Im}(z)\leq 0 \}~$ and $D_0= R_0 \setminus D_0$
• If you have a followup to someone else's answer, you should not post that followup as an answer to your question. Instead you can post your followup as a comment to that person's answer. Or, you can edit your question by adding something like "The answer of Joe Schmo leads me to ask a refined question ..." – Lee Mosher Jan 29 '16 at 21:36
$\text{Log}$ is usually used for the principal branch of the logarithm, where the argument is in the interval $(-\pi,\pi)$. That has branch cut on the negative real axis, and if $U$ intersects that branch cut your $f(z)$ is certainly not $\text{Log}(z)$. Indeed, you could have a simply connected region like this:
where the imaginary part of $f(z)$ is forced to have an arbitrarily large range. As you go around the spiral counterclockwise, the argument increases by $2 \pi$ on every turn.
EDIT: If $f(4) = \ln(4)$, then \eqalign{f(5 i) &= \ln(5) + i \pi/2\cr f(-6) &= \ln(6) + i \pi\cr f(-7i) &= \ln(7)+ 3 i \pi/2\cr f(8) &= \ln(8) + 2 i \pi\cr f(9i) &= \ln(9) + 5 i \pi/2\cr f(-10) &= \ln(10) + 3 i \pi\cr &etc.}
• Beautiful picture, that! Maybe you could write down the values of $f(-5i), f(6), f(15i), f(-17)$ for the sake of explicitness ? – Georges Elencwajg Jan 29 '16 at 20:26 | 915 | 2,418 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-21 | latest | en | 0.778585 |
http://sciencedocbox.com/Physics/73836135-In-particular-if-a-is-a-square-matrix-and-l-is-one-of-its-eigenvalues-then-we-can-find-a-non-zero-column-vector-x-with.html | 1,548,008,140,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583728901.52/warc/CC-MAIN-20190120163942-20190120185942-00409.warc.gz | 207,506,756 | 28,233 | # In particular, if A is a square matrix and λ is one of its eigenvalues, then we can find a non-zero column vector X with
Save this PDF as:
Size: px
Start display at page:
Download "In particular, if A is a square matrix and λ is one of its eigenvalues, then we can find a non-zero column vector X with"
## Transcription
1 Appendix: Matrix Estimates and the Perron-Frobenius Theorem. This Appendix will first present some well known estimates. For any m n matrix A = [a ij ] over the real or complex numbers, it will be convenient to use the notation A = i,j For any product matrix AB, note that AB = a ij b jk i,k j i,j,j,k a ij. a ij b j k = A B. (A : 1) In particular, if A is a square matrix and λ is one of its eigenvalues, then we can find a non-zero column vector X with λx = AX, hence λ k X = A k X and λ k X A k X. Dividing by X, it follows that λ k A k. (Here λ and X may be complex, even if A is real.) Define the spectral radius λ max of an n n matrix A to be the largest of the absolute values λ 1,... λ n of its n eigenvalues. Then it follows that λ k max A k. (A : 2) Lemma A.1. The sequence of powers A k converges uniformly to zero if and only if the spectral radius satisfies λ max < 1. (However, examples such as A = show that the convergence may be very slow.) [ ] Proof of A.1. If λ max 1, then A k 1 for all k by (A : 2). The proof for the case λ max < 1 will be by induction on n. Evidently the statement is clear when n = 1. For n > 1, we can always find a non-singular complex matrix S so that SAS 1 has zeros below the diagonal. We can then write SAS 1 = [ ] P Q, 0 R where P and R are square matrices. Since each eigenvalue of P or R is also an eigenvalue of A, we may assume inductively that P k 0 and R k 0 as k. Now set where SA k S 1 = [ P Q 0 R ] k = [ P k ] Q(k) 0 R k Q(k + l) = P k Q(l) + Q(k)R l. Let q m be the maximum of Q(k) over the range 2 m k < 2 m+1. Choosing m large enough so that P k < 1/4 and R k < 1/4 for k 2 m, it follows easily that A-1,
2 APPENDIX q m+1 q m /2, and more generally, q m+r q m /2 r, tending to zero. It follows that Q(k) tends to zero as k, and hence that A k does also. Corollary A.2 Let f(z) = a 0 + a 1 z + a 2 z 2 + be a complex power series with radius of convergence 0 < r. If A is a real or complex n n matrix with spectral radius satisfying λ max < r, then the corresponding series converges absolutely. f(a) = a 0 I + a 1 A + a 2 A 2 +, Proof. Choose η with λ max < η < r. Then the powers of A/η converge to zero by A.1, and hence are bounded, A k /η k c for some uniform constant c. On the other hand, the series a k η k converges, and hence the series a k A k also converges. a k A k c a k η k, Example. If the spectral radius of A satisfies λ max < 1, then it follows that the series A k converges absolutely, and the identity (I A) 1 = I + A + A 2 + A 3 + (A : 3) can be verified by multiplying both sides of this equation by I A. We next prove two formulas for the spectral radius of a matrix A = [a ij ] in terms of the powers A k. We will also make use of the trace trace A = a 11 + a a nn of the matrix A, as well as the traces of its powers A k. Theorem A.3. The spectral radius of any real or complex n n matrix A is given by λ max = lim k Ak 1/k = lim sup trace (A k ) 1/k. k Here it is essential to work with the lim sup, since the limit may not exist. For example, if { A = then trace(a k 3 if k 0 (mod 3) ) = 0 otherwise, so that lim sup trace A k 1/k = +1 but lim inf trace A k 1/k = 0. (The eigenvalues of this matrix are the three cube roots of unity.) Proof of A.3. Since A k λ k max by (A : 2), it certainly follows that A k 1/k λ max. On the other hand, if c = λ max + ɛ, then it follows from A.1 that the powers of A/c converge to zero. Hence we can choose some constant c so that A k c k c for all k. It follows that A k 1/k c k c which converges to c = λ max + ɛ as k. A-2
3 PERRON-FROBENIUS THEOREM Since ɛ can be arbitrarily small, this proves that A k 1/k λ max as k. The inequality lim sup trace(a k ) 1/k λ max k follows immediately, since trace(a k ) A k. To obtain a lower bound for this lim sup, let λ 1,..., λ n be the eigenvalues of A, and note that trace(a k ) = λ k λ k n. For each non-zero λ j, let µ j = λ j / λ j. We will need the following. Lemma A.4. Let µ 1,..., µ q ɛ > 0 there exist infinitely many values of k such that µ j k Re (µ k j ) > 1 ɛ for every µ j. be complex numbers with µ j = 1. For any 1 < ɛ and hence Proof. If p > 2π/ɛ, then we can cover the unit circle by p intervals J 1,..., J p of length less than ɛ. Hence we can cover the torus T q = S 1 S 1 q by p q boxes J i(1) J i(q). Consider the sequence of points µ h = (µ 1 h,..., µ q h ) on the torus T q. Given any p q + 1 of these points, there must be at least two, say µ h and µ l, belonging to the same box. Setting k = h l, it follows that µ j k 1 = µ h j µ j l < ɛ for all j, as required. Since ɛ can be arbitrarily small, we can construct infinitely many such integers k by this procedure. The proof of A.3 continues as follows. Taking ɛ = 1/2, we can choose k as above, and conclude that the real part satisfies Re ((λ j / λ j ) k ) > 1/2 whenever λ j 0. Multiplying by λ j k, it follows that Re (λj k ) 1 2 λ j k for every eigenvalue λ j. Thus trace(a k ) Re (trace(a k )) = j for infinitely many values of k. The required inequality follows immediately. lim sup trace(a k ) 1/k k Re (λj k ) 1 2 λ j k 1 2 λ max k j λ max The Perron-Frobenius Theorem. We say that an m n matrix is non-negative if all of its entries are real and non-negative, and strictly positive if all of the entries are strictly greater than zero. The following helps to show the special properties of such matrices. Lemma A.5. If the non-negative n n matrix A has an eigenvector X which is strictly positive, then the corresponding eigenvalue λ ( where AX = λx ) is equal to the spectral radius λ max. Furthermore λ > 0 unless A is the zero matrix. Proof. If A is not the zero matrix, then since X is strictly positive, it follows that the vector AX has at least one strictly positive component. The equation AX = λx then implies that λ > 0. A-3
4 APPENDIX Now since A k X = λ k X, it follows that A k X = λ k X. If X has components x j c > 0, then A k X c A k, hence λ k X c A k. Taking the k-th root and passing to the limit as k, using A.3, we get as required. λ lim A k 1/k = λ max, and hence λ = λ max, Definition. We will say that a non-negative n n matrix A is transitive if for every i and j there exists an integer k > 0 so that the ij -th entry a (k) ij of the k-th power matrix A k is non-zero. (Some authors use the term irreducible for such matrices.) We can interpret this condition graphically as follows. Let G(A) be the graph with one vertex v i for eacch integer between 1 and n, and with a directed edge from v i to v j if and only if a ij > 0. Then A is transitive if and only if we can get from any vertex to any other vertex by following these directed edges. Since such a path can always be chosen so as to pass through each vertex at most once, we may assume that its length is strictly less than n. Hence a completely equivalent condition would be that the matrix sum is strictly positive. A + A 2 + A A n 1 Theorem A.6. (O. Perron and G. Frobenius). Every non-negative n n matrix A has a non-negative eigenvector, AY = λy, with the property that the associated eigenvalue λ is equal to the spectral radius λ max. If the matrix A is transitive, then there is only one non-negative eigenvector Y up to multiplication by a positive constant, and this eigenvector is strictly positive: y i > 0 for all i. Proof in the transitive case. We will make use of the Brouwer Fixed Point Theorem, which asserts that any continuous mapping f : from a closed simplex to itself must have at least one fixed point, X = f(x). In fact let be the standard (n 1)-dimensional simplex consisting of all non-negative column vectors Y with Y = y y n = 1. If A is transitive and Y, note that AY 0. For if AY were the zero vector, then A k Y would be zero for all k > 0. But choosing some component y j > 0, for any i we can choose some power of A k so that the (i, j)-th component of A k is non-zero, and hence so that the i-th component of A k Y is non-zero. A similar argument shows that any eigenvalue AY = λy must be strictly positive. We can now define the required map f : by the formula Thus there exists at least one fixed point f(y ) = AY/ AY. Y = f(y ) = AY/ AY. Taking λ = AY > 0, it follows that AY = λy where, as noted above, Y must be strictly positive, with λ > 0. It then follows from A.5 that λ = λ max. A-4
5 PERRON-FROBENIUS THEOREM Now suppose that there were two distinct eigenvectors Y and Y in, both necessarily with eigenvalue equal to λ max. Then every point on the straight line joining Y and Y would also be an eigenvector. But such a line would contain boundary points of, contradicting the assertion that any such eigenvector must be strictly positive. This proves A.6 in the transitive case. Proof in the non-transitive case. We will give a constructive procedure for reducing the non-transitive case to the transitive one. The statement of A.6 is trivially true when n = 1, so we may assume that A is a non-transitive n n matrix with n 2. Then there exist indices i j so that the (i, j)-th entry a (k) ij of A k is zero for all k. Let us say that the index j is accessible from i if a (k) > 0 for some k > 0. Evidently this is a transitive relation between indices. After conjugating A by a permutation matrix, we may assume that i = 1, and that all of the indices which are accessible from i are listed first, with all of the remaining indices at the end. It follows that A can be written as a block matrix of the form [ P ] 0 Q R where P and R are square matrices. Here we may assume by induction on n that Theorem A.5 is true for P and R. Since the eigenvalues of such a block matrix are just the eigenvalues of P together with the eigenvalues of R, it certainly follows that the spectral radius λ max is itself an eigenvalue. To complete the proof, we must solve the equation [ ] [ ] [ ] P 0 Y Y Q R Y = λ Y, with λ = λ max, and with Y and Y non-negative. If λ is an eigenvalue of R, we can simply take Y to be the appropriate eigenvector for R, with Y = 0. Otherwise, we can assume that the spectral radius of R is strictly less than λ = λ max. Let P Y = λy be a non-negative eigenvalue for P. Then we must find a non-negative vector Y which satisfies the equation QY + RY = λy, or in other words (λi R)Y = QY or Y = (I R/λ) 1 QY/λ where I is the identity matrix of appropriate size. Since QY is certainly non-negative, it only remains to show that (I R/λ) 1 is non-negative. But according to (A : 3) this inverse matrix can be expressed as the sum of a convergentpower series (I R/λ) 1 = I + (R/λ) + (R/λ) 2 +, where all summands on the right are non-negative. This completes the proof of A.6. We conclude with two problems. Problem A-a. Iteration of Y AY. For any real or complex n n matrix A, show that there is a lower dimensional subspace V in the vector space of n 1 column vectors so that, for any fixed Y V. we have λ max = lim k Ak Y 1/k. (To prove this in the complex case, it is convenient to conjugate A to an appropriate normal form. It is then necessary to check that the real case follows.) A-5 ij,
6 APPENDIX Problem A-b. Simplicity of the largest eigenvector, and computation. Suppose that A is an n n matrix with a left eigenvector ˆXA = λ ˆX and a right eigenvector AŶ = µŷ such that ˆXŶ = 1. Show that λ = µ. If λ 0, then setting B = A/λ we have ˆXB = ˆX and BŶ = Ŷ with ˆXŶ = 1. If ˆX is the hyperplane consisting of all column vectors Y with ˆXY = 0, show that the space of all column vectors splits as a dirct sum n = ˆX Ŷ. Show that the linear map Y BY carries the hyperplane ˆX into itself and fixes each point of the eigenspace Ŷ. Show that every orbit under this map Y BY converges towards some vector in Ŷ if and only if all but one of the n eigenvalues of B belong to the open unit disk, λ j < 1. Now suppose that A is strictly positive. Let be the simplex consisting of all nonnegative Y with ˆXY = 1 Show that the associated transformation Y BY = AY/ λ max carries into its interior. Conclude that all orbits in converge to Ŷ. For example we can prove this by constructing a kind of norm function N on which is linear on each line segment joining Ŷ to the boundary of, taking the value 0 at Ŷ and the value +1 on. It is then easy to check that N(BY ) cn(y ) for some constant c < 1. Conclude that all but one of the n eigenvalues of A must satisfy λ j < λ max. Next suppose only that T is a non-negative transitive matrix. Applying the discussion above to the strictly positive matrix A = T + T T n 1, we see that all but one of the eigenvalues of A satisfies λ j < λ max. Conclude that the spectral radius of T must be a simple eigenvalue. That is, all but one of the eigenvalues of T must be different from the spectral radius of T. Show that the positive eigenvector for a transitive matrix T can always be located by iterating the associated map Y AY/ AY, or even the map Y (I + T )Y/ (I + T )Y. A-6
### Spectral radius, symmetric and positive matrices
Spectral radius, symmetric and positive matrices Zdeněk Dvořák April 28, 2016 1 Spectral radius Definition 1. The spectral radius of a square matrix A is ρ(a) = max{ λ : λ is an eigenvalue of A}. For an
### FIXED POINT ITERATIONS
FIXED POINT ITERATIONS MARKUS GRASMAIR 1. Fixed Point Iteration for Non-linear Equations Our goal is the solution of an equation (1) F (x) = 0, where F : R n R n is a continuous vector valued mapping in
### Optimization Theory. A Concise Introduction. Jiongmin Yong
October 11, 017 16:5 ws-book9x6 Book Title Optimization Theory 017-08-Lecture Notes page 1 1 Optimization Theory A Concise Introduction Jiongmin Yong Optimization Theory 017-08-Lecture Notes page Optimization
### NONCOMMUTATIVE POLYNOMIAL EQUATIONS. Edward S. Letzter. Introduction
NONCOMMUTATIVE POLYNOMIAL EQUATIONS Edward S Letzter Introduction My aim in these notes is twofold: First, to briefly review some linear algebra Second, to provide you with some new tools and techniques
### AN ELEMENTARY PROOF OF THE SPECTRAL RADIUS FORMULA FOR MATRICES
AN ELEMENTARY PROOF OF THE SPECTRAL RADIUS FORMULA FOR MATRICES JOEL A. TROPP Abstract. We present an elementary proof that the spectral radius of a matrix A may be obtained using the formula ρ(a) lim
### The Eigenvalue Problem: Perturbation Theory
Jim Lambers MAT 610 Summer Session 2009-10 Lecture 13 Notes These notes correspond to Sections 7.2 and 8.1 in the text. The Eigenvalue Problem: Perturbation Theory The Unsymmetric Eigenvalue Problem Just
### w T 1 w T 2. w T n 0 if i j 1 if i = j
Lyapunov Operator Let A F n n be given, and define a linear operator L A : C n n C n n as L A (X) := A X + XA Suppose A is diagonalizable (what follows can be generalized even if this is not possible -
### A Perron-type theorem on the principal eigenvalue of nonsymmetric elliptic operators
A Perron-type theorem on the principal eigenvalue of nonsymmetric elliptic operators Lei Ni And I cherish more than anything else the Analogies, my most trustworthy masters. They know all the secrets of
### Computational math: Assignment 1
Computational math: Assignment 1 Thanks Ting Gao for her Latex file 11 Let B be a 4 4 matrix to which we apply the following operations: 1double column 1, halve row 3, 3add row 3 to row 1, 4interchange
### In English, this means that if we travel on a straight line between any two points in C, then we never leave C.
Convex sets In this section, we will be introduced to some of the mathematical fundamentals of convex sets. In order to motivate some of the definitions, we will look at the closest point problem from
### Invertibility and stability. Irreducibly diagonally dominant. Invertibility and stability, stronger result. Reducible matrices
Geršgorin circles Lecture 8: Outline Chapter 6 + Appendix D: Location and perturbation of eigenvalues Some other results on perturbed eigenvalue problems Chapter 8: Nonnegative matrices Geršgorin s Thm:
### NATIONAL BOARD FOR HIGHER MATHEMATICS. M. A. and M.Sc. Scholarship Test. September 25, Time Allowed: 150 Minutes Maximum Marks: 30
NATIONAL BOARD FOR HIGHER MATHEMATICS M. A. and M.Sc. Scholarship Test September 25, 2010 Time Allowed: 150 Minutes Maximum Marks: 30 Please read, carefully, the instructions on the following page 1 INSTRUCTIONS
### 4. Linear transformations as a vector space 17
4 Linear transformations as a vector space 17 d) 1 2 0 0 1 2 0 0 1 0 0 0 1 2 3 4 32 Let a linear transformation in R 2 be the reflection in the line = x 2 Find its matrix 33 For each linear transformation
### Convexity in R n. The following lemma will be needed in a while. Lemma 1 Let x E, u R n. If τ I(x, u), τ 0, define. f(x + τu) f(x). τ.
Convexity in R n Let E be a convex subset of R n. A function f : E (, ] is convex iff f(tx + (1 t)y) (1 t)f(x) + tf(y) x, y E, t [0, 1]. A similar definition holds in any vector space. A topology is needed
### 4. Ergodicity and mixing
4. Ergodicity and mixing 4. Introduction In the previous lecture we defined what is meant by an invariant measure. In this lecture, we define what is meant by an ergodic measure. The primary motivation
### Symmetric Matrices and Eigendecomposition
Symmetric Matrices and Eigendecomposition Robert M. Freund January, 2014 c 2014 Massachusetts Institute of Technology. All rights reserved. 1 2 1 Symmetric Matrices and Convexity of Quadratic Functions
### Jordan Normal Form. Chapter Minimal Polynomials
Chapter 8 Jordan Normal Form 81 Minimal Polynomials Recall p A (x) =det(xi A) is called the characteristic polynomial of the matrix A Theorem 811 Let A M n Then there exists a unique monic polynomial q
### Linear Algebra Massoud Malek
CSUEB Linear Algebra Massoud Malek Inner Product and Normed Space In all that follows, the n n identity matrix is denoted by I n, the n n zero matrix by Z n, and the zero vector by θ n An inner product
### Properties of Matrices and Operations on Matrices
Properties of Matrices and Operations on Matrices A common data structure for statistical analysis is a rectangular array or matris. Rows represent individual observational units, or just observations,
### 18.S34 linear algebra problems (2007)
18.S34 linear algebra problems (2007) Useful ideas for evaluating determinants 1. Row reduction, expanding by minors, or combinations thereof; sometimes these are useful in combination with an induction
### Topics in linear algebra
Chapter 6 Topics in linear algebra 6.1 Change of basis I want to remind you of one of the basic ideas in linear algebra: change of basis. Let F be a field, V and W be finite dimensional vector spaces over
### LECTURE 12 A MODEL OF LIQUIDITY PREFERENCE (CONCLUDED)
LECTURE 12 A MODEL OF LIQUIDITY PREFERENCE (CONCLUDED) JACOB T. SCHWARTZ EDITED BY KENNETH R. DRIESSEL Abstract. We prove Theorem 11.6 and Theorem 11.9 of the previous lecture. 1. Proof of Theorem 11.6
### Lecture Notes: Eigenvalues and Eigenvectors. 1 Definitions. 2 Finding All Eigenvalues
Lecture Notes: Eigenvalues and Eigenvectors Yufei Tao Department of Computer Science and Engineering Chinese University of Hong Kong taoyf@cse.cuhk.edu.hk 1 Definitions Let A be an n n matrix. If there
### Mathematical Methods wk 2: Linear Operators
John Magorrian, magog@thphysoxacuk These are work-in-progress notes for the second-year course on mathematical methods The most up-to-date version is available from http://www-thphysphysicsoxacuk/people/johnmagorrian/mm
### Math Camp Lecture 4: Linear Algebra. Xiao Yu Wang. Aug 2010 MIT. Xiao Yu Wang (MIT) Math Camp /10 1 / 88
Math Camp 2010 Lecture 4: Linear Algebra Xiao Yu Wang MIT Aug 2010 Xiao Yu Wang (MIT) Math Camp 2010 08/10 1 / 88 Linear Algebra Game Plan Vector Spaces Linear Transformations and Matrices Determinant
### Chapter 7 Iterative Techniques in Matrix Algebra
Chapter 7 Iterative Techniques in Matrix Algebra Per-Olof Persson persson@berkeley.edu Department of Mathematics University of California, Berkeley Math 128B Numerical Analysis Vector Norms Definition
### Some notes on Coxeter groups
Some notes on Coxeter groups Brooks Roberts November 28, 2017 CONTENTS 1 Contents 1 Sources 2 2 Reflections 3 3 The orthogonal group 7 4 Finite subgroups in two dimensions 9 5 Finite subgroups in three
### A proof of the Jordan normal form theorem
A proof of the Jordan normal form theorem Jordan normal form theorem states that any matrix is similar to a blockdiagonal matrix with Jordan blocks on the diagonal. To prove it, we first reformulate it
### Scaling of Symmetric Matrices by Positive Diagonal Congruence
Scaling of Symmetric Matrices by Positive Diagonal Congruence Abstract Charles R. Johnson a and Robert Reams b a Department of Mathematics, College of William and Mary, P.O. Box 8795, Williamsburg, VA
### SPRING 2006 PRELIMINARY EXAMINATION SOLUTIONS
SPRING 006 PRELIMINARY EXAMINATION SOLUTIONS 1A. Let G be the subgroup of the free abelian group Z 4 consisting of all integer vectors (x, y, z, w) such that x + 3y + 5z + 7w = 0. (a) Determine a linearly
### Eigenvalues and Eigenvectors
Chapter 1 Eigenvalues and Eigenvectors Among problems in numerical linear algebra, the determination of the eigenvalues and eigenvectors of matrices is second in importance only to the solution of linear
### Problem 1A. Suppose that f is a continuous real function on [0, 1]. Prove that
Problem 1A. Suppose that f is a continuous real function on [, 1]. Prove that lim α α + x α 1 f(x)dx = f(). Solution: This is obvious for f a constant, so by subtracting f() from both sides we can assume
### Symmetric matrices and dot products
Symmetric matrices and dot products Proposition An n n matrix A is symmetric iff, for all x, y in R n, (Ax) y = x (Ay). Proof. If A is symmetric, then (Ax) y = x T A T y = x T Ay = x (Ay). If equality
### Ir O D = D = ( ) Section 2.6 Example 1. (Bottom of page 119) dim(v ) = dim(l(v, W )) = dim(v ) dim(f ) = dim(v )
Section 3.2 Theorem 3.6. Let A be an m n matrix of rank r. Then r m, r n, and, by means of a finite number of elementary row and column operations, A can be transformed into the matrix ( ) Ir O D = 1 O
### Real symmetric matrices/1. 1 Eigenvalues and eigenvectors
Real symmetric matrices 1 Eigenvalues and eigenvectors We use the convention that vectors are row vectors and matrices act on the right. Let A be a square matrix with entries in a field F; suppose that
### On the mathematical background of Google PageRank algorithm
Working Paper Series Department of Economics University of Verona On the mathematical background of Google PageRank algorithm Alberto Peretti, Alberto Roveda WP Number: 25 December 2014 ISSN: 2036-2919
### JUST THE MATHS UNIT NUMBER 9.8. MATRICES 8 (Characteristic properties) & (Similarity transformations) A.J.Hobson
JUST THE MATHS UNIT NUMBER 9.8 MATRICES 8 (Characteristic properties) & (Similarity transformations) by A.J.Hobson 9.8. Properties of eigenvalues and eigenvectors 9.8. Similar matrices 9.8.3 Exercises
### IMPORTANT DEFINITIONS AND THEOREMS REFERENCE SHEET
IMPORTANT DEFINITIONS AND THEOREMS REFERENCE SHEET This is a (not quite comprehensive) list of definitions and theorems given in Math 1553. Pay particular attention to the ones in red. Study Tip For each
### 2. Metric Spaces. 2.1 Definitions etc.
2. Metric Spaces 2.1 Definitions etc. The procedure in Section for regarding R as a topological space may be generalized to many other sets in which there is some kind of distance (formally, sets with
### General Convexity, General Concavity, Fixed Points, Geometry, and Min-Max Points
Mathematica Moravica Vol. 12-1 2008), 63 109 General Convexity, General Concavity, Fixed Points, Geometry, and Min-Max Points Milan R. Tasković Abstract. This paper continues the study of general convexity
### Spectrally arbitrary star sign patterns
Linear Algebra and its Applications 400 (2005) 99 119 wwwelseviercom/locate/laa Spectrally arbitrary star sign patterns G MacGillivray, RM Tifenbach, P van den Driessche Department of Mathematics and Statistics,
### The third smallest eigenvalue of the Laplacian matrix
Electronic Journal of Linear Algebra Volume 8 ELA Volume 8 (001) Article 11 001 The third smallest eigenvalue of the Laplacian matrix Sukanta Pati pati@iitg.ernet.in Follow this and additional works at:
MTH50 Spring 07 HW Assignment 7 {From [FIS0]}: Sec 44 #4a h 6; Sec 5 #ad ac 4ae 4 7 The due date for this assignment is 04/05/7 Sec 44 #4a h Evaluate the erminant of the following matrices by any legitimate
IMC 05, Blagoevgrad, Bulgaria Day, July 9, 05 Problem. For any integer n and two n n matrices with real entries, B that satisfy the equation + B ( + B prove that det( det(b. Does the same conclusion follow
### Up to this point, our main theoretical tools for finding eigenvalues without using det{a λi} = 0 have been the trace and determinant formulas
Finding Eigenvalues Up to this point, our main theoretical tools for finding eigenvalues without using det{a λi} = 0 have been the trace and determinant formulas plus the facts that det{a} = λ λ λ n, Tr{A}
### A Note on Eigenvalues of Perturbed Hermitian Matrices
A Note on Eigenvalues of Perturbed Hermitian Matrices Chi-Kwong Li Ren-Cang Li July 2004 Let ( H1 E A = E H 2 Abstract and à = ( H1 H 2 be Hermitian matrices with eigenvalues λ 1 λ k and λ 1 λ k, respectively.
### Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University
Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University February 7, 2007 2 Contents 1 Metric Spaces 1 1.1 Basic definitions...........................
### 1 Topology Definition of a topology Basis (Base) of a topology The subspace topology & the product topology on X Y 3
Index Page 1 Topology 2 1.1 Definition of a topology 2 1.2 Basis (Base) of a topology 2 1.3 The subspace topology & the product topology on X Y 3 1.4 Basic topology concepts: limit points, closed sets,
### Chapter Two Elements of Linear Algebra
Chapter Two Elements of Linear Algebra Previously, in chapter one, we have considered single first order differential equations involving a single unknown function. In the next chapter we will begin to
### 1.10 Matrix Representation of Graphs
42 Basic Concepts of Graphs 1.10 Matrix Representation of Graphs Definitions: In this section, we introduce two kinds of matrix representations of a graph, that is, the adjacency matrix and incidence matrix
### Convex Sets. Prof. Dan A. Simovici UMB
Convex Sets Prof. Dan A. Simovici UMB 1 / 57 Outline 1 Closures, Interiors, Borders of Sets in R n 2 Segments and Convex Sets 3 Properties of the Class of Convex Sets 4 Closure and Interior Points of Convex
### 1 Positive definiteness and semidefiniteness
Positive definiteness and semidefiniteness Zdeněk Dvořák May 9, 205 For integers a, b, and c, let D(a, b, c) be the diagonal matrix with + for i =,..., a, D i,i = for i = a +,..., a + b,. 0 for i = a +
### Generalizations of M-matrices which may not have a nonnegative inverse
Available online at www.sciencedirect.com Linear Algebra and its Applications 429 (2008) 2435 2450 www.elsevier.com/locate/laa Generalizations of M-matrices which may not have a nonnegative inverse Abed
### 1 T 1 = where 1 is the all-ones vector. For the upper bound, let v 1 be the eigenvector corresponding. u:(u,v) E v 1(u)
CME 305: Discrete Mathematics and Algorithms Instructor: Reza Zadeh (rezab@stanford.edu) Final Review Session 03/20/17 1. Let G = (V, E) be an unweighted, undirected graph. Let λ 1 be the maximum eigenvalue
### Competitive and Cooperative Differential Equations
Chapter 3 Competitive and Cooperative Differential Equations 0. Introduction This chapter and the next one focus on ordinary differential equations in IR n. A natural partial ordering on IR n is generated
### Chapter 3. Matrices. 3.1 Matrices
40 Chapter 3 Matrices 3.1 Matrices Definition 3.1 Matrix) A matrix A is a rectangular array of m n real numbers {a ij } written as a 11 a 12 a 1n a 21 a 22 a 2n A =.... a m1 a m2 a mn The array has m rows
### Stat 159/259: Linear Algebra Notes
Stat 159/259: Linear Algebra Notes Jarrod Millman November 16, 2015 Abstract These notes assume you ve taken a semester of undergraduate linear algebra. In particular, I assume you are familiar with the
### c 1995 Society for Industrial and Applied Mathematics Vol. 37, No. 1, pp , March
SIAM REVIEW. c 1995 Society for Industrial and Applied Mathematics Vol. 37, No. 1, pp. 93 97, March 1995 008 A UNIFIED PROOF FOR THE CONVERGENCE OF JACOBI AND GAUSS-SEIDEL METHODS * ROBERTO BAGNARA Abstract.
### Affine iterations on nonnegative vectors
Affine iterations on nonnegative vectors V. Blondel L. Ninove P. Van Dooren CESAME Université catholique de Louvain Av. G. Lemaître 4 B-348 Louvain-la-Neuve Belgium Introduction In this paper we consider
### Epilogue: Quivers. Gabriel s Theorem
Epilogue: Quivers Gabriel s Theorem A figure consisting of several points connected by edges is called a graph. More precisely, a graph is a purely combinatorial object, which is considered given, if a
### Linear Algebra Primer
Linear Algebra Primer David Doria daviddoria@gmail.com Wednesday 3 rd December, 2008 Contents Why is it called Linear Algebra? 4 2 What is a Matrix? 4 2. Input and Output.....................................
### BASIC ALGORITHMS IN LINEAR ALGEBRA. Matrices and Applications of Gaussian Elimination. A 2 x. A T m x. A 1 x A T 1. A m x
BASIC ALGORITHMS IN LINEAR ALGEBRA STEVEN DALE CUTKOSKY Matrices and Applications of Gaussian Elimination Systems of Equations Suppose that A is an n n matrix with coefficents in a field F, and x = (x,,
### Chapter 1. Optimality Conditions: Unconstrained Optimization. 1.1 Differentiable Problems
Chapter 1 Optimality Conditions: Unconstrained Optimization 1.1 Differentiable Problems Consider the problem of minimizing the function f : R n R where f is twice continuously differentiable on R n : P
### NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA4247 Complex Analysis II Lecture Notes Part II
NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA4247 Complex Analysis II Lecture Notes Part II Chapter 2 Further properties of analytic functions 21 Local/Global behavior of analytic functions;
### Jordan normal form notes (version date: 11/21/07)
Jordan normal form notes (version date: /2/7) If A has an eigenbasis {u,, u n }, ie a basis made up of eigenvectors, so that Au j = λ j u j, then A is diagonal with respect to that basis To see this, let
LECTURE NOTES AMRITANSHU PRASAD Let K be a field. 1. Basic definitions Definition 1.1. A K-algebra is a K-vector space together with an associative product A A A which is K-linear, with respect to which
### PROBLEMS. (b) (Polarization Identity) Show that in any inner product space
1 Professor Carl Cowen Math 54600 Fall 09 PROBLEMS 1. (Geometry in Inner Product Spaces) (a) (Parallelogram Law) Show that in any inner product space x + y 2 + x y 2 = 2( x 2 + y 2 ). (b) (Polarization
### (January 14, 2009) q n 1 q d 1. D = q n = q + d
(January 14, 2009) [10.1] Prove that a finite division ring D (a not-necessarily commutative ring with 1 in which any non-zero element has a multiplicative inverse) is commutative. (This is due to Wedderburn.)
### Continuity of convex functions in normed spaces
Continuity of convex functions in normed spaces In this chapter, we consider continuity properties of real-valued convex functions defined on open convex sets in normed spaces. Recall that every infinitedimensional
### Topics in Tensors III Nonnegative tensors
Topics in Tensors III Nonnegative tensors A Summer School by Shmuel Friedland 1 July 6-8, 2011 given in Department of Mathematics University of Coimbra, Portugal 1 Department of Mathematics, Statistics
### Math 443/543 Graph Theory Notes 5: Graphs as matrices, spectral graph theory, and PageRank
Math 443/543 Graph Theory Notes 5: Graphs as matrices, spectral graph theory, and PageRank David Glickenstein November 3, 4 Representing graphs as matrices It will sometimes be useful to represent graphs
### 2016 EF Exam Texas A&M High School Students Contest Solutions October 22, 2016
6 EF Exam Texas A&M High School Students Contest Solutions October, 6. Assume that p and q are real numbers such that the polynomial x + is divisible by x + px + q. Find q. p Answer Solution (without knowledge
### The Symmetric Space for SL n (R)
The Symmetric Space for SL n (R) Rich Schwartz November 27, 2013 The purpose of these notes is to discuss the symmetric space X on which SL n (R) acts. Here, as usual, SL n (R) denotes the group of n n
### Most Continuous Functions are Nowhere Differentiable
Most Continuous Functions are Nowhere Differentiable Spring 2004 The Space of Continuous Functions Let K = [0, 1] and let C(K) be the set of all continuous functions f : K R. Definition 1 For f C(K) we
### The proximal mapping
The proximal mapping http://bicmr.pku.edu.cn/~wenzw/opt-2016-fall.html Acknowledgement: this slides is based on Prof. Lieven Vandenberghes lecture notes Outline 2/37 1 closed function 2 Conjugate function
### A matrix is a rectangular array of. objects arranged in rows and columns. The objects are called the entries. is called the size of the matrix, and
Section 5.5. Matrices and Vectors A matrix is a rectangular array of objects arranged in rows and columns. The objects are called the entries. A matrix with m rows and n columns is called an m n matrix.
### Diagonalisierung. Eigenwerte, Eigenvektoren, Mathematische Methoden der Physik I. Vorlesungsnotizen zu
Eigenwerte, Eigenvektoren, Diagonalisierung Vorlesungsnotizen zu Mathematische Methoden der Physik I J. Mark Heinzle Gravitational Physics, Faculty of Physics University of Vienna Version /6/29 2 version
### NOTES (1) FOR MATH 375, FALL 2012
NOTES 1) FOR MATH 375, FALL 2012 1 Vector Spaces 11 Axioms Linear algebra grows out of the problem of solving simultaneous systems of linear equations such as 3x + 2y = 5, 111) x 3y = 9, or 2x + 3y z =
### HOMEWORK PROBLEMS FROM STRANG S LINEAR ALGEBRA AND ITS APPLICATIONS (4TH EDITION)
HOMEWORK PROBLEMS FROM STRANG S LINEAR ALGEBRA AND ITS APPLICATIONS (4TH EDITION) PROFESSOR STEVEN MILLER: BROWN UNIVERSITY: SPRING 2007 1. CHAPTER 1: MATRICES AND GAUSSIAN ELIMINATION Page 9, # 3: Describe
### Majorizations for the Eigenvectors of Graph-Adjacency Matrices: A Tool for Complex Network Design
Majorizations for the Eigenvectors of Graph-Adjacency Matrices: A Tool for Complex Network Design Rahul Dhal Electrical Engineering and Computer Science Washington State University Pullman, WA rdhal@eecs.wsu.edu
### CHUN-HUA GUO. Key words. matrix equations, minimal nonnegative solution, Markov chains, cyclic reduction, iterative methods, convergence rate
CONVERGENCE ANALYSIS OF THE LATOUCHE-RAMASWAMI ALGORITHM FOR NULL RECURRENT QUASI-BIRTH-DEATH PROCESSES CHUN-HUA GUO Abstract The minimal nonnegative solution G of the matrix equation G = A 0 + A 1 G +
### Tangent spaces, normals and extrema
Chapter 3 Tangent spaces, normals and extrema If S is a surface in 3-space, with a point a S where S looks smooth, i.e., without any fold or cusp or self-crossing, we can intuitively define the tangent
### The Nearest Doubly Stochastic Matrix to a Real Matrix with the same First Moment
he Nearest Doubly Stochastic Matrix to a Real Matrix with the same First Moment William Glunt 1, homas L. Hayden 2 and Robert Reams 2 1 Department of Mathematics and Computer Science, Austin Peay State
### (x 1, y 1 ) = (x 2, y 2 ) if and only if x 1 = x 2 and y 1 = y 2.
1. Complex numbers A complex number z is defined as an ordered pair z = (x, y), where x and y are a pair of real numbers. In usual notation, we write z = x + iy, where i is a symbol. The operations of
### HW1 solutions. 1. α Ef(x) β, where Ef(x) is the expected value of f(x), i.e., Ef(x) = n. i=1 p if(a i ). (The function f : R R is given.
HW1 solutions Exercise 1 (Some sets of probability distributions.) Let x be a real-valued random variable with Prob(x = a i ) = p i, i = 1,..., n, where a 1 < a 2 < < a n. Of course p R n lies in the standard
### Lecture 11: Eigenvalues and Eigenvectors
Lecture : Eigenvalues and Eigenvectors De nition.. Let A be a square matrix (or linear transformation). A number λ is called an eigenvalue of A if there exists a non-zero vector u such that A u λ u. ()
### Weierstraß-Institut. für Angewandte Analysis und Stochastik. im Forschungsverbund Berlin e.v. Preprint ISSN
Weierstraß-Institut für Angewandte Analysis und Stochastik im Forschungsverbund Berlin e.v. Preprint ISSN 0946 8633 Proof of a Counterexample to the Finiteness Conjecture in the Spirit of the Theory of
### 5 Quiver Representations
5 Quiver Representations 5. Problems Problem 5.. Field embeddings. Recall that k(y,..., y m ) denotes the field of rational functions of y,..., y m over a field k. Let f : k[x,..., x n ] k(y,..., y m )
### Real Analysis Problems
Real Analysis Problems Cristian E. Gutiérrez September 14, 29 1 1 CONTINUITY 1 Continuity Problem 1.1 Let r n be the sequence of rational numbers and Prove that f(x) = 1. f is continuous on the irrationals.
### Integral Similarity and Commutators of Integral Matrices Thomas J. Laffey and Robert Reams (University College Dublin, Ireland) Abstract
Integral Similarity and Commutators of Integral Matrices Thomas J. Laffey and Robert Reams (University College Dublin, Ireland) Abstract Let F be a field, M n (F ) the algebra of n n matrices over F and
### Problem set 1, Real Analysis I, Spring, 2015.
Problem set 1, Real Analysis I, Spring, 015. (1) Let f n : D R be a sequence of functions with domain D R n. Recall that f n f uniformly if and only if for all ɛ > 0, there is an N = N(ɛ) so that if n
### Math 123 Homework Assignment #2 Due Monday, April 21, 2008
Math 123 Homework Assignment #2 Due Monday, April 21, 2008 Part I: 1. Suppose that A is a C -algebra. (a) Suppose that e A satisfies xe = x for all x A. Show that e = e and that e = 1. Conclude that e
### Linear Algebra, Theory And Applications. Kenneth Kuttler
Linear Algebra, Theory And Applications Kenneth Kuttler September 3, 26 2 Contents Preliminaries Sets And Set Notation 2 Functions 2 3 The Number Line And Algebra Of The Real Numbers 2 4 Ordered fields
### Boolean Inner-Product Spaces and Boolean Matrices
Boolean Inner-Product Spaces and Boolean Matrices Stan Gudder Department of Mathematics, University of Denver, Denver CO 80208 Frédéric Latrémolière Department of Mathematics, University of Denver, Denver
### System theory and system identification of compartmental systems Hof, Jacoba Marchiena van den
University of Groningen System theory and system identification of compartmental systems Hof, Jacoba Marchiena van den IMPORTANT NOTE: You are advised to consult the publisher's version (publisher's PDF)
### THE MINIMAL POLYNOMIAL AND SOME APPLICATIONS
THE MINIMAL POLYNOMIAL AND SOME APPLICATIONS KEITH CONRAD. Introduction The easiest matrices to compute with are the diagonal ones. The sum and product of diagonal matrices can be computed componentwise
### . The following is a 3 3 orthogonal matrix: 2/3 1/3 2/3 2/3 2/3 1/3 1/3 2/3 2/3
Lecture Notes: Orthogonal and Symmetric Matrices Yufei Tao Department of Computer Science and Engineering Chinese University of Hong Kong taoyf@cse.cuhk.edu.hk Orthogonal Matrix Definition. An n n matrix
### Optimality, Duality, Complementarity for Constrained Optimization
Optimality, Duality, Complementarity for Constrained Optimization Stephen Wright University of Wisconsin-Madison May 2014 Wright (UW-Madison) Optimality, Duality, Complementarity May 2014 1 / 41 Linear | 10,855 | 39,956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2019-04 | longest | en | 0.847703 |
https://www.construct.net/en/forum/construct-2/how-do-i-18/mathematics-waves-sin-cos-angl-68675 | 1,539,864,580,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511806.8/warc/CC-MAIN-20181018105742-20181018131242-00466.warc.gz | 906,272,461 | 14,271 | # Mathematics, Waves, Sin, Cos, Angles and HTML5
0 Favourites
• 19 posts
• Hi Everybody,
I ask for the communty help, especially to those stronger in mathematics to solve a problem:
I've making tests in a program that should be able to represent in real time a sinus wave depending on a slider value. The idea is to have something similar to a Osciloscope made with a canvas plugin.
https://dl.dropboxusercontent.com/u/87172782/test.JPG
So that I try to do so and, as you can see in the example, I'm just near to pick up a solution, but not completely. The line represented in the canvas is not responding to the changes made on the slider. It only takes the initial values of X and Y.
The aim would be to make possible to represent a y=cos(x) for example, where X is given by the slider.
Here you would find an image and a example.
https://dl.dropboxusercontent.com/u/87172782/test.capx
Any idea or comment? Is my approach to the solution a good one? Perhaps it is not possible to be done with a canvas plugin?
• If you try to do something similar to this, it is totally possible and relatively easy as well. The link leads to an app I once made with CC, but the graphic representation was done with the canvas plugin, and nothing else was used than "line to". Both is also available in C2 using ROJOhound's Canvas plugin for C2.
The basic approach of drawing a waveform (be it a sinus or anything else) is to split it into segments, tiny parts of the complete wave. Those segments represent the resolution, the wave is drawn at.
If you are not afraid of installing CC, I could give you the source the the link above. But maybe somebody else already working on a demo for C2
• You are setting Y to a constant: 4*cos(3.141516)^2
It needs to be a function of X, something like: 4*cos(X * _yourfrequencyvalue_)^2
• wouldn't using the curve feature in the canvas plugin be easier?
• I haven't loaded your example but if blackhornet copied from your event:??Doesn't Construct 2 use degree instead of radian in sin()???Also better use pi instead of 3.141516
• Wow! That too much of course. But this is the kind of control of the waves that I seek in my app.
What I do is
Every tick
-----> Loop For 1 to BoxBound : Drawing the points.
The problem is this only represents the first values, no changes.
• Y is a constant in the example.
Oh, I'm going to change this inmediately
• wouldn't using the curve feature in the canvas plugin be easier?
I have never used it before, I'll check it out
• ## Try Construct 3
Develop games in your browser. Powerful, performant & highly capable.
Construct 3 users don't see these ads
• Here you can find the project in it's last state:
As you can see, the problem is that it seems no way to represent the changes in real time
• easy with the canvas plugin:
<img src="http://i.imgur.com/3MvPTqe.png" border="0" />
<img src="http://i.imgur.com/fNVeCj6.png" border="0" />
• Thank you mindfaQ
That is exactly what I was talking about.
@Heptagono
mindfaQ's example might need some explanation for your understanding. 'Amplitude' is Y and 'Period' (or Frequency) is what you called 'distance pp' in your graphic.
• Wonderful solution!
Now I can see that the appoach that I was using was not the best way to pick up a proper result. I'm goint to test It and tell you how it has been.
Thanks and cheers!
• I probably use too many points with rising new_period. I think canvas.width*newperiod/4 or even /8 will be enough and stay fast enough if the new_period number gets large. Also what I called period in my example is the frequency instead of period in (sinwaves/360 pixel) as unit.
and instead of adding sin to y, substract it, as it should point upwards when sin is 1, not downwards
• mindfaQ, there is something that I want to ask you.
Can I use an Amplitude and a Period value to draw a wave with the code you have showed me?
• Not sure what you mean with that. The code already does that, no? The slider is not a necessity. | 981 | 3,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-43 | latest | en | 0.93392 |
https://www.physicsforums.com/threads/green-function-question.381283/ | 1,606,646,612,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00689.warc.gz | 819,333,207 | 14,821 | # Green function question
I was reading up on Classical Mechanics and the general method for solving for an undamped harmonic oscillator was given as
$$\frac{d^{2}q}{dt^{2}} + \omega^{2}q = F(t)$$
was solved using the Green function, G, to the equation
$$\frac{d^{2}G}{dt^{2}} + \omega^{2}q = \delta(t-t')$$
and then integrating via the normal procedure.
Then the next case considered was the damped harmonic oscillator which had a damping term proportional to $$\frac{\omega}{Q}$$ times the velocity. The equation has the form
$$\frac{d^{2}q}{dt^{2}} + \frac{\omega}{Q}\frac{dq}{dt} + \omega^{2}q = F(t)$$
Now, the author wants to find the particular solution to this equation and says that his Green function is of the form
G = $$A e^{\frac{{\omega(t-t')}}{2Q} + i\omega(t-t')}$$ plus this term's complex conjugate. This is a solution to STEADY STATE ONLY but does not consider the transient part.
BUT, he also claims that the above Green function is a solution the equation
$$\frac{d^{2}G}{dt^{2}} + \frac{\omega}{Q}\frac{dG}{dt} + \omega^{2}G = \delta(t-t')$$
How can he do this? The Green function which will solve the above equation will also have a transient part apart from the steady state function which he has considered. G will be of the form above PLUS a dying out part. What happens to that?
Last edited:
## Answers and Replies
Related Differential Equations News on Phys.org | 386 | 1,404 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-50 | latest | en | 0.915492 |
https://www.dcode.fr/alphabetic-transcription | 1,726,309,250,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00141.warc.gz | 663,100,848 | 6,824 | Search for a tool
Alphabetic Transcription
Tool to convert text composed of symbols into letters of the alphabet (Transcription usually for the purpose of alphabetical substitutions).
Results
Alphabetic Transcription -
Tag(s) : Substitution Cipher
Share
dCode and more
dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!
A suggestion ? a feedback ? a bug ? an idea ? Write to dCode!
Please, check our dCode Discord community for help requests!
NB: for encrypted messages, test our automatic cipher identifier!
Feedback and suggestions are welcome so that dCode offers the best 'Alphabetic Transcription' tool for free! Thank you!
# Alphabetic Transcription
## Symbols to Letter Conversion
Spaces andLine return are Removed Kept as is Treated as character to substitute
Substitution alphabet Classic (ABCDEF…) Match ngrams with English Frequencies Random
### What is alphabetical transcription? (Definition)
Alphabetical transcription is the process of converting text composed of symbols into letters of the alphabet.
### How to transcribe a message into letters?
To transcribe a message into letters, the user must associate each symbol in the message with a letter of the alphabet using a correspondence table.
To create a correspondence table, the user can draw up a list of symbols to be transcribed and assign them a letter of the alphabet in front of each symbol.
To go faster, identify the distinct characters (or ngrams) in the text, and associate to each one, a letter of the alphabet from A to Z taking care not to associate the same letter several times with a symbol and not to associate several symbols with the same letter.
Example: &%##& will be transcribed into ABCCA
If there are more than 26 distinct characters (or ngrams), possibly use the 10 digits to obtain up to 36 characters, and if there are more than 36 characters, it is still possible to go up to 62 by distinguishing between upper and lower case.
### Why transcribe a message in letters?
The transcription of the messages in letters of the Latin alphabet makes it possible to preserve the properties of the frequencies of the characters of the message in order to use more easily the tools of cryptanalysis (on dCode among others) as the mono-alphabetical substitution.
It is possible that the original message is a homophonic substitution (uses several symbols for a single letter), even in this case transcription can help. This was the case for the deciphering of famous messages like that of the zodiac killer.
## Source code
dCode retains ownership of the "Alphabetic Transcription" source code. Except explicit open source licence (indicated Creative Commons / free), the "Alphabetic Transcription" algorithm, the applet or snippet (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, breaker, translator), or the "Alphabetic Transcription" functions (calculate, convert, solve, decrypt / encrypt, decipher / cipher, decode / encode, translate) written in any informatic language (Python, Java, PHP, C#, Javascript, Matlab, etc.) and all data download, script, or API access for "Alphabetic Transcription" are not public, same for offline use on PC, mobile, tablet, iPhone or Android app!
Reminder : dCode is free to use.
## Cite dCode
The copy-paste of the page "Alphabetic Transcription" or any of its results, is allowed (even for commercial purposes) as long as you credit dCode!
Exporting results as a .csv or .txt file is free by clicking on the export icon
Cite as source (bibliography):
Alphabetic Transcription on dCode.fr [online website], retrieved on 2024-09-14, https://www.dcode.fr/alphabetic-transcription
## Need Help ?
Please, check our dCode Discord community for help requests!
NB: for encrypted messages, test our automatic cipher identifier! | 840 | 3,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.871013 |
https://www.physicsforums.com/threads/can-anyone-help-me-slove-this-question.598277/ | 1,537,306,471,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155702.33/warc/CC-MAIN-20180918205149-20180918225149-00097.warc.gz | 838,681,827 | 12,256 | # Homework Help: Can anyone help me slove this question?
1. Apr 19, 2012
### dangkykodc
Calculate the terminal speed for a bacterium in water. Use the expression for the drag force in the equation
F drag = −Crv .
Assume the bacterium has a mass m = 4 10−15 kg, a drag coefficient
C = 0.02 N • s/m2 and can be approximated as a sphere of radius r = 1 μm.
a. Draw a force diagram for the bacterium. Is it in equilibrium? Why?
b.Find the speed (size of the velocity) of the bacterium
c.How long does it take a bacterium to fall from the top of a lake of depth 2.5 m to the bottom if it falls at this speed?
2. Apr 19, 2012
### Steely Dan
Hi dangkykodc, welcome to PF. The way this homework forum works is that if you show us your work and where you got stuck, we'll be happy to help you. That is why we have the template :) | 239 | 826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-39 | latest | en | 0.865745 |
http://nationalpost.com/health/seniors/grey-matters-a-little-math-can-add-up-in-your-retirement-savings | 1,529,578,353,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864139.22/warc/CC-MAIN-20180621094633-20180621114633-00227.warc.gz | 224,267,567 | 31,699 | # Grey Matters: A little math can add up in your retirement savings
Wanda Morris
When I was little, my dad used to sit my two sisters and me at the kitchen table and play the Smarties game. He would ask us questions and we’d get a Smartie when we got one right. When we got one wrong, he’d get a Smartie — or two or three. The Smarties game taught me my times tables and combined two loves that shaped my life: math and chocolate.
Because of this early positive exposure, I always saw math as fun. In my University of Calgary days when I was, admittedly, somewhat obnoxious, I took great joy in pointing out math errors in textbooks and leaving exams early to show how easy they were.
I understand that many people find no joy in math at all. That’s too bad. Not only are they missing out on the fun, but understanding math is fundamental to our financial security. To miss the first is regrettable; to miss the second can endanger your retirement.
The good news is that there is one single specific mathematical concept that anyone can learn to greatly help increase financial security: compounding.
Imagine you receive a bequest of \$100,000 and decide to invest it for 25 years then use it to fund your retirement. You don’t know how much of a return you’ll make on your windfall, but let’s say either four per cent, six per cent, eight per cent or 10 per cent.
You decide you need a financial adviser to draw up a financial plan (an excellent idea) and make and monitor your mutual fund investments. You find one you like who will do all this by investing you in mutual funds which include an embedded commission of two per cent of your assets a year (mutual fund fees typically run between 1.5 per cent and 2.5 per cent annually).
Depending on the gross return on your mutual fund investment, after 25 years, your \$100,000 would be worth:
• Mutual funds earning four per cent will grow to \$164,000
• Mutual funds earning six per cent will grow to \$267,000
• Mutual funds earning eight per cent will grow to \$429,000
• Mutual funds earning 10 per cent will grow to \$685,000
Imagine that instead you visit a fee-for-service financial planner who provides you with a financial plan; you then invest your money into index funds. Let’s assume you pay \$2,500 for your financial plan and your index fund costs you one per cent a year. You start out with \$2,500 less but pay one per cent less per year in fees every year. What happens to your investment over time? Again it depends on the return you earn. Here are the returns after 25 years on \$97,500 at different rates of return:
• Index funds earning two per cent will grow to \$204,000
• Index funds earning six per cent will grow to \$330,000
• Index funds earning eight per cent will grow to \$529,000
• Index funds earning 10 per cent will grow to \$841,000
The extra return, whether \$40,000 for an investment earning four per cent or \$100,000 for one earning eight per cent, is the compounded effect of the one per cent difference in investment fees. (Note that many exchange-traded funds have an even lower investment cost, but they typically require a minimum investment and attract trading costs, so I’ve used index funds for this example).
Now you may be thinking, nothing I’d love better than an extra \$40,000 in my retirement fund, unless of course it’s an extra \$100,000, but how do I know if this applies to my situation?
The good news is that CARP has partnered with ex-banker Larry Bates to give everyone this information through a tool he developed to help you understand how compounding affects you.
Go to carp.ca/TREX to get your own T-Rex score (an investor’s total return efficiency index) and find out how big a bite fees and commissions are taking out of your savings.
And when you’re done, why not treat yourself to some Smarties? Because math can save you money and be fun, too.
25 Years On
What’s the difference in value between an investment of \$100,000 in mutual funds (two per cent annual fees) and an investment of \$97,500 in index funds (one per cent annual fees)?
• At a four per cent return, the savings from the lower commission index funds are \$40,000.
• At a six per cent return, the savings from the lower commission index funds are \$63,000.
• At an eight per cent return, the savings from the lower commission index funds are \$100,000.
• At a 10 per cent return, the savings from the lower commission index funds are \$156,000.
Grey Matters is a weekly column by Wanda Morris, the VP of Advocacy for CARP, a 300,000 member national, non-partisan, non-profit organization that advocates for financial security, improved health-care for Canadians as we age. Missed a week? Past columns by Wanda and other key CARP contributors can be found at carp.ca/blogs
#### WatchAre the unknown victims of alleged serial killer Bruce McArthur hidden among these 22 cold cases?
The Post combed through 186 Toronto cold cases and countless newspaper reports between 1975 and 1997, searching for a pattern
#### Watch'Unthinkable, immoral act': Woman ordered to pay \$375,000 for sabotaging boyfriend's musical career
When McGill student Eric Abramovitz won a scholarship to study with a prestigious L.A. professor, Jennifer Lee began a ‘despicable’ scheme of fake emails to stop him
#### WatchHow quickly could Canada build an atomic bomb?
We’ve got the uranium, the know-how and a sudden desire to be respected by our nearest neighbour
#### Keep calm, and legislate on: Bill Blair is arguably the reason Canada is on the cusp of cannabis legalization
Former cop turned Liberal MP has the life experience to run the delicate file
This Week's Flyers | 1,267 | 5,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-26 | latest | en | 0.944861 |
https://www.convertunits.com/from/gibibyte/to/kilobit | 1,620,928,804,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991943.36/warc/CC-MAIN-20210513173321-20210513203321-00298.warc.gz | 747,223,888 | 16,756 | ## ››Convert gibibyte to kilobit
gibibyte kilobit
How many gibibyte in 1 kilobit? The answer is 1.1368683772162E-7.
We assume you are converting between gibibyte and kilobit.
You can view more details on each measurement unit:
gibibyte or kilobit
The main non-SI unit for computer data storage is the byte.
1 byte is equal to 9.3132257461548E-10 gibibyte, or 0.008192 kilobit.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between gibibytes and kilobits.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of gibibyte to kilobit
1 gibibyte to kilobit = 8796093.02221 kilobit
2 gibibyte to kilobit = 17592186.04442 kilobit
3 gibibyte to kilobit = 26388279.06662 kilobit
4 gibibyte to kilobit = 35184372.08883 kilobit
5 gibibyte to kilobit = 43980465.11104 kilobit
6 gibibyte to kilobit = 52776558.13325 kilobit
7 gibibyte to kilobit = 61572651.15546 kilobit
8 gibibyte to kilobit = 70368744.17766 kilobit
9 gibibyte to kilobit = 79164837.19987 kilobit
10 gibibyte to kilobit = 87960930.22208 kilobit
## ››Want other units?
You can do the reverse unit conversion from kilobit to gibibyte, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Gibibyte
The gibibyte (a contraction of giga binary byte) is a unit of digital information storage. It is equal to 1,024 mebibytes.
## ››Definition: Kilobit
The SI prefix "kilo" represents a factor of 103, or in exponential notation, 1E3.
So 1 kilobit = 103 bits.
The definition of a bit is as follows:
A bit is a binary digit, taking a value of either 0 or 1. The bit is also a unit of measurement, the information capacity of one binary digit. It has the symbol bit, or b. There are 8 bits in 1 byte.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 689 | 2,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-21 | latest | en | 0.703138 |
https://chemistry.stackexchange.com/questions/109962/how-much-of-ag-has-formed-on-the-cu-tile/109984#109984 | 1,632,752,027,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058450.44/warc/CC-MAIN-20210927120736-20210927150736-00221.warc.gz | 213,011,646 | 41,972 | # How much of Ag has formed on the Cu tile?
$$\pu{9.547 g}$$ $$\ce{Cu}$$ tile is added to an $$\ce{AgNO3}$$ solution. After some time, $$\ce{Cu}$$ was taken out of the solution, washed, dried and weighed. The mass appeared to be $$\pu{9.983 g}$$. How much of $$\ce{Ag}$$ has formed on the $$\ce{Cu}$$ tile?
The right answer to be found is supposed to be: $$\pu{1.08 g}$$.
My imaginary equations (with unlikely scenarios):
\begin{align} \ce{Cu + 2AgNO3 &-> Cu(NO3)2 + 2Ag}\\ \ce{Cu + AgNO3 &-> CuNO3 + 2Ag} \end{align}
I cannot solve this problem. There are a few others, but this is bugging me the most. How does one solve it? I believe this is something I wasn't taught before or there is again a mistake in the references given. I've tried calculating in ways I thought was right and anyhow, I cannot find the answer.
After discussions, I have come to the conclusion that I am taking $$\pu{0.618 g}$$ as the solution to this problem and will look forward into it tomorrow with the teacher as well.
This is the only closest answer I seem to arrive at by myself and others, as we get $$\pu{0.618 g}$$ when using the $$\Delta m_\mathrm{p}=\pu{0.436 g}$$ and $$\Delta m_\mathrm{t}=\pu{22.853 g}$$ differences. However, the problem regarding the other exercises still persists, which are of similar nature. I am starting to question my teacher's logic and abilities.
There is a close answer to the referenced answer, which assumes the $$\ce{Cu}$$ having a charge of $$+1$$, thus $$\ce{Cu + Ag+ -> Cu+ + Ag}$$ and $$\Delta m_\mathrm{p}=\pu{0.436 g}$$, $$\Delta m_\mathrm{t}=\pu{16.2 g}$$. However, I assume that to be a very unlikely scenario considering the current circumstances. This scenario lets you acquire \pu{1.0617 g} as the answer.
• Could you please edit what you've tried so far into the question? Feb 23 '19 at 13:39
• To me it seems like there is a typo somewhere. For some reason this question is all over Runet (Russian internet) and the answer is 0.62 g (which to me is the correct one). Feb 23 '19 at 14:02
• Could you provide a link for that? I can't seen to find anything regarding this anywhere Feb 23 '19 at 14:55
• @Alchimista This is my first post on stack, I got extremely surprised by the hold put on the topic, as well as it being treated as off-topic. I even put the homework tag beforehand, yet it was removed by the moderator Feb 24 '19 at 19:40
• I have reopened the question, but I'd still think it would have been a much better question, if the actual calculation that you did were included. In its current form it only discusses results, not the way to get them, therefore it doesn't point to the actual question you have. You should also explain the abbreviations you use. This question was probably closed, because there was a certain lack of attempt in previous versions. the edits then have not reached the critical number to reopen. (Please note that the homework tag is deprecated, it says that in all caps in its description.) Mar 2 '19 at 14:01
In this experiment the tile lose mass due to oxidation and dissolution as Cu dication (Cu++) and acquire mass due to the reduction and deposition of Ag.
This is what is measured by weighting before and after.
This difference in mass correspond to 2n moles of Ag minus n moles of Cu so is just matter of solving a first degree equation.
Quickly, without being consistent in digits and truncation., solving it for the current data gave about 0.6 g of Ag.
All others outcomes seems to me to assume an unlike stoichiometry or refuses perpetuating around.
A way to get into an "effective stoichiometry" of 1:1 is to consider formation of Cu+ followed by disproportionation AND assuming that all metal copper finely precipitates without depositing back to the tile from which originally came from. While, partially occurring, this scenario would explain discrepancies found performing a real experiment, it seems unlikely that is intended as to give a fixed result as for it requires the assumption in italic above. Doing that the amount of silver is now about 1.06 g.
• Thank you for the answer. Can you possibly provide how you got the answer and why 1.08g is not a viable answer? I wish to discuss this with the teacher whether she got the wrong answer herself Feb 23 '19 at 21:09
• You get it exactly as I wrote assuming that tge reaction is Cu and 2 Ag+ giving 2 Ag and Cu++. It seems the likely one. You can play with 1 to 1 stoichiometry but I guess you did and does not give the answer. If not retry following the same reasoning as in my answer. It would be funny if it turns to be the wanted 1.08 g :)) but what I and and others have proposed (Cu++) is the one that makes sense. Feb 23 '19 at 21:29
• It seems given by an author that thought of that. But it could be a mistake. You could add another exercise of that sort to see if it is the same problem. Read again my answer as I added an end. But my opinion is unchanged. Feb 24 '19 at 0:17
• Another exercise: 73g of Zn put in NiSO4 solution, which has a mass of 240g. After some time, the Zn tile appeared to be 71.8g. Find ZnSO4 mass percent in the solution (Answer: 13.35%) Feb 24 '19 at 0:19
• Thank you. If you can, could you provide the steps if you manage to get the answer? (possibly in chat or somewhere) As well as for the previous exercise where you get 1.06g, as that seems to be the most accurate answer I've yet to see and never gotten, as I am completely frustrated with just these exercises Feb 24 '19 at 0:22
Well, let's work backwards... Given:
9.547g Cu tile
9.983g tile + Ag
1.08 g Ag
then the Cu that is left on the tile is
9.983 - 1.080 = 8.903 g
and the Cu that dissolved must be
9.547 - 8.903 = 0.644 g
Using the atomic masses:
AM(Cu) = 63.546
AM(Ag) = 107.8682
we can calculate the moles of Ag
1.08/107.8682 = 0.010012
and the moles of Cu
0.644/63.546 = 0.010134
So the poser of the question was assuming the chemical reaction:
$$\ce{Cu + Ag^+ -> Cu^+ + Ag}$$
• Copper(I) in aqueous solution? No way. Feb 23 '19 at 15:08
• Didn't say that I agreed, just what poser of the question must have done...
– MaxW
Feb 23 '19 at 15:09
• Charge balance is fubar for that one. Feb 23 '19 at 15:29
• All in all I don't know how the poser messed the problem up. Old problems get reworked to form new problems. Perhaps in the original problem there was some sort of metal complexes that gave the overall 1:1 reaction.
– MaxW
Feb 23 '19 at 15:29
• @I'mPatrick - Lacking any other information I'd expect an acidic solution and that the copper would be oxidized to $\ce{Cu^{2+}}$. However to get the given answer you have to reason that the copper goes to $\ce{Cu^+}$ and assume some small rounding errors.
– MaxW
Feb 23 '19 at 17:20 | 1,828 | 6,714 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-39 | latest | en | 0.95927 |
https://www.gamedev.net/forums/topic/98385-simulated-air-friction-and-wind/ | 1,511,589,955,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809419.96/warc/CC-MAIN-20171125051513-20171125071513-00707.warc.gz | 820,068,831 | 25,883 | #### Archived
This topic is now archived and is closed to further replies.
# simulated air friction and wind
## Recommended Posts
I am simply curious if anyone has any thoughts on a fairly convincing way to simulate air resistance, specifically in cases of free fall. I''ve looked over some of the physics for it and found a couple different formulas for different speeds of the ojects. In a perfect world, it would be neat to actually use these equations with drag coefficients and exact values to mimic reality, but from what I''ve read, it seems like a very difficult process to analyze specifics about a body experiencing air resistance. The bodies I was mainly thinking about were leaves. Is there a realistic (and more to the point moderately easy) way to simulate objects falling that are highly succeptible to air resistance? Also, one other question, it''s somewhat related but not really. Is there any set rule to invoke to determine how far a gust of wind will travel before diminishing to nothing and how powerful it will be at any point from source to destination? Anyway, I''m sure I''ve wasted enough of your time and mine, so thanks in advance. Elijah ----------------------------- here are a few things I like, maybe you like them too: dew colored ponies, crisp apple strudel, doorbells, sleighbells, schnitzel with noodle, wild geese that fly with the moon on their wings
##### Share on other sites
I''ve seen some fairly realistic falling paper simulations. I don''t recall exactly where, though. It may have been a downloadable demo from havok.com or mathengine.com or one of the other (now defunct or acquired) game physics engine companies. Some games do a decent job with leaves, but it may be artistic animation and not real-time simulation. See Soul Calibur and DOA3 for examples.
There are certainly ways to approximate the variation in drag coefficient (and other so-called "stability derivatives") for arbitrary shaped objects. You can simulate the aerodynamic forces on an airplane wing, or baseball, or even quite weird-shaped object, quite easily using fairly simple formulas and lookup tables.
The difficulty with very light-weight objects (especially thin sheets such as paper/leaves) that are flexible and not "aerodynamically sound" is that the aerodynamic forces are highly nonlinear and difficult to simulate. But I think you realized that and that''s why you asked the question!
That said, obviously some folks have come up with solutions to this that are realistic enough for games, and work in real-time. I actually had to develop a flight simulator for the World Book Multimedia Encyclopedia a few years ago (if you have a copy of the 1997 or later CD-ROM, look for the "glider" simulation---my company did that simulation, and I wrote the flight simulator. Actually, we did all of the simulations in the CD-ROM.) The simulator was for a hand-thrown toy glider that does have some behaviors similar to a sheet of paper. The glider happens to be longitudinally stable while a sheet of paper is not, so the paper will kind of rock back and forth or flutter down, while the glider follows an oscillatory pattern of climb, stall, descent to accelerate, then start over.
I don''t have any references that I can just point to, but if I think of any specific advice, I''ll let you know. You might search the archives, because I have given some fairly detailed comments on drag calculation in the past, and those comments could be useful. I may have mentioned lift and moment calculations as well.
As for the distance wind travels before dying down, I think you could come up with a fairly realistic approximation using the formula for a potential vortex, which is basically an exponential decay of velocity away from some vortex core. So, basically, the velocity would decay away from the strongest gust point with the formula: v = V_max * exp(-a*r), where v is the velocity at a point, V_max is the maximum gust, a is a positive coefficient to be tweaked until things look/feel right, and r is the distance away from the point of maximum gust. Wind gusts are more complex in reality, but using that formula, and by animating the center of the gust and magnitude of the gust, you should be able to get a fairly realistic overall behavior. At least its based on real physical models of reality.
Your post is actually quite interesting to me personally, because my educational background is aerospace engineering and low and high-speed aerodynamics specifically.
Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.
##### Share on other sites
Well, if you don''t want to delve into the physics of it, you can always just make something that looks right. If you watch a leave fall to the ground, it tends to spin about an axis normal to the ground, and rock back and forth. When it''s mostly parallel to the ground it falls slowly...when it''s mostly perpendicular to the ground it falls faster. We the face of the leaf is perpendicular to the wind direction it is more likely to be caught in the gust and blown in that direction. So really you should be able to use a few simple rules with a basic wind engine (generate randomly changing wind gusts) to create a "leaf-in-the-wind" in a game without giving yourself a headache over the physics.
##### Share on other sites
bob_the_third is right.
The role of a game is not to simulate something with infinitesimal precision. The goal is to have it look right.
If it means you have to give your object additional parameters besides mass and vertices... well you do it this way. Else you will end up calculating all single air molecules interacting with your object.
What you could do is add a balancing and spining value to your object. As for the direction of the fall, simply use common drag force formulas... they take resistant surface into acount.
• ### Forum Statistics
• Total Topics
628401
• Total Posts
2982463
• 9
• 10
• 9
• 19
• 24 | 1,261 | 5,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-47 | longest | en | 0.949372 |
https://servantboy.com/temperature-measurement/ | 1,716,419,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00425.warc.gz | 442,191,839 | 18,304 | Physic Tutorials
# What Is Temperature And Temperature Measurement In Physics
The idea of temperature is rooted in qualitative ideas of “hot” and “cold” based on the sense of touch. As we know that an object that feels hot usually has a higher temperature than when it feels cold.
We are used to the idea that a thermometer shows the temperature of an object with which is in contact with.
What I will be discussing in this tutorial is listed below
1. What is temperature or how temperature is defined in physics?
2. What are the different units of temperature?
3. Conversion of different temperature scales
4. Temperature measurement
5. Temperature examples
6. Types of thermometer
## What is temperature or how temperature is defined in physics?
Temperature is a measure of degree of hotness or coldness of an object on some scale. You can also say it tells us the direction in which energy flows (energy flow from a region of higher temperature to the region of lower temperature).
When two objects in contact have the same temperature, there will be no transfer of thermal energy between them. This is referred to as thermal equilibrium.
## What are the different units of temperature?
The unit of temperature includes Celsius, Kelvin, and Fahrenheit. However, the S.I base unit of temperature is Kelvin. Measuring temperature in Kelvin scale is better than that of Celsius scale in that one of its fixed points, absolute zero has much significance than the higher or lower fixed points of Celsius scale.
Note: it is impossible to have a temperature lower than zero Kelvin (0 K is absolute zero).
## Conversion of different temperature scales
To convert from Celsius to Kelvin = temperature in Celsius + 273.15
To convert from Kelvin to Celsius = temperature in Kelvin – 273.15
Example
Convert -23 degree Celsius to Kelvin
Solution
-23 + 273.15 = 250.15 K
Note: the triple point of water is the temperature at which ice, water, and water vapour can co-exist. And it is defined as 273.16 K (273.16 – 273.15 = 0.01 C)
To convert from Celsius to Fahrenheit or vice-versa
C/5 = (F-32)/9
C is Celsius scale while F is Fahrenheit scale
Example
Convert 50 degree Celsius to Fahrenheit
50/5 = (F-32)/9
10 = (F-32)/9
10 X 9 = F-32
90 = F -32
F = 90 + 32
F = 122
## Temperature measurement
Temperature of an object is measured with a thermometer. Types of thermometer
Recommended
Heat Capacity
Linear Expansivity
### Bolarinwa Olajire
A tutor with a demonstrated history of working in the education industry. Skilled in analytical skills. Strong education professional with a M. SC focused in condensed matter. You can follow me on Twitter by clicking on the icon below to ask questions. | 611 | 2,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-22 | latest | en | 0.911962 |
https://www.msuperl.org/wikis/pcubed/doku.php?id=183_notes:constantf | 1,709,270,846,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00234.warc.gz | 904,966,178 | 10,817 | 183_notes:constantf
Section 2.5 in Matter and Interactions (4th edition)
You read previously how to (separately) predict the final momentum and final location of a system. In these notes, you will read how to put those two ideas together for a system where the net force is a constant vector (unchanging magnitude and direction) to be able to predict the motion of such a system.
A system that experiences a constant net force may be subject to one or more individual forces. What matters is that the sum of all the forces acting on the object result in a net force that has a constant magnitude and direction. A system which experiences such a force only changes its momentum in the direction of that net force.
Depending on how you select your coordinate system, it might mean that more than one component of the momentum vector changes. Often, it is convenient to select a coordinate system where the net force is aligned with a coordinate direction, then only one momentum vector component changes in time.
# Predicting the Motion
Consider a fan cart that is released on a low-friction track. Here's a video of the situation.
Notice that the fan cart's position changes more rapidly near the end of the video. The fan cart experiences (to a good approximation) a constant net force. The sum of all the forces acting on the fan cart give (roughly) a net force of constant magnitude and direction. Furthermore, the motion is constrained to a single dimension (namely, the horizontal direction).
With this setup, you can predict the position of the fan cart given only information about its initial position, velocity (or momentum), and the net force acting on it.
If you choose the horizontal direction to be the x-direction, we have the following equations to describe the motion.
$$p_{fx} = p_{ix} + F_{net,x} \Delta t$$ $$x_{f} = x_{i} + v_{avg,x} \Delta t$$
For this system, the momentum and, thus, the velocity change linearly in time. So the arithmetic average velocity and average velocity are equivalent. Hence, we can determine the final location of the system exactly.1)
Starting with the Update Form of the Momentum Principle, you determine the velocity of the object after a time $\Delta t$,
$$p_{fx} = p_{ix} + F_{net,x} \Delta t$$ $$mv_{fx} = mv_{ix} + F_{net,x} \Delta t$$ $$v_{fx} = v_{ix} + \dfrac{F_{net,x}}{m} \Delta t$$
From this equation, you can determine the arithmetic average velocity, which in this case is equal to the average velocity. $$v_{avg,x} = \dfrac{v_{ix} + v_{fx}}{2} = \dfrac{ v_{ix} + v_{ix} + \dfrac{F_{net,x}}{m} \Delta t}{2} = \dfrac{2v_{ix}}{2}+ \dfrac{\dfrac{F_{net,x}}{m} \Delta t}{2} = v_{ix}+ \dfrac{1}{2}\dfrac{F_{net,x}}{m} \Delta t$$
By using this average velocity in the position update formula, you obtain the final expression that predicts the location of the system given only information about its initial position, velocity, and the force acting on it.
$$x_{f} = x_{i} + v_{avg,x} \Delta t = x_{i} + v_{ix} \Delta t + \dfrac{1}{2}\dfrac{F_{net,x}}{m} \Delta t^2$$
In physics, the information about the system prior to predicting its motion is called the “initial state” of the system. The starting values of these properties (position, velocity, net force) are called the “initial conditions” of the system.
As you will read, the motion of systems can also be predicted or explained by using the energy principle in addition to or, as an alternative, to using the momentum principle. You will find that using energy, you can often think about the initial and final states of the system's motion and not how that motion evolves (e.g., over what time the motion occurs).
For constant force motion in one dimension (e.g., x-direction), you could solve the two motion prediction equations above (i.e., combining them into a single equation that removes the time variable). The resulting equation predicts the final speed of a system given its initial speed, the net force acting on the system, and the displacement of the system,
$$v_{xf}^2 = v_{xi}^2 + 2\dfrac{F_{net,x}}{m}\Delta x$$
Again, as you will read, this equation can also be derived from the relationship between kinetic energy and work.
The relationship between force and acceleration (even for a variable net force): $\vec{F}_{net}=m\vec{a}$ OR $\vec{a}=\frac{\vec{F}_{net}}{m}$.
The following 1D equations are valid ONLY if the net force (and therefore, the acceleration) is constant. These equations are commonly known as kinematic equations: $$x_{f} = x_{i} + v_{avg,x} \Delta t$$ $$v_{fx} = v_{ix} + \dfrac{F_{net,x}}{m} \Delta t$$ $$v_{avg,x} = \dfrac{v_{ix} + v_{fx}}{2} = v_{ix}+ \dfrac{1}{2}\dfrac{F_{net,x}}{m} \Delta t$$ $$x_{f} = x_{i} + v_{ix} \Delta t + \dfrac{1}{2}\dfrac{F_{net,x}}{m} \Delta t^2$$ $$v_{xf}^2 = v_{xi}^2 + 2\dfrac{F_{net,x}}{m}\Delta x$$
The derivation for each dimension is similar (so long as the force is constant in each direction). The result is the following general equation,
$$\vec{r}_{f} = \vec{r}_{i} + \vec{v}_{i} \Delta t + \dfrac{1}{2}\dfrac{\vec{F}_{net}}{m} \Delta t^2$$
1)
This is why so many physics courses spend so much time working with constant force motion. You can predict precisely where the system will be at any time.
• 183_notes/constantf.txt | 1,407 | 5,244 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-10 | latest | en | 0.901079 |
http://www.waseian.com/2019/06/ | 1,571,575,636,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986707990.49/warc/CC-MAIN-20191020105426-20191020132926-00442.warc.gz | 338,512,151 | 40,487 | Network Security - Comprehensive Paper Solution
Note: This is a previous year comprehensive solutions for your reference, feel free to provide solutions by navigating Submit Question/Answer tab in case you have latest solutions.
1) Jira's password is made of up 6 alphanumeric characters only. One password attempt takes 1 millisecond, What is the time to crack it in days?
ii) If password is case- insensitive.
i) Case sensitive
total chars = 26 + 26 + 10 = 62
possible combinations = 62 ^ 6 passwords
total time taken = 62 ^ 6 . 1 ms
approx. 62.62 =~ 3600 sec = 1 hour
Total time taken = 62. 62. 62. 62 / 1000 hours
=~ 360. 36 hours = 360.36/24 days
=~ 360.3/2= 180.3 = 540 days
Note: 62^4/1000/24 =~ 615 days
So an approx. answer between 520 to 620 days is good enough.
ii) Case insensitive
total chars = 26 + 10 = 36
possible combinations = 36 ^ 6 passwords
total time taken = 36 ^ 6 . 1 ms
= 36.36.36./ 1000 . 36 ^3
=~ 36. 36 ^ 3 seconds
= 36. 36. 36. 36 / 3600 hours
= 466 hours
=~ 19 days
So an approx. answer between 18 to 20 days is good enough.
2) What is a self-signed SSL certificate ? Detail on the security perspective when a website is using a self-signed SSL certificate.
A certificate not-signed by a Publicly trusted CA, but signed by a locally setup CA server is a self-signed certificate. Any entity/website/server using a self-signed SSL certificate cannot be trusted and very commonly used in phising attacks. I can setup a server to act as gmail.com fradulently by creating a self-signed certificate for www.gmail.com and deploying it in the server.
Most standard browsers – Firefox, Google Chrome, Safari, etc. throw errors when trying to browse to websites having self-signed certificates.
3) How to avoid man-in-the-middle attack in SSH sessions? Show passwordless SSH logins at work.
Man-in-the-middle attack is at-work when a client C logs in to a server M thinking it is server S and the client C is unable to detect it. In this case, the server M has successfully duped the client C and has forged a man-in-the-middle attack. So it can be a passive two-way data forwarder between client C and the actual server S, or an active data-mangler.
Every host server in SSH have their public keys sent to the client in the Key Exchange., which gets stored in client’s .ssh/known_hosts file. So the next time, client connects to the host, the server sent public key is matched with the client’s .ssh/known_hosts file and if there is a mismatch, SSH does not connect. So this SSH behavior effectively thwarts a middle server M trying to pose as actual server S.
Consider Client C connecting to Server S. In server S side, in file .ssh/authorized_keys, there should be entry containing client C’s public key. Then the server S will use it to exchange - encrypting/signing initial key exchange material with the client C. Because of the property that any data encrypted with public key can only be decrypted using the matching private key, this mechanism automatically authenticates the client C as only client C holds the private key. Client’s private key file are usually stored in file .ssh/id_rsa or .ssh/id_dsa depending on the public key algorithm chosen.
Discrete Structures for Computer Science - MCQS
Note: We have tried to upload as much as we can, all the question and answers might be shuffled - Please find the answer below each question, some answers might be wrong please review on the last date(some answers might be changed) if you find any wrong answer please comment down below.
Question:
Let m be an integer with m > 1. R on the set of integers is an equivalence relation if
Select one:
a. {(a, b) | a ≡ b (mod m)}
b. {(a, b) | a ≡ a (mod b)}
c. {(a, b) | b ≡ b (mod a)}
d. {(a, b) | b ≡ a (mod m)}
The Correct answer is: {(a, b) | a ≡ b (mod m)}
Question:
Consider the statement: x, y E Z if both xy and x + y are even, then
Select one:
a. both x and y are odd
b. both x and y are even
c. x is even and y is odd
d. x is odd and y is even
The Correct answer is:both x and y are even
Question:
If A = {0, 1}, B = {1, 2}, and C = {0, 1, 2} then what of the following isn’t in A × B × C ?
Select one:
a. (1, 1, 0)
b. (2, 2, 0)
c. (1, 1, 1)
d. (1, 2, 2)
The Correct answer is: (2, 2, 0)
Question:
Let A, B, and C be sets. Identify the the correct one among the following
Select one:
a. None of these
b. A ∩ (B ∩ C) = (C B) A
c. A (B C) = (C B) B
d. A ∩ (B C) = (C B) A
The Correct answer is: None of these
Question:
Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) = x − x2. What are
the functions f1 + f2 and f1 f2?
Select one:
a. x2 and x4 – x3
b. x and x3 – x2
c. x3 and x2 – x3
d. x and x3 − x4
The Correct answer is: x and x3 − x4
Question:
Determine for what of the following for “f” is not a one-to-one function:
Select one:
a. f for f (x) = x + 1 from the set of real numbers to itself
b. f from {a, b, c, d} to {1, 2, 3, 4, 5} with f (a) = 4, f (b) = 5, f (c) = 1, and f (d) = 3
c. f for f (x) = x2 from Z to the set of integers
d. f for f (x) = x2 from Z+ to the set of integers
The Correct answer is: f for f (x) = x2 from Z to the set of integers
Question:
Which of the following is not a logical equivalence for bi-conditional representation ?
Select one:
a. (p → q) ≡ p q
b. p ↔ q ≡ (p q) (p q)
c. p ↔ q ≡ p ↔q)
d. All mentioned
The Correct answer is: (p → q) ≡ p q
Question:
By the second law of de-Morgan (r s) is equivalent to
Select one:
a. r s
b. (r s)
c. r s
d. r s
Question:
Let P(x) be the statement “x + 1 > x.” For the real number domain, qualify the statement for truth value;
Select one:
a. xP(x) is true
b. xP(x) is true
c. xP(x) is true
d. xP(x) is true
The Correct answer is: xP(x) is true
Question:
For {Z+: Z+ < 5} verify if xP(x) holds good for P(x) is x2 < 10
Select one:
a. xP(x) is a conjunction
b. None of these
c. xP(x) is false
d. xP(x) holds good
The Correct answer is: xP(x) is false
Question:
Identify correct statement/s among the following
Select one:
a. The relation"Union of sets" is reflexive,but not symmetric
b. The relation"parallel of lines" is always an equivalence relation
c. The relation "Division" is Symmetric
The Correct answer is: The relation"parallel of lines" is always an equivalence relation
Question:
Which of the following is/are true ?
Select one or more:
a. p p is always a contradiction
b. p p is always a tautology
c. p p is always a contradiction
d. p p is always a tautology
The Correct answer is: p is always a tautology, p is always a contradiction
Question:
What of the following expressions does not imply the negation of the proposition, “there is an honest politician” if h(x) represents honesty function:
Select one:
a. xH(x)
b. xH(x)
c. xH(x)
d. All of these
Question:
“The sum of two positive integers is always positive” into a logical expression
Select one:
a. BOTH :xy((x > 0) (y > 0) (x +y > 0)) and xy(x +y > 0)
b. xy((x > 0) (y > 0) (x +y > 0))
c. Either xy((x > 0) (y > 0) (x +y > 0)) or xy(x +y > 0)
d. xy(x +y > 0)
The Correct answer is: BOTH :xy((x > 0) (y > 0) (x +y > 0)) and xy(x +y > 0)
Question:
Which of the following is/are statement/s
Select one or more:
a. How hot the day is
b. None of these are statements
c. The temperature is 40 degrees.
d. It is raining in the summer.
The Correct answer is: The temperature is 40 degrees., It is raining in the summer.
Question:
Let P be “you can take the flight,”
Let Q be “you buy a ticket.”
What of the following notates “you can take the flight if and only if you buy a ticket”
Select one:
a. None of these
b. Q → P
c. P → Q
d. P ↔ Q
The Correct answer is: P ↔ Q
Digital Electronics and Microprocessors - MCQS
Note: We have tried to upload as much as we can, all the question and answers might be shuffled - Please find the answer below each question, some answers might be wrong please review on the last date(some answers might be changed) if you find any of the answer is wrong please comment down below.
Question
How many memory chips of (64 x 2) are needed to provide a memory capacity of 2048 x 8?
A 32
B 128
C 4
D 16
Select one:
a. A
b. D
c. B
d. C
The correct answer is : 128
Question
In 2's complement binary representation what is the magnitude of these two numbers 1001 & 11001
A -6 & 6
B 9 & 25
C -6 & -6
D 9 & -6
Select one:
a. C
b. D
c. B
d. A
The correct answer is : -6 & -6
Question
Minimum no. of two input NAND gate required to implement a Ex-OR function is
(A)2 (B)3
(C)4 (D)5
Select one:
a. A
b. D
c. B
d. C
The correct answer is : 4
Question
What is the logic function implemented by the 2X1 mux shown below
A AND logic
B OR logic
C NAND logic
D Nor Logic
Select one:
a. d
b. A
c. c
d. b
The correct answer is : OR logic
Linear Algebra and Optimization- MCQS
Note : We have tried upload as much as we can, all the question and answers might be shuffled - Answers are marked in green. Thanks for Murthy,Abirami who helped us on this quiz.
Questions:
1. In R^3 the four vectors (1,2,5) (5,9,3) (329,431,731) and (21, -13,58) are
Select one:
A. Linearly dependent
B. Every linear combination is zero
C. Linearly Independent
D. Forms a basis for R^3
2.
Select one:
A. 45
B. 0
C. 10
D. 3
3. The elements along principal diagonal of a Hermitian matrix are all
Select one:
A. either zero or purely imaginary
B. real
C. 0
D. imaginary
4. For the linearly dependent vectors X=[3,-9,12] and Y=[-4,12,-16]:
Select one:
A. X-2Y=0
B. 4X+3Y=0
C. 4X-3Y=0
D. 3X+4Y=0
5. If A and B are two orthogonal matrices, each of order n then AB and BA are
Select one:
A. Orthogonal
B. asymmetric
C. symmetric
D. Hermitian
6. A set of single non-zero vector is
Select one:
A. Basis
B. None
C. Linearly dependent
D. Linearly Independent
7.Which one of the following is false :
Select one:
A. Matrices is an arrangement while determinant is a value of square matrices.
B. If Rank of A= Rank of B
C. Row equivalent matrices have the same rank.
D. A linearly independent set in v consisting on a maximum possible no of vectors in V is called a basis for V
The correct answer: If Rank of A= Rank of B
8. The Gauss-Elimination method, the augmented matrix reduces to…..................matrix
Select one:
a. Upper triangular
b. Diagonal
c. unit
d. None of these
9. The Characteristics roots of an orthogonal matrix are of ------ modulus
Select one:
A. Unit
B. Two
C. Three
D. Zero
10.Which of the following is not a elementary transformation?
Select one:
b. Squaring all the elements of the matrix
d. Multiplying a row by a non-zero number
The correct answer: Squaring all the elements of the matrix
11.
Select one:
A.
B.
C.
D.
12. If rank = number of unknowns and x = y = z = 0 the equation have only -------- solution
Select one:
A. Non trivial
B. Infinite
C. Unique
D. Trivial
13.
Select one:
A. 3
B. 9
C. 1
D. -3
14.If r is the rank of the matrix [A] of order m x n then r is
Select one:
A. r is less than or equal to minimum of (m, n)
B. r is greater than or equal to 'm'
C. r is greater than 'n'
D. r is less than or equal to 'n'
The correct answer: r is less than or equal to 'n'
15. The standard basis of an inner product space R x R x R (R) forms an
Select one:
A. Orthonormal set
B. Orthogonal set and Orthonormal set
C. Orthogonal set
D. Neither Orthogonal set nor Orthonormal set
The correct answer: Orthogonal set and Orthonormal set
16. Let A be a matrix having rank ‘r ' & B be the equivalent matrix obtained from A. Then rank of matrix B is…
Select one:
A. r + 2
B. r -1
C. r
D. r + 1
17. State : which one is false :
Select one:
A. If A and B are Hermitian: then AB-BA is skew Hermitian
B.
C. If A is a skew Hermitian : then iA is Hermitian
D.
18.
Select one:
A. 6,6
B. 6,-12
C. 12,6
D. -6,12.
19. The linear transformation Y = AX is regular if
Select one:
A. I A I#0
B. I A I=1
C. I A I=-1
D. I A I=0
The correct answer: I A I#0
20. A square matrix A of order 3 has 3 linearly independent Eigen vectors then a matrix P can be found such that P-1AP is a
Select one:
A. Diagonal Matrix
B. Symmetric matrix
C. Singular matrix
D. Unit matrix | 3,772 | 12,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-43 | longest | en | 0.907782 |
https://www.essay-writing.com/exp19_excel_appcapstone_intro_collection-computer-science-homework-help/ | 1,695,405,194,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506421.14/warc/CC-MAIN-20230922170343-20230922200343-00024.warc.gz | 854,566,275 | 16,615 | # Exp19_excel_appcapstone_intro_collection | Computer Science homework help
## Exp19_Excel_AppCapstone_Intro_Collection
### Project Description:
You are the financial manager for Judi’s Art Gallery. One of your clients, Raymond Chancellor, has a large collection of limited edition signed art by James C. Christensen (1942-2017). Over the years, you have helped Raymond maintain a list of his collection, including the title of the art, medium, issue date, issue price, the price he paid, and current market values. Most of the art is sold out from the publisher, which increases the value of the art. Other art is still readily available or is available in limited quantities. You want to update the list so that he can properly insure his collection.
### Steps to Perform:
Step
Instructions
Points Possible
1
0
2
Select the range A1:A6 on the Christensen worksheet, merge the cells, and apply Middle Align vertical alignment.
2
3
Change the width of column K to 17.00, select the range K1:K3, and apply Thick Outside Borders.
2
4
Click cell C9, and freeze panes so that rows 1 through 8 and columns A and B are frozen.
1
5
Select the range E9:E54 and apply the Mar-12 date format.
2
6
Find all occurrences of Retired and replace them with Sold Out.
2
7
Click cell H9 on the Christensen worksheet, and insert a formula that calculates the percentage Raymond paid of the issue price by dividing the amount Paid by the Issue Price. Copy the formula from cell H9 to the range H10:H54.
3
8
Click cell J9, and insert a formula that calculates the percentage change in value by subtracting the Issue Price from the Current Value and then dividing that result by the Issue Price. Copy the formula from cell J9 to the range J10:J54.
3
9
Apply Percent Style with one decimal place to the ranges H9:H54 and J9:J54.
2
10
Insert in the Current Values section at the top of the worksheet summary functions that use the range I9:I54. In cell I2, calculate the total of all the Current Values. In cell I3, calculate the average current value. in cell I4, calculate the lowest current value. In cell I5, calculate the highest current value.
12
11
Click cell C9 and insert a VLOOKUP function that looks up the code in cell B9, compares it to the codes and types of art in the range B2:C6, and returns the type of art. Copy the function in cell C9 to the range C9:C54. Hide column B that contains the codes.
5
12
Click cell K9 and insert an IF function that determines if the Issue Price is equal to the Current Value. If the values are the same, display Same as Issue (using the cell reference K2); otherwise, display Increased in Value (using the cell reference K3). Copy the function from cell K9 to the range K10:K54.
4
13
Display the Purchase worksheet, insert a row above Monthly Payment. Type Monthly Payments in 1 Year in cell A5 and type 12 in cell B5.
2
14
In cell B6 in the Purchasing sheet, insert the payment function to calculate the monthly payment using cell references, not numbers, in the function, and make sure the function displays a positive result. Apply Accounting Number Format and the Output cell style to cell B6.
5
15
Display the Christensen worksheet, select the range C1:E6, and create a clustered column chart.
4
16
Cut the chart and paste it in cell A57, change the height to 4″, and change the width to 6.5″. Add Alt Text The column chart compares total issue prices to total current values by type of art. (include the period).
4
17
Type Raymond’s Art Collection for the chart title, apply bold, and Black, Text 1 font color. Place the legend at the top of the chart. Add Primary Minor Horizontal gridlines.
4
18
Create a pie chart using the ranges C2:C6 and E2:E6, and then move the chart to a new chart sheet named Current Values. Move the Current Values sheet to the right of the Purchase sheet.
5
19
Type Percentage of Total Current Value as the chart title and change the font size to 18 pt. Choose Colorful Palette 3 for the chart colors. Add this description for Alt Text: The pie chart shows each art type by percentage of total current value. (include the period).
4
20
Hide the legend. Add data labels for categories and percentages; remove value data labels. Change the font size to 16 pt, bold, and Black, Text 1 font color for the data labels.
4
21
Explode the Masterwork Anniversary Edition slice by 15% and change the fill color to Light Blue.
1
22
Display the Christensen worksheet, click in any cell within the dataset, convert the data to a table, assign a table name Collection, and apply Green, Table Style Light 14.
4
23
Apply a conditional format to the range J9:J54 that highlights cells where the value is greater than 200% with Green Fill with Dark Green Text.
3
24
Sort the dataset by Type of Art in alphabetical order and then within Type of Art, sort by Current Value from largest to smallest.
4
25
Set a filter to display art that equals Sold Out as the status.
3
26
Add a total row to display the sum of the Issue Price, Paid, and Current Values columns. Remove the total for the Note column.
3
27
Select the Purchase sheet, set 2″ top margin, and center horizontally on page.
2
28
Select the Christensen sheet, select Landscape orientation, Legal paper size, set 0.2″ left and right margins, 0.5″ top and bottom margins, and set row 8 to repeat at the top of pages.
5
29
On the Christensen worksheet, change the width of column A to 27, the width of column D to 11, the width of column J to 12. Wrap text in cell A1 and cell J8.
3
30
Create a footer with Exploring Series on the left side, the sheet tab code in the center, and file name code on the right side on the Christensen sheet.
2
31
Save and close Exp19_Excel_AppCapstone_Intro_Collection.xlsx. Exit Excel. Submit the file as directed.
Get FREE Essay Price Quote
Pages (550 words)
Approximate price: - | 1,545 | 6,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-40 | latest | en | 0.866006 |
https://study.com/academy/topic/limits.html | 1,553,610,207,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205534.99/warc/CC-MAIN-20190326135436-20190326161436-00059.warc.gz | 619,626,495 | 20,826 | # Ch 7: Limits
Watch online video lessons on limits and learn about asymptotes, infinity, one-sided limits, continuity and more. Each lesson is accompanied by a short multiple-choice quiz you can use to check your understanding of these math topics.
## Limits
Join us on a cross-country road trip in this video lesson series on limits. As our car travels at various speeds over time, we'll learn how to describe, understand and determine the limits of functions.
In a function, limits are used to described what happens as a variable approaches a certain number. For instance, if we hop on the interstate and start cruising along at highway speed, after we drive for 10 minutes, we could say that the limit of our function as we approach 10 minutes is 60 miles per hour. As we speed up or slow down, the limits as we approach different times will either increase or decrease. Let's hope we don't get pulled over for speeding!
Building on your understanding of limits and how they apply to the real world, you'll learn how to graph limits using proper notation. You'll learn how to 'divide and conquer' with the properties of limits, which include addition, subtraction, multiplication and division properties.
Finally, you'll understand more complex aspects of limits, such as what happens in discontinuous functions (one-sided limits), how functions relate to each other through the Squeeze Theorem and how to graph limits for asymptotes and infinity. To illustrate these concepts, we'll visit the rivers of the fictional Kingdom of Rimonn and encounter an earthquake on the way to grandma's house. So, fasten your seat belts, and let's explore the limits of functions!
7 Lessons in Chapter 7: Limits
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
### Earning College Credit
Did you know… We have over 160 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 479 | 2,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2019-13 | longest | en | 0.934967 |
https://www.bybloggers.net/249184-slope-intercept-form-into-point-slope-form-seven-ugly-truth-about-slope-intercept-form-into-point-slope-form.html | 1,582,152,487,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00056.warc.gz | 702,663,540 | 5,960 | # Slope Intercept Form Into Point Slope Form Seven Ugly Truth About Slope Intercept Form Into Point Slope Form
Slope Intercept Form Into Point Slope Form Seven Ugly Truth About Slope Intercept Form Into Point Slope Form – slope intercept form into point slope form
| Delightful to my own website, in this particular time period I am going to demonstrate about keyword. And from now on, this is the 1st impression:
Slope-Intercept and Point-Slope Forms of a Linear Equation .. | slope intercept form into point slope form
Why don’t you consider photograph above? is usually that will incredible???. if you’re more dedicated consequently, I’l d teach you many photograph once again under:
Here you are at our site, articleabove (Slope Intercept Form Into Point Slope Form Seven Ugly Truth About Slope Intercept Form Into Point Slope Form) published . At this time we’re pleased to declare we have discovered an extremelyinteresting topicto be pointed out, that is (Slope Intercept Form Into Point Slope Form Seven Ugly Truth About Slope Intercept Form Into Point Slope Form) Many people trying to find info about(Slope Intercept Form Into Point Slope Form Seven Ugly Truth About Slope Intercept Form Into Point Slope Form) and of course one of these is you, is not it?
Algebra – Graphing using Point-Slope Form – slope intercept form into point slope form | slope intercept form into point slope form
Changing Linear Forms – from point-slope form to slope-intercept & standard form – slope intercept form into point slope form | slope intercept form into point slope form
Advanced Algebra 13. Slope-Intercept Form Point-Slope Form .. | slope intercept form into point slope form
How Do You Write The Equation Of Line In Point Slope Form .. | slope intercept form into point slope form
Point Slope Form Into Slope Intercept Form Calculator Five .. | slope intercept form into point slope form
standard form equations – akasharyans | slope intercept form into point slope form
How Do You Put an Equation in Point-Slope Form Into Standard .. | slope intercept form into point slope form
changing standard form to slope intercept form worksheet .. | slope intercept form into point slope form
Point-slope form | Algebra (practice) | Khan Academy – slope intercept form into point slope form | slope intercept form into point slope form
How Do You Put an Equation in Standard Form Into Slope .. | slope intercept form into point slope form
13 Use The Slope Intercept Form In Algebra Step Version .. | slope intercept form into point slope form
How Do You Put an Equation in Slope-Intercept Form Into .. | slope intercept form into point slope form
Last Updated: December 2nd, 2019 by
Essay Outline Template Microsoft Word Most Effective Ways To Overcome Essay Outline Template Microsoft Word’s Problem 9 Line Medevac Card Ten Great Seven Signs You’re In Love With 9 Line Medevac Card Ten Great Financial Statement Format In Uae The Modern Rules Of Financial Statement Format In Uae Self Employment Invoice Template Self Employment Invoice Template Will Be A Thing Of The Past And Here’s Why Metlife Claim Form The Biggest Contribution Of Metlife Claim Form To Humanity Insurance Damage Report Template Insurance Damage Report Template Will Be A Thing Of The Past And Here’s Why Slope Intercept Form Using Two Points Calculator Understanding The Background Of Slope Intercept Form Using Two Points Calculator Disney Customer Service Powerpoint Five Great Disney Customer Service Powerpoint Ideas That You Can Share With Your Friends Standard Form Ks13 The Death Of Standard Form Ks13 | 739 | 3,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-10 | latest | en | 0.833081 |
https://www.extendoffice.com/documents/excel/5088-excel-calculate-bond-price | 1,723,261,769,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00080.warc.gz | 608,001,869 | 23,611 | ## How to calculate bond price in Excel?
#### Calculate price of a zero coupon bond in Excel
For example there is 10-years bond, its face value is \$1000, and the interest rate is 5.00%. Before the maturity date, the bondholder cannot get any coupon as below screenshot shown. You can calculate the price of this zero coupon bond as follows:
Select the cell you will place the calculated result at, type the formula =PV(B4,B3,0,B2) into it, and press the Enter key. See screenshot:
Note: In above formula, B4 is the interest rate, B3 is the maturity year, 0 means no coupon, B2 is the face value, and you can change them as you need.
#### Calculate price of an annual coupon bond in Excel
Let’s say there is a annul coupon bond, by which bondholders can get a coupon every year as below screenshot shown. You can calculate the price of this annual coupon bond as follows:
Select the cell you will place the calculated result at, type the formula =PV(B11,B12,(B10*B13),B10), and press the Enter key. See screenshot:
Note: In above formula, B11 is the interest rate, B12 is the maturity year, B10 is the face value, B10*B13 is the coupon you will get every year, and you can change them as you need.
#### Calculate price of a semi-annual coupon bond in Excel
Sometimes, bondholders can get coupons twice in a year from a bond. In this condition, you can calculate the price of the semi-annual coupon bond as follows:
Select the cell you will place the calculated price at, type the formula =PV(B20/2,B22,B19*B23/2,B19), and press the Enter key.
Note: In above formula, B20 is the annual interest rate, B22 is the number of actual periods, B19*B23/2 gets the coupon, B19 is the face value, and you can change them as you need.
### Best Office Productivity Tools
🤖 Kutools AI Aide: Revolutionize data analysis based on: Intelligent Execution | Generate Code | Create Custom Formulas | Analyze Data and Generate Charts | Invoke Kutools Functions… Popular Features: Find, Highlight or Identify Duplicates | Delete Blank Rows | Combine Columns or Cells without Losing Data | Round without Formula ... Super Lookup: Multiple Criteria VLookup | Multiple Value VLookup | VLookup Across Multiple Sheets | Fuzzy Lookup .... Advanced Drop-down List: Quickly Create Drop Down List | Dependent Drop Down List | Multi-select Drop Down List .... Column Manager: Add a Specific Number of Columns | Move Columns | Toggle Visibility Status of Hidden Columns | Compare Ranges & Columns ... Featured Features: Grid Focus | Design View | Big Formula Bar | Workbook & Sheet Manager | Resource Library (Auto Text) | Date Picker | Combine Worksheets | Encrypt/Decrypt Cells | Send Emails by List | Super Filter | Special Filter (filter bold/italic/strikethrough...) ... Top 15 Toolsets: 12 Text Tools (Add Text, Remove Characters, ...) | 50+ Chart Types (Gantt Chart, ...) | 40+ Practical Formulas (Calculate age based on birthday, ...) | 19 Insertion Tools (Insert QR Code, Insert Picture from Path, ...) | 12 Conversion Tools (Numbers to Words, Currency Conversion, ...) | 7 Merge & Split Tools (Advanced Combine Rows, Split Cells, ...) | ... and more
Supercharge Your Excel Skills with Kutools for Excel, and Experience Efficiency Like Never Before. Kutools for Excel Offers Over 300 Advanced Features to Boost Productivity and Save Time. Click Here to Get The Feature You Need The Most...
#### Office Tab Brings Tabbed interface to Office, and Make Your Work Much Easier
• Enable tabbed editing and reading in Word, Excel, PowerPoint, Publisher, Access, Visio and Project.
• Open and create multiple documents in new tabs of the same window, rather than in new windows.
• Increases your productivity by 50%, and reduces hundreds of mouse clicks for you every day! | 957 | 3,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-33 | latest | en | 0.85348 |
https://www.teacherspayteachers.com/Browse/Search:skittles%20graphing%20activity | 1,537,537,078,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157203.39/warc/CC-MAIN-20180921131727-20180921152127-00460.warc.gz | 894,090,425 | 48,911 | showing 1-24 of 373 results
This math activity is perfect for wrapping up your unit on graphing. Included in this 7 page unit is a data recording sheet, "skittles" to color, and 3 different types of graphs including a bar graph, pictograph, and line plot for the students to create. Also included is a page of questions to help
Subjects:
Types:
\$2.00
42 Ratings
4.0
PDF (108.01 KB)
Students will learn about tallies, column graphs and picture graphs. Students are given a set number of skittles. I recommnend that you buy a big bag of skittles and divide them up into zip-lock bags. (This way you can control the number of skittles for the worksheet or if you would like to use th
Subjects:
Types:
\$4.50
11 Ratings
4.0
PDF (2.94 MB)
If your class is anything like mine, they can not stop talking about the upcoming Super Bowl! This activity was inspired by Marshawn Lynch's love for Skittles. What's better than Lynch, Skittles, and Common Core Standards all mixed into one activity? This activity is appropriate for grades 3-5 and
Subjects:
\$2.00
1 Rating
3.8
PDF (789.08 KB)
Use this item to teach the following common core aligned objective: Students will collect and record data of classmates favorite flavor of skittles candy, draw a bar graph to represent the data set. They will also be able to answer several questions using the information in their bar graph. Include
Subjects:
\$3.00
4 Ratings
4.0
PDF (354.24 KB)
Students use skittles to make a tally chart and graph. Students also answer questions about the graphs and label the graphs. The rubric is also included on the sheet for the teacher and/or students to evaluate the assignment. This work is licensed under a Creative Commons Attribution-NoDerivs 3.0 U
Subjects:
Types:
\$1.00
4 Ratings
4.0
PDF (43.07 KB)
After students collect their data on a tally chart, they will graph their data using a bar graph and pictograph. After, there are ten questions they answer using their data that require addition and subtraction. After reading a problem, students must write an equation and solve using their data. For
Subjects:
Types:
CCSS:
\$2.00
4 Ratings
3.8
PDF (101.77 KB)
This graphing activity was created for use with a unit on graphing. It can be used as an activity or as a final assessment for the unit. **NOTE: I have also posted a rubric that can be used to grade this activity. Please look for it in my TpT store. This work is licensed under a Creative Comm
Subjects:
Types:
\$2.00
3 Ratings
4.0
PDF (177.15 KB)
If your class is anything like mine, they can not stop talking about the upcoming Super Bowl! This activity was inspired by Marshawn Lynch's love for Skittles. What's better than Lynch, Skittles, and Common Core Standards all mixed into one activity? This activity is appropriate for grades K-2 (s
Subjects:
Types:
CCSS:
\$2.00
2 Ratings
4.0
PDF (781.97 KB)
Students practice converting fractions to decimals and percents. Students practice creating bar and circle graphs.
Subjects:
CCSS:
\$1.00
1 Rating
4.0
PDF (90.13 KB)
Students each get a small bag of Skittles. They count the total and the total of each color. Then they create a picture graph and bar graph. Students will answer questions about their data. They can also create a graph using class data.
Subjects:
\$2.00
not yet rated
N/A
DOCX (78.44 KB)
Students make 3 different types of graphs using skittles. The graphs are already made for you all the students have to do is fill in the data they gather!
Subjects:
\$1.00
not yet rated
N/A
PDF (784.82 KB)
Did this activity in a 5th grade neurological support class, they really enjoyed this hands on activity.
\$3.00
not yet rated
N/A
DOC (396.5 KB)
This file contains the instructions, worksheets, and a rubric for a FUN sorting and graphing activity. Each child gets a small snack pack of skilltes. They are first asked to answer a few general questions about what they THINK will be in the bag when they open it. Then, they will sort out the co
Subjects:
Types:
FREE
10 Ratings
4.0
ZIP (537.39 KB)
Use this activity with original skittles to teach even the youngest learners how to graph! Learning is always better with food, right? Please remember to leave feedback and visit my blog to see how I used this activity in my classroom! droppinknowledge2.blogspot.com Thank you!
Subjects:
Types:
FREE
6 Ratings
4.0
PDF (2.64 MB)
Students use candy to convert fractions, decimals, and percents. They also use the information to create a double bar graph.
Subjects:
Types:
FREE
1 Rating
N/A
ZIP (460.25 KB)
These skittles labs are a fun way to have students learn about the scientific method. Lab contents: 1. Skittle Colors Lab: A. Question: Which color skittle is most common in a package of skittles? B. Developing a hypothesis C. Procedure D. Data Tables E. Graphing F. Post-Lab Questions 2. Skittles
Subjects:
Types:
\$3.25
507 Ratings
4.0
PDF (116.07 KB)
Graphing - Sorting - Patterns - Tallying - Sight Words - Measuring Students love using candy to practice skills! This pack gives you some great activities and ideas using Skittles. These are great to use during the spring or around St. Patrick's Day since they are rainbow colors! Includes: Estima
Subjects:
Types:
\$3.50
150 Ratings
4.0
PDF (1.85 MB)
A fun and interactive way to teach graphing skills with the use of Skittles. Let kids "taste the rainbow" as they learn how to use tally marks, pictographs, and bar graphs. Great activity to do in March around St. Patricks day!
Subjects:
Types:
FREE
76 Ratings
4.0
PDF (241.69 KB)
An 14 page packet with activities all that can be done with Skittles! Activities include bar graphs, pictographs, line graphs, tally charts, fractions, addition, subtraction, money, sorting, estimation, place value, number sense, coordinate grids, working on a hundreds chart, and multiplication.
Subjects:
Types:
\$5.00
\$2.50
82 Ratings
4.0
PDF (452.07 KB)
Who knew a bag of fun sized skittles could mean some big time learning!?! Your students can enjoy a fun and tasty graphing project when learning about graphing, at the end of the year, or for some anytime fun! Skittles Graphing includes: An estimation page. A sorting sheet. A bar graph. A questio
Subjects:
\$1.00
71 Ratings
4.0
PDF (3.8 MB)
Here is a tasty way to end (or begin!) your graphing unit. Using personal size bags of Skittles, students will graph the color of their Skittles, collect data from peers about their favorite flavor and color of their Skittles, and create their own bar graph or pictographs. There are 2 graphing optio
Subjects:
Types:
\$2.50
51 Ratings
4.0
PDF (2.83 MB)
Add this hands-on graphing, multiplying, and fraction mini unit to your St. Patrick's Day planning. Students will sort skittles and create a scaled bar graph and pictograph and more. What you get: ~Sorting mat and tally chart. ~Scaled bar graph and pictograph with data sheet. 3 different data s
Subjects:
Types:
\$2.50
25 Ratings
4.0
PDF (4.31 MB)
My students are learning comparing numbers in Math this week and I wanted to create something fun and easy to start with. I love using food and thought skittles would be great to use since there are so many different colors. I hope your students enjoy it! :)
Subjects:
Types:
\$1.00
20 Ratings
4.0
PDF (457.28 KB)
This is a fun graphing activity that can be used with or without Skittles. It can also be used with a color printer or without. On each activity page, there is a graphing element, and a follow-up question section. It covers Common Core State Standard K.MD.B.3. Contents: Cover (Page 1) Activity 1 -
Subjects:
Types:
\$2.00
20 Ratings
4.0
PDF (742.49 KB)
showing 1-24 of 373 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 1,986 | 7,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-39 | latest | en | 0.929842 |
http://nrich.maths.org/public/leg.php?code=-396&cl=4&cldcmpid=5840 | 1,484,756,547,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00153-ip-10-171-10-70.ec2.internal.warc.gz | 212,046,008 | 9,337 | # Search by Topic
#### Resources tagged with chemistry similar to More Bridge Building:
Filter by: Content type:
Stage:
Challenge level:
### There are 75 results
Broad Topics > Applications > chemistry
### Big and Small Numbers in Chemistry
##### Stage: 4 Challenge Level:
Get some practice using big and small numbers in chemistry.
### Constantly Changing
##### Stage: 4 Challenge Level:
Many physical constants are only known to a certain accuracy. Explore the numerical error bounds in the mass of water and its constituents.
### Diamonds Aren't Forever
##### Stage: 5 Challenge Level:
Ever wondered what it would be like to vaporise a diamond? Find out inside...
### The Power of Dimensional Analysis
##### Stage: 4 and 5
An introduction to a useful tool to check the validity of an equation.
### Eudiometry
##### Stage: 5 Challenge Level:
When a mixture of gases burn, will the volume change?
##### Stage: 5
Read about the mathematics behind the measuring devices used in quantitative chemistry
### Ideal Axes
##### Stage: 5 Challenge Level:
Explore how can changing the axes for a plot of an equation can lead to different shaped graphs emerging
### Reaction Types
##### Stage: 5 Challenge Level:
Explore the rates of growth of the sorts of simple polynomials often used in mathematical modelling.
### Lennard Jones Potential
##### Stage: 5 Challenge Level:
Investigate why the Lennard-Jones potential gives a good approximate explanation for the behaviour of atoms at close ranges
### What Salt?
##### Stage: 5 Challenge Level:
Can you deduce why common salt isn't NaCl_2?
### Core Scientific Mathematics
##### Stage: 4 and 5 Challenge Level:
This is the area of the advanced stemNRICH site devoted to the core applied mathematics underlying the sciences.
### Smoke and Daggers
##### Stage: 5 Challenge Level:
We all know that smoking poses a long term health risk and has the potential to cause cancer. But what actually happens when you light up a cigarette, place it to your mouth, take a tidal breath. . . .
### Clear as Crystal
##### Stage: 5 Challenge Level:
Unearth the beautiful mathematics of symmetry whilst investigating the properties of crystal lattices
### The Real Hydrogen Atom
##### Stage: 5 Challenge Level:
Dip your toe into the world of quantum mechanics by looking at the Schrodinger equation for hydrogen atoms
### The Amazing Properties of Water
##### Stage: 4 and 5 Challenge Level:
Find out why water is one of the most amazing compounds in the universe and why it is essential for life. - UNDER DEVELOPMENT
### Striking Gold
##### Stage: 5 Challenge Level:
Investigate some of the issues raised by Geiger and Marsden's famous scattering experiment in which they fired alpha particles at a sheet of gold.
### Maths in the Undergraduate Physical Sciences
##### Stage: 5
An article about the kind of maths a first year undergraduate in physics, engineering and other physical sciences courses might encounter. The aim is to highlight the link between particular maths. . . .
### Reaction Rates
##### Stage: 5 Challenge Level:
Explore the possibilities for reaction rates versus concentrations with this non-linear differential equation
### Big and Small Numbers in Physics
##### Stage: 4 Challenge Level:
Work out the numerical values for these physical quantities.
### Blood Buffers
##### Stage: 5 Challenge Level:
Investigate the mathematics behind blood buffers and derive the form of a titration curve.
### A Question of Scale
##### Stage: 4 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
### Stats Statements
##### Stage: 5 Challenge Level:
Are these statistical statements sometimes, always or never true? Or it is impossible to say?
##### Stage: 4 and 5 Challenge Level:
Advanced problems in the mathematical sciences.
### Extreme Dissociation
##### Stage: 5 Challenge Level:
In this question we push the pH formula to its theoretical limits.
### Mixing Ph
##### Stage: 5 Challenge Level:
Use the logarithm to work out these pH values
### Ph Temperature
##### Stage: 5 Challenge Level:
At what temperature is the pH of water exactly 7?
### Concentrations
##### Stage: 5 Challenge Level:
Use the interactivity to practise your skills with concentrations and molarity.
### Real-life Equations
##### Stage: 5 Challenge Level:
Here are several equations from real life. Can you work out which measurements are possible from each equation?
### Approximately Certain
##### Stage: 4 and 5 Challenge Level:
Estimate these curious quantities sufficiently accurately that you can rank them in order of size
### Mathematical Issues for Chemists
##### Stage: 5
A brief outline of the mathematical issues faced by chemistry students.
### Cobalt Decay
##### Stage: 5 Challenge Level:
Investigate the effects of the half-lifes of the isotopes of cobalt on the mass of a mystery lump of the element.
### Spectrometry Detective
##### Stage: 5 Challenge Level:
From the atomic masses recorded in a mass spectrometry analysis can you deduce the possible form of these compounds?
### Chemnrich
##### Stage: 4 and 5 Challenge Level:
chemNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of chemistry, designed to help develop the mathematics required to get the most from your study. . . .
### CSI: Chemical Scene Investigation
##### Stage: 5 Challenge Level:
There has been a murder on the Stevenson estate. Use your analytical chemistry skills to assess the crime scene and identify the cause of death...
### Conversion Sorter
##### Stage: 4 Challenge Level:
Can you break down this conversion process into logical steps?
### Dilution Series Calculator
##### Stage: 4 Challenge Level:
Which dilutions can you make using 10ml pipettes and 100ml measuring cylinders?
### Investigating the Dilution Series
##### Stage: 4 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### Mixed up Mixture
##### Stage: 4 Challenge Level:
Can you fill in the mixed up numbers in this dilution calculation?
### Reaction Rates!
##### Stage: 5
Fancy learning a bit more about rates of reaction, but don't know where to look? Come inside and find out more...
### Catalyse That!
##### Stage: 5 Challenge Level:
Can you work out how to produce the right amount of chemical in a temperature-dependent reaction?
### Exact Dilutions
##### Stage: 4 Challenge Level:
Which exact dilution ratios can you make using only 2 dilutions?
### Heavy Hydrocarbons
##### Stage: 4 and 5 Challenge Level:
Explore the distribution of molecular masses for various hydrocarbons
### Bond Angles
##### Stage: 5 Challenge Level:
Think about the bond angles occurring in a simple tetrahedral molecule and ammonia.
### Reductant Ratios
##### Stage: 5 Challenge Level:
What does the empirical formula of this mixture of iron oxides tell you about its consituents?
### Molecular Sequencer
##### Stage: 4 and 5 Challenge Level:
Investigate the molecular masses in this sequence of molecules and deduce which molecule has been analysed in the mass spectrometer.
### Coordinated Crystals
##### Stage: 5 Challenge Level:
Explore the lattice and vector structure of this crystal.
### Gassy Information
##### Stage: 5 Challenge Level:
Do each of these scenarios allow you fully to deduce the required facts about the reactants?
### Stemnrich - Applied Mathematics
##### Stage: 3 and 4 Challenge Level:
This is the area of the stemNRICH site devoted to the core applied mathematics underlying the sciences.
### Stereoisomers
##### Stage: 5 Challenge Level:
Put your visualisation skills to the test by seeing which of these molecules can be rotated onto each other.
### Bent Out of Shape
##### Stage: 4 and 5 Challenge Level:
An introduction to bond angle geometry. | 1,690 | 8,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-04 | longest | en | 0.837149 |
https://www.beatthegmat.com/mba/2012/03/17/the-next-gen-gmat-two-part-analysis | 1,618,395,911,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077810.20/warc/CC-MAIN-20210414095300-20210414125300-00449.warc.gz | 748,265,997 | 10,708 | # The Next-Gen GMAT: Two-Part Analysis
by , Mar 17, 2012
The launch of the Integrated Reasoning (IR) section is getting close! The last administration of the old version of the GMAT will be on 2 June; the next-generation GMAT will launch on 5 June. Its not too late to study for the old version, but its also not too early to start thinking about studying for the next-gen test, including IR.
So lets talk about one of the four IR question categories: the Two-Part Analysis. IR in general is a mix of quant and logical reasoning, so expect to bring your critical reasoning and reading comp skills into play on this section.
Before we dive in, just a note: a new Official Guide (13th edition!) will be published next month (April) with an IR section and the test makers will also launch additional online IR resources. I would guess that most test prep companies will also be releasing their IR study materials next month (we certainly are!).
## Try the problem
Lets try out the question:here it is. Just in case that link changes, you can also click on this link to go to the next-gen GMAT website, and then, about halfway down the page, click on the Two-Part Analysis link. Were going to try the 2nd of the 5 questions.
Note: when you are done, do NOT click the next button. Just leave it up on the screen and come back here.
Set your timer for 2 minutes 30 seconds and go! (Note: we have an average of 2 minutes and 30 seconds for each IR question in the section, but some question types are more complicated than others. I recommend trying this one for 2 minutes initially, but take the full 2 minutes 30 seconds if you think you need it. Just note that, as you study, youre going to have to determine your strengths and weaknesses so you can learn to balance your time appropriately.)
## A morefa? Huh?
Oh. They told us that this morefa word is made up. Make sure you read the little opening line at the top, if there is one. The beginning of the sentence (The following excerpt) makes it seem unimportant, but if they bothered to include the sentence, theres a reason. In this case, Ive learned that morefa is a made-up word and also that it is some type of location.
## Orient yourself
If there is quite a bit of material to work through before getting to the question (a whole paragraph of text, a complicated table or graph), then preview the question before diving into the actual text or table.
In this case, we do have a long paragraph to read, so lets read the question first. Lets see
Based on the definition. that can be inferred from the previous paragraph
Apparently, the paragraph is going to define this word morefa for me. So now I know that I need to figure out what that definition is when Im reading the paragraph.
Keep reading. Remember how the opening line told us that the morefa is a type of location? The question is asking us what activities do take place within a species location (I dont know what morefa means yet) and what activities dont take place there. When I read the paragraph, then, I need to make sure that I (a) understand the definition of a morefa, and (b) know which activities do and dont take place, by definition, in a specific morefa.
## Dive in and take notes
Great, Im ready to dive into the paragraph. Im going to take notes here Ive got a space on my scrap paper for the definition of the morefa and Ive also drawn a quick table with two columns. One says IN m. and the other says NOT in m.
Below, Ill show in italics what Im thinking, and then Ill show what Id write on my scrap paper.
(Reading sentence 1) Hmm. Scientists are observing birds in the morefa during breeding season. (Sentence 2) By doing this, they can study courtship displays and social hierarchy, so this things must occur in the morefa, whatever that is. (Sentence 3) Some species keep returning I guess that means maybe some of them leave when its not breeding season? Or they just sometimes leave who knows when or why. (Sentence 4) They also want to observe the morefa when the birds arent there, such as when oh, this is interesting, such as when theyre off somewhere else building nests. So thats not happening in the morefa.
Okay, so the morefa is like a habitat or a place for the birds to live, though they dont necessarily live there all the time. The focus seems to be on breeding in the morefa.
Def: habitat; place to live, breed
Now the question makes a little more sense. I need to find an answer that represents something that always occurs in the morefa and something else that never occurs in the morefa.
Here are the options:
Sleeping: I dont see anything in my notes about sleeping. This isnt it.
Occupying the location multiple times: they did mention that some birds leave, yes but I dont think they said that all of the birds leave and come back. Ill leave this one for now.
Establishing nests: Oh, yes, the last sentence specifically said they leave to make nests elsewhere. This is the answer to the second part.
Gathering together with members of their own species: Lets see, theyre both courting and breeding in the morefa, and animals have to breed with members of their own species. So this one is sounding pretty good for the first part.
Territorial competition with members of different species: I dont have anything in my notes about competition or other species. Nope.
Okay, Im confident that the answer to the second part is establishing nests. For the first part, I just want to examine the text again where it talks about leaving and coming back. Okay, sentence 3 says that some species repeatedly return which means that not all species keep going back to the same morefa. Therefore, this isnt something that absolutely has to happen. The first sentence does link the morefa to breeding season, though, and obviously the birds need to be with members of their own species in order to breed.
The answer to the first part is: gathering together with members of their own species. The answer to the second part is: establishing nests.
## Key Takeaways for Two-Part Analysis questions:
(1) The Two-Part questions will always contain two parts. (Surprise, surprise.) The two halves may be dependent upon each other or may be determined independently; either way, we have to answer both halves correctly in order to get any credit.
(2) On the more critical-reasoning-like questions, the details are going to matter. Preview the question stem so that you have a good idea of what youll be asked to do. That will help you to read the paragraph and take notes with a purpose, focusing on the ideas and details that are most applicable to the specific question asked.
(3) Be aware that, though this particular question was more like a critical reasoning question, these two-part questions can also be much more quant focused.
* All quotes copyright and courtesy of the Graduate Management Admissions Council. Usage of this material does not imply endorsement by GMAC. | 1,520 | 6,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-17 | latest | en | 0.957327 |
http://blog.longpengfei.com/post/48-rotate-image-%E6%97%8B%E8%BD%AC%E5%9B%BE%E5%83%8F/ | 1,566,357,144,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315750.62/warc/CC-MAIN-20190821022901-20190821044901-00093.warc.gz | 31,271,899 | 3,999 | # 48. Rotate Image 旋转图像
Medium
You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Note:
You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.
`````` Example 1:
Given input matrix =
[
[1,2,3],
[4,5,6],
[7,8,9]
],
rotate the input matrix in-place such that it becomes:
[
[7,4,1],
[8,5,2],
[9,6,3]
]
Example 2:
Given input matrix =
[
[ 5, 1, 9,11], i=0,j=1,n=4
[ 2, 4, 8,10],
[13, 3, 6, 7],
[15,14,12,16]
],
rotate the input matrix in-place such that it becomes:
[
[15,13, 2, 5],
[14, 3, 4, 1],
[12, 6, 8, 9],
[16, 7,10,11]
]
``````
``````/*
i=0,j=1,n=4时,
[ [
[ 5, 1, 9,11], [15,13, 2, 5],
[ 2, 4, 8,10], ----> [14, 3, 4, 1],
[13, 3, 6, 7], [12, 6, 8, 9],
[15,14,12,16] [16, 7,10,11]
], ]
*/
void rotate2(vector<vector<int>> &matrix){
int n = matrix.size();
for (int i = 0; i<n/2; i++) {
for (int j = i; j<n-1-i; j++) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[n-1-j][i];
matrix[n-1-j][i] = matrix[n-1-i][n-1-j];
matrix[n-1-i][n-1-j] = matrix[j][n-1-i];
matrix[j][n-1-i] = tmp;
}
}
}
``````
``````void rotate(vector<vector<int>>& matrix) {
int n = matrix.size();
for (int i=0; i<n; i++) {
for (int j=i+1; j<n; j++) {
swap(matrix[i][j], matrix[j][i]);
}
reverse(matrix[i].begin(), matrix[i].end());
}
}
`````` | 595 | 1,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-35 | latest | en | 0.573576 |
http://www.slideshare.net/PabloCasMaz/solucionario-resnick-5-ed-vol-2http-galeanoingwordpresscom | 1,394,228,263,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999651148/warc/CC-MAIN-20140305060731-00020-ip-10-183-142-35.ec2.internal.warc.gz | 519,917,389 | 205,913 | Like this document? Why not share!
# Solucionario resnick 5° ed - vol 2(http- -galeanoing.wordpress.com_)
## by Pablo CasMaz, Working at Real Madrid C.F. on Feb 26, 2013
• 1,636 views
### Views
Total Views
1,636
Views on SlideShare
1,636
Embed Views
0
Likes
1
Downloads
27
Comments
0
No embeds
### Upload Details
Uploaded via SlideShare as Adobe PDF
### Usage Rights
© All Rights Reserved
### Report content
Edit your comment
## Solucionario resnick 5° ed - vol 2(http- -galeanoing.wordpress.com_) Document Transcript
• Instructor Solutions Manual for Physics byHalliday, Resnick, and Krane Paul Stanley Beloit College Volume 2
• A Note To The Instructor... The solutions here are somewhat brief, as they are designed for the instructor, not for the student.Check with the publishers before electronically posting any part of these solutions; website, ftp, orserver access must be restricted to your students. I have been somewhat casual about subscripts whenever it is obvious that a problem is onedimensional, or that the choice of the coordinate system is irrelevant to the numerical solution.Although this does not change the validity of the answer, it will sometimes obfuscate the approachif viewed by a novice. There are some traditional formula, such as 2 2 vx = v0x + 2ax x,which are not used in the text. The worked solutions use only material from the text, so there maybe times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know aneasier approach existed. But if it was not in the text, I did not use it here. I also tried to avoid reinventing the wheel. There are some exercises and problems in the textwhich build upon previous exercises and problems. Instead of rederiving expressions, I simply referyou to the previous solution. I adopt a different approach for rounding of significant figures than previous authors; in partic-ular, I usually round intermediate answers. As such, some of my answers will differ from those inthe back of the book. Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manualwith considerably more detail and, when appropriate, include discussion on any physical implicationsof the answer. These student solutions carefully discuss the steps required for solving problems, pointout the relevant equation numbers, or even specify where in the text additional information can befound. When two almost equivalent methods of solution exist, often both are presented. You areencouraged to refer students to the Student’s Solution Manual for these exercises and problems.However, the material from the Student’s Solution Manual must not be copied. Paul Stanley Beloit College stanley@clunet.edu 1
• E25-1 The charge transferred is Q = (2.5 × 104 C/s)(20 × 10−6 s) = 5.0 × 10−1 C.E25-2 Use Eq. 25-4: (8.99×109 N·m2 /C2 )(26.3×10−6 C)(47.1×10−6 C) r= = 1.40 m (5.66 N)E25-3 Use Eq. 25-4: (8.99×109 N·m2 /C2 )(3.12×10−6 C)(1.48×10−6 C) F = = 2.74 N. (0.123 m)2E25-4 (a) The forces are equal, so m1 a1 = m2 a2 , or m2 = (6.31×10−7 kg)(7.22 m/s2 )/(9.16 m/s2 ) = 4.97×10−7 kg. (b) Use Eq. 25-4: (6.31×10−7 kg)(7.22 m/s2 )(3.20×10−3 m)2 q= = 7.20×10−11 C (8.99×109 N·m2 /C2 )E25-5 (a) Use Eq. 25-4, 1 q1 q2 1 (21.3 µC)(21.3 µC) F = 2 = −12 C2 /N · m2 ) = 1.77 N 4π 0 r12 4π(8.85×10 (1.52 m)2 (b) In part (a) we found F12 ; to solve part (b) we need to first find F13 . Since q3 = q2 andr13 = r12 , we can immediately conclude that F13 = F12 . We must assess the direction of the force of q3 on q1 ; it will be directed along the line whichconnects the two charges, and will be directed away from q3 . The diagram below shows the directions. F 23 F 12 θ F F 12 23 F net From this diagram we want to find the magnitude of the net force on q1 . The cosine law isappropriate here: F net 2 = 2 2 F12 + F13 − 2F12 F13 cos θ, = (1.77 N) + (1.77 N)2 − 2(1.77 N)(1.77 N) cos(120◦ ), 2 = 9.40 N2 , F net = 3.07 N. 2
• E25-6 Originally F0 = CQ2 = 0.088 N, where C is a constant. When sphere 3 touches 1 the 0charge on both becomes Q0 /2. When sphere 3 the touches sphere 2 the charge on each becomes(Q0 + Q0 /2)/2 = 3Q0 /4. The force between sphere 1 and 2 is then F = C(Q0 /2)(3Q0 /4) = (3/8)CQ2 = (3/8)F0 = 0.033 N. 0 E25-7 The forces on q3 are F31 and F32 . These forces are given by the vector form of Coulomb’sLaw, Eq. 25-5, 1 q3 q1 1 q3 q1 F31 = ˆ 2 r31 = 4π ˆ , r 2 31 4π 0 r31 0 (2d) 1 q3 q2 1 q3 q2 F32 = ˆ 2 r32 = 4π ˆ . r 2 32 4π 0 r32 0 (d)These two forces are the only forces which act on q3 , so in order to have q3 in equilibrium the forcesmust be equal in magnitude, but opposite in direction. In short, F31 = −F32 , 1 q3 q1 1 q3 q2 ˆ31 r = − ˆ32 , r 4π 0 (2d)2 4π 0 (d)2 q1 q2 ˆ31 r = − ˆ32 . r 4 1Note that ˆ31 and ˆ32 both point in the same direction and are both of unit length. We then get r r q1 = −4q2 .E25-8 The horizontal and vertical contributions from the upper left charge and lower right chargeare straightforward to find. The contributions from the upper left charge require slightly more work. √ √The diagonal distance is 2a; the components will be weighted by cos 45◦ = 2/2. The diagonalcharge will contribute √ √ 1 (q)(2q) 2ˆ 2 q2 ˆ Fx = √ i= i, 4π 0 ( 2a)2 2 8π 0 a2 √ √ 1 (q)(2q) 2 ˆ 2 q2 ˆ Fy = √ j= j. 4π 0 ( 2a)2 2 8π 0 a2 (a) The horizontal component of the net force is then √ 1 (2q)(2q)ˆ 2 q2 ˆ Fx = i+ i, 4π 0 a2 8π 0 a2 √ 4 + 2/2 q 2 ˆ = i, 4π 0 a2 = (4.707)(8.99×109 N · m2 /C2 )(1.13×10−6 C)2 /(0.152 m)2ˆ = 2.34 N ˆ i i. (b) The vertical component of the net force is then √ 1 (q)(2q) ˆ 2 q2 ˆ Fy = − 2 j+ j, 4π 0 a 8π 0 a2 √ −2 + 2/2 q 2 ˆ = j, 8π 0 a2 = (−1.293)(8.99×109 N · m2 /C2 )(1.13×10−6 C)2 /(0.152 m)2ˆ = −0.642 N ˆ j j. 3
• E25-9 The magnitude of the force on the negative charge from each positive charge is F = (8.99×109 N · m2 /C2 )(4.18×10−6 C)(6.36×10−6 C)/(0.13 m)2 = 14.1 N.The force from each positive charge is directed along the side of the triangle; but from symmetryonly the component along the bisector is of interest. This means that we need to weight the aboveanswer by a factor of 2 cos(30◦ ) = 1.73. The net force is then 24.5 N.E25-10 Let the charge on one sphere be q, then the charge on the other sphere is Q = (52.6 ×10−6 C) − q. Then 1 qQ = F, 4π 0 r2 (8.99×109 N·m2 /C2 )q(52.6×10−6 C − q) = (1.19 N)(1.94 m)2 .Solve this quadratic expression for q and get answers q1 = 4.02×10−5 C and q2 = 1.24×10−6 N. E25-11 This problem is similar to Ex. 25-7. There are some additional issues, however. It iseasy enough to write expressions for the forces on the third charge 1 q3 q1 F31 = ˆ 2 r31 , 4π 0 r31 1 q3 q2 F32 = ˆ 2 r32 . 4π 0 r32Then F31 = −F32 , 1 q3 q1 1 q3 q2 ˆ 2 r31 = − ˆ 2 r32 , 4π 0 r31 4π 0 r32 q1 q2 ˆ 2 r31 = − 2 ˆ32 . r r31 r32The only way to satisfy the vector nature of the above expression is to have ˆ31 = ±ˆ32 ; this means r rthat q3 must be collinear with q1 and q2 . q3 could be between q1 and q2 , or it could be on eitherside. Let’s resolve this issue now by putting the values for q1 and q2 into the expression: (1.07 µC) (−3.28 µC) 2 ˆ31 r = − 2 ˆ32 , r r31 r32 2 2 r32 ˆ31 r = (3.07)r31 ˆ32 . rSince squared quantities are positive, we can only get this to work if ˆ31 = ˆ32 , so q3 is not between r rq1 and q2 . We are then left with 2 2 r32 = (3.07)r31 ,so that q3 is closer to q1 than it is to q2 . Then r32 = r31 + r12 = r31 + 0.618 m, and if we take thesquare root of both sides of the above expression, r31 + (0.618 m) = (3.07)r31 , (0.618 m) = (3.07)r31 − r31 , (0.618 m) = 0.752r31 , 0.822 m = r31 4
• E25-12 The magnitude of the magnetic force between any two charges is kq 2 /a2 , where a =0.153 m. The force between each charge is directed along the side of the triangle; but from symmetryonly the component along the bisector is of interest. This means that we need to weight the aboveanswer by a factor of 2 cos(30◦ ) = 1.73. The net force on any charge is then 1.73kq 2 /a2 . The length of the angle bisector, d, is given by d = a cos(30◦ ). The distance from any charge to the center of the equilateral triangle is x, given by x2 =(a/2)2 + (d − x)2 . Then x = a2 /8d + d/2 = 0.644a. The angle between the strings and the plane of the charges is θ, given by sin θ = x/(1.17 m) = (0.644)(0.153 m)/(1.17 m) = 0.0842,or θ = 4.83◦ . The force of gravity on each ball is directed vertically and the electric force is directed horizontally.The two must then be related by tan θ = F E /F G ,so 1.73(8.99×109 N · m2 /C2 )q 2 /(0.153 m)2 = (0.0133 kg)(9.81 m/s2 ) tan(4.83◦ ),or q = 1.29×10−7 C.E25-13 On any corner charge there are seven forces; one from each of the other seven charges.The net force will be the sum. Since all eight charges are the same all of the forces will be repulsive.We need to sketch a diagram to show how the charges are labeled. 2 1 6 4 7 3 8 5 The magnitude of the force of charge 2 on charge 1 is 1 q2 F12 = 2 , 4π 0 r12where r12 = a, the length of a side. Since both charges are the same we wrote q 2 . By symmetry weexpect that the magnitudes of F12 , F13 , and F14 will all be the same and they will all be at rightangles to each other directed along the edges of the cube. Written in terms of vectors the forces 5
• would be 1 q2 ˆ F12 = i, 4π 0 a2 1 q2 ˆ F13 = j, 4π 0 a2 1 q2 ˆ F14 = k. 4π 0 a2 The force from charge 5 is 1 q2 F15 = 2 , 4π 0 r15and is directed along the side diagonal away from charge 5. The distance r15 is also the side diagonaldistance, and can be found from r15 = a2 + a2 = 2a2 , 2then 1 q2 F15 = . 4π 0 2a2By symmetry we expect that the magnitudes of F15 , F16 , and F17 will all be the same and they willall be directed along the diagonals of the faces of the cube. In terms of components we would have 1 q2 √ √ F15 = ˆ 2 + k/ 2 , j/ ˆ 4π 0 2a2 1 q2 √ √ F16 = ˆ 2 + k/ 2 , i/ ˆ 4π 0 2a2 1 q2 √ √ F17 = ˆ 2 +ˆ 2 . i/ j/ 4π 0 2a2 The last force is the force from charge 8 on charge 1, and is given by 1 q2 F18 = 2 , 4π 0 r18and is directed along the cube diagonal away from charge 8. The distance r18 is also the cubediagonal distance, and can be found from r18 = a2 + a2 + a2 = 3a2 , 2then in term of components 1 q2 ˆ √ √ √ F18 = 2 i/ 3 + ˆ 3 + k/ 3 . j/ ˆ 4π 0 3a We can add the components together. By symmetry we expect the same answer for each com-ponents, so we’ll just do one. How about ˆ This component has contributions from charge 2, 6, 7, i.and 8: 1 q2 1 2 1 2 + √ + √ , 4π 0 a 1 2 2 3 3or 1 q2 (1.90) 4π 0 a2 √The three components add according to Pythagoras to pick up a final factor of 3, so q2 F net = (0.262) 2 . 0a 6
• E25-14 (a) Yes. Changing the sign of y will change the sign of Fy ; since this is equivalent toputting the charge q0 on the “other” side, we would expect the force to also push in the “other”direction. (b) The equation should look Eq. 25-15, except all y’s should be replaced by x’s. Then 1 q0 q Fx = . 4π 0 x x2 + L2 /4 (c) Setting the particle a distance d away should give a force with the same magnitude as 1 q0 q F = . 4π 0 d d2 + L2 /4This force is directed along the 45◦ line, so Fx = F cos 45◦ and Fy = F sin 45◦ . (d) Let the distance be d = x2 + y 2 , and then use the fact that Fx /F = cos θ = x/d. Then x 1 x q0 q Fx = F = 2 + y 2 + L2 /4)3/2 . d 4π 0 (xand y 1 y q0 q Fy = F = 2 + y 2 + L2 /4)3/2 . d 4π 0 (xE25-15 (a) The equation is valid for both positive and negative z, so in vector form it would read ˆ 1 q0 q z ˆ F = Fz k = k. 4π 0 (z 2 + R2 )3/2 (b) The equation is not valid for both positive and negative√ Reversing the sign of z should z.reverse the sign of Fz , and one way to fix this is to write 1 = z/ z 2 . Then ˆ 1 2q0 qz 1 1 ˆ F = Fz k = √ −√ k. 4π 0 R2 z 2 z2E25-16 Divide the rod into small differential lengths dr, each with charge dQ = (Q/L)dr. Eachdifferential length contributes a differential force 1 q dQ 1 qQ dF = 2 = dr. 4π 0 r 4π 0 r2 LIntegrate: x+L 1 qQ F = dF = dr, x 4π 0 r2 L 1 qQ 1 1 = − 4π 0 L x x + LE25-17 You must solve Ex. 16 before solving this problem! q0 refers to the charge that had beencalled q in that problem. In either case the distance from q0 will be the same regardless of the signof q; if q = Q then q will be on the right, while if q = −Q then q will be on the left. Setting the forces equal to each other one gets 1 qQ 1 1 1 qQ − = , 4π 0 L x x+L 4π 0 r2or r= x(x + L). 7
• E25-18 You must solve Ex. 16 and Ex. 17 before solving this problem. If all charges are positive then moving q0 off axis will result in a net force away from the axis.That’s unstable. If q = −Q then both q and Q are on the same side of q0 . Moving q0 closer to q will result in theattractive force growing faster than the repulsive force, so q0 will move away from equilibrium. E25-19 We can start with the work that was done for us on Page 577, except since we areconcerned with sin θ = z/r we would have 1 q0 λ dz z dFx = dF sin θ = . 4π 0 (y 2 + z 2 ) y2 + z2We will need to take into consideration that λ changes sign for the two halves of the rod. Then 0 L/2 q0 λ −z dz +z dz Fx = + , 4π 0 −L/2 (y 2 + z 2 )3/2 0 (y 2 + z 2 )3/2 L/2 q0 λ z dz = , 2π 0 0 (y 2 + z 2 )3/2 L/2 q0 λ −1 = , 2π 0 y2 + z2 0 q0 λ 1 1 = − . 2π 0 y y2 + (L/2)2E25-20 Use Eq. 25-15 to find the magnitude of the force from any one rod, but write it as 1 qQ F = , 4π 0 r r2 + L2 /4where r2 = z 2 + L2 /4. The component of this along the z axis is Fz = F z/r. Since there are 4 rods,we have 1 qQz 1 qQz F = ,= , π 0r r 2 2 + L2 /4 π 0 (z 2 + L2 /4) z 2 + L2 /2Equating the electric force with the force of gravity and solving for Q, π 0 mg 2 Q= (z + L2 /4) z 2 + L2 /2; qzputting in the numbers,π(8.85×10−12 C2 /N·m2 )(3.46×10−7 kg)(9.8m/s2 ) ((0.214m)2+(0.25m)2 /4) (0.214m)2 +(0.25m)2 /2 (2.45×10−12 C)(0.214 m)so Q = 3.07×10−6 C. E25-21 In each case we conserve charge by making sure that the total number of protons is thesame on both sides of the expression. We also need to conserve the number of neutrons. (a) Hydrogen has one proton, Beryllium has four, so X must have five protons. Then X must beBoron, B. (b) Carbon has six protons, Hydrogen has one, so X must have seven. Then X is Nitrogen, N. (c) Nitrogen has seven protons, Hydrogen has one, but Helium has two, so X has 7 + 1 − 2 = 6protons. This means X is Carbon, C. 8
• E25-22 (a) Use Eq. 25-4: (8.99×109 N·m2 /C2 )(2)(90)(1.60×10−19 C)2 F = = 290 N. (12×10−15 m)2 (b) a = (290 N)/(4)(1.66×10−27 kg) = 4.4×1028 m/s2 .E25-23 Use Eq. 25-4: (8.99×109 N·m2 /C2 )(1.60×10−19 C)2 F = = 2.89×10−9 N. (282×10−12 m)2E25-24 (a) Use Eq. 25-4: (3.7×10−9 N)(5.0×10−10 m)2 q= = 3.20×10−19 C. (8.99×109 N·m2 /C2 ) (b) N = (3.20×10−19 C)/(1.60×10−19 C) = 2.E25-25 Use Eq. 25-4, 1 q1 q2 ( 1 1.6 × 10−19 C)( 1 1.6 × 10−19 C) 3 3 F = 2 = = 3.8 N. 4π 0 r12 4π(8.85 × 10−12 C2 /N · m2 )(2.6 × 10−15 m)2E25-26 (a) N = (1.15×10−7 C)/(1.60×10−19 C) = 7.19×1011 . (b) The penny has enough electrons to make a total charge of −1.37×105 C. The fraction is then (1.15×10−7 C)/(1.37×105 C) = 8.40×10−13 .E25-27 Equate the magnitudes of the forces: 1 q2 = mg, 4π 0 r2so (8.99×109 N·m2 /C2 )(1.60×10−19 C)2 r= = 5.07 m (9.11×10−31 kg)(9.81 m/s2 )E25-28 Q = (75.0 kg)(−1.60×10−19 C)/(9.11×10−31 kg) = −1.3×1013 C. 3E25-29 The mass of water is (250 cm3 )(1.00 g/cm ) = 250 g. The number of moles of water is(250 g)/(18.0 g/mol) = 13.9 mol. The number of water molecules is (13.9 mol)(6.02×1023 mol−1 ) =8.37×1024 . Each molecule has ten protons, so the total positive charge is Q = (8.37×1024 )(10)(1.60×10−19 C) = 1.34×107 C.E25-30 The total positive charge in 0.250 kg of water is 1.34×107 C. Mary’s imbalance is then q1 = (52.0)(4)(1.34×107 C)(0.0001) = 2.79×105 C,while John’s imbalance is q2 = (90.7)(4)(1.34×107 C)(0.0001) = 4.86×105 C,The electrostatic force of attraction is then 1 q1 q2 (2.79×105 )(4.86×105 ) F = = (8.99×109 N · m2 /C2 ) = 1.6×1018 N. 4π 0 r2 (28.0 m)2 9
• E25-31 (a) The gravitational force of attraction between the Moon and the Earth is GM E M M FG = , R2where R is the distance between them. If both the Earth and the moon are provided a charge q,then the electrostatic repulsion would be 1 q2 FE = . 4π 0 R2Setting these two expression equal to each other, q2 = GM E M M , 4π 0which has solution q = 4π 0 GM E M M , = 4π(8.85×10−12 C2/Nm2 )(6.67×10−11 Nm2/kg2 )(5.98×1024 kg)(7.36×1022 kg), = 5.71 × 1013 C. (b) We need (5.71 × 1013 C)/(1.60 × 10−19 C) = 3.57 × 1032protons on each body. The mass of protons needed is then (3.57 × 1032 )(1.67 × 10−27 kg) = 5.97 × 1065 kg.Ignoring the mass of the electron (why not?) we can assume that hydrogen is all protons, so weneed that much hydrogen. P25-1 Assume that the spheres initially have charges q1 and q2 . The force of attraction betweenthem is 1 q1 q2 F1 = 2 = −0.108 N, 4π 0 r12where r12 = 0.500 m. The net charge is q1 + q2 , and after the conducting wire is connected eachsphere will get half of the total. The spheres will have the same charge, and repel with a force of 1 1 (q1 + q2 ) 1 (q1 + q2 ) 2 2 F2 = 2 = 0.0360 N. 4π 0 r12Since we know the separation of the spheres we can find q1 + q2 quickly, 2 q1 + q2 = 2 4π 0 r12 (0.0360 N) = 2.00 µCWe’ll put this back into the first expression and solve for q2 . 1 (2.00 µC − q2 )q2 −0.108 N = 2 , 4π 0 r12 −3.00 × 10−12 C2 = (2.00 µC − q2 )q2 , 0 = −q2 + (2.00 µC)q2 + (1.73 µC)2 . 2The solution is q2 = 3.0 µC or q2 = −1.0 µC. Then q1 = −1.0 µC or q1 = 3.0 µC. 10
• P25-2 The electrostatic force on Q from each q has magnitude qQ/4π 0 a2 , where a is the lengthof the side of the square. The magnitude of the vertical (horizontal) component of the force of Q on √Q is 2Q2 /16π 0 a2 . (a) In order to have a zero net force on Q the magnitudes of the two contributions must balance,so √ 2 2Q qQ 2 = , 16π 0 a 4π 0 a2 √or q = 2Q/4. The charges must actually have opposite charge. (b) No.P25-3 (a) The third charge, q3 , will be between the first two. The net force on the third chargewill be zero if 1 q q3 1 4q q3 2 = 4π 4π 0 r31 2 , 0 r32which will occur if 1 2 = r31 r32The total distance is L, so r31 + r32 = L, or r31 = L/3 and r32 = 2L/3. Now that we have found the position of the third charge we need to find the magnitude. Thesecond and third charges both exert a force on the first charge; we want this net force on the firstcharge to be zero, so 1 q q3 1 q 4q 2 = 4π 4π 0 r13 2 , 0 r12or q3 4q = 2, (L/3)2 Lwhich has solution q3 = −4q/9. The negative sign is because the force between the first and secondcharge must be in the opposite direction to the force between the first and third charge. (b) Consider what happens to the net force on the middle charge if is is displaced a small distancez. If the charge 3 is moved toward charge 1 then the force of attraction with charge 1 will increase.But moving charge 3 closer to charge 1 means moving charge 3 away from charge 2, so the force ofattraction between charge 3 and charge 2 will decrease. So charge 3 experiences more attraction toward the charge that it moves toward, and less attraction to the charge it moves away from. Soundsunstable to me.P25-4 (a) The electrostatic force on the charge on the right has magnitude q2 F = , 4π 0 x2The weight of the ball is W = mg, and the two forces are related by F/W = tan θ ≈ sin θ = x/2L.Combining, 2Lq 2 = 4π 0 mgx3 , so 1/3 q2 L x= . 2π 0 (b) Rearrange and solve for q, 2π(8.85×10−12 C2 /N · m2 )(0.0112 kg)(9.81 m/s2 )(4.70×10−2 m)3 q= = 2.28×10−8 C. (1.22 m) 11
• P25-5 (a) Originally the balls would not repel, so they would move together and touch; aftertouching the balls would “split” the charge ending up with q/2 each. They would then repel again. (b) The new equilibrium separation is 1/3 1/3 (q/2)2 L 1 x = = x = 2.96 cm. 2π 0 mg 4P25-6 Take the time derivative of the expression in Problem 25-4. Then dx 2 x dq 2 (4.70×10−2 m) = = (−1.20×10−9 C/s) = 1.65×10−3 m/s. dt 3 q dt 3 (2.28×10−8 C)P25-7 The force between the two charges is 1 (Q − q)q F = 2 . 4π 0 r12We want to maximize this force with respect to variation in q, this means finding dF/dq and settingit equal to 0. Then dF d 1 (Q − q)q 1 Q − 2q = 2 = 2 . dq dq 4π 0 r12 4π 0 r12This will vanish if Q − 2q = 0, or q = 1 Q. 2P25-8 Displace the charge q a distance y. The net restoring force on q will be approximately qQ 1 y qQ 16 F ≈2 = y. 4π 0 (d/2)2 (d/2) 4π 0 d3Since F/y is effectively a force constant, the period of oscillation is 3 3 1/2 m 0 mπ d T = 2π = . k qQP25-9 Displace the charge q a distance x toward one of the positive charges Q. The net restoringforce on q will be qQ 1 1 F = 2 − , 4π 0 (d/2 − x) (d/2 + x)2 qQ 32 ≈ x. 4π 0 d3Since F/x is effectively a force constant, the period of oscillation is 3 3 1/2 m 0 mπ d T = 2π = . k 2qQ 12
• P25-10 (a) Zero, by symmetry. (b) Removing a positive Cesium ion is equivalent to adding a singly charged negative ion at thatsame location. The net force is then F = e2 /4π 0 r2 ,where r is the distance between the Chloride ion and the newly placed negative ion, or r= 3(0.20×10−9 m)2The force is then (1.6×10−19 C)2 F = = 1.92×10−9 N. 4π(8.85×10−12 C2 /N · m2 )3(0.20×10−9 m)2P25-11 We can pretend that this problem is in a single plane containing all three charges. Themagnitude of the force on the test charge q0 from the charge q on the left is 1 q q0 Fl = . 4π 0 (a2 + R2 )A force of identical magnitude exists from the charge on the right. we need to add these two forcesas vectors. Only the components along R will survive, and each force will contribute an amount R F l sin θ = F l √ , R2+ a2so the net force on the test particle will be 2 q q0 R √ . 4π 0 (a2 + R2 ) R2 + a2We want to find the maximum value as a function of R. This means take the derivative, and set itequal to zero. The derivative is 2q q0 1 3R2 − 2 , 4π 0 (a2 +R 2 )3/2 (a + R2 )5/2which will vanish when a2 + R2 = 3R2 , √a simple quadratic equation with solutions R = ±a/ 2. 13
• E26-1 E = F/q = ma/q. Then E = (9.11×10−31 kg)(1.84×109 m/s2 )/(1.60×10−19 C) = 1.05×10−2 N/C.E26-2 The answers to (a) and (b) are the same! F = Eq = (3.0×106 N/C)(1.60×10−19 C) = 4.8×10−13 N.E26-3 F = W , or Eq = mg, so mg (6.64 × 10−27 kg)(9.81 m/s2 ) E= = = 2.03 × 10−7 N/C. q 2(1.60 × 10−19 C)The alpha particle has a positive charge, this means that it will experience an electric force whichis in the same direction as the electric field. Since the gravitational force is down, the electric force,and consequently the electric field, must be directed up.E26-4 (a) E = F/q = (3.0×10−6 N)/(2.0×10−9 C) = 1.5×103 N/C. (b) F = Eq = (1.5×103 N/C)(1.60×10−19 C) = 2.4×10−16 N. (c) F = mg = (1.67×10−27 kg)(9.81 m/s2 ) = 1.6×10−26 N. (d) (2.4×10−16 N)/(1.6×10−26 N) = 1.5×1010 .E26-5 Rearrange E = q/4π 0 r2 , q = 4π(8.85×10−12 C2 /N · m2 )(0.750 m)2 (2.30 N/C) = 1.44×10−10 C.E26-6 p = qd = (1.60×10−19 C)(4.30×10−9 ) = 6.88×10−28 C · m.E26-7 Use Eq. 26-12 for points along the perpendicular bisector. Then 1 p (3.56 × 10−29 C · m) E= = (8.99 × 109 N · m2 /C2 ) = 1.95 × 104 N/C. 4π 0 x3 (25.4 × 10−9 m)3E26-8 If the charges on the line x = a where +q and −q instead of +2q and −2q then at thecenter of the square E = 0 by symmetry. This simplifies the problem into finding E for a charge +qat (a, 0) and −q at (a, a). This is a dipole, and the field is given by Eq. 26-11. For this exercise wehave x = a/2 and d = a, so 1 qa E= , 4π 0 [2(a/2)2 ]3/2or, putting in the numbers, E = 1.11×105 N/C.E26-9 The charges at 1 and 7 are opposite and can be effectively replaced with a single charge of−6q at 7. The same is true for 2 and 8, 3 and 9, on up to 6 and 12. By symmetry we expect thefield to point along a line so that three charges are above and three below. That would mean 9:30.E26-10 If both charges are positive then Eq. 26-10 would read E = 2E+ sin θ, and Eq. 26-11would look like 1 q x E = 2 2 + (d/2)2 , 4π 0 x x2 + (d/2)2 1 q x ≈ 2 √ 4π 0 x2 x2when x d. This can be simplified to E = 2q/4π 0 x2 . 14
• E26-11 Treat the two charges on the left as one dipole and treat the two charges on the right asa second dipole. Point P is on the perpendicular bisector of both dipoles, so we can use Eq. 26-12to find the two fields. For the dipole on the left p = 2aq and the electric field due to this dipole at P has magnitude 1 2aq El = 4π 0 (x + a)3and is directed up. For the dipole on the right p = 2aq and the electric field due to this dipole at P has magnitude 1 2aq Er = 4π 0 (x − a)3and is directed down. The net electric field at P is the sum of these two fields, but since the two component fields pointin opposite directions we must actually subtract these values, E = Er − El, 2aq 1 1 = 3 − , 4π 0 (x − a) (x + a)3 aq 1 1 1 = 3 3 − . 2π 0 x (1 − a/x) (1 + a/x)3We can use the binomial expansion on the terms containing 1 ± a/x, aq 1 E ≈ ((1 + 3a/x) − (1 − 3a/x)) , 2π 0 x3 aq 1 = (6a/x) , 2π 0 x3 3(2qa2 ) = . 2π 0 x4E26-12 Do a series expansion on the part in the parentheses 1 1 R2 R2 1− ≈1− 1− = . 1 + R2 /z 2 2 z2 2z 2Substitute this in, σ R2 π Q Ez ≈ 2 π = . 2 0 2z 4π 0 z 2E26-13 At the surface z = 0 and Ez = σ/2 0 . Half of this value occurs when z is given by 1 z =1− √ , 2 z 2 + R2 √which can be written as z 2 + R2 = (2z)2 . Solve this, and z = R/ 3.E26-14 Look at Eq. 26-18. The electric field will be a maximum when z/(z 2 + R2 )3/2 is amaximum. Take the derivative of this with respect to z, and get 1 3 2z 2 z 2 + R2 − 3z 2 − = . (z 2 + R2 )3/2 2 (z 2 + R2 )5/2 (z 2 + R2 )5/2 √This will vanish when the numerator vanishes, or when z = R/ 2. 15
• E26-15 (a) The electric field strength just above the center surface of a charged disk is given byEq. 26-19, but with z = 0, σ E= 2 0The surface charge density is σ = q/A = q/(πR2 ). Combining, q = 2 0 πR2 E = 2(8.85 × 10−12 C2 /N · m2 )π(2.5 × 10−2 m)2 (3 × 106 N/C) = 1.04 × 10−7 C.Notice we used an electric field strength of E = 3 × 106 N/C, which is the field at air breaks downand sparks happen. (b) We want to find out how many atoms are on the surface; if a is the cross sectional area ofone atom, and N the number of atoms, then A = N a is the surface area of the disk. The numberof atoms is A π(0.0250 m)2 N= = = 1.31 × 1017 a (0.015 × 10−18 m2 ) (c) The total charge on the disk is 1.04 × 10−7 C, this corresponds to (1.04 × 10−7 C)/(1.6 × 10−19 C) = 6.5 × 1011electrons. (We are ignoring the sign of the charge here.) If each surface atom can have at most oneexcess electron, then the fraction of atoms which are charged is (6.5 × 1011 )/(1.31 × 1017 ) = 4.96 × 10−6 ,which isn’t very many.E26-16 Imagine switching the positive and negative charges. The electric field would also needto switch directions. By symmetry, then, the electric field can only point vertically down. Keepingonly that component, π/2 1 λdθ E = 2 sin θ, 0 4π 0 r2 2 λ = . 4π 0 r2But λ = q/(π/2), so E = q/π 2 0 r2 .E26-17 We want to fit the data to Eq. 26-19, σ z Ez = 1− √ . 2 0 z2 + R2There are only two variables, R and q, with q = σπR2 . We can find σ very easily if we assume that the measurements have no error because then at thesurface (where z = 0), the expression for the electric field simplifies to σ E= . 2 0Then σ = 2 0 E = 2(8.854 × 10−12 C2 /N · m2 )(2.043 × 107 N/C) = 3.618 × 10−4 C/m2 . Finding the radius will take a little more work. We can choose one point, and make that thereference point, and then solve for R. Starting with σ z Ez = 1− √ , 2 0 z2 + R2 16
• and then rearranging, 2 0 Ez z = 1− √ , σ z 2 + R2 2 0 Ez 1 = 1− , σ 1 + (R/z)2 1 2 0 Ez = 1− , 1 + (R/z)2 σ 1 1 + (R/z)2 = 2, (1 − 2 0 Ez /σ) R 1 = 2 − 1. z (1 − 2 0 Ez /σ)Using z = 0.03 m and Ez = 1.187 × 107 N/C, along with our value of σ = 3.618 × 10−4 C/m2 , wefind R 1 = 2 − 1, z (1 − 2(8.854×10−12 C2/Nm2 )(1.187×107 N/C)/(3.618×10−4 C/m2 )) R = 2.167(0.03 m) = 0.065 m. (b) And now find the charge from the charge density and the radius, q = πR2 σ = π(0.065 m)2 (3.618 × 10−4 C/m2 ) = 4.80 µC.E26-18 (a) λ = −q/L. (b) Integrate: L+a 1 E = λ dxx2 , a 4π 0 λ 1 1 = − , 4π 0 a L + a q 1 = , 4π 0 a(L + a)since λ = q/L. (c) If a L then L can be replaced with 0 in the above expression.E26-19 A sketch of the field looks like this. 17
• E26-20 (a) F = Eq = (40 N/C)(1.60×10−19 C) = 6.4×10−18 N (b) Lines are twice as far apart, so the field is half as large, or E = 20N/C.E26-21 Consider a view of the disk on edge.E26-22 A sketch of the field looks like this. 18
• E26-23 To the right.E26-24 (a) The electric field is zero nearer to the smaller charge; since the charges have oppositesigns it must be to the right of the +2q charge. Equating the magnitudes of the two fields, 2q 5q = , 4π 0 x2 4π 0 (x + a)2or √ √ 5x = 2(x + a),which has solution √ 2a x= √ √ = 2.72a. 5− 2E26-25 This can be done quickly with a spreadsheet. E x dE26-26 (a) At point A, 1 q −2q 1 −q E= − − = , 4π 0 d2 (2d)2 4π 0 2d2 19
• where the negative sign indicates that E is directed to the left. At point B, 1 q −2q 1 6q E= 2 − 2 = , 4π 0 (d/2) (d/2) 4π 0 d2where the positive sign indicates that E is directed to the right. At point C, 1 q −2q 1 −7q E= 2 + 2 = , 4π 0 (2d) d 4π 0 4d2where the negative sign indicates that E is directed to the left. E26-27 (a) The electric field does (negative) work on the electron. The magnitude of this workis W = F d, where F = Eq is the magnitude of the electric force on the electron and d is the distancethrough which the electron moves. Combining, W = F · d = q E · d,which gives the work done by the electric field on the electron. The electron originally possessed a 1kinetic energy of K = 2 mv 2 , since we want to bring the electron to a rest, the work done must benegative. The charge q of the electron is negative, so E and d are pointing in the same direction,and E · d = Ed. By the work energy theorem, 1 W = ∆K = 0 − mv 2 . 2We put all of this together and find d, W −mv 2 −(9.11×10−31 kg)(4.86 × 106 m/s)2 d= = = = 0.0653 m. qE 2qE 2(−1.60×10−19 C)(1030 N/C) (b) Eq = ma gives the magnitude of the acceleration, and v f = v i + at gives the time. Butv f = 0. Combining these expressions, mv i (9.11×10−31 kg)(4.86 × 106 m/s) t=− =− = 2.69×10−8 s. Eq (1030 N/C)(−1.60×10−19 C) (c) We will apply the work energy theorem again, except now we don’t assume the final kineticenergy is zero. Instead, W = ∆K = K f − K i ,and dividing through by the initial kinetic energy to get the fraction lost, W Kf − Ki = = fractional change of kinetic energy. Ki KiBut K i = 1 mv 2 , and W = qEd, so the fractional change is 2 W qEd (−1.60×10−19 C)(1030 N/C)(7.88×10−3 m) = 1 2 = 1 −31 kg)(4.86 × 106 m/s)2 = −12.1%. Ki 2 mv 2 (9.11×10E26-28 (a) a = Eq/m = (2.16×104 N/C)(1.60×10−19 C)/(1.67×10−27 kg) = 2.07×1012 m/s2 . √ (b) v = 2ax = 2(2.07×1012 m/s2 )(1.22×10−2 m) = 2.25×105 m/s. 20
• E26-29 (a) E = 2q/4π 0 r2 , or (1.88×10−7 C) E= = 5.85×105 N/C. 2π(8.85×10−12 C2 /N · m2 )(0.152 m/2)2 (b) F = Eq = (5.85×105 N/C)(1.60×10−19 C) = 9.36×10−14 N.E26-30 (a) The average speed between the plates is (1.95×10−2 m)/(14.7×10−9 s) = 1.33×106 m/s.The speed with which the electron hits the plate is twice this, or 2.65×106 m/s. (b) The acceleration is a = (2.65×106 m/s)/(14.7×10−9 s) = 1.80×1014 m/s2 . The electric fieldthen has magnitude E = ma/q, or E = (9.11×10−31 kg)(1.80×1014 m/s2 )/(1.60×10−19 C) = 1.03×103 N/C.E26-31 The drop is balanced if the electric force is equal to the force of gravity, or Eq = mg.The mass of the drop is given in terms of the density by 4 m = ρV = ρ πr3 . 3Combining, mg 4πρr3 g 4π(851 kg/m3 )(1.64×10−6 m)3 (9.81 m/s2 ) q= = = = 8.11×10−19 C. E 3E 3(1.92×105 N/C)We want the charge in terms of e, so we divide, and get q (8.11×10−19 C) = = 5.07 ≈ 5. e (1.60×10−19 C)E26-32 (b) F = (8.99×109 N · m2 /C2 )(2.16×10−6 C)(85.3×10−9 C)/(0.117m)2 = 0.121 N. (a) E2 = F/q1 = (0.121 N)/(2.16×10−6 C) = 5.60×104 N/C. E1 = F/q2 = (0.121 N)/(85.3×10−9 C) = 1.42×106 N/C. E26-33 If each value of q measured by Millikan was a multiple of e, then the difference betweenany two values of q must also be a multiple of q. The smallest difference would be the smallestmultiple, and this multiple might be unity. The differences are 1.641, 1.63, 1.60, 1.63, 3.30, 3.35,3.18, 3.24, all times 10−19 C. This is a pretty clear indication that the fundamental charge is on theorder of 1.6 × 10−19 C. If so, the likely number of fundamental charges on each of the drops is shownbelow in a table arranged like the one in the book: 4 8 12 5 10 14 7 11 16The total number of charges is 87, while the total charge is 142.69 × 10−19 C, so the average chargeper quanta is 1.64 × 10−19 C. 21
• E26-34 Because of the electric field the acceleration toward the ground of a charged particle isnot g, but g ± Eq/m, where the sign depends on the direction of the electric field. (a) If the lower plate is positively charged then a = g − Eq/m. Replace g in the pendulum periodexpression by this, and then L T = 2π . g − Eq/m (b) If the lower plate is negatively charged then a = g + Eq/m. Replace g in the pendulumperiod expression by this, and then L T = 2π . g + Eq/mE26-35 The ink drop travels an additional time t = d/vx , where d is the additional horizontaldistance between the plates and the paper. During this time it travels an additional vertical distancey = vy t , where vy = at = 2y/t = 2yvx /L. Combining, 2yvx t 2yd 2(6.4×10−4 m)(6.8×10−3 m) y = = = = 5.44×10−4 m, L L (1.6×10−2 m)so the total deflection is y + y = 1.18×10−3 m.E26-36 (a) p = (1.48×10−9 C)(6.23×10−6 m) = 9.22×10−15 C · m. (b) ∆U = 2pE = 2(9.22×10−15 C · m)(1100 N/C) = 2.03×10−11 J. E26-37 Use τ = pE sin θ, where θ is the angle between p and E. For this dipole p = qd = 2edor p = 2(1.6 × 10−19 C)(0.78 × 10−9 m) = 2.5 × 10−28 C · m. For all three cases pE = (2.5 × 10−28 C · m)(3.4 × 106 N/C) = 8.5 × 10−22 N · m.The only thing we care about is the angle. (a) For the parallel case θ = 0, so sin θ = 0, and τ = 0. (b) For the perpendicular case θ = 90◦ , so sin θ = 1, and τ = 8.5 × 10−22 N · m.. (c) For the anti-parallel case θ = 180◦ , so sin θ = 0, and τ = 0.E26-38 (c) Equal and opposite, or 5.22×10−16 N. (d) Use Eq. 26-12 and F = Eq. Then 4π 0 x3 F p = , q 4π(8.85×10−12 C2 /N · m2 )(0.285m)3 (5.22×10−16 N) = , (3.16×10−6 C) = 4.25×10−22 C · m.E26-39 The point-like nucleus contributes an electric field 1 Ze E+ = , 4π 0 r2while the uniform sphere of negatively charged electron cloud of radius R contributes an electricfield given by Eq. 26-24, 1 −Zer E− = . 4π 0 R3 22
• The net electric field is just the sum, Ze 1 r E= − 3 4π 0 r2 RE26-40 The shell theorem first described for gravitation in chapter 14 is applicable here since bothelectric forces and gravitational forces fall off as 1/r2 . The net positive charge inside the sphere ofradius d/2 is given by Q = 2e(d/2)3 /R3 = ed3 /4R3 . The net force on either electron will be zero when e2 eQ 4e2 d3 e2 d = = 2 = 3, d2 (d/2)2 d 4R3 Rwhich has solution d = R. P26-1 (a) Let the positive charge be located closer to the point in question, then the electricfield from the positive charge is 1 q E+ = 4π 0 (x − d/2)2and is directed away from the dipole. The negative charge is located farther from the point in question, so 1 q E− = 4π 0 (x + d/2)2and is directed toward the dipole. The net electric field is the sum of these two fields, but since the two component fields point inopposite direction we must actually subtract these values, E = E + − E− , 1 q 1 q = − , 4π 0 (z − d/2)2 4π 0 (z + d/2)2 1 q 1 1 = − 4π 0 z 2 (1 − d/2z)2 (1 + d/2z)2We can use the binomial expansion on the terms containing 1 ± d/2z, 1 q E ≈ ((1 + d/z) − (1 − d/z)) , 4π 0 z 2 1 qd = 2π 0 z 3 (b) The electric field is directed away from the positive charge when you are closer to the positivecharge; the electric field is directed toward the negative charge when you are closer to the negativecharge. In short, along the axis the electric field is directed in the same direction as the dipolemoment.P26-2 The key to this problem will be the expansion of 1 1 3 zd ≈ 2 1 . (x2 + (z ± d/2)2 )3/2 (x + z 2 )3/2 2 x2 + z 2 23
• √for d x2 + z 2 . Far from the charges the electric field of the positive charge has magnitude 1 q E+ = , 4π 0 x2 + (z − d/2)2the components of this are 1 q x Ex,+ = 2 + z2 , 4π 0 x x2+ (z − d/2)2 1 q (z − d/2) Ez,+ = . 4π 0 x2 + z 2 x2 + (z − d/2)2Expand both according to the first sentence, then 1 xq 3 zd Ex,+ ≈ 1+ , 4π 0 (x2 + z 2 )3/2 2 x2 + z 2 1 (z − d/2)q 3 zd Ez,+ = 1+ . 4π 0 (x2 + z 2 )3/2 2 x2 + z 2Similar expression exist for the negative charge, except we must replace q with −q and the + in theparentheses with a −, and z − d/2 with z + d/2 in the Ez expression. All that is left is to add theexpressions. Then 1 xq 3 zd 1 −xq 3 zd Ex = 1+ + 1− , 4π 0 (x2 + z 2 )3/2 2 x2 + z 2 4π 0 (x2 + z 2 )3/2 2 x2 + z 2 1 3xqzd = , 4π 0 (x2 + z 2 )5/2 1 (z − d/2)q 3 zd 1 −(z + d/2)q 3 zd Ez = 1+ 2 + z2 + 1− , 4π 0 (x2 + z 2 )3/2 2x 4π 0 (x2 + z 2 )3/2 2 x2 + z 2 1 3z 2 dq 1 dq = 2 + z 2 )5/2 − , 4π 0 (x 4π 0 (x2 + z 2 )3/2 1 (2z 2 − x2 )dq = . 4π 0 (x2 + z 2 )5/2 √P26-3 (a) Each point on the ring is a distance z 2 + R2 from the point on the axis in question.Since all points are equal distant and subtend the same angle from the axis then the top half of thering contributes q1 z E1z = 2 + R2 ) √ , 4π 0 (x z 2 + R2while the bottom half contributes a similar expression. Add, and q1 + q2 z q z Ez = = , 4π 0 (z 2 + R2 )3/2 4π 0 (z 2 + R2 )3/2which is identical to Eq. 26-18. (b) The perpendicular component would be zero if q1 = q2 . It isn’t, so it must be the differenceq1 − q2 which is of interest. Assume this charge difference is evenly distributed on the top half ofthe ring. If it is a positive difference, then E⊥ must point down. We are only interested then in thevertical component as we integrate around the top half of the ring. Then π 1 (q1 − q2 )/π E⊥ = cos θ dθ, 0 4π 0 z 2 + R2 q1 − q2 1 = 2 2 + R2 . 2π 0 z 24
• P26-4 Use the approximation 1/(z ± d)2 ≈ (1/z 2 )(1 2d/z + 3d2 /z 2 ). Add the contributions: 1 q 2q q E = 2 − 2 + , 4π 0 (z + d) z (z − d)2 q 2d 3d2 2d 3d2 ≈ 2 1− + 2 −2+1+ + 2 , 4π 0 z z z z z 2 q 6d 3Q = = , 4π 0 z 2 z 2 4π 0 z 4where Q = 2qd2 . P26-5 A monopole field falls off as 1/r2 . A dipole field falls off as 1/r3 , and consists of twooppositely charge monopoles close together. A quadrupole field (see Exercise 11 above or readProblem 4) falls off as 1/r4 and (can) consist of two otherwise identical dipoles arranged with anti-parallel dipole moments. Just taking a leap of faith it seems as if we can construct a 1/r6 fieldbehavior by extending the reasoning. First we need an octopole which is constructed from a quadrupole. We want to keep things assimple as possible, so the construction steps are 1. The monopole is a charge +q at x = 0. 2. The dipole is a charge +q at x = 0 and a charge −q at x = a. We’ll call this a dipole at x = a/2 3. The quadrupole is the dipole at x = a/2, and a second dipole pointing the other way at x = −a/2. The charges are then −q at x = −a, +2q at x = 0, and −q at x = a. 4. The octopole will be two stacked, offset quadrupoles. There will be −q at x = −a, +3q at x = 0, −3q at x = a, and +q at x = 2a. 5. Finally, our distribution with a far field behavior of 1/r6 . There will be +q at x = 2a, −4q at x = −a, +6q at x = 0, −4q at x = a, and +q at x = 2a.P26-6 The vertical component of E is simply half of Eq. 26-17. The horizontal component isgiven by a variation of the work required to derive Eq. 26-16, 1 λ dz z dEz = dE sin θ = , 4π 0 y 2 + z 2 y2 + z2which integrates to zero if the limits are −∞ to +∞, but in this case, ∞ λ 1 Ez = dEz = . 0 4π 0 zSince the vertical and horizontal components are equal then E makes an angle of 45◦ .P26-7 (a) Swap all positive and negative charges in the problem and the electric field must reversedirection. But this is the same as flipping the problem over; consequently, the electric field mustpoint parallel to the rod. This only holds true at point P , because point P doesn’t move when youflip the rod. 25
• (b) We are only interested in the vertical component of the field as contributed from each pointon the rod. We can integrate only half of the rod and double the answer, so we want to evaluate L/2 1 λ dz z Ez = 2 , 0 4π 0 y 2 + z 2 y2 + z2 2λ (L/2)2 + y 2 − y = . 4π 0 y (L/2)2 + y 2 (c) The previous expression is exact. If y L, then the expression simplifies with a Taylorexpansion to λ L2 Ez = , 4π 0 y 3which looks similar to a dipole.P26-8 Evaluate R 1 z dq E= , 0 4π 0 (z 2 + r2 )3/2where r is the radius of the ring, z the distance to the plane of the ring, and dq the differentialcharge on the ring. But r2 + z 2 = R2 , and dq = σ(2πr dr), where σ = q/2πR2 . Then R √ q R2 − r2 r dr E = , 0 4π 0 R5 q 1 = . 4π 0 3R2 P26-9 The key statement is the second italicized paragraph on page 595; the number of fieldlines through a unit cross-sectional area is proportional to the electric field strength. If the exponentis n, then the electric field strength a distance r from a point charge is kq E= , rnand the total cross sectional area at a distance r is the area of a spherical shell, 4πr2 . Then thenumber of field lines through the shell is proportional to kq EA = 4πr2 = 4πkqr2−n . rnNote that the number of field lines varies with r if n = 2. This means that as we go farther fromthe point charge we need more and more field lines (or fewer and fewer). But the field lines can onlystart on charges, and we don’t have any except for the point charge. We have a problem; we reallydo need n = 2 if we want workable field lines.P26-10 The distance traveled by the electron will be d1 = a1 t2 /2; the distance traveled by theproton will be d2 = a2 t2 /2. a1 and a2 are related by m1 a1 = m2 a2 , since the electric force is thesame (same charge magnitude). Then d1 + d2 = (a1 + a2 )t2 /2 is the 5.00 cm distance. Divide bythe proton distance, and then d1 + d2 a1 + a2 m2 = = + 1. d2 a2 m1Then d2 = (5.00×10−2 m)/(1.67×10−27 /9.11×10−31 + 1) = 2.73×10−5 m. 26
• P26-11 This is merely a fancy projectile motion problem. vx = v0 cos θ while vy,0 = v0 sin θ. Thex and y positions are x = vx t and 1 2 ax2 y= at + vy,0 t = 2 + x tan θ. 2 2v0 cos2 θ The acceleration of the electron is vertically down and has a magnitude of F Eq (1870 N/C)(1.6×10−19 C) a= = = = 3.284×1014 m/s2 . m m (9.11×10−31 kg) We want to find out how the vertical velocity of the electron at the location of the top plate. Ifwe get an imaginary answer, then the electron doesn’t get as high as the top plate. vy = vy,0 2 + 2a∆y, = (5.83×106 m/s)2 sin(39◦ )2 + 2(−3.284×1014 m/s2 )(1.97×10−2 m), = 7.226×105 m/s.This is a real answer, so this means the electron either hits the top plate, or it misses both plates.The time taken to reach the height of the top plate is ∆vy (7.226×105 m/s) − (5.83×106 m/s) sin(39◦ ) t= = = 8.972×10−9 s. a (−3.284×1014 m/s2 )In this time the electron has moved a horizontal distance of x = (5.83×106 m/s) cos(39◦ )(8.972×10−9 s) = 4.065×10−2 m.This is clearly on the upper plate.P26-12 Near the center of the ring z R, so a Taylor expansion yields λ z E= . 2 0 R2The force on the electron is F = Ee, so the effective “spring” constant is k = eλ/2 0 R2 . This means k eλ eq ω= = = . m 2 0 mR2 4π 0 mR3P26-13 U = −pE cos θ, so the work required to flip the dipole is W = −pE [cos(θ0 + π) − cos θ0 ] = 2pE cos θ0 .P26-14 If the torque on a system is given by |τ | = κθ, where κ is a constant, then the frequencyof oscillation of the system is f = κ/I/2π. In this case τ = pE sin θ ≈ pEθ, so f= pE/I/2π. 27
• P26-15 Use the a variation of the exact result from Problem 26-1. The two charge are positive,but since we will eventually focus on the area between the charges we must subtract the two fieldcontributions, since they point in opposite directions. Then q 1 1 Ez = − 4π 0 (z − a/2)2 (z + a/2)2and then take the derivative, dEz q 1 1 =− − . dz 2π 0 (z − a/2)3 (z + a/2)3Applying the binomial expansion for points z a, dEz 8q 1 1 1 = − 3 3 − , dz 2π 0 a (2z/a − 1) (2z/a + 1)3 8q 1 ≈ − (−(1 + 6z/a) − (1 − 6z/a)) , 2π 0 a3 8q 1 = . π 0 a3There were some fancy sign flips in the second line, so review those steps carefully! (b) The electrostatic force on a dipole is the difference in the magnitudes of the electrostaticforces on the two charges that make up the dipole. Near the center of the above charge arrangementthe electric field behaves as dEz Ez ≈ Ez (0) + z + higher ordered terms. dz z=0The net force on a dipole is dEz dEz F+ − F− = q(E+ − E− ) = q Ez (0) + z+ − Ez (0) − z− dz z=0 dz z=0where the “+” and “-” subscripts refer to the locations of the positive and negative charges. Thislast line can be simplified to yield dEz dEz q (z+ − z− ) = qd . dz z=0 dz z=0 28
• E27-1 ΦE = (1800 N/C)(3.2×10−3 m)2 cos(145◦ ) = −7.8×10−3 N · m2 /C.E27-2 The right face has an area element given by A = (1.4 m)2ˆ j. 2 ˆ ˆ = 0. (a) ΦE = A · E = (2.0 m )j · (6 N/C)i (b) ΦE = (2.0 m2 )ˆ · (−2 N/C)ˆ = −4N · m2 /C. j j (c) ΦE = (2.0 m2 )ˆ · [(−3 N/C)ˆ + (4 N/C)k] = 0. j i ˆ (d) In each case the field is uniform so we can simply evaluate ΦE = E · A, where A has sixparts, one for every face. The faces, however, have the same size but are organized in pairs withopposite directions. These will cancel, so the total flux is zero in all three cases.E27-3 (a) The flat base is easy enough, since according to Eq. 27-7, ΦE = E · dA.There are two important facts to consider in order to integrate this expression. E is parallel to theaxis of the hemisphere, E points inward while dA points outward on the flat base. E is uniform, soit is everywhere the same on the flat base. Since E is anti-parallel to dA, E · dA = −E dA, then ΦE = E · dA = − E dA.Since E is uniform we can simplify this as ΦE = − E dA = −E dA = −EA = −πR2 E.The last steps are just substituting the area of a circle for the flat side of the hemisphere. (b) We must first sort out the dot product E dA R θ φ We can simplify the vector part of the problem with E · dA = cos θE dA, so ΦE = E · dA = cos θE dAOnce again, E is uniform, so we can take it out of the integral, ΦE = cos θE dA = E cos θ dAFinally, dA = (R dθ)(R sin θ dφ) on the surface of a sphere centered on R = 0. 29
• We’ll integrate φ around the axis, from 0 to 2π. We’ll integrate θ from the axis to the equator,from 0 to π/2. Then 2π π/2 ΦE = E cos θ dA = E R2 cos θ sin θ dθ dφ. 0 0Pulling out the constants, doing the φ integration, and then writing 2 cos θ sin θ as sin(2θ), π/2 π/2 ΦE = 2πR2 E cos θ sin θ dθ = πR2 E sin(2θ) dθ, 0 0Change variables and let β = 2θ, then we have π 1 ΦE = πR2 E sin β dβ = πR2 E. 0 2E27-4 Through S1 , ΦE = q/ 0 . Through S2 , ΦE = −q/ 0 . Through S3 , ΦE = q/ 0 . Through S4 ,ΦE = 0. Through S5 , ΦE = q/ 0 .E27-5 By Eq. 27-8, q (1.84 µC) ΦE = = = 2.08×105 N · m2 /C. 0 (8.85×10−12 C2 /N · m2 )E27-6 The total flux through the sphere is ΦE = (−1 + 2 − 3 + 4 − 5 + 6)(×103 N · m2 /C) = 3×103 N · m2 /C.The charge inside the die is (8.85×10−12 C2 /N · m2 )(3×103 N · m2 /C) = 2.66×10−8 C.E27-7 The total flux through a cube would be q/ 0 . Since the charge is in the center of the cubewe expect that the flux through any side would be the same, or 1/6 of the total flux. Hence the fluxthrough the square surface is q/6 0 .E27-8 If the electric field is uniform then there are no free charges near (or inside) the net. Theflux through the netting must be equal to, but opposite in sign, from the flux through the opening.The flux through the opening is Eπa2 , so the flux through the netting is −Eπa2 .E27-9 There is no flux through the sides of the cube. The flux through the top of the cube is(−58 N/C)(100 m)2 = −5.8×105 N · m2 /C. The flux through the bottom of the cube is (110 N/C)(100 m)2 = 1.1×106 N · m2 /C.The total flux is the sum, so the charge contained in the cube is q = (8.85×10−12 C2 /N · m2 )(5.2×105 N · m2 /C) = 4.60×10−6 C.E27-10 (a) There is only a flux through the right and left faces. Through the right face ΦR = (2.0 m2 )ˆ · (3 N/C · m)(1.4 m)ˆ = 8.4 N · m2 /C. j jThe flux through the left face is zero because y = 0. 30
• E27-11 There are eight cubes which can be “wrapped” around the charge. Each cube has threeexternal faces that are indistinguishable for a total of twenty-four faces, each with the same fluxΦE . The total flux is q/ 0 , so the flux through one face is ΦE = q/24 0 . Note that this is the fluxthrough faces opposite the charge; for faces which touch the charge the electric field is parallel tothe surface, so the flux would be zero.E27-12 Use Eq. 27-11, λ = 2π 0 rE = 2π(8.85×10−12 C2 /N · m2 )(1.96 m)(4.52×104 N/C) = 4.93×10−6 C/m.E27-13 (a) q = σA = (2.0×10−6 C/m2 )π(0.12 m)(0.42 m) = 3.17×10−7 C. (b) The charge density will be the same! q = σA = (2.0 × 10−6 C/m2 )π(0.08 m)(0.28 m) =1.41×10−7 C.E27-14 The electric field from the sheet on the left is of magnitude E l = σ/2 0 , and points directlyaway from the sheet. The magnitude of the electric field from the sheet on the right is the same,but it points directly away from the sheet on the right. (a) To the left of the sheets the two fields add since they point in the same direction. This meansthat the electric field is E = −(σ/ 0 )ˆi. (b) Between the sheets the two electric fields cancel, so E = 0. (c) To the right of the sheets the two fields add since they point in the same direction. Thismeans that the electric field is E = (σ/ 0 )ˆi.E27-15 The electric field from the plate on the left is of magnitude E l = σ/2 0 , and points directlytoward the plate. The magnitude of the electric field from the plate on the right is the same, but itpoints directly away from the plate on the right. (a) To the left of the plates the two fields cancel since they point in the opposite directions. Thismeans that the electric field is E = 0. (b) Between the plates the two electric fields add since they point in the same direction. Thismeans that the electric field is E = −(σ/ 0 )ˆ i. (c) To the right of the plates the two fields cancel since they point in the opposite directions.This means that the electric field is E = 0.E27-16 The magnitude of the electric field is E = mg/q. The surface charge density on the platesis σ = 0 E = 0 mg/q, or (8.85×10−12 C2 /N · m2 )(9.11×10−31 kg)(9.81 m/s2 ) σ= = 4.94×10−22 C/m2 . (1.60×10−19 C) E27-17 We don’t really need to write an integral, we just need the charge per unit length in thecylinder to be equal to zero. This means that the positive charge in cylinder must be +3.60nC/m.This positive charge is uniformly distributed in a circle of radius R = 1.50 cm, so 3.60nC/m 3.60nC/m ρ= = = 5.09µC/m3 . πR2 π(0.0150 m)2 31
• E27-18 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell.The E field will be perpendicular to the surface, so Gauss’ law will simplify to q enc / 0 = E · dA = E dA = E dA = 4πr2 E. (a) For point P1 the charge enclosed is q enc = 1.26×10−7 C, so (1.26×10−7 C) E= = 3.38×106 N/C. 4π(8.85×10−12 C2 /N · m2 )(1.83×10−2 m)2 (b) Inside a conductor E = 0. E27-19 The proton orbits with a speed v, so the centripetal force on the proton is FC = mv 2 /r.This centripetal force is from the electrostatic attraction with the sphere; so long as the proton isoutside the sphere the electric field is equivalent to that of a point charge Q (Eq. 27-15), 1 Q E= . 4π 0 r2If q is the charge on the proton we can write F = Eq, or mv 2 1 Q =q r 4π 0 r2Solving for Q, 4π 0 mv 2 r Q = , q 4π(8.85×10−12 C2 /N · m2 )(1.67×10−27 kg)(294×103 m/s)2 (0.0113 m) = , (1.60×10−19 C) = −1.13×10−9 C.E27-20 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell.The E field will be perpendicular to the surface, so Gauss’ law will simplify to q enc / 0 = E · dA = E dA = E dA = 4πr2 E. (a) At r = 0.120 m q enc = 4.06×10−8 C. Then (4.06×10−8 C) E= = 2.54×104 N/C. 4π(8.85×10−12 C2 /N · m2 )(1.20×10−1 m)2 (b) At r = 0.220 m q enc = 5.99×10−8 C. Then (5.99×10−8 C) E= = 1.11×104 N/C. 4π(8.85×10−12 C2 /N · m2 )(2.20×10−1 m)2 (c) At r = 0.0818 m q enc = 0 C. Then E = 0. 32
• E27-21 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindricalshell. The E field will be perpendicular to the curved surface and parallel to the end surfaces, soGauss’ law will simplify to q enc / 0 = E · dA = E dA = E dA = 2πrLE,where L is the length of the cylinder. Note that σ = q/2πrL represents a surface charge density. (a) r = 0.0410 m is between the two cylinders. Then (24.1×10−6 C/m2 )(0.0322 m) E= = 2.14×106 N/C. (8.85×10−12 C2 /N · m2 )(0.0410 m)It points outward. (b) r = 0.0820 m is outside the two cylinders. Then (24.1×10−6 C/m2 )(0.0322 m) + (−18.0×10−6 C/m2 )(0.0618 m) E= = −4.64×105 N/C. (8.85×10−12 C2 /N · m2 )(0.0820 m)The negative sign is because it is pointing inward.E27-22 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindricalshell. The E field will be perpendicular to the curved surface and parallel to the end surfaces, soGauss’ law will simplify to q enc / 0 = E · dA = E dA = E dA = 2πrLE,where L is the length of the cylinder. The charge enclosed is q enc = ρdV = ρπL r2 − R2The electric field is given by ρπL r2 − R2 ρ r 2 − R2 E= = . 2π 0 rL 2 0rAt the surface, ρ (2R)2 − R2 3ρR Es = = . 2 0 2R 4 0Solve for r when E is half of this: 3R r 2 − R2 = , 8 2r 3rR = 4r2 − 4R2 , 0 = 4r2 − 3rR − 4R2 .The solution is r = 1.443R. That’s (2R − 1.443R) = 0.557R beneath the surface.E27-23 The electric field must do work on the electron to stop it. The electric field is given byE = σ/2 0 . The work done is W = F d = Eqd. d is the distance in question, so 2 0K 2(8.85×10−12 C2 /N · m2 )(1.15×105 eV) d= = = 0.979 m σq (2.08×10−6 C/m2 )e 33
• E27-24 Let the spherical Gaussian surface have a radius of R and be centered on the origin.Choose the orientation of the axis so that the infinite line of charge is along the z axis. The electricfield is then directed radially outward from the z axis with magnitude E = λ/2π 0 ρ, where ρ is theperpendicular distance from the z axis. Now we want to evaluate ΦE = E · dA,over the surface of the sphere. In spherical coordinates, dA = R2 sin θ dθ dφ, ρ = R sin θ, andE · dA = EA sin θ. Then λ 2λR ΦE = sin θR dθ dφ = . 2π 0 0 E27-25 (a) The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindricalshell. The E field will be perpendicular to the curved surface and parallel to the end surfaces, soGauss’ law will simplify to q enc / 0 = E · dA = E dA = E dA = 2πrLE,where L is the length of the cylinder. Now for the q enc part. If the (uniform) volume charge densityis ρ, then the charge enclosed in the Gaussian cylinder is q enc = ρdV = ρ dV = ρV = πr2 Lρ.Combining, πr2 Lρ/ 0 = E2πrL or E = ρr/2 0 . (b) Outside the charged cylinder the charge enclosed in the Gaussian surface is just the chargein the cylinder. Then q enc = ρdV = ρ dV = ρV = πR2 Lρ.and πR2 Lρ/ 0 = E2πrL,and then finally R2 ρ E= . 2 0rE27-26 (a) q = 4π(1.22 m)2 (8.13×10−6 C/m2 ) = 1.52×10−4 C. (b) ΦE = q/ 0 = (1.52×10−4 C)/(8.85×10−12 C2 /N · m2 ) = 1.72×107 N · m2 /C. (c) E = σ/ 0 = (8.13×10−6 C/m2 )/(8.85×10−12 C2 /N · m2 ) = 9.19×105 N/CE27-27 (a) σ = (2.4×10−6 C)/4π(0.65 m)2 = 4.52×10−7 C/m2 . (b) E = σ/ 0 = (4.52×10−7 C/m2 )/(8.85×10−12 C2 /N · m2 ) = 5.11×104 N/C.E27-28 E = σ/ 0 = q/4πr2 0 .E27-29 (a) The near field is given by Eq. 27-12, E = σ/2 0 , so (6.0×10−6 C)/(8.0×10−2 m)2 E≈ = 5.3×107 N/C. 2(8.85×10−12 C2 /N · m2 ) (b) Very far from any object a point charge approximation is valid. Then 1 q 1 (6.0×10−6 C) E= = = 60N/C. 4π 0 r2 4π(8.85×10−12 C2 /N · m2 ) (30 m)2 34
• P27-1 For a spherically symmetric mass distribution choose a spherical Gaussian shell. Then g · dA = g dA = g dA = 4πr2 g.Then Φg gr2 = = −m, 4πG Gor Gm g=− . r2The negative sign indicates the direction; g point toward the mass center.P27-2 (a) The flux through all surfaces except the right and left faces will be zero. Through theleft face, √ Φl = −Ey A = −b aa2 .Through the right face, √ Φr = Ey A = b 2aa2 .The net flux is then √ √ Φ = ba5/2 ( 2 − 1) = (8830 N/C · m1/2 )(0.130 m)5/2 ( 2 − 1) = 22.3 N · m2 /C. (b) The charge enclosed is q = (8.85×10−12 C2 /N · m2 )(22.3 N · m2 /C) = 1.97×10−10 C.P27-3 The net force on the small sphere is zero; this force is the vector sum of the force of gravityW , the electric force FE , and the tension T . θ T FE W These forces are related by Eq = mg tan θ. We also have E = σ/2 0 , so 2 0 mg tan θ σ = , q 2(8.85×10−12 C2 /N · m2 )(1.12×10−6 kg)(9.81 m/s2 ) tan(27.4◦ ) = , (19.7×10−9 C) = 5.11×10−9 C/m2 . 35
• P27-4 The materials are conducting, so all charge will reside on the surfaces. The electric fieldinside any conductor is zero. The problem has spherical symmetry, so use a Gaussian surface whichis a spherical shell. The E field will be perpendicular to the surface, so Gauss’ law will simplify to q enc / 0 = E · dA = E dA = E dA = 4πr2 E.Consequently, E = q enc /4π 0 r2 . (a) Within the sphere E = 0. (b) Between the sphere and the shell q enc = q. Then E = q/4π 0 r2 . (c) Within the shell E = 0. (d) Outside the shell q enc = +q − q = 0. Then E = 0. (e) Since E = 0 inside the shell, q enc = 0, this requires that −q reside on the inside surface. Thisis no charge on the outside surface.P27-5 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindricalshell. The E field will be perpendicular to the curved surface and parallel to the end surfaces, soGauss’ law will simplify to q enc / 0 = E · dA = E dA = E dA = 2πrLE,where L is the length of the cylinder. Consequently, E = q enc /2π 0 rL. (a) Outside the conducting shell q enc = +q − 2q = −q. Then E = −q/2π 0 rL. The negative signindicates that the field is pointing inward toward the axis of the cylinder. (b) Since E = 0 inside the conducting shell, q enc = 0, which means a charge of −q is on theinside surface of the shell. The remaining −q must reside on the outside surface of the shell. (c) In the region between the cylinders q enc = +q. Then E = +q/2π 0 rL. The positive signindicates that the field is pointing outward from the axis of the cylinder.P27-6 Subtract Eq. 26-19 from Eq. 26-20. Then σ z E= √ . 2 0 z 2 + R2 P27-7 This problem is closely related to Ex. 27-25, except for the part concerning q enc . We’llset up the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered onthe axis of the physical cylinder. For Gaussian surfaces of radius r < R, there is no charge enclosedwhile for Gaussian surfaces of radius r > R, q enc = λl. We’ve already worked out the integral E · dA = 2πrlE, tubefor the cylinder, and then from Gauss’ law, q enc = 0 E · dA = 2π 0 rlE. tube (a) When r < R there is no enclosed charge, so the left hand vanishes and consequently E = 0inside the physical cylinder. (b) When r > R there is a charge λl enclosed, so λ E= . 2π 0 r 36
• P27-8 This problem is closely related to Ex. 27-25, except for the part concerning q enc . We’ll setup the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered on theaxis of the physical cylinders. For Gaussian surfaces of radius r < a, there is no charge enclosedwhile for Gaussian surfaces of radius b > r > a, q enc = λl. We’ve already worked out the integral E · dA = 2πrlE, tubefor the cylinder, and then from Gauss’ law, q enc = 0 E · dA = 2π 0 rlE. tube (a) When r < a there is no enclosed charge, so the left hand vanishes and consequently E = 0inside the inner cylinder. (b) When b > r > a there is a charge λl enclosed, so λ E= . 2π 0 rP27-9 Uniform circular orbits require a constant net force towards the center, so F = Eq =λq/2π 0 r. The speed of the positron is given by F = mv 2 /r; the kinetic energy is K = mv 2 /2 =F r/2. Combining, λq K = , 4π 0 (30×10−9 C/m)(1.6×10−19 C) = , 4π((8.85 × 10−12 C2 /N · m2 ) = 4.31×10−17 J = 270 eV.P27-10 λ = 2π 0 rE, so q = 2π(8.85×10−12 C2 /N · m2 )(0.014 m)(0.16 m)(2.9×104 N/C) = 3.6×10−9 C. P27-11 (a) Put a spherical Gaussian surface inside the shell centered on the point charge. Gauss’law states q enc E · dA = . 0Since there is spherical symmetry the electric field is normal to the spherical Gaussian surface, andit is everywhere the same on this surface. The dot product simplifies to E · dA = E dA, while sinceE is a constant on the surface we can pull it out of the integral, and we end up with q E dA = , 0where q is the point charge in the center. Now dA = 4πr2 , where r is the radius of the Gaussiansurface, so q E= . 4π 0 r2 (b) Repeat the above steps, except put the Gaussian surface outside the conducting shell. Keepit centered on the charge. Two things are different from the above derivation: (1) r is bigger, and 37
• (2) there is an uncharged spherical conducting shell inside the Gaussian surface. Neither change willaffect the surface integral or q enc , so the electric field outside the shell is still q E= , 4π 0 r2 (c) This is a subtle question. With all the symmetry here it appears as if the shell has no effect;the field just looks like a point charge field. If, however, the charge were moved off center the fieldinside the shell would become distorted, and we wouldn’t be able to use Gauss’ law to find it. Sothe shell does make a difference. Outside the shell, however, we can’t tell what is going on inside the shell. So the electric fieldoutside the shell looks like a point charge field originating from the center of the shell regardless ofwhere inside the shell the point charge is placed! (d) Yes, q induces surface charges on the shell. There will be a charge −q on the inside surfaceand a charge q on the outside surface. (e) Yes, as there is an electric field from the shell, isn’t there? (f) No, as the electric field from the outside charge won’t make it through a conducting shell.The conductor acts as a shield. (g) No, this is not a contradiction, because the outside charge never experienced any electrostaticattraction or repulsion from the inside charge. The force is between the shell and the outside charge.P27-12 The repulsive electrostatic forces must exactly balance the attractive gravitational forces.Then 1 q2 m2 =G 2 , 4π 0 r2 r √or m = q/ 4π 0 G. Numerically, (1.60×10−19 C) m= = 1.86×10−9 kg. 2 4π(8.85×10−12 C2 /N · m2 )(6.67×10−11 N · m2 /kg )P27-13 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell.The E field will be perpendicular to the surface, so Gauss’ law will simplify to q enc / 0 = E · dA = E dA = E dA = 4πr2 E.Consequently, E = q enc /4π 0 r2 . q enc = q + 4π ρ r2 dr, or r q enc = q + 4π Ar dr = q + 2πA(r2 − a2 ). aThe electric field will be constant if q enc behaves as r2 , which requires q = 2πAa2 , or A = q/2πa2 .P27-14 (a) The problem has spherical symmetry, so use a Gaussian surface which is a sphericalshell. The E field will be perpendicular to the surface, so Gauss’ law will simplify to q enc / 0 = E · dA = E dA = E dA = 4πr2 E.Consequently, E = q enc /4π 0 r2 . q enc = 4π ρ r2 dr = 4πρr3 /3, so E = ρr/3 0 38
• and is directed radially out from the center. Then E = ρr/3 0 . (b) The electric field in the hole is given by Eh = E − Eb , where E is the field from part (a) andEb is the field that would be produced by the matter that would have been in the hole had the holenot been there. Then Eb = ρb/3 0 ,where b is a vector pointing from the center of the hole. Then ρr ρb ρ Eh = − = (r − b). 3 0 3 0 3 0But r − b = a, so Eh = ρa/3 0 . P27-15 If a point is an equilibrium point then the electric field at that point should be zero.If it is a stable point then moving the test charge (assumed positive) a small distance from theequilibrium point should result in a restoring force directed back toward the equilibrium point. Inother words, there will be a point where the electric field is zero, and around this point there will bean electric field pointing inward. Applying Gauss’ law to a small surface surrounding our point P ,we have a net inward flux, so there must be a negative charge inside the surface. But there shouldbe nothing inside the surface except an empty point P , so we have a contradiction.P27-16 (a) Follow the example on Page 618. By symmetry E = 0 along the median plane. Thecharge enclosed between the median plane and a surface a distance x from the plane is q = ρAx.Then E = ρAx/ 0 A = ρA/ 0 . (b) Outside the slab the charge enclosed between the median plane and a surface a distance xfrom the plane is is q = ρAd/2, regardless of x. The E = ρAd/2/ 0 A = ρd/2 0 .P27-17 (a) The total charge is the volume integral over the whole sphere, Q= ρ dV.This is actually a three dimensional integral, and dV = A dr, where A = 4πr2 . Then Q = ρ dV, R ρS r = 4πr2 dr, 0 R 4πρS 1 4 = R , R 4 = πρS R3 . (b) Put a spherical Gaussian surface inside the sphere centered on the center. We can use Gauss’law here because there is spherical symmetry in the entire problem, both inside and outside theGaussian surface. Gauss’ law states q enc E · dA = . 0 39
• Since there is spherical symmetry the electric field is normal to the spherical Gaussian surface, andit is everywhere the same on this surface. The dot product simplifies to E · dA = E dA, while sinceE is a constant on the surface we can pull it out of the integral, and we end up with q enc E dA = , 0Now dA = 4πr2 , where r is the radius of the Gaussian surface, so q enc E= . 4π 0 r2We aren’t done yet, because the charge enclosed depends on the radius of the Gaussian surface. Weneed to do part (a) again, except this time we don’t want to do the whole volume of the sphere, weonly want to go out as far as the Gaussian surface. Then q enc = ρ dV, r ρS r = 4πr2 dr, 0 R 4πρS 1 4 = r , R 4 r4 = πρS . R Combine these last two results and πρS r4 E = , 4π 0 r2 R πρS r2 = , 4π 0 R Q r2 = . 4π 0 R4In the last line we used the results of part (a) to eliminate ρS from the expression.P27-18 (a) Inside the conductor E = 0, so a Gaussian surface which is embedded in the conductorbut containing the hole must have a net enclosed charge of zero. The cavity wall must then have acharge of −3.0 µC. (b) The net charge on the conductor is +10.0 µC; the charge on the outer surface must then be+13.0 µC.P27-19 (a) Inside the shell E = 0, so the net charge inside a Gaussian surface embedded in theshell must be zero, so the inside surface has a charge −Q. (b) Still −Q; the outside has nothing to do with the inside. (c) −(Q + q); see reason (a). (d) Yes. 40
• Throughout this chapter we will use the convention that V (∞) = 0 unless explicitly stated otherwise. Then the potential in the vicinity of a point charge will be given by Eq. 28-18, V = q/4π 0 r.E28-1 (a) Let U12 be the potential energy of the interaction between the two “up” quarks. Then (2/3)2 e(1.60×10−19 C) U12 = (8.99×109 N · m2 /C2 ) = 4.84×105 eV. (1.32×10−15 m) (b) Let U13 be the potential energy of the interaction between an “up” quark and a “down”quark. Then (−1/3)(2/3)e(1.60×10−19 C) U13 = (8.99×109 N · m2 /C2 ) = −2.42×105 eV (1.32×10−15 m)Note that U13 = U23 . The total electric potential energy is the sum of these three terms, or zero.E28-2 There are six interaction terms, one for every charge pair. Number the charges clockwisefrom the upper left hand corner. Then U12 = −q 2 /4π 0 a, U23 = −q 2 /4π 0 a, U34 = −q 2 /4π 0 a, U41 = −q 2 /4π 0 a, 2 √ U13 = (−q) /4π 0 ( 2a), √ U24 = q 2 /4π 0 ( 2a).Add these terms and get 2 q2 q2 U= √ −4 = −0.206 2 4π 0 a 0aThe amount of work required is W = U . E28-3 (a) We build the electron one part at a time; each part has a charge q = e/3. Moving thefirst part from infinity to the location where we want to construct the electron is easy and takes nowork at all. Moving the second part in requires work to change the potential energy to 1 q1 q2 U12 = , 4π 0 rwhich is basically Eq. 28-7. The separation r = 2.82 × 10−15 m. Bringing in the third part requires work against the force of repulsion between the third chargeand both of the other two charges. Potential energy then exists in the form U13 and U23 , where allthree charges are the same, and all three separations are the same. Then U12 = U13 = U12 , so thetotal potential energy of the system is 1 (e/3)2 3 (1.60×10−19 C/3)2 U =3 = = 2.72×10−14 J 4π 0 r 4π(8.85×10−12 C2 /N · m2 ) (2.82×10−15 m) (b) Dividing our answer by the speed of light squared to find the mass, 2.72 × 10−14 J m= = 3.02 × 10−31 kg. (3.00 × 108 m/s)2 41
• E28-4 There are three interaction terms, one for every charge pair. Number the charges from theleft; let a = 0.146 m. Then (25.5×10−9 C)(17.2×10−9 C) U12 = , 4π 0 a (25.5×10−9 C)(−19.2×10−9 C) U13 = , 4π 0 (a + x) (17.2×10−9 C)(−19.2×10−9 C) U23 = . 4π 0 xAdd these and set it equal to zero. Then (25.5)(17.2) (25.5)(19.2) (17.2)(19.2) = + , a a+x xwhich has solution x = 1.405a = 0.205 m.E28-5 The volume of the nuclear material is 4πa3 /3, where a = 8.0×10−15 m. Upon dividing in √half each part will have a radius r where 4πr3 /3 = 4πa3 /6. Consequently, r = a/ 3 2 = 6.35×10−15 m.Each fragment will have a charge of +46e. (a) The force of repulsion is (46)2 (1.60×10−19 C)2 F = = 3000 N 4π(8.85×10−12 C2 /N · m2 )[2(6.35×10−15 m)]2 (b) The potential energy is (46)2 e(1.60×10−19 C) U= = 2.4×108 eV 4π(8.85×10−12 C2 /N · m2 )2(6.35×10−15 m) 1 2E28-6 This is a work/kinetic energy problem: 2 mv0 = q∆V . Then 2(1.60×10−19 C)(10.3×103 V) v0 = = 6.0×107 m/s. (9.11×10−31 kg) E28-7 (a) The energy released is equal to the charges times the potential through which thecharge was moved. Then ∆U = q∆V = (30 C)(1.0 × 109 V) = 3.0 × 1010 J. (b) Although the problem mentions acceleration, we want to focus on energy. The energy willchange the kinetic energy of the car from 0 to K f = 3.0 × 1010 J. The speed of the car is then 2K 2(3.0 × 1010 J) v= = = 7100 m/s. m (1200 kg) (c) The energy required to melt ice is given by Q = mL, where L is the latent heat of fusion.Then Q (3.0 × 1010 J) m= = = 90, 100kg. L (3.33×105 J/kg) 42
• E28-8 (a) ∆U = (1.60×10−19 C)(1.23×109 V) = 1.97×10−10 J. (b) ∆U = e(1.23×109 V) = 1.23×109 eV. 1 2E28-9 This is an energy conservation problem: 2 mv = q∆V ; ∆V = q/4π 0 (1/r1 − 1/r2 ). Com-bining, q2 1 1 v = − , 2π 0 m r1 r2 (3.1×10−6 C)2 1 1 = − , 2π(8.85×10−12 C2 /N · m2 )(18×10−6 kg) (0.90×10−3 m) (2.5×10−3 m) = 2600 m/s.E28-10 This is an energy conservation problem: 1 q2 1 m(2v)2 − = mv 2 . 2 4π 0 r 2Rearrange, q2 r = , 6π 0 mv 2 (1.60×10−19 C)2 = = 1.42×10−9 m. 6π(8.85×10−12 C2 /N · m2 )(9.11×10−31 kg)(3.44×105 m/s)2 )E28-11 (a) V = (1.60×10−19 C)/4π(8.85×10−12 C2 /N · m2 )(5.29×10−11 m) = 27.2 V. (b) U = qV = (−e)(27.2 V) = −27.2 eV. (c) For uniform circular orbits F = mv 2 /r; the force is electrical, or F = e2 /4π 0 r2 . Kineticenergy is K = mv 2 /2 = F r/2, so e2 (1.60×10−19 C) K= = −12 C2 /N · m2 )(5.29×10−11 m) = 13.6 eV. 8π 0 r 8π(8.85×10 (d) The ionization energy is −(K + U ), or E ion = −[(13.6 eV) + (−27.2 eV)] = 13.6 eV.E28-12 (a) The electric potential at A is 1 q1 q2 (−5.0×10−6 C) (2.0×10−6 C) VA = + = (8.99×109 N · m2 /C) + = 6.0×104 V. 4π 0 r1 r2 (0.15 m) (0.05 m)The electric potential at B is 1 q1 q2 (−5.0×10−6 C) (2.0×10−6 C)VB = + = (8.99×109 N · m2 /C) + = −7.8×105 V. 4π 0 r2 r1 (0.05 m) (0.15 m) (b) W = q∆V = (3.0×10−6 C)(6.0×104 V − −7.8×105 V) = 2.5 J. (c) Since work is positive then external work is converted to electrostatic potential energy. 43
• E28-13 (a) The magnitude of the electric field would be found from F (3.90 × 10−15 N) E= = = 2.44 × 104 N/C. q (1.60 × 10−19 C) (b) The potential difference between the plates is found by evaluating Eq. 28-15, b ∆V = − E · ds. aThe electric field between two parallel plates is uniform and perpendicular to the plates. ThenE · ds = E ds along this path, and since E is uniform, b b b ∆V = − E · ds = − E ds = −E ds = E∆x, a a awhere ∆x is the separation between the plates. Finally, ∆V = (2.44 × 104 N/C)(0.120 m) = 2930 V.E28-14 ∆V = E∆x, so 2 0 2(8.85×10−12 C2 /N · m2 ) ∆x = ∆V = (48 V) = 7.1×10−3 m σ (0.12×10−6 C/m2 )E28-15 The electric field around an infinitely long straight wire is given by E = λ/2π 0 r. Thepotential difference between the inner wire and the outer cylinder is given by b ∆V = − (λ/2π 0 r) dr = (λ/2π 0 ) ln(a/b). aThe electric field near the surface of the wire is then given by λ ∆V (−855 V) E= = = = 1.32×108 V/m. 2π 0 a a ln(a/b) (6.70×10−7 m) ln(6.70×10−7 m/1.05×10−2 m)The electric field near the surface of the cylinder is then given by λ ∆V (−855 V) E= = = −2 m) ln(6.70×10−7 m/1.05×10−2 m) = 8.43×103 V/m. 2π 0 a a ln(a/b) (1.05×10E28-16 ∆V = E∆x = (1.92×105 N/C)(1.50×10−2 m) = 2.88×103 V.E28-17 (a) This is an energy conservation problem: 1 (2)(79)e2 (2)(79)e(1.60×10−19 C) K= = (8.99×109 N · m2 /C) = 3.2×107 eV 4π 0 r (7.0×10−15 m) (b) The alpha particles used by Rutherford never came close to hitting the gold nuclei.E28-18 This is an energy conservation problem: mv 2 /2 = eq/4π 0 r, or (1.60×10−19 C)(1.76×10−15 C) v= = 2.13×104 m/s 2π(8.85×10−12 C2 /N · m2 )(1.22×10−2 m)(9.11×10−31 kg) 44
• E28-19 (a) We evaluate VA and VB individually, and then find the difference. 1 q 1 (1.16µC) VA = = −12 C2 /N · m2 ) (2.06 m) = 5060 V, 4π 0 r 4π(8.85 × 10and 1 q 1 (1.16µC) VB = = = 8910 V, 4π 0 r 4π(8.85 × 10−12 C2 /N · m2 ) (1.17 m) The difference is then VA − VB = −3850 V. (b) The answer is the same, since when concerning ourselves with electric potential we only careabout distances, and not directions.E28-20 The number of “excess” electrons on each grain is 4π 0 rV 4π(8.85×10−12 C2 /N · m)(1.0×10−6 m)(−400 V) n= = = 2.8×105 e (−1.60×10−19 C)E28-21 The excess charge on the shuttle is q = 4π 0 rV = 4π(8.85×10−12 C2 /N · m)(10 m)(−1.0 V) = −1.1×10−9 CE28-22 q = 1.37×105 C, so (1.37×105 C) V = (8.99×109 N · m2 /C2 ) = 1.93×108 V. (6.37×106 m)E28-23 The ratio of the electric potential to the electric field strength is V 1 q 1 q = / = r. E 4π 0 r 4π 0 r2In this problem r is the radius of the Earth, so at the surface of the Earth the potential is V = Er = (100 V/m)(6.38×106 m) = 6.38×108 V.E28-24 Use Eq. 28-22: (1.47)(3.34×10−30 C · m) V = (8.99×109 N · m2 /C2 ) = 1.63×10−5 V. (52.0×10−9 m)2 E28-25 (a) When finding VA we need to consider the contribution from both the positive andthe negative charge, so 1 −q VA = qa + 4π 0 a+dThere will be a similar expression for VB , 1 q VB = −qa + . 4π 0 a+d 45
• Now to evaluate the difference. 1 −q 1 q V A − VB = qa + − −qa + , 4π 0 a+d 4π 0 a+d q 1 1 = − , 2π 0 a a + d q a+d a = − , 2π 0 a(a + d) a(a + d) q d = . 2π 0 a(a + d) (b) Does it do what we expect when d = 0? I expect it the difference to go to zero as thetwo points A and B get closer together. The numerator will go to zero as d gets smaller. Thedenominator, however, stays finite, which is a good thing. So yes, Va − VB → 0 as d → 0.E28-26 (a) Since both charges are positive the electric potential from both charges will be positive.There will be no finite points where V = 0, since two positives can’t add to zero. (b) Between the charges the electric field from each charge points toward the√ √ other, so E willvanish when q/x2 = 2q/(d − x)2 . This happens when d − x = 2x, or x = d/(1 + 2). √E28-27 The distance from C to either charge is 2d/2 = 1.39×10−2 m. (a) V at C is 2(2.13×10−6 C) V = (8.99×109 N · m2 /C2 ) = 2.76×106 V (1.39×10−2 m) (b) W = qδV = (1.91×10−6 C)(2.76×106 V) = 5.27 J. (c) Don’t forget about the potential energy of the original two charges! (2.13×10−6 C)2 U0 = (8.99×109 N · m2 /C2 ) = 2.08 J (1.96×10−2 m)Add this to the answer from part (b) to get 7.35 J.E28-28 The potential is given by Eq. 28-32; at the surface V s = σR/2 0 , half of this occurs when R2 + z 2 − z = R/2, R2 + z 2 = R2 /4 + Rz + z 2 , 3R/4 = z.E28-29 We can find the linear charge density by dividing the charge by the circumference, Q λ= , 2πRwhere Q refers to the charge on the ring. The work done to move a charge q from a point x to theorigin will be given by W = q∆V, W = q (V (0) − V (x)) , 1 Q 1 Q = q √ − √ , 4π 0 R2 4π 0 R 2 + x2 qQ 1 1 = −√ . 4π 0 R R 2 + x2 46
• Putting in the numbers, (−5.93×10−12 C)(−9.12×10−9 C) 1 1 − = 1.86×10−10 J. 4π(8.85×10−12 C2 /N · m2 ) 1.48m (1.48m)2 + (3.07m)2E28-30 (a) The electric field strength is greatest where the gradient of V is greatest. That isbetween d and e. (b) The least absolute value occurs where the gradient is zero, which is between b and c andagain between e and f .E28-31 The potential on the positive plate is 2(5.52 V) = 11.0 V; the electric field between theplates is E = (11.0 V)/(1.48×10−2 m) = 743 V/m.E28-32 Take the derivative: E = −∂V /∂z. E28-33 The radial potential gradient is just the magnitude of the radial component of the electricfield, ∂V Er = − ∂rThen ∂V 1 q = − , ∂r 4π 0 r2 1 79(1.60 × 10−19 C) = , 4π(8.85 × 10−12 C2 /N · m2 ) (7.0 × 10−15 m)2 = −2.32×1021 V/m.E28-34 Evaluate ∂V /∂r, and Ze −1 r E=− 2 +2 3 . 4π 0 r 2RE28-35 Ex = −∂V /∂x = −2(1530 V/m2 )x. At the point in question, E = −2(1530 V/m2 )(1.28×10−2 m) = 39.2 V/m.E28-36 Draw the wires so that they are perpendicular to the plane of the page; they will then“come out of” the page. The equipotential surfaces are then lines where they intersect the page,and they look like 47
• E28-37 (a) |VB − VA | = |W/q| = |(3.94 × 10−19 J)/(1.60 × 10−19 C)| = 2.46 V. The electric fielddid work on the electron, so the electron was moving from a region of low potential to a region ofhigh potential; or VB > VA . Consequently, VB − VA = 2.46 V. (b) VC is at the same potential as VB (both points are on the same equipotential line), soVC − VA = VB − VA = 2.46 V. (c) VC is at the same potential as VB (both points are on the same equipotential line), soVC − VB = 0 V.E28-38 (a) For point charges r = q/4π 0 V , so r = (8.99×109 N · m2 /C2 )(1.5×10−8 C)/(30 V) = 4.5 m (b) No, since V ∝ 1/r.E28-39 The dotted lines are equipotential lines, the solid arrows are electric field lines. Note thatthere are twice as many electric field lines from the larger charge! 48
• E28-40 The dotted lines are equipotential lines, the solid arrows are electric field lines.E28-41 This can easily be done with a spreadsheet. The following is a sketch; the electric field isthe bold curve, the potential is the thin curve. 49
• sphere radius rE28-42 Originally V = q/4π 0 r, where r is the radius of the smaller sphere. (a) Connecting the spheres will bring them to the same potential, or V1 = V2 . (b) q1 + q2 = q; V1 = q1 /4π 0 r and V2 = q2 /4π 0 2r; combining all of the above q2 = 2q1 andq1 = q/3 and q2 = 2q/3.E28-43 (a) q = 4πR2 σ, so V = q/4π 0 R = σR/ 0 , or V = (−1.60×10−19 C/m2 )(6.37×106 m)/(8.85×10−12 C2 /N · m2 ) = 0.115 V (b) Pretend the Earth is a conductor, then E = σ/epsilon0 , so E = (−1.60×10−19 C/m2 )/(8.85×10−12 C2 /N · m2 ) = 1.81×10−8 V/m.E28-44 V = q/4π 0 R, so V = (8.99×109 N · m2 /C2 )(15×10−9 C)/(0.16 m) = 850 V.E28-45 (a) q = 4π 0 RV = 4π(8.85×10−12 C2 /N · m2 )(0.152 m)(215 V) = 3.63×10−9 C (b) σ = q/4πR2 = (3.63×10−9 C)/4π(0.152 m)2 = 1.25×10−8 C/m2 .E28-46 The dotted lines are equipotential lines, the solid arrows are electric field lines. 50
• E28-47 (a) The total charge (Q = 57.2nC) will be divided up between the two spheres so thatthey are at the same potential. If q1 is the charge on one sphere, then q2 = Q − q1 is the charge onthe other. Consequently V1 = V2 , 1 q1 1 Q − q1 = , 4π 0 r1 4π 0 r2 q 1 r2 = (Q − q1 )r1 , Qr2 q1 = . r2 + r1Putting in the numbers, we find Qr1 (57.2 nC)(12.2 cm) q1 = = = 38.6 nC, r2 + r1 (5.88 cm) + (12.2 cm)and q2 = Q − q1 = (57.2 nC) − (38.6 nC) = 18.6 nC. (b) The potential on each sphere should be the same, so we only need to solve one. Then 1 q1 1 (38.6 nC) = −12 C2 /N · m2 ) (12.2 cm) = 2850 V. 4π 0 r1 4π(8.85 × 10E28-48 (a) V = (8.99×109 N · m2 /C2 )(31.5×10−9 C)/(0.162 m) = 1.75×103 V. (b) V = q/4π 0 r, so r = q/4π 0 V , and then r = (8.99×109 N · m2 /C2 )(31.5×10−9 C)/(1.20×103 V) = 0.236 m.That is (0.236 m) − (0.162 m) = 0.074 m above the surface. 51
• E28-49 (a) Apply the point charge formula, but solve for the charge. Then 1 q = V, 4π 0 r q = 4π 0 rV, q = 4π(8.85 × 10−12 C2 /N · m2 )(1 m)(106 V) = 0.11 mC.Now that’s a fairly small charge. But if the radius were decreased by a factor of 100, so wouldthe charge (1.10 µC). Consequently, smaller metal balls can be raised to higher potentials with lesscharge. (b) The electric field near the surface of the ball is a function of the surface charge density,E = σ/ 0 . But surface charge density depends on the area, and varies as r−2 . For a given potential,the electric field near the surface would then be given by σ q V E= = = . 0 4π 0 r2 rNote that the electric field grows as the ball gets smaller. This means that the break down field ismore likely to be exceeded with a low voltage small ball; you’ll get sparking.E28-50 A “Volt” is a Joule per Coulomb. The power required by the drive belt is the product(3.41×106 V)(2.83×10−3 C/s) = 9650 W.P28-1 (a) According to Newtonian mechanics we want K = 1 mv 2 to be equal to W = q∆V 2which means mv 2 (0.511 MeV) ∆V = = = 256 kV. 2q 2emc2 is the rest mass energy of an electron. (b) Let’s do some rearranging first. 1 K = mc2 −1 , 1 − β2 K 1 = − 1, mc2 1 − β2 K 1 +1 = , mc2 1 − β2 1 K = 1 − β2, mc2 + 1 1 2 = 1 − β2, K mc2 + 1and finally, 1 β= 1− 2 K mc2 +1Putting in the numbers, 1 1− 2 = 0.746, (256 keV) + 1 (511 keV)so v = 0.746c. 52
• P28-2 (a) The potential of the hollow sphere is V = q/4π 0 r. The work required to increase thecharge by an amount dq is dW = V /, dq. Integrating, e q e2 W = dq = . 0 4π 0 r 8π 0 rThis corresponds to an electric potential energy of e(1.60×10−19 C) W = = 2.55×105 eV = 4.08×10−14 J. 8π(8.85×10−12 C2 /N · m2 )(2.82×10−15 m) (b) This would be a mass of m = (4.08×10−14 J)/(3.00×108 m/s)2 = 4.53×10−31 kg.P28-3 The negative charge is held in orbit by electrostatic attraction, or mv 2 qQ = . r 4π 0 r2The kinetic energy of the charge is 1 qQ K= mv 2 = . 2 8π 0 rThe electrostatic potential energy is qQ U =− , 4π 0 rso the total energy is qQ E=− . 8π 0 rThe work required to change orbit is then qQ 1 1 W = − . 8π 0 r1 r2P28-4 (a) V = − E dr, so r qr qr2 V =− 3 dr = − . 0 4π 0 R 8π 0 R3 (b) ∆V = q/8π 0 R. (c) If instead of V = 0 at r = 0 as was done in part (a) we take V = 0 at r = ∞, thenV = q/4π 0 R on the surface of the sphere. The new expression for the potential inside the spherewill look like V = V + Vs , where V is the answer from part (a) and Vs is a constant so that thesurface potential is correct. Then q qR2 3qR2 Vs = + = , 4π 0 R 8π 0 R3 8π 0 R3and then qr2 3qR2 q(3R2 − r2 ) V =− 3 + 3 = . 8π 0 R 8π 0 R 8π 0 R3 53
• P28-5 The total electric potential energy of the system is the sum of the three interaction pairs.One of these pairs does not change during the process, so it can be ignored when finding the changein potential energy. The change in electrical potential energy is then q2 q2 q2 1 1 ∆U = 2 −2 = − . 4π 0 rf 4π 0 ri 2π 0 rf riIn this case ri = 1.72 m, while rf = 0.86 m. The change in potential energy is then 1 1 ∆U = 2(8.99×109 N · m2 /C2 )(0.122 C)2 − = 1.56×108 J (0.86 m) (1.72 m)The time required is t = (1.56×108 )/(831 W) = 1.87×105 s = 2.17 days.P28-6 (a) Apply conservation of energy: qQ qQ K= , or d = , 4π 0 d 4π 0 Kwhere d is the distance of closest approach. (b) Apply conservation of energy: qQ 1 K= + mv 2 , 4π 0 (2d) 2so, combining with the results in part (a), v = K/m.P28-7 (a) First apply Eq. 28-18, but solve for r. Then q (32.0 × 10−12 C) r= = = 562 µm. 4π 0 V 4π(8.85 × 10−12 C2 /N · m2 )(512 V) (b) If two such drops join together the charge doubles, and the volume of water doubles, but the √radius of the new drop only increases by a factor of 3 2 = 1.26 because volume is proportional tothe radius cubed. The potential on the surface of the new drop will be 1 q new V new = , 4π 0 rnew 1 2q old = √ , 4π 0 3 2 rold 1 q old = (2)2/3 = (2)2/3 V old . 4π 0 roldThe new potential is 813 V.P28-8 (a) The work done is W = −F z = −Eqz = −qσz/2 0 . (b) Since W = q∆V , ∆V = −σz/2 0 , so V = V0 − (σ/2 0 )z. 54
• P28-9 (a) The potential at any point will be the sum of the contribution from each charge, 1 q1 1 q2 V = + , 4π 0 r1 4π 0 r2where r1 is the distance the point in question from q1 and r2 is the distance the point in questionfrom q2 . Pick a point, call it (x, y). Since q1 is at the origin, r1 = x2 + y 2 .Since q2 is at (d, 0), where d = 9.60 nm, r2 = (x − d)2 + y 2 .Define the “Stanley Number” as S = 4π 0 V . Equipotential surfaces are also equi-Stanley surfaces.In particular, when V = 0, so does S. We can then write the potential expression in a sightlysimplified form q1 q2 S= + . r1 r2 If S = 0 we can rearrange and square this expression. q1 q2 = − , r1 r2 2 2 r1 r2 2 = 2, q1 q2 x2 + y 2 (x − d)2 + y 2 2 = 2 , q1 q2Let α = q2 /q1 , then we can write α2 x2 + y 2 = (x − d)2 + y 2 , α2 x2 + α2 y 2 = x2 − 2xd + d2 + y 2 , (α − 1)x + 2xd + (α2 − 1)y 2 2 2 = d2 .We complete the square for the (α2 − 1)x2 + 2xd term by adding d2 /(α2 − 1) to both sides of theequation. Then 2 d 1 (α2 − 1) x + 2 + y 2 = d2 1 + 2 . α −1 α −1 The center of the circle is at d (9.60 nm) − = = −5.4 nm. α2 −1 (−10/6)2 − 1 (b) The radius of the circle is 1 1+ α2 −1 d2 , α2 − 1which can be simplified to α |(−10/6)| d = (9.6 nm) = 9.00 nm. α2 − 1 (−10/6)2 − 1 (c) No. 55
• P28-10 An annulus is composed of differential rings of varying radii r and width dr; the chargeon any ring is the product of the area of the ring, dA = 2πr dr, and the surface charge density, or k 2πk dq = σ dA = 3 2πr dr = 2 dr. r rThe potential at the center can be found by adding up the contributions from each ring. Since weare at the center, the contributions will each be dV = dq/4π 0 r. Then b k dr k 1 1 k b2 − a2 V = 3 = 2 − 2 .= . a 2 0r 4 0 a b 4 0 b2 a2The total charge on the annulus is b 2πk 1 1 b−a Q= dr = 2πk − = 2πk . a r2 a b baCombining, Q a+b V = . 8π 0 abP28-11 Add the three contributions, and then do a series expansion for d r. q −1 1 1 V = + + , 4π 0 r + d r r−d q −1 1 = +1+ , 4π 0 r 1 + d/r 1 − d/r q d d ≈ −1 + + 1 + 1 + , 4π 0 r r r q 2d ≈ 1+ . 4π 0 r rP28-12 (a) Add the contributions from each differential charge: dq = λ dy. Then y+L λ λ y+L V = dy = ln . y 4π 0 y 4π 0 y (b) Take the derivative: ∂V λ y −L λ L Ey = − =− = . ∂y 4π 0 y + L y 2 4π 0 y(y + L) (c) By symmetry it must be zero, since the system is invariant under rotations about the axisof the rod. Note that we can’t determine E⊥ from derivatives because we don’t have the functionalform of V for points off-axis!P28-13 (a) We follow the work done in Section 28-6 for a uniform line of charge, starting withEq. 28-26, 1 λ dx dV = , 4π 0 x2 + y 2 L 1 kx dx dV = , 4π 0 0 x2 + y 2 56
• k L = x2 + y 2 , 4π 0 0 k = L2 + y 2 − y . 4π 0 (b) The y component of the electric field can be found from ∂V Ey = − , ∂ywhich (using a computer-aided math program) is k y Ey = 1− . 4π 0 L2 + y2 (c) We could find Ex if we knew the x variation of V . But we don’t; we only found the values ofV along a fixed value of x. (d) We want to find y such that the ratio k k L2 + y 2 − y / (L) 4π 0 4π 0is one-half. Simplifying, L2 + y 2 − y = L/2, which can be written as L2 + y 2 = L2 /4 + Ly + y 2 ,or 3L2 /4 = Ly, with solution y = 3L/4.P28-14 The spheres are small compared to the separation distance. Assuming only one sphere ata potential of 1500 V, the charge would be q = 4π 0 rV = 4π(8.85×10−12 C2 /N · m)(0.150 m)(1500 V) = 2.50×10−8 C.The potential from the sphere at a distance of 10.0 m would be (0.150 m) V = (1500 V) = 22.5 V. (10.0 m)This is small compared to 1500 V, so we will treat it as a perturbation. This means that we canassume that the spheres have charges of q = 4π 0 rV = 4π(8.85×10−12 C2 /N · m)(0.150 m)(1500 V + 22.5 V) = 2.54×10−8 C. P28-15 Calculating the fraction of excess electrons is the same as calculating the fraction ofexcess charge, so we’ll skip counting the electrons. This problem is effectively the same as Exercise28-47; we have a total charge that is divided between two unequal size spheres which are at the samepotential on the surface. Using the result from that exercise we have Qr1 q1 = , r2 + r1where Q = −6.2 nC is the total charge available, and q1 is the charge left on the sphere. r1 is theradius of the small ball, r2 is the radius of Earth. Since the fraction of charge remaining is q1 /Q,we can write q1 r1 r1 = ≈ = 2.0 × 10−8 . Q r 2 + r1 r2 57
• P28-16 The positive charge on the sphere would be q = 4π 0 rV = 4π(8.85×10−12 C2 /N · m2 )(1.08×10−2 m)(1000 V) = 1.20×10−9 C.The number of decays required to build up this charge is n = 2(1.20×10−9 C)/(1.60×10−19 C) = 1.50×1010 .The extra factor of two is because only half of the decays result in an increase in charge. The timerequired is t = (1.50×1010 )/(3.70×108 s−1 ) = 40.6 s.P28-17 (a) None. (b) None. (c) None. (d) None. (e) No.P28-18 (a) Outside of an isolated charged spherical object E = q/4π 0 r2 and V = q/4π 0 r.Then E = V /r. Consequently, the sphere must have a radius larger than r = (9.15×106 V)/(100×106 V/m) = 9.15×10−2 m. (b) The power required is (320×10−6 C/s)(9.15×106 V) = 2930 W. (c) σwv = (320×10−6 C/s), so (320×10−6 C/s) σ= = 2.00×10−5 C/m2 . (0.485 m)(33.0 m/s) 58
• E29-1 (a) The charge which flows through a cross sectional surface area in a time t is given byq = it, where i is the current. For this exercise we have q = (4.82 A)(4.60 × 60 s) = 1330 Cas the charge which passes through a cross section of this resistor. (b) The number of electrons is given by (1330 C)/(1.60 × 10−19 C) = 8.31 × 1021 electrons.E29-2 Q/t = (200×10−6 A/s)(60s/min)/(1.60×10−19 C) = 7.5×1016 electrons per minute.E29-3 (a) j = nqv = (2.10×1014 /m3 )2(1.60×10−19 C)(1.40×105 m/s) = 9.41 A/m2 . Since the ionshave positive charge then the current density is in the same direction as the velocity. (b) We need an area to calculate the current.E29-4 (a) j = i/A = (123×10−12 A)/π(1.23×10−3 m)2 = 2.59×10−5 A/m2 . (b) v d = j/ne = (2.59×10−5 A/m2 )/(8.49×1028 /m3 )(1.60×10−19 C) = 1.91×10−15 m/s.E29-5 The current rating of a fuse of cross sectional area A would be imax = (440 A/cm2 )A,and if the fuse wire is cylindrical A = πd2 /4. Then 4 (0.552 A) d= = 4.00×10−2 cm. π (440 A/m2 )E29-6 Current density is current divided by cross section of wire, so the graph would look like: 4 I (A/mil^2 x10^−3) 3 2 1 50 100 150 200 d(mils) 59
• E29-7 The current is in the direction of the motion of the positive charges. The magnitude of thecurrent is i = (3.1×1018 /s + 1.1×1018 /s)(1.60×10−19 C) = 0.672 A.E29-8 (a) The total current is i = (3.50×1015 /s + 2.25×1015 /s)(1.60×10−19 C) = 9.20×10−4 A. (b) The current density is j = (9.20×10−4 A)/π(0.165×10−3 m)2 = 1.08×104 A/m2 .E29-9 (a) j = (8.70×106 /m3 )(1.60×10−19 C)(470×103 m/s) = 6.54×10−7 A/m2 . (b) i = (6.54×10−7 A/m2 )π(6.37×106 m)2 = 8.34×107 A.E29-10 i = σwv, so σ = (95.0×10−6 A)/(0.520 m)(28.0 m/s) = 6.52×10−6 C/m2 .E29-11 The drift velocity is given by Eq. 29-6, j i (115 A) vd = = = = 2.71×10−4 m/s. ne Ane (31.2×10−6 m2 )(8.49×1028 /m3 )(1.60×10−19 C)The time it takes for the electrons to get to the starter motor is x (0.855 m) t= = = 3.26×103 s. v (2.71×10−4 m/s)That’s about 54 minutes.E29-12 ∆V = iR = (50×10−3 A)(1800 Ω) = 90 V.E29-13 The resistance of an object with constant cross section is given by Eq. 29-13, L (11, 000 m) R=ρ = (3.0 × 10−7 Ω · m) = 0.59 Ω. A (0.0056 m2 )E29-14 The slope is approximately [(8.2 − 1.7)/1000]µΩ · cm/◦ C, so 1 α= 6.5×10−3 µΩ · cm/◦ C ≈ 4×10−3 /C◦ 1.7µΩ · cmE29-15 (a) i = ∆V /R = (23 V)/(15×10−3 Ω) = 1500 A. (b) j = i/A = (1500 A)/π(3.0×10−3 m)2 = 5.3×107 A/m2 . (c) ρ = RA/L = (15×10−3 Ω)π(3.0×10−3 m)2 /(4.0 m) = 1.1×10−7 Ω · m. The material is possiblyplatinum.E29-16 Use the equation from Exercise 29-17. ∆R = 8 Ω; then ∆T = (8 Ω)/(50 Ω)(4.3×10−3 /C◦ ) = 37 C◦ .The final temperature is then 57◦ C. 60
• E29-17 Start with Eq. 29-16, ρ − ρ0 = ρ0 αav (T − T0 ),and multiply through by L/A, L L (ρ − ρ0 ) = ρ0 αav (T − T0 ), A Ato get R − R0 = R0 αav (T − T0 ).E29-18 The wire has a length L = (250)2π(0.122 m) = 192 m. The diameter is 0.129 inches; thecross sectional area is then A = π(0.129 × 0.0254 m)2 /4 = 8.43×10−6 m2 .The resistance is R = ρL/A = (1.69×10−8 Ω · m)(192 m)/(8.43×10−6 m2 ) = 0.385 Ω.E29-19 If the length of each conductor is L and has resistivity ρ, then L 4L RA = ρ =ρ πD2 /4 πD2and L 4L RB = ρ =ρ . (π4D2 /4 − πD2 /4) 3πD2The ratio of the resistances is then RA = 3. RBE29-20 R = R, so ρ1 L1 /π(d1 /2)2 = ρ2 L2 /π(d2 /2)2 . Simplifying, ρ1 /d2 = ρ2 /d2 . Then 1 2 d2 = (1.19×10−3 m) (9.68×10−8 Ω · m)/(1.69×10−8 Ω · m) = 2.85×10−3 m.E29-21 (a) (750×10−3 A)/(125) = 6.00×10−3 A. (b) ∆V = iR = (6.00×10−3 A)(2.65×10−6 Ω) = 1.59×10−8 V. (c) R = ∆V /i = (1.59×10−8 V)/(750×10−3 A) = 2.12×10−8 Ω.E29-22 Since ∆V = iR, then if ∆V and i are the same, then R must be the same. 2 2 2 2 (a) Since R = R, ρ1 L1 /πr1 = ρ2 L2 /πr2 , or ρ1 /r1 = ρ2 /r2 . Then riron /rcopper = (9.68×10−8 Ω · m)(1.69×10−8 Ω · m) = 2.39. (b) Start with the definition of current density: i ∆V ∆V j= = = . A RA ρLSince ∆V and L is the same, but ρ is different, then the current densities will be different. 61
• E29-23 Conductivity is given by Eq. 29-8, j = σ E. If the wire is long and thin, then themagnitude of the electric field in the wire will be given by E ≈ ∆V /L = (115 V)/(9.66 m) = 11.9 V/m.We can now find the conductivity, j (1.42×104 A/m2 ) σ= = = 1.19×103 (Ω · m)−1 . E (11.9 V/m)E29-24 (a) v d = j/en = σE/en. Thenv d = (2.70×10−14 /Ω · m)(120 V/m)/(1.60×10−19 C)(620×106 /m3 + 550×106 /m3 ) = 1.73×10−2 m/s. (b) j = σE = (2.70×10−14 /Ω · m)(120 V/m) = 3.24×10−14 A/m2 .E29-25 (a) R/L = ρ/A, so j = i/A = (R/L)i/ρ. For copper, j = (0.152×10−3 Ω/m)(62.3 A)/(1.69×10−8 Ω · m) = 5.60×105 A/m2 ;for aluminum, j = (0.152×10−3 Ω/m)(62.3 A)/(2.75×10−8 Ω · m) = 3.44×105 A/m2 . (b) A = ρL/R; if δ is density, then m = δlA = lδρ/(R/L). For copper, m = (1.0 m)(8960 kg/m3 )(1.69×10−8 Ω · m)/(0.152×10−3 Ω/m) = 0.996 kg;for aluminum, m = (1.0 m)(2700 kg/m3 )(2.75×10−8 Ω · m)/(0.152×10−3 Ω/m) = 0.488 kg.E29-26 The resistance for potential differences less than 1.5 V are beyond the scale. 10 8 R (Kilo−ohms) 6 4 2 1 2 3 4 V(Volts) 62
• E29-27 (a) The resistance is defined as ∆V (3.55 × 106 V/A2 )i2 R= = = (3.55 × 106 V/A2 )i. i iWhen i = 2.40 mA the resistance would be R = (3.55 × 106 V/A2 )(2.40 × 10−3 A) = 8.52 kΩ. (b) Invert the above expression, and i = R/(3.55 × 106 V/A2 ) = (16.0 Ω)/(3.55 × 106 V/A2 ) = 4.51 µA.E29-28 First, n = 3(6.02×1023 )(2700 kg/m3 )(27.0×10−3 kg) = 1.81×1029 /m3 . Then m (9.11×10−31 kg) τ= 2ρ = 29 /m3 )(1.60×10−19 C)2 (2.75×10−8 Ω · m) = 7.15×10−15 s. ne (1.81×10E29-29 (a) E = E0 /κe = q/4π 0 κe R2 , so (1.00×10−6 C) E= = 4π(8.85×10−12 C2 /N · m2 )(4.7)(0.10 m)2 (b) E = E0 = q/4π 0 R2 , so (1.00×10−6 C) E= = 4π(8.85×10−12 C2 /N · m2 )(0.10 m)2 (c) σ ind = 0 (E0 − E) = q(1 − 1/κe )/4πR2 . Then (1.00×10−6 C) 1 σ ind = 1− = 6.23×10−6 C/m2 . 4π(0.10 m)2 (4.7)E29-30 Midway between the charges E = q/π 0 d, so q = π(8.85×10−12 C2 /N · m2 )(0.10 m)(3×106 V/m) = 8.3×10−6 C. E29-31 (a) At the surface of a conductor of radius R with charge Q the magnitude of the electricfield is given by 1 E= QR2 , 4π 0while the potential (assuming V = 0 at infinity) is given by 1 V = QR. 4π 0The ratio is V /E = R. The potential on the sphere that would result in “sparking” is V = ER = (3×106 N/C)R. (b) It is “easier” to get a spark off of a sphere with a smaller radius, because any potential onthe sphere will result in a larger electric field. (c) The points of a lighting rod are like small hemispheres; the electric field will be large nearthese points so that this will be the likely place for sparks to form and lightning bolts to strike. 63
• P29-1 If there is more current flowing into the sphere than is flowing out then there must be achange in the net charge on the sphere. The net current is the difference, or 2 µA. The potential onthe surface of the sphere will be given by the point-charge expression, 1 q V = , 4π 0 rand the charge will be related to the current by q = it. Combining, 1 it V = , 4π 0 ror 4π 0 V r 4π(8.85 × 10−12 C2 /N · m2 )(980 V)(0.13 m) t= = = 7.1 ms. i (2 µA)P29-2 The net current density is in the direction of the positive charges, which is to the east. Thereare two electrons for every alpha particle, and each alpha particle has a charge equal in magnitudeto two electrons. The current density is then j = q e ne v e + qα + nα vα , = (−1.6×10−19 C)(5.6×1021 /m3 )(−88 m/s) + (3.2×10−19 C)(2.8×1021 /m3 )(25 m/s), = 1.0×105 C/m2 .P29-3 (a) The resistance of the segment of the wire is R = ρL/A = (1.69×10−8 Ω · m)(4.0×10−2 m)/π(2.6×10−3 m)2 = 3.18×10−5 Ω.The potential difference across the segment is ∆V = iR = (12 A)(3.18×10−5 Ω) = 3.8×10−4 V. (b) The tail is negative. (c) The drift speed is v = j/en = i/Aen, so v = (12 A)/π(2.6×10−3 m)2 (1.6×10−19 C)(8.49×1028 /m3 ) = 4.16×10−5 m/s.The electrons will move 1 cm in (1.0×10−2 m)/(4.16×10−5 m/s) = 240 s.P29-4 (a) N = it/q = (250×10−9 A)(2.9 s)/(3.2×10−19 C) = 2.27×1012 . (b) The speed of the particles in the beam is given by v = 2K/m, so v= 2(22.4 MeV)/4(932 MeV/c2 ) = 0.110c.It takes (0.180 m)/(0.110)(3.00×108 m/s) = 5.45×10−9 s for the beam to travel 18.0 cm. The numberof charges is then N = it/q = (250×10−9 A)(5.45×10−9 s)/(3.2×10−19 C) = 4260. (c) W = q∆V , so ∆V = (22.4 MeV)/2e = 11.2 MV. 64
• P29-5 (a) The time it takes to complete one turn is t = (250 m)/c. The total charge is q = it = (30.0 A)(950 m)/(3.00×108 m/s) = 9.50×10−5 C. (b) The number of charges is N = q/e, the total energy absorbed by the block is then ∆U = (28.0×109 eV)(9.50×10−5 C)/e = 2.66×106 J.This will raise the temperature of the block by ∆T = ∆U/mC = (2.66×106 J)/(43.5 kg)(385J/kgC◦ ) = 159 C◦ .P29-6 (a) i = j dA = 2π jr dr; i = 2π −0R j0 (1 − r/R)r dr = 2πj0 (R2 /2 − R3 /3R) = πj0 R2 /6. (b) Integrate, again: i = 2π −0R j0 (r/R)r dr = 2πj0 (R3 /3R) = πj0 R2 /3.P29-7 (a) Solve 2ρ0 = ρ0 [1 + α(T − 20◦ C)], or T = 20◦ C + 1/(4.3×10−3 /C◦ ) = 250◦ C. (b) Yes, ignoring changes in the physical dimensions of the resistor.P29-8 The resistance when on is (2.90 V)/(0.310 A) = 9.35 Ω. The temperature is given by T = 20◦ C + (9.35 Ω − 1.12 Ω)/(1.12 Ω)(4.5×10−3 /◦ C) = 1650◦ C. P29-9 Originally we have a resistance R1 made out of a wire of length l1 and cross sectional areaA1 . The volume of this wire is V1 = A1 l1 . When the wire is drawn out to the new length we havel2 = 3l1 , but the volume of the wire should be constant so A2 l2 = A1 l1 , A2 (3l1 ) = A1 l1 , A2 = A1 /3.The original resistance is l1 R1 = ρ . A1The new resistance is l2 3l1 R2 = ρ =ρ = 9R1 , A2 A1 /3or R2 = 54 Ω.P29-10 (a) i = (35.8 V)/(935 Ω) = 3.83×10−2 A. (b) j = i/A = (3.83×10−2 A)/(3.50×10−4 m2 ) = 109 A/m2 . (c) v = (109 A/m2 )/(1.6×10−19 C)(5.33×1022 /m3 ) = 1.28×10−2 m/s. (d) E = (35.8 V)/(0.158 m) = 227 V/m. 65
• P29-11 (a) ρ = (1.09×10−3 Ω)π(5.5×10−3 m)2 /4(1.6 m) = 1.62×10−8 Ω · m. This is possibly silver. (b) R = (1.62×10−8 Ω · m)(1.35×10−3 m)4/π(2.14×10−2 m)2 = 6.08×10−8 Ω.P29-12 (a) ∆L/L = 1.7×10−5 for a temperature change of 1.0 C◦ . Area changes are twice this,or ∆A/A = 3.4×10−5 . Take the differential of RA = ρL: R dA+A dR = ρ dL+L dρ, or dR = ρ dL/A+L dρ/A−R dA/A.For finite changes this can be written as ∆R ∆L ∆ρ ∆A = + − . R L ρ A∆ρ/ρ = 4.3×10−3 . Since this term is so much larger than the other two it is the only significanteffect.P29-13 We will use the results of Exercise 29-17, R − R0 = R0 αav (T − T0 ).To save on subscripts we will drop the “av” notation, and just specify whether it is carbon “c” oriron “i”. The disks will be effectively in series, so we will add the resistances to get the total. Lookingonly at one disk pair, we have Rc + Ri = R0,c (αc (T − T0 ) + 1) + R0,i (αi (T − T0 ) + 1) , = R0,c + R0,i + (R0,c αc + R0,i αi ) (T − T0 ).This last equation will only be constant if the coefficient for the term (T − T0 ) vanishes. Then R0,c αc + R0,i αi = 0,but R = ρL/A, and the disks have the same cross sectional area, so Lc ρc αc + Li ρi αi = 0,or Lc ρi α i (9.68×10−8 Ω · m)(6.5×10−3 /C◦ ) =− =− = 0.036. Li ρc α c (3500×10−8 Ω · m)(−0.50×10−3 /C◦ )P29-14 The current entering the cone is i. The current density as a function of distance x fromthe left end is then i j= . π[a + x(b − a)/L]2The electric field is given by E = ρj. The potential difference between the ends is then L L iρ iρL ∆V = E dx = 2 dx = 0 0 π[a + x(b − a)/L] πabThe resistance is R = ∆V /i = ρL/πab. 66
• P29-15 The current is found from Eq. 29-5, i= j · dA,where the region of integration is over a spherical shell concentric with the two conducting shellsbut between them. The current density is given by Eq. 29-10, j = E/ρ,and we will have an electric field which is perpendicular to the spherical shell. Consequently, 1 1 i= E · dA = E dA ρ ρBy symmetry we expect the electric field to have the same magnitude anywhere on a spherical shellwhich is concentric with the two conducting shells, so we can bring it out of the integral sign, andthen 1 4πr2 E i = E dA = , ρ ρwhere E is the magnitude of the electric field on the shell, which has radius r such that b > r > a. The above expression can be inverted to give the electric field as a function of radial distance,since the current is a constant in the above expression. Then E = iρ/4πr2 The potential is given by a ∆V = − E · ds, bwe will integrate along a radial line, which is parallel to the electric field, so a ∆V = − E dr, b a iρ = − dr, b 4πr2 iρ a dr = − , 4π b r iρ 1 1 = − . 4π a bWe divide this expression by the current to get the resistance. Then ρ 1 1 R= − 4π a bP29-16 Since τ =√ d , √ ∝ v d . For an ideal gas the kinetic energy is proportional to the λ/v ρtemperature, so ρ ∝ K ∝ T . 67
• E30-1 We apply Eq. 30-1, q = C∆V = (50 × 10−12 F)(0.15 V) = 7.5 × 10−12 C;E30-2 (a) C = ∆V /q = (73.0×10−12 C)/(19.2 V) = 3.80×10−12 F. (b) The capacitance doesn’t change! (c) ∆V = q/C = (210×10−12 C)/(3.80×10−12 F) = 55.3 V.E30-3 q = C∆V = (26.0×10−6 F)(125 V) = 3.25×10−3 C.E30-4 (a) C = 0 A/d = (8.85×10−12 F/m)π(8.22×10−2 m)2 /(1.31×10−3 m) = 1.43×10−10 F. (b) q = C∆V = (1.43×10−10 F)(116 V) = 1.66×10−8 C.E30-5 Eq. 30-11 gives the capacitance of a cylinder, L (0.0238 m) C = 2π 0 = 2π(8.85×10−12 F/m) = 5.46×10−13 F. ln(b/a) ln((9.15mm)/(0.81mm))E30-6 (a) A = Cd/ 0 = (9.70×10−12 F)(1.20×10−3 m)/(8.85×10−12 F/m) = 1.32×10−3 m2 . (b) C = C0 d0 /d = (9.70×10−12 F)(1.20×10−3 m)/(1.10×10−3 m) = 1.06×10−11 F. (c) ∆V = q0 /C = [∆V ]0 C0 /C = [∆V ]0 d/d0 . Using this formula, the new potential differencewould be [∆V ]0 = (13.0 V)(1.10×10−3 m)/(1.20×10−3 m) = 11.9 V. The potential energy has changedby (11.9 V) − (30.0 V) = −1.1 V.E30-7 (a) From Eq. 30-8, (0.040 m)(0.038 m) C = 4π(8.85×10−12 F/m) = 8.45×10−11 F. (0.040 m) − (0.038 m) (b) A = Cd/ 0 = (8.45×10−11 F)(2.00×10−3 m)/(8.85×10−12 F/m) = 1.91×10−2 m2 .E30-8 Let a = b + d, where d is the small separation between the shells. Then ab (b + d)b C = 4π 0 = 4π 0 , a−b d b2 ≈ 4π 0 = 0 A/d. dE30-9 The potential difference across each capacitor in parallel is the same; it is equal to 110 V.The charge on each of the capacitors is then q = C∆V = (1.00 × 10−6 F)(110 V) = 1.10 × 10−4 C.If there are N capacitors, then the total charge will be N q, and we want this total charge to be1.00 C. Then (1.00 C) (1.00 C) N= = = 9090. q (1.10 × 10−4 C) 68
• E30-10 First find the equivalent capacitance of the parallel part: C eq = C1 + C2 = (10.3×10−6 F) + (4.80×10−6 F) = 15.1×10−6 F.Then find the equivalent capacitance of the series part: 1 1 1 = −6 F) + = 3.23×105 F−1 . C eq (15.1×10 (3.90×10−6 F)Then the equivalent capacitance of the entire arrangement is 3.10×10−6 F.E30-11 First find the equivalent capacitance of the series part: 1 1 1 = + = 3.05×105 F−1 . C eq (10.3×10−6 F) (4.80×10−6 F)The equivalent capacitance is 3.28×10−6 F. Then find the equivalent capacitance of the parallel part: C eq = C1 + C2 = (3.28×10−6 F) + (3.90×10−6 F) = 7.18×10−6 F.This is the equivalent capacitance for the entire arrangement.E30-12 For one capacitor q = C∆V = (25.0×10−6 F)(4200 V) = 0.105 C. There are three capaci-tors, so the total charge to pass through the ammeter is 0.315 C.E30-13 (a) The equivalent capacitance is given by Eq. 30-21, 1 1 1 1 1 5 = + = + = C eq C1 C2 (4.0µF) (6.0µF) (12.0µF)or C eq = 2.40µF. (b) The charge on the equivalent capacitor is q = C∆V = (2.40µF)(200 V) = 0.480 mC. Forseries capacitors, the charge on the equivalent capacitor is the same as the charge on each of thecapacitors. This statement is wrong in the Student Solutions! (c) The potential difference across the equivalent capacitor is not the same as the potentialdifference across each of the individual capacitors. We need to apply q = C∆V to each capacitorusing the charge from part (b). Then for the 4.0µF capacitor, q (0.480 mC) ∆V = = = 120 V; C (4.0µF)and for the 6.0µF capacitor, q (0.480 mC) ∆V = = = 80 V. C (6.0µF)Note that the sum of the potential differences across each of the capacitors is equal to the potentialdifference across the equivalent capacitor.E30-14 (a) The equivalent capacitance is C eq = C1 + C2 = (4.0µF) + (6.0µF) = (10.0µF). (c) For parallel capacitors, the potential difference across the equivalent capacitor is the same asthe potential difference across either of the capacitors. (b) For the 4.0µF capacitor, q = C∆V = (4.0µF)(200 V) = 8.0×10−4 C;and for the 6.0µF capacitor, q = C∆V = (6.0µF)(200 V) = 12.0×10−4 C. 69
• E30-15 (a) C eq = C + C + C = 3C; 0A 0A d deq = = = . C eq 3C 3 (b) 1/C eq = 1/C + 1/C + 1/C = 3/C; 0A 0A deq = = = 3d. C eq C/3E30-16 (a) The maximum potential across any individual capacitor is 200 V; so there must beat least (1000 V)/(200 V) = 5 series capacitors in any parallel branch. This branch would have anequivalent capacitance of C eq = C/5 = (2.0×10−6 F)/5 = 0.40×10−6 F. (b) For parallel branches we add, which means we need (1.2×10−6 F)/(0.40×10−6 F) = 3 parallelbranches of the combination found in part (a).E30-17 Look back at the solution to Ex. 30-10. If C3 breaks down electrically then the circuit iseffectively two capacitors in parallel. (b) ∆V = 115 V after the breakdown. (a) q1 = (10.3×10−6 F)(115 V) = 1.18×10−3 C.E30-18 The 108µF capacitor originally has a charge of q = (108×10−6 F)(52.4 V) = 5.66×10−3 C.After it is connected to the second capacitor the 108µF capacitor has a charge of q = (108 ×10−6 F)(35.8 V) = 3.87×10−3 C. The difference in charge must reside on the second capacitor, so thecapacitance is C = (1.79×10−3 C)/(35.8 V) = 5.00×10−5 F. E30-19 Consider any junction other than A or B. Call this junction point 0; label the four nearestjunctions to this as points 1, 2, 3, and 4. The charge on the capacitor that links point 0 to point 1 isq1 = C∆V01 , where ∆V01 is the potential difference across the capacitor, so ∆V01 = V0 − V1 , whereV0 is the potential at the junction 0, and V1 is the potential at the junction 1. Similar expressionsexist for the other three capacitors. For the junction 0 the net charge must be zero; there is no way for charge to cross the plates ofthe capacitors. Then q1 + q2 + q3 + q4 = 0, and this means C∆V01 + C∆V02 + C∆V03 + C∆V04 = 0or ∆V01 + ∆V02 + ∆V03 + ∆V04 = 0.Let ∆V0i = V0 − Vi , and then rearrange, 4V0 = V1 + V2 + V3 + V4 ,or 1 V0 = (V1 + V2 + V3 + V4 ) . 4 2E30-20 U = uV = 0E V /2, where V is the volume. Then 1 U= (8.85×10−12 F/m)(150 V/m)2 (2.0 m3 ) = 1.99×10−7 J. 2 70
• E30-21 The total capacitance is (2100)(5.0×10−6 F) = 1.05×10−2 F. The total energy stored is 1 1 U= C(∆V )2 = (1.05×10−2 F)(55×103 V)2 = 1.59×107 J. 2 2The cost is \$0.03 (1.59×107 J) = \$0.133. 3600×103 JE30-22 (a) U = 2 C(∆V )2 = 1 (0.061 F)(1.0×104 V)2 = 3.05×106 J. 1 2 (b) (3.05×10 J)/(3600×103 J/kW · h) = 0.847kW · h. 6E30-23 (a) The capacitance of an air filled parallel-plate capacitor is given by Eq. 30-5, 0A (8.85×10−12 F/m)(42.0 × 10−4 m2 ) C= = = 2.86×10−11 F. d (1.30 × 10−3 m) (b) The magnitude of the charge on each plate is given by q = C∆V = (2.86×10−11 F)(625 V) = 1.79×10−8 C. (c) The stored energy in a capacitor is given by Eq. 30-25, regardless of the type or shape of thecapacitor, so 1 1 U = C(∆V )2 = (2.86×10−11 F)(625 V)2 = 5.59 µJ. 2 2 (d) Assuming a parallel plate arrangement with no fringing effects, the magnitude of the electricfield between the plates is given by Ed = ∆V , where d is the separation between the plates. Then E = ∆V /d = (625 V)/(0.00130 m) = 4.81×105 V/m. (e) The energy density is Eq. 30-28, 1 2 1 −12 u= 0 E = ((8.85×10 F/m))(4.81×105 V/m)2 = 1.02 J/m3 . 2 2E30-24 The equivalent capacitance is given by 1/C eq = 1/(2.12×10−6 F) + 1/(3.88×10−6 F) = 1/(1.37×10−6 F).The energy stored is U = 1 (1.37×10−6 F)(328 V)2 = 7.37×10−2 J. 2E30-25 V /r = q/4π 0 r2 = E, so that if V is the potential of the sphere then E = V /r is theelectric field on the surface. Then the energy density of the electric field near the surface is 2 1 (8.85×10−12 F/m) (8150 V) u= 2 0E = = 7.41×10−2 J/m3 . 2 2 (0.063 m)E30-26 The charge on C3 can be found from considering the equivalent capacitance. q3 = (3.10×10−6 F)(112 V) = 3.47×10−4 C. The potential across C3 is given by [∆V ]3 = (3.47×10−4 C)/(3.90×10−6 F) = 89.0 V. The potential across the parallel segment is then (112 V) − (89.0 V) = 23.0 V. So [∆V ]1 =[∆V ]2 = 23.0 V. Then q1 = (10.3×10−6 F)(23.0 V) = 2.37×10−4 C and q2 = (4.80×10−6 F)(23.0 V) = 1.10×10−4 C.. 71
• E30-27 There is enough work on this problem without deriving once again the electric fieldbetween charged cylinders. I will instead refer you back to Section 26-4, and state 1 q E= , 2π 0 Lrwhere q is the magnitude of the charge on a cylinder and L is the length of the cylinders. The energy density as a function of radial distance is found from Eq. 30-28, 1 2 1 q2 u= 0E = 2 8π 2 2 2 0 L r The total energy stored in the electric field is given by Eq. 30-24, 1 q2 q 2 ln(b/a) U= = , 2C 2 2π 0 Lwhere we substituted into the last part Eq. 30-11, the capacitance of a cylindrical capacitor. √ We want to show that integrating a volume integral from r = a to r = ab over the energydensity function will yield U/2. Since we want to do this problem the hard way, we will pretend wedon’t know the answer, and integrate from r = a to r = c, and then find out what c is. Then 1 U = u dV, 2 c 2π L 1 q2 = r dr dφ dz, a 0 0 8π 2 2 2 0 L r 2 c 2π L q dr = dφ dz, 8π 2 0 L2 a 0 0 r c q2 dr = , 4π 0 L a r q2 c = ln . 4π 0 L aNow we equate this to the value for U that we found above, and we solve for c. 1 q 2 ln(b/a) q2 c = ln , 2 2 2π 0 L 4π 0 L a ln(b/a) = 2 ln(c/a), (b/a) = (c/a)2 , √ ab = c.E30-28 (a) d = 0 A/C, or d = (8.85×10−12 F/m)(0.350 m2 )/(51.3×10−12 F) = 6.04×10−3 m. (b) C = (5.60)(51.3×10−12 F) = 2.87×10−10 F.E30-29 Originally, C1 = 0 A/d1 . After the changes, C2 = κ 0 A/d2 . Dividing C2 by C1 yieldsC2 /C1 = κd1 /d2 , so κ = d2 C2 /d1 C1 = (2)(2.57×10−12 F)/(1.32×10−12 F) = 3.89. 72
• E30-30 The required capacitance is found from U = 1 C(∆V )2 , or 2 C = 2(6.61×10−6 J)/(630 V)2 = 3.33×10−11 F.The dielectric constant required is κ = (3.33×10−11 F)/(7.40×10−12 F) = 4.50. Try transformer oil.E30-31 Capacitance with dielectric media is given by Eq. 30-31, κe 0 A C= . dThe various sheets have different dielectric constants and different thicknesses, and we want tomaximize C, which means maximizing κe /d. For mica this ratio is 54 mm−1 , for glass this ratio is35 mm−1 , and for paraffin this ratio is 0.20 mm−1 . Mica wins.E30-32 The minimum plate separation is given by d = (4.13×103 V)/(18.2×106 V/m) = 2.27×10−4 m.The minimum plate area is then dC (2.27×10−4 m)(68.4×10−9 F) A= = = 0.627 m2 . κ 0 (2.80)(8.85×10−12 F/m)E30-33 The capacitance of a cylindrical capacitor is given by Eq. 30-11, 1.0×103 m C = 2π(8.85×10−12 F/m)(2.6) = 8.63×10−8 F. ln(0.588/0.11)E30-34 (a) U = C (∆V )2 /2, C = κe 0 A/d, and ∆V /d is less than or equal to the dielectricstrength (which we will call S). Then ∆V = Sd and 1 U= κe 0 AdS 2 , 2so the volume is given by V = 2U/κe 0 S 2 .This quantity is a minimum for mica, so V = 2(250×103 J)/(5.4)(8.85×10−12 F/m)(160×106 V/m)2 = 0.41 m3 . 2 (b) κe = 2U/V 0S , so κe = 2(250×103 J)/(0.087m3 )(8.85×10−12 F/m)(160×106 V/m)2 = 25.E30-35 (a) The capacitance of a cylindrical capacitor is given by Eq. 30-11, L C = 2π 0 κe . ln(b/a)The factor of κe is introduced because there is now a dielectric (the Pyrex drinking glass) betweenthe plates. We can look back to Table 29-2 to get the dielectric properties of Pyrex. The capacitanceof our “glass” is then (0.15 m) C = 2π(8.85×10−12 F/m)(4.7) = 7.3×10−10 F. ln((3.8 cm)/(3.6 cm) (b) The breakdown potential is (14 kV/mm)(2 mm) = 28 kV. 73
• E30-36 (a) C = κe C = (6.5)(13.5×10−12 F) = 8.8×10−11 F. (b) Q = C ∆V = (8.8×10−11 F)(12.5 V) = 1.1×10−9 C. (c) E = ∆V /d, but we don’t know d. (d) E = E/κe , but we couldn’t find E. E30-37 (a) Insert the slab so that it is a distance a above the lower plate. Then the distancebetween the slab and the upper plate is d − a − b. Inserting the slab has the same effect as havingtwo capacitors wired in series; the separation of the bottom capacitor is a, while that of the topcapacitor is d − a − b. The bottom capacitor has a capacitance of C1 = 0 A/a, while the top capacitor has a capacitanceof C2 = 0 A/(d − a − b). Adding these in series, 1 1 1 = + , C eq C1 C2 a d−a−b = + , 0A 0A d−b = . 0A So the capacitance of the system after putting the copper slab in is C = 0 A/(d − b). (b) The energy stored in the system before the slab is inserted is q2 q2 d Ui = = 2C i 2 0Awhile the energy stored after the slab is inserted is q2 q2 d − b Uf = = 2C f 2 0AThe ratio is U i /U f = d/(d − b). (c) Since there was more energy before the slab was inserted, then the slab must have gone inwillingly, it was pulled in!. To get the slab back out we will need to do work on the slab equal tothe energy difference. q2 d q2 d − b q2 b Ui − Uf = − = . 2 0A 2 0A 2 0AE30-38 (a) Insert the slab so that it is a distance a above the lower plate. Then the distancebetween the slab and the upper plate is d − a − b. Inserting the slab has the same effect as havingtwo capacitors wired in series; the separation of the bottom capacitor is a, while that of the topcapacitor is d − a − b. The bottom capacitor has a capacitance of C1 = 0 A/a, while the top capacitor has a capacitanceof C2 = 0 A/(d − a − b). Adding these in series, 1 1 1 = + , C eq C1 C2 a d−a−b = + , 0A 0A d−b = . 0A So the capacitance of the system after putting the copper slab in is C = 0 A/(d − b). 74
• (b) The energy stored in the system before the slab is inserted is C i (∆V )2 (∆V )2 0 A Ui = = 2 2 dwhile the energy stored after the slab is inserted is C f (∆V )2 (∆V )2 0 A Uf = = 2 2 d−bThe ratio is U i /U f = (d − b)/d. (c) Since there was more energy after the slab was inserted, then the slab must not have gone inwillingly, it was being repelled!. To get the slab in we will need to do work on the slab equal to theenergy difference. (∆V )2 0 A (∆V )2 0 A (∆V )2 0 Ab Uf − Ui = − = . 2 d−b 2 d 2 d(d − b)E30-39 C = κe 0 A/d, so d = κe 0 A/C. (a) E = ∆V /d = C∆V /κe 0 A, or (112×10−12 F)(55.0 V) E= = 13400 V/m. (5.4)(8.85×10−12 F/m)(96.5×10−4 m2 ) (b) Q = C∆V = (112×10−12 F)(55.0 V) = 6.16×10−9 C.. (c) Q = Q(1 − 1/κe ) = (6.16×10−9 C)(1 − 1/(5.4)) = 5.02×10−9 C.E30-40 (a) E = q/κe 0 A, so (890×10−9 C) κe = = 6.53 (1.40×106 V/m)(8.85×10−12 F/m)(110×10−4 m2 ) (b) q = q(1 − 1/κe ) = (890×10−9 C)(1 − 1/(6.53)) = 7.54×10−7 C.P30-1 The capacitance of the cylindrical capacitor is from Eq. 30-11, 2π 0 L C= . ln(b/a)If the cylinders are very close together we can write b = a + d, where d, the separation between thecylinders, is a small number, so 2π 0 L 2π 0 L C= = . ln ((a + d)/a) ln (1 + d/a)Expanding according to the hint, 2π 0 L 2πa 0 L C≈ = d/a dNow 2πa is the circumference of the cylinder, and L is the length, so 2πaL is the area of a cylindricalplate. Hence, for small separation between the cylinders we have 0A C≈ , dwhich is the expression for the parallel plates. 75
• P30-2 (a) C = 0 A/x; take the derivative and dC 0 dA 0 A dx = − 2 , dT x dT x dT 1 dA 1 dx = C − . A dT x dT (b) Since (1/A)dA/dT = 2αa and (1/x)dx/dT = αs , we need αs = 2αa = 2(23×10−6 /C◦ ) = 46×10−6 /C◦ . P30-3 Insert the slab so that it is a distance d above the lower plate. Then the distance betweenthe slab and the upper plate is a−b−d. Inserting the slab has the same effect as having two capacitorswired in series; the separation of the bottom capacitor is d, while that of the top capacitor is a−b−d. The bottom capacitor has a capacitance of C1 = 0 A/d, while the top capacitor has a capacitanceof C2 = 0 A/(a − b − d). Adding these in series, 1 1 1 = + , C eq C1 C2 d a−b−d = + , 0A 0A a−b = . 0A So the capacitance of the system after putting the slab in is C = 0 A/(a − b).P30-4 The potential difference between any two adjacent plates is ∆V . Each interior plate has acharge q on each surface; the exterior plate (one pink, one gray) has a charge of q on the interiorsurface only. The capacitance of one pink/gray plate pair is C = 0 A/d. There are n plates, but only n − 1plate pairs, so the total charge is (n − 1)q. This means the total capacitance is C = 0 (n − 1)A/d. P30-5 Let ∆V0 = 96.6 V. As far as point e is concerned point a looks like it is originally positively charged, and point d isoriginally negatively charged. It is then convenient to define the charges on the capacitors in termsof the charges on the top sides, so the original charge on C1 is q 1,i = C1 ∆V0 while the original chargeon C2 is q 2,i = −C2 ∆V0 . Note the negative sign reflecting the opposite polarity of C2 . (a) Conservation of charge requires q 1,i + q 2,i = q 1,f + q 2,f ,but since q = C∆V and the two capacitors will be at the same potential after the switches are closedwe can write C1 ∆V0 − C2 ∆V0 = C1 ∆V + C2 ∆V, (C1 − C2 ) ∆V0 = (C1 + C2 ) ∆V, C1 − C2 ∆V0 = ∆V. C1 + C2With numbers, (1.16 µF) − (3.22 µF) ∆V = (96.6 V) = −45.4 V. (1.16 µF) + (3.22 µF) 76
• The negative sign means that the top sides of both capacitor will be negatively charged after theswitches are closed. (b) The charge on C1 is C1 ∆V = (1.16 µF)(45.4 V) = 52.7µC. (c) The charge on C2 is C2 ∆V = (3.22 µF)(45.4 V) = 146µC.P30-6 C2 and C3 form an effective capacitor with equivalent capacitance Ca = C2 C3 /(C2 + C3 ).The charge on C1 is originally q0 = C1 ∆V0 . After throwing the switch the potential across C1is given by q1 = C1 ∆V1 . The same potential is across Ca ; q2 = q3 , so q2 = Ca ∆V1 . Charge isconserved, so q1 + q2 = q0 . Combining some of the above, q0 C1 ∆V1 = = ∆V0 , C1 + Ca C1 + Caand then 2 2 C1 C1 (C2 + C3 ) q1 = ∆V0 = ∆V0 . C1 + Ca C1 C2 + C1 C3 + C2 C3Similarly, −1 Ca C1 1 1 1 q2 = ∆V0 = + + ∆V0 . C1 + Ca C1 C2 C3q3 = q2 because they are in series. P30-7 (a) If terminal a is more positive than terminal b then current can flow that will charge thecapacitor on the left, the current can flow through the diode on the top, and the current can chargethe capacitor on the right. Current will not flow through the diode on the left. The capacitors areeffectively in series. Since the capacitors are identical and series capacitors have the same charge, we expect thecapacitors to have the same potential difference across them. But the total potential differenceacross both capacitors is equal to 100 V, so the potential difference across either capacitor is 50 V. The output pins are connected to the capacitor on the right, so the potential difference acrossthe output is 50 V. (b) If terminal b is more positive than terminal a the current can flow through the diode on theleft. If we assume the diode is resistanceless in this configuration then the potential difference acrossit will be zero. The net result is that the potential difference across the output pins is 0 V. In real life the potential difference across the diode would not be zero, even if forward biased. Itwill be somewhere around 0.5 Volts.P30-8 Divide the strip of width a into N segments, each of width ∆x = a/N . The capacitance ofeach strip is ∆C = 0 a∆x/y. If θ is small then 1 1 1 d = ≈ ≈ 1 − xθ/d). y d + x sin θ d + xθ (Since parallel capacitances add, a 2 0a 0a aθ C= ∆C = (1 − xθ/d)dx = 1− . d 0 d 2d 77
• P30-9 (a) When S2 is open the circuit acts as two parallel capacitors. The branch on the left hasan effective capacitance given by 1 1 1 1 = + = , Cl (1.0×10−6 F) (3.0×10−6 F) 7.5×10−7 Fwhile the branch on the right has an effective capacitance given by 1 1 1 1 = + = . Cl (2.0×10−6 F) (4.0×10−6 F) 1.33×10−6 FThe charge on either capacitor in the branch on the left is q = (7.5×10−7 F)(12 V) = 9.0×10−6 C,while the charge on either capacitor in the branch on the right is q = (1.33×10−6 F)(12 V) = 1.6×10−5 C. (b) After closing S2 the circuit is effectively two capacitors in series. The top part has an effectivecapacitance of C t = (1.0×10−6 F) + (2.0×10−6 F) = (3.0×10−6 F),while the effective capacitance of the bottom part is C b = (3.0×10−6 F) + (4.0×10−6 F) = (7.0×10−6 F).The effective capacitance of the series combination is given by 1 1 1 1 = −6 F) + −6 F) = . C eq (3.0×10 (7.0×10 2.1×10−6 FThe charge on each part is q = (2.1×10−6 F)(12 V) = 2.52×10−5 C. The potential difference acrossthe top part is ∆V t = (2.52×10−5 C)/(3.0×10−6 F) = 8.4 V,and then the charge on the top two capacitors is q1 = (1.0 × 10−6 F)(8.4 V) = 8.4 × 10−6 C andq2 = (2.0×10−6 F)(8.4 V) = 1.68×10−5 C. The potential difference across the bottom part is ∆V t = (2.52×10−5 C)/(7.0×10−6 F) = 3.6 V,and then the charge on the top two capacitors is q1 = (3.0 × 10−6 F)(3.6 V) = 1.08 × 10−5 C andq2 = (4.0×10−6 F)(3.6 V) = 1.44×10−5 C.P30-10 Let ∆V = ∆Vxy . By symmetry ∆V2 = 0 and ∆V1 = ∆V4 = ∆V5 = ∆V3 = ∆V /2.Suddenly the problem is very easy. The charges on each capacitor is q1 , except for q2 = 0. Then theequivalent capacitance of the circuit is q q1 + q4 C eq = = = C1 = 4.0×10−6 F. ∆V 2∆V1 78
• P30-11 (a) The charge on the capacitor with stored energy U0 = 4.0 J is q0 , where 2 q0 U0 = . 2CWhen this capacitor is connected to an identical uncharged capacitor the charge is shared equally,so that the charge on either capacitor is now q = q0 /2. The stored energy in one capacitor is then q2 q 2 /4 1 U= = 0 = U0 . 2C 2C 4But there are two capacitors, so the total energy stored is 2U = U0 /2 = 2.0 J. (b) Good question. Current had to flow through the connecting wires to get the charge from onecapacitor to the other. Originally the second capacitor was uncharged, so the potential differenceacross that capacitor would have been zero, which means the potential difference across the con-necting wires would have been equal to that of the first capacitor, and there would then have beenenergy dissipation in the wires according to P = i2 R.That’s where the missing energy went.P30-12 R = ρL/A and C = 0 A/L. Combining, R = ρ 0 /C, or R = (9.40 Ω · m)(8.85×10−12 F/m)/(110×10−12 F) = 0.756 Ω. 1P30-13 (a) u = 2 0 E 2 = e2 /32π 2 0 r4 . (b) U = u dV where dV = 4πr2 dr. Then ∞ e2 e2 1 U = 4pi 2 r4 r2 dr = . R 32π 0 8π 0 R (c) R = e2 /8π 0 mc2 , or (1.60×10−19 C)2 R= = 1.40×10−15 m. 8π(8.85×10−12 F/m)(9.11×10−31 kg)(3.00×108 m/s)2 1P30-14 U = 2 q 2 /C = q 2 x/2A 0 . F = dU/dx = q 2 /2A 0 .P30-15 According to Problem 14, the force on a plate of a parallel plate capacitor is q2 F = . 2 0AThe force per unit area is then F q2 σ2 = = , A 2 0 A2 2 0where σ = q/A is the surface charge density. But we know that the electric field near the surface ofa conductor is given by E = σ/ 0 , so F 1 = 0E2. A 2 79
• P30-16 A small surface area element dA carries a charge dq = q dA/4πR2 . There are three forceson the elements which balance, so p(V0 /V )dA + q dq/4π 0 R2 = p dA,or pR0 + q 2 /16π 2 0 R = pR3 . 3This can be rearranged as q 2 = 16π 2 0 pR(R3 − R0 ). 3P30-17 The magnitude of the electric field in the cylindrical region is given by E = λ/2π 0 r,where λ is the linear charge density on the anode. The potential difference is given by ∆V =(λ/2π 0 ) ln(b/a), where a is the radius of the anode b the radius of the cathode. Combining, E =∆V /r ln(b/a), this will be a maximum when r = a, so ∆V = (0.180×10−3 m) ln[(11.0×10−3 m)/(0.180×10−3 m)](2.20×106 V/m) = 1630 V.P30-18 This is effectively two capacitors in parallel, each with an area of A/2. Then 0 A/2 0 A/2 0A κe1 + κe2 C eq = κe1 + κe2 = . d d d 2 P30-19 We will treat the system as two capacitors in series by pretending there is an infinitesi-mally thin conductor between them. The slabs are (I assume) the same thickness. The capacitanceof one of the slabs is then given by Eq. 30-31, κe1 0 A C1 = , d/2where d/2 is the thickness of the slab. There would be a similar expression for the other slab. Theequivalent series capacitance would be given by Eq. 30-21, 1 1 1 = + , C eq C1 C2 d/2 d/2 = + , κe1 0 A κe2 0 A d κe2 + κe1 = , 2 0 A κe1 κe2 2 0 A κe1 κe2 C eq = . d κe2 + κe1P30-20 Treat this as three capacitors. Find the equivalent capacitance of the series combinationon the right, and then add on the parallel part on the left. The right hand side is 1 d d 2d κe2 + κe3 = + = . C eq κe2 0 A/2 κe3 0 A/2 0A κe2 κe3Add this to the left hand side, and κe1 0 A/2 0A κe2 κe3 C eq = + , 2d 2d κe2 + κe3 0 A κe1 κe2 κe3 = + . 2d 2 κe2 + κe3 80
• P30-21 (a) q doesn’t change, but C = C/2. Then ∆V = q/C = 2∆V . (b) U = C(∆V )2 /2 = 0 A(∆V )2 /2d. U = C (∆V )2 /2 = 0 A(2∆V )2 /4d = 2U . (c) W = U − U = 2U − U = U = 0 A(∆V )2 /2d.P30-22 The total energy is U = qδV /2 = (7.02×10−10 C)(52.3 V)/2 = 1.84×10−8 J. (a) In the air gap we have 2 0 E0 V (8.85×10−12 F/m)(6.9×103 V/m)2 (1.15×10−2 m2 )(4.6×10−3 m) Ua = = = 1.11×10−8 J. 2 2That is (1.11/1.85) = 60% of the total. (b) The remaining 40% is in the slab.P30-23 (a) C = 0 A/d = (8.85×10−12 F/m)(0.118 m2 )/(1.22×10−2 m) = 8.56×10−11 F. (b) Use the results of Problem 30-24. (4.8)(8.85×10−12 F/m)(0.118 m2 ) C = = 1.19×10−10 F (4.8)(1.22×10−2 m) − (4.3×10−3 m)(4.8 − 1) (c) q = C∆V = (8.56×10−11 F)(120 V) = 1.03×10−8 C; since the battery is disconnected q = q. (d) E = q/ 0 A = (1.03×10−8 C)/(8.85×10−12 F/m)(0.118 m2 ) = 9860 V/m in the space betweenthe plates. (e) E = E/κe = (9860 V/m)/(4.8) = 2050 V/m in the dielectric. (f) ∆V = q/C = (1.03×10−8 C)/(1.19×10−10 F) = 86.6 V. (g) W = U − U = q 2 (1/C − 1/C )/2, or (1.03×10−8 C)2 W = [1/(8.56×10−11 F) − 1/(1.19×10−10 F)] = 1.73×10−7 J. 2P30-24 The result is effectively three capacitors in series. Two are air filled with thicknesses ofx and d − b − x, the third is dielectric filled with thickness b. All have an area A. The effectivecapacitance is given by 1 x d−b−x b = + + , C 0A 0A κe 0 A 1 b = (d − b) + , 0 A κe 0A C = , d − b + b/κe κe 0 A = . κe − b(κe − 1) 81
• E31-1 (5.12 A)(6.00 V)(5.75 min)(60 s/min) = 1.06×104 J.E31-2 (a) (12.0 V)(1.60×10−19 C) = 1.92×10−18 J. (b) (1.92×10−18 J)(3.40×1018 /s) = 6.53 W.E31-3 If the energy is delivered at a rate of 110 W, then the current through the battery is P (110 W) i= = = 9.17 A. ∆V (12 V) Current is the flow of charge in some period of time, so ∆q (125 A · h) ∆t = = = 13.6 h, i (9.2 A)which is the same as 13 hours and 36 minutes.E31-4 (100 W)(8 h) = 800 W · h. (a) (800 W · h)/(2.0 W · h) = 400 batteries, at a cost of (400)(\$0.80) = \$320. (b) (800 W · h)(\$0.12×10−3 W · h) = \$0.096. E31-5 Go all of the way around the circuit. It is a simple one loop circuit, and although it doesnot matter which way we go around, we will follow the direction of the larger emf. Then (150 V) − i(2.0 Ω) − (50 V) − i(3.0 Ω) = 0,where i is positive if it is counterclockwise. Rearranging, 100 V = i(5.0 Ω),or i = 20 A. Assuming the potential at P is VP = 100 V, then the potential at Q will be given by VQ = VP − (50 V) − i(3.0 Ω) = (100 V) − (50 V) − (20 A)(3.0 Ω) = −10 V.E31-6 (a) Req = (10 Ω) + (140 Ω) = 150 Ω. i = (12.0 V)/(150 Ω) = 0.080 A. (b) Req = (10 Ω) + (80 Ω) = 90 Ω. i = (12.0 V)/(90 Ω) = 0.133 A. (c) Req = (10 Ω) + (20 Ω) = 30 Ω. i = (12.0 V)/(30 Ω) = 0.400 A.E31-7 (a) Req = (3.0 V − 2.0 V)/(0.050 A) = 20 Ω. Then R = (20 Ω) − (3.0 Ω) − (3.0 Ω) = 14 Ω. (b) P = i∆V = i2 R = (0.050 A)2 (14 Ω) = 3.5×10−2 W.E31-8 (5.0 A)R1 = ∆V . (4.0 A)(R1 +2.0 Ω) = ∆V . Combining, 5R1 = 4R1 +8.0 Ω, or R1 = 8.0 Ω.E31-9 (a) (53.0 W)/(1.20 A) = 44.2 V. (b) (1.20 A)(19.0 Ω) = 22.8 V is the potential difference across R. Then an additional potentialdifference of (44.2 V) − (22.8 V) = 21.4 V must exist across C. (c) The left side is positive; it is a reverse emf.E31-10 (a) The current in the resistor is (9.88 W)/(0.108 Ω) = 9.56 A. The total resistance ofthe circuit is (1.50 V)/(9.56 A) = 0.157 Ω. The internal resistance of the battery is then (0.157 Ω) −(0.108 Ω) = 0.049 Ω. (b) (9.88 W)/(9.56 A) = 1.03 V. 82
• E31-11 We assign directions to the currents through the four resistors as shown in the figure. 1 2 a b 3 4 Since the ammeter has no resistance the potential at a is the same as the potential at b. Con-sequently the potential difference (∆V b ) across both of the bottom resistors is the same, and thepotential difference (∆V t ) across the two top resistors is also the same (but different from thebottom). We then have the following relationships: ∆V t + ∆V b = E, i1 + i2 = i3 + i4 , ∆Vj = i j Rj ,where the j subscript in the last line refers to resistor 1, 2, 3, or 4. For the top resistors, ∆V1 = ∆V2 implies 2i1 = i2 ;while for the bottom resistors, ∆V3 = ∆V4 implies i3 = i4 .Then the junction rule requires i4 = 3i1 /2, and the loop rule requires (i1 )(2R) + (3i1 /2)(R) = E or i1 = 2E/(7R).The current that flows through the ammeter is the difference between i2 and i4 , or 4E/(7R) −3E/(7R) = E/(7R).E31-12 (a) Define the current i1 as moving to the left through r1 and the current i2 as movingto the left through r2 . i3 = i1 + i2 is moving to the right through R. Then there are two loopequations: E1 = i1 r1 + i3 R, E2 = (i3 − i1 )r2 + i3 R.Multiply the top equation by r2 and the bottom by r1 and then add: r2 E1 + r1 E2 = i3 r1 r2 + i3 R(r1 + r2 ),which can be rearranged as r2 E1 + r1 E2 i3 = . r1 r2 + Rr1 + Rr2 (b) There is only one current, so E1 + E2 = i(r1 + r2 + R),or E1 + E2 i= . r1 + r 2 + R 83
• E31-13 (a) Assume that the current flows through each source of emf in the same direction asthe emf. The the loop rule will give us three equations E1 − i1 R1 + i2 R2 − E2 − i1 R1 = 0, E2 − i2 R2 + i3 R1 − E3 + i3 R1 = 0, E1 − i1 R1 + i3 R1 − E3 + i3 R1 − i1 R1 = 0.The junction rule (looks at point a) gives us i1 + i2 + i3 = 0. Use this to eliminate i2 from the secondloop equation, E2 + i1 R2 + i3 R2 + 2i3 R1 − E3 = 0,and then combine this with the the third equation to eliminate i3 , 2 E1 R2 − E3 R2 + 2i3 R1 R2 + 2E2 R1 + 2i3 R1 R2 + 4i3 R1 − 2E3 R1 = 0,or 2E3 R1 + E3 R2 − E1 R2 − 2E2 R1 i3 = 2 = 0.582 A. 4R1 R2 + 4R1Then we can find i1 from E3 − E2 − i3 R2 − 2i3 R1 i1 = = −0.668 A, R2where the negative sign indicates the current is down. Finally, we can find i2 = −(i1 + i3 ) = 0.0854 A. (b) Start at a and go to b (final minus initial!), +i2 R2 − E2 = −3.60 V.E31-14 (a) The current through the circuit is i = E/(r + R). The power delivered to R is thenP = i∆V = i2 R = E 2 R/(r + R)2 . Evaluate dP/dR and set it equal to zero to find the maximum.Then dP r−R 0= = E 2R , dR (r + R)3which has the solution r = R. (b) When r = R the power is 1 E2 P = E 2R = . (R + R)2 4r E31-15 (a) We first use P = F v to find the power output by the electric motor. Then P =(2.0 N)(0.50 m/s) = 1.0 W. The potential difference across the motor is ∆V m = E − ir. The power output from the motor isthe rate of energy dissipation, so P m = ∆V m i. Combining these two expressions, Pm = (E − ir) i, = Ei − i2 r, 0 = −i2 r + Ei − P m , 0 = (0.50 Ω)i2 − (2.0 V)i + (1.0 W).Rearrange and solve for i, (2.0 V) ± (2.0 V)2 − 4(0.50 Ω)(1.0 W) i= , 2(0.50 Ω) 84
• which has solutions i = 3.4 A and i = 0.59 A. (b) The potential difference across the terminals of the motor is ∆V m = E − ir which if i = 3.4 Ayields ∆V m = 0.3 V, but if i = 0.59 A yields ∆V m = 1.7 V. The battery provides an emf of 2.0 V; itisn’t possible for the potential difference across the motor to be larger than this, but both solutionsseem to satisfy this constraint, so we will move to the next part and see what happens. (c) So what is the significance of the two possible solutions? It is a consequence of the fact thatpower is related to the current squared, and with any quadratics we expect two solutions. Bothare possible, but it might be that only one is stable, or even that neither is stable, and a smallperturbation to the friction involved in turning the motor will cause the system to break down. Wewill learn in a later chapter that the effective resistance of an electric motor depends on the speedat which it is spinning, and although that won’t affect the problem here as worded, it will affect thephysical problem that provided the numbers in this problem!E31-16 req = 4r = 4(18 Ω) = 72 Ω. The current is i = (27 V)/(72 Ω) = 0.375 A. E31-17 In parallel connections of two resistors the effective resistance is less than the smallerresistance but larger than half the smaller resistance. In series connections of two resistors theeffective resistance is greater than the larger resistance but less than twice the larger resistance. Since the effective resistance of the parallel combination is less than either single resistanceand the effective resistance of the series combinations is larger than either single resistance we canconclude that 3.0 Ω must have been the parallel combination and 16 Ω must have been the seriescombination. The resistors are then 4.0 Ω and 12 Ω resistors.E31-18 Points B and C are effectively the same point! (a) The three resistors are in parallel. Then req = R/3. (b) See (a). (c) 0, since there is no resistance between B and C.E31-19 Focus on the loop through the battery, the 3.0 Ω, and the 5.0 Ω resistors. The loop ruleyields (12.0 V) = i[(3.0 Ω) + (5.0 Ω)] = i(8.0 Ω).The potential difference across the 5.0 Ω resistor is then ∆V = i(5.0 Ω) = (5.0 Ω)(12.0 V)/(8.0 Ω) = 7.5 V.E31-20 Each lamp draws a current of (500 W)/(120 V) = 4.17 A. Furthermore, the fuse cansupport (15 A)/(4.17 A) = 3.60 lamps. That is a maximum of 3.E31-21 The current in the series combination is is = E/(R1 + R2 ). The power dissipated isP s = iE = E 2 /(R1 + R2 ). In a parallel arrangement R1 dissipates P1 = i1 E = E 2 /R1 . A similar expression exists for R2 ,so the total power dissipated is P p = E 2 (1/R1 + 1/R2 ). The ratio is 5, so 5 = P p /P s = (1/R1 + 1/R2 )(R1 + R2 ), or 5R1 R2 = (R1 + R2 )2 . Solving forR2 yields 2.618R1 or 0.382R1 . Then R2 = 262 Ω or R2 = 38.2 Ω. 85
• E31-22 Combining n identical resistors in series results in an equivalent resistance of req = nR.Combining n identical resistors in parallel results in an equivalent resistance of req = R/n. If theresistors are arranged in a square array consisting of n parallel branches of n series resistors, thenthe effective resistance is R. Each will dissipate a power P , together they will dissipate n2 P . So we want nine resistors, since four would be too small. E31-23 (a) Work through the circuit one step at a time. We first “add” R2 , R3 , and R4 inparallel: 1 1 1 1 1 = + + = Reff 42.0 Ω 61.6 Ω 75.0 Ω 18.7 ΩWe then “add” this resistance in series with R1 , Reff = (112 Ω) + (18.7 Ω) = 131 Ω. (b) The current through the battery is i = E/R = (6.22 V)/(131 Ω) = 47.5 mA. This is also thecurrent through R1 , since all the current through the battery must also go through R1 . The potential difference across R1 is ∆V1 = (47.5 mA)(112 Ω) = 5.32 V. The potential differenceacross each of the three remaining resistors is 6.22 V − 5.32 V = 0.90 V. The current through each resistor is then i2 = (0.90 V)/(42.0 Ω) = 21.4 mA, i3 = (0.90 V)/(61.6 Ω) = 14.6 mA, i4 = (0.90 V)/(75.0 Ω) = 12.0 mA.E31-24 The equivalent resistance of the parallel part is r = R2 R/(R2 + R). The equivalentresistance for the circuit is r = R1 + R2 R/(R2 + R). The current through the circuit is i = E/r.The potential difference across R is ∆V = E − i R1 , or ∆V = E(1 − R1 /r), R2 + R = E 1 − R1 , R1 R2 + R1 R + RR2 RR2 = E . R1 R2 + R1 R + RR2Since P = i∆V = (∆V )2 /R, 2 RR2 P = E2 . (R1 R2 + R1 R + RR2 )2Set dP/dR = 0, the solution is R = R1 R2 /(R1 + R2 ). E31-25 (a) First “add” the left two resistors in series; the effective resistance of that branch is2R. Then “add” the right two resistors in series; the effective resistance of that branch is also 2R. Now we combine the three parallel branches and find the effective resistance to be 1 1 1 1 4 = + + = , Reff 2R R 2R 2Ror Reff = R/2. (b) First we “add” the right two resistors in series; the effective resistance of that branch is 2R.We then combine this branch with the resistor which connects points F and H. This is a parallelconnection, so the effective resistance is 1 1 1 3 = + = , Reff 2R R 2R 86
• or 2R/3. This value is effectively in series with the resistor which connects G and H, so the “total” is5R/3. Finally, we can combine this value in parallel with the resistor that directly connects F and Gaccording to 1 1 3 8 = + = , Reff R 5R 5Ror Reff = 5R/8.E31-26 The resistance of the second resistor is r2 = (2.4 V)/(0.001 A) = 2400 Ω. The potentialdifference across the first resistor is (12 V) − (2.4 V) = 9.6 V. The resistance of the first resistor is(9.6 V)/(0.001 A) = 9600 Ω.E31-27 See Exercise 31-26. The resistance ratio is r1 (0.95 ± 0.1 V) = , r 1 + r2 (1.50 V)or r2 (1.50 V) = − 1. r1 (0.95 ± 0.1 V)The allowed range for the ratio r2 /r1 is between 0.5625 and 0.5957. We can choose any standard resistors we want, and we could use any tolerance, but then wewill need to check our results. 22Ω and 39Ω would work; as would 27Ω and 47Ω. There are otherchoices. E31-28 Consider any junction other than A or B. Call this junction point 0; label the fournearest junctions to this as points 1, 2, 3, and 4. The current through the resistor that linkspoint 0 to point 1 is i1 = ∆V01 /R, where ∆V01 is the potential difference across the resistor, so∆V01 = V0 − V1 , where V0 is the potential at the junction 0, and V1 is the potential at the junction1. Similar expressions exist for the other three resistor. For the junction 0 the net current must be zero; there is no way for charge to accumulate on thejunction. Then i1 + i2 + i3 + i4 = 0, and this means ∆V01 /R + ∆V02 /R + ∆V03 /R + ∆V04 /R = 0or ∆V01 + ∆V02 + ∆V03 + ∆V04 = 0.Let ∆V0i = V0 − Vi , and then rearrange, 4V0 = V1 + V2 + V3 + V4 ,or 1 V0 = (V1 + V2 + V3 + V4 ) . 4 E31-29 The current through the radio is i = P/∆V = (7.5 W)/(9.0 V) = 0.83 A. The radiowas left one for 6 hours, or 2.16×104 s. The total charge to flow through the radio in that time is(0.83 A)(2.16×104 s) = 1.8×104 C.E31-30 The power dissipated by the headlights is (9.7 A)(12.0 V) = 116 W. The power requiredby the engine is (116 W)/(0.82) = 142 W, which is equivalent to 0.190 hp. 87
• E31-31 (a) P = (120 V)(120 V)/(14.0 Ω) = 1030 W. (b) W = (1030 W)(6.42 h) = 6.61 kW · h. The cost is \$0.345.E31-32E31-33 We want to apply either Eq. 31-21, PR = i2 R,or Eq. 31-22, PR = (∆VR )2 /R,depending on whether we are in series (the current is the same through each bulb), or in parallel(the potential difference across each bulb is the same. The brightness of a bulb will be measured byP , even though P is not necessarily a measure of the rate radiant energy is emitted from the bulb. (b) If the bulbs are in parallel then PR = (∆VR )2 /R is how we want to compare the brightness.The potential difference across each bulb is the same, so the bulb with the smaller resistance isbrighter. (b) If the bulbs are in series then PR = i2 R is how we want to compare the brightness. Bothbulbs have the same current, so the larger value of R results in the brighter bulb. One direct consequence of this can be tried at home. Wire up a 60 W, 120 V bulb and a 100 W,120 V bulb in series. Which is brighter? You should observe that the 60 W bulb will be brighter. 2E31-34 (a) j = i/A = (25 A)/π(0.05 in) = 3180 A/in = 4.93×106 A/m2 . (b) E = ρj = (1.69×10−8 Ω · m)(4.93×106 A/m2 ) = 8.33×10−2 V/m. (c) ∆V = Ed = (8.33×10−2 V/m)(305 m) = 25 V. (d) P = i∆V = (25 A)(25 V) = 625 W.E31-35 (a) The bulb is on for 744 hours. The energy consumed is (100 W)(744 h) = 74.4 kW · h,at a cost of (74.4)(0.06) = \$4.46. (b) r = V 2 /P = (120 V)2 /(100 W) = 144 Ω. (c) i = P/V = (100 W)/(120 V) = 0.83 A.E31-36 P = (∆V )2 /r and r = r0 (1 + α∆T ). Then P0 (500 W) P = = = 660 W 1 + α∆T 1 + (4.0×10−4 /C◦ )(−600C◦ )E31-37 (a) n = q/e = it/e, so n = (485×10−3 A)(95×10−9 s)/(1.6×10−19 C) = 2.88×1011 . (b) iav = (520/s)(485×10−3 A)(95×10−9 s) = 2.4×10−5 A. (c) P p = ip ∆V = (485×10−3 A)(47.7×106 V) = 2.3×106 W; while P a = ia ∆V = (2.4×10−5 A)(47.7×10 V) = 1.14×103 W. 6E31-38 r = ρL/A = (3.5×10−5 Ω · m)(1.96×10−2 m)/π(5.12×10−3 m)2 = 8.33×10−3 Ω. (a) i = P/r = (1.55 W)/(8.33×10−3 Ω) = 13.6 A, so j = i/A = (13.6 A)/π(5.12×10−3 m)2 = 1.66×105 A/m2 . √ (b) ∆V = Pr = (1.55 W)(8.33×10−3 Ω) = 0.114 V. 88
• E31-39 (a) The current through the wire is i = P/∆V = (4800 W)/(75 V) = 64 A,The resistance of the wire is R = ∆V /i = (75 V)/(64 A) = 1.17 Ω.The length of the wire is then found from RA (1.17 Ω)(2.6×10−6 m2 ) L= = = 6.1 m. ρ (5.0×10−7 Ωm)One could easily wind this much nichrome to make a toaster oven. Of course allowing 64 Amps tobe drawn through household wiring will likely blow a fuse. (b) We want to combine the above calculations into one formula, so RA A∆V /i A(∆V )2 L= = = , ρ ρ Pρthen (2.6×10−6 m2 )(110 V)2 L= = 13 m. (4800 W)(5.0×10−7 Ωm)Hmm. We need more wire if the potential difference is increased? Does this make sense? Yes, itdoes. We need more wire because we need more resistance to decrease the current so that the samepower output occurs.E31-40 (a) The energy required to bring the water to boiling is Q = mC∆T . The time requiredis Q (2.1 kg)(4200 J/kg)(100◦ C − 18.5◦ C) t= = = 2.22×103 s 0.77P 0.77(420 W) (b) The additional time required to boil half of the water away is mL/2 (2.1 kg)(2.26×106 J/kg)/2 t= = = 7340 s. 0.77P 0.77(420 W)E31-41 (a) Integrate both sides of Eq. 31-26; q t dq dt = − , 0 q − EC 0 RC t q t ln(q − EC)|0 = − , RC 0 q − EC t ln = − , −EC RC q − EC = e−t/RC , −EC q = EC 1 − e−t/RC .That wasn’t so bad, was it? 89
• (b) Rearrange Eq. 31-26 in order to get q terms on the left and t terms on the right, thenintegrate; q t dq dt = − , q0 q 0 RC t q t ln q|q0 = − , RC 0 q t ln = − , q0 RC q = e−t/RC , q0 q = q0 e−t/RC .That wasn’t so bad either, was it?E31-42 (a) τC = RC = (1.42×106 Ω)(1.80×10−6 F) = 2.56 s. (b) q0 = C∆V = (1.80×10−6 F)(11.0 V) = 1.98×10−5 C. (c) t = −τC ln(1 − q/q0 ), so t = −(2.56 s) ln(1 − 15.5×10−6 C/1.98×10−5 C) = 3.91 s.E31-43 Solve n = t/τC = − ln(1 − 0.99) = 4.61.E31-44 (a) ∆V = E(1 − e−t/τC ), so τC = −(1.28×10−6 s)/ ln(1 − 5.00 V/13.0 V) = 2.64×10−6 s (b) C = τC /R = (2.64×10−6 s)/(15.2×103 Ω) = 1.73×10−10 FE31-45 (a) ∆V = Ee−t/τC , so τC = −(10.0 s)/ ln(1.06 V/100 V) = 2.20 s (b) ∆V = (100 V)e−17 s/2.20 s = 4.4×10−2 V.E31-46 ∆V = Ee−t/τC and τC = RC, so t t t R=− =− = . C ln(∆V /∆V0 ) (220×10−9 F) ln(0.8 V/5 V) 4.03×10−7 FIf t is between 10.0 µs and 6.0 ms, then R is between R = (10×10−6 s)/(4.03×10−7 F) = 24.8Ω,and R = (6×10−3 s)/(4.03×10−7 F) = 14.9×103 Ω. E31-47 The charge on the capacitor needs to build up to a point where the potential across thecapacitor is VL = 72 V, and this needs to happen within 0.5 seconds. This means that we want tosolve C∆VL = CE 1 − eT /RCfor R knowing that T = 0.5 s. This expression can be written as T (0.5 s) R=− =− = 2.35×106 Ω. C ln(1 − VL /E) (0.15 µC) ln(1 − (72 V)/(95 V)) 90
• √E31-48 (a) q0 = 2U C = 2(0.50 J)(1.0×10−6 F) = 1×10−3 C. (b) i0 = ∆V0 /R = q0 /RC = (1×10−3 C)/(1.0×106 Ω)(1.0×10−6 F) = 1×10−3 A. (c) ∆VC = ∆V0 e−t/τC , so (1×10−3 C) −t/(1.0×106 Ω)(1.0×10−6 F) ∆VC = e = (1000 V)e−t/(1.0 s) (1.0×10−6 F)Note that ∆VR = ∆VC . (d) PR = (∆VR )2 /R, so PR = (1000 V)2 e−2t/(1.0 s) /(1×106 ω) = (1 W)e−2t/(1.0 s) .E31-49 (a) i = dq/dt = Ee−t/τC /R, so (4.0 V) 6 −6 i= e−(1.0 s)/(3.0×10 Ω)(1.0×10 F) = 9.55×10−7 A. (3.0×106 Ω) (b) PC = i∆V = (E 2 /R)e−t/τC (1 − e−t/τC ), so (4.0 V)2 −(1.0 s)/(3.0×106 Ω)(1.0×10−6 F) 6 −6 PC = e 1 − e−(1.0 s)/(3.0×10 Ω)(1.0×10 F) = 1.08×10−6 W. (3.0×106 Ω) (c) PR = i2 R = (E 2 /R)e−2t/τC , so (4.0 V)2 −2(1.0 s)/(3.0×106 Ω)(1.0×10−6 F) PR = e = 2.74×10−6 W. (3.0×106 Ω) (d) P = PR + PC , or P = 2.74×10−6 W + 1.08×10−6 W = 3.82×10−6 WE31-50 The rate of energy dissipation in the resistor is PR = i2 R = (E 2 /R)e−2t/τC .Evaluating ∞ E 2 ∞ −2t/RC E2 PR dt = e dt = C, 0 R 0 2but that is the original energy stored in the capacitor.P31-1 The terminal voltage of the battery is given by V = E − ir, so the internal resistance is E −V (12.0 V) − (11.4 V) r= = = 0.012 Ω, i (50 A)so the battery appears within specs. The resistance of the wire is given by ∆V (3.0 V) R= = = 0.06 Ω, i (50 A)so the cable appears to be bad. What about the motor? Trying it, ∆V (11.4 V) − (3.0 V) R= = = 0.168 Ω, i (50 A)so it appears to be within spec. 91
• P31-2 Traversing the circuit we have E − ir1 + E − ir2 − iR = 0,so i = 2E/(r1 + r2 + R). The potential difference across the first battery is then 2r1 r2 − r 1 + R ∆V1 = E − ir1 = E 1 − =E r 1 + r2 + R r1 + r 2 + RThis quantity will only vanish if r2 − r1 + R = 0, or r1 = R + r2 . Since r1 > r2 this is actuallypossible; R = r1 − r2 .P31-3 ∆V = E − iri and i = E/(ri + R), so R ∆V = E , ri + RThere are then two simultaneous equations: (0.10 V)(500 Ω) + (0.10 V)ri = E(500 Ω)and (0.16 V)(1000 Ω) + (0.16 V)ri = E(1000 Ω),with solution (a) ri = 1.5×103 Ω and (b) E = 0.400 V. 2 (c) The cell receives energy from the sun at a rate (2.0 mW/cm )(5.0 cm2 ) = 0.010 W. The cell 2 2converts energy at a rate of V /R = (0.16 V) /(1000 Ω) = 0.26 %P31-4 (a) The emf of the battery can be found from E = iri + ∆V l = (10 A)(0.05 Ω) + (12 V) = 12.5 V (b) Assume that resistance is not a function of temperature. The resistance of the headlights isthen rl = (12.0 V)/(10.0 A) = 1.2 Ω.The potential difference across the lights when the starter motor is on is ∆V l = (8.0 A)(1.2 Ω) = 9.6 V,and this is also the potential difference across the terminals of the battery. The current through thebattery is then E − ∆V (12.5 V) − (9.6 V) i= = = 58 A, ri (0.05 Ω)so the current through the motor is 50 Amps.P31-5 (a) The resistivities are ρA = rA A/L = (76.2×10−6 Ω)(91.0×10−4 m2 )/(42.6 m) = 1.63×10−8 Ω · m,and ρB = rB A/L = (35.0×10−6 Ω)(91.0×10−4 m2 )/(42.6 m) = 7.48×10−9 Ω · m. (b) The current is i = ∆V /(rA + rB ) = (630 V)/(111.2 µΩ) = 5.67×106 A. The current densityis then j = (5.67×106 A)/(91.0×10−4 m2 ) = 6.23×108 A/m2 . (c) EA = ρA j = (1.63×10−8 Ω · m)(6.23×108 A/m2 ) = 10.2 V/m and EB = ρB j = (7.48×10−9 Ω ·m)(6.23×108 A/m2 ) = 4.66 V/m. (d) ∆VA = EA L = (10.2 V/m)(42.6 m) = 435 V and ∆VB = EB L = (4.66 V/m)(42.6 m) = 198 V. 92
• P31-6 Set up the problem with the traditional presentation of the Wheatstone bridge problem.Then the symmetry of the problem (flip it over on the line between x and y) implies that there is nocurrent through r. As such, the problem is equivalent to two identical parallel branches each withtwo identical series resistances. Each branch has resistance R + R = 2R, so the overall circuit has resistance 1 1 1 1 = + = , Req 2R 2R Rso Req = R.P31-7P31-8 (a) The loop through R1 is trivial: i1 = E2 /R1 = (5.0 V)/(100 Ω) = 0.05 A. The loopthrough R2 is only slightly harder: i2 = (E2 + E3 − E1 )/R2 = 0.06 A. (b) ∆Vab = E3 + E2 = (5.0 V) + (4.0 V) = 9.0 V. P31-9 (a) The three way light-bulb has two filaments (or so we are told in the question). Thereare four ways for these two filaments to be wired: either one alone, both in series, or both inparallel. Wiring the filaments in series will have the largest total resistance, and since P = V 2 /Rthis arrangement would result in the dimmest light. But we are told the light still operates at thelowest setting, and if a filament burned out in a series arrangement the light would go out. We then conclude that the lowest setting is one filament, the middle setting is another filament,and the brightest setting is both filaments in parallel. (b) The beauty of parallel settings is that then power is additive (it is also addictive, but that’sa different field.) One filament dissipates 100 W at 120 V; the other filament (the one that burnsout) dissipates 200 W at 120 V, and both together dissipate 300 W at 120 V. The resistance of one filament is then (∆V )2 (120 V)2 R= = = 144 Ω. P (100 W)The resistance of the other filament is (∆V )2 (120 V)2 R= = = 72 Ω. P (200 W)P31-10 We can assume that R “contains” all of the resistance of the resistor, the battery and theammeter, then R = (1.50 V)/(1.0 m/A) = 1500 Ω.For each of the following parts we apply R + r = ∆V /i, so (a) r = (1.5 V)/(0.1 mA) − (1500 Ω) = 1.35×104 Ω, (b) r = (1.5 V)/(0.5 mA) − (1500 Ω) = 1.5×103 Ω, (c) r = (1.5 V)/(0.9 mA) − (1500 Ω) = 167Ω. (d) R = (1500 Ω) − (18.5 Ω) = 1482 ΩP31-11 (a) The effective resistance of the parallel branches on the middle and the right is R2 R 3 . R2 + R3 93
• The effective resistance of the circuit as seen by the battery is then R2 R3 R1 R 2 + R1 R3 + R2 R3 R1 + = , R2 + R3 R 2 + R3The current through the battery is R 2 + R3 i=E , R1 R 2 + R1 R3 + R2 R3The potential difference across R1 is then R2 + R 3 ∆V1 = E R1 , R 1 R2 + R 1 R 3 + R 2 R 3while ∆V3 = E − ∆V1 , or R 2 R3 ∆V3 = E , R1 R2 + R 1 R3 + R 2 R 3so the current through the ammeter is ∆V3 R2 i3 = =E , R3 R1 R2 + R 1 R3 + R 2 R 3or (4 Ω) i3 = (5.0 V) = 0.45 A. (2 Ω)(4 Ω) + (2 Ω)(6 Ω) + (4 Ω)(6 Ω) (b) Changing the locations of the battery and the ammeter is equivalent to swapping R1 andR3 . But since the expression for the current doesn’t change, then the current is the same.P31-12 ∆V1 + ∆V2 = ∆VS + ∆VX ; if Va = Vb , then ∆V1 = ∆VS . Using the first expression, ia (R1 + R2 ) = ib (RS + RX ),using the second, i a R1 = i b R2 .Dividing the first by the second, 1 + R2 /R1 = 1 + RX /RS ,or RX = RS (R2 /R1 ).P31-13P31-14 Lv = ∆Q/∆m and ∆Q/∆t = P = i∆V , so i∆V (5.2 A)(12 V) Lv = = = 2.97×106 J/kg. ∆m/∆t (21×10−6 kg/s)P31-15 P = i2 R. W = p∆V , where V is volume. p = mg/A and V = Ay, where y is the heightof the piston. Then P = dW/dt = mgv. Combining all of this, i2 R (0.240 A)2 (550 Ω) v= = = 0.274 m/s. mg (11.8 kg)(9.8 m/s2 ) 94
• P31-16 (a) Since q = CV , then q = (32×10−6 F) (6 V) + (4 V/s)(0.5 s) − (2 V/s2 )(0.5 s)2 = 2.4×10−4 C. (b) Since i = dq/dt = C dV /dt, then i = (32×10−6 F) (4 V/s) − 2(2 V/s2 )(0.5 s) = 6.4×10−5 A. (c) Since P = iV , P = [ (4 V/s) − 2(2 V/s2 )(0.5 s) (6 V) + (4 V/s)(0.5 s) − (2 V/s2 )(0.5 s)2 = 4.8×10−4 W.P31-17 (a) We have P = 30P0 and i = 4i0 . Then P 30P0 30 R= = = R0 . i2 (4i0 )2 16We don’t really care what happened with the potential difference, since knowing the change inresistance of the wire should give all the information we need. The volume of the wire is a constant, even upon drawing the wire out, so LA = L0 A0 ; theproduct of the length and the cross sectional area must be a constant. Resistance is given by R = ρL/A, but A = L0 A0 /L, so the length of the wire is A0 L0 R 30 A0 L0 R0 L= = = 1.37L0 . ρ 16 ρ (b) We know that A = L0 A0 /L, so L A0 A= A0 = = 0.73A0 . L0 1.37P31-18 (a) The capacitor charge as a function of time is given by Eq. 31-27, q = CE 1 − e−t/RC ,while the current through the circuit (and the resistor) is given by Eq. 31-28, E −t/RC i= e . RThe energy supplied by the emf is U= Ei dt = E dq = Eq;but the energy in the capacitor is UC = q∆V /2 = Eq/2. (b) Integrating, E2 E2 Eq UR = i2 Rdt = e−2t/RC dt = = . R 2C 2 95
• P31-19 The capacitor charge as a function of time is given by Eq. 31-27, q = CE 1 − e−t/RC ,while the current through the circuit (and the resistor) is given by Eq. 31-28, E −t/RC i= e . R The energy stored in the capacitor is given by q2 U= , 2Cso the rate that energy is being stored in the capacitor is dU q dq q PC = = = i. dt C dt CThe rate of energy dissipation in the resistor is PR = i2 R,so the time at which the rate of energy dissipation in the resistor is equal to the rate of energystorage in the capacitor can be found by solving PC = PR , 2 q i R = i, C iRC = q, ECe−t/RC = CE 1 − e−t/RC , e−t/RC = 1/2, t = RC ln 2. 96
• E32-1 Apply Eq. 32-3, F = qv× B. All of the paths which involve left hand turns are positive particles (path 1); those paths whichinvolve right hand turns are negative particle (path 2 and path 4); and those paths which don’t turninvolve neutral particles (path 3).E32-2 (a) The greatest magnitude of force is F = qvB = (1.6×10−19 C)(7.2×106 m/s)(83×10−3 T) =9.6×10−14 N. The least magnitude of force is 0. (b) The force on the electron is F = ma; the angle between the velocity and the magnetic fieldis θ, given by ma = qvB sin θ. Then (9.1×10−31 kg)(4.9×1016 m/s2 ) θ = arcsin = 28◦ . (1.6×10−19 C)(7.2×106 m/s)(83×10−3 T)E32-3 (a) v = E/B = (1.5×103 V/m)/(0.44 T) = 3.4×103 m/s.E32-4 (a) v = F/qB sin θ = (6.48×10−17 N/(1.60×10−19 C)(2.63×10−3 T) sin(23.0◦ ) = 3.94×105 m/s. (b) K = mv 2 /2 = (938 MeV/c2 )(3.94×105 m/s)2 /2 = 809 eV.E32-5 The magnetic force on the proton is FB = qvB = (1.6×10−19 C)(2.8×107 m/s)(30eex−6 T) = 1.3×10−16 N.The gravitational force on the proton is mg = (1.7×10−27 kg)(9.8 m/s2 ) = 1.7×10−26 N.The ratio is then 7.6×109 . If, however, you carry the number of significant digits for the intermediateanswers farther you will get the answer which is in the back of the book.E32-6 The speed of the electron is given by v = 2q∆V /m, or v= 2(1000 eV)/(5.1×105 eV/c2 ) = 0.063c.The electric field between the plates is E = (100 V)/(0.020 m) = 5000 V/m. The required magneticfield is then B = E/v = (5000 V/m)/(0.063c) = 2.6×10−4 T.E32-7 Both have the same velocity. Then K p /K e = mp v 2 /me v 2 = mp /me =.E32-8 The speed of the ion is given by v = 2q∆V /m, or v= 2(10.8 keV)/(6.01)(932 MeV/c2 ) = 1.96×10−3 c.The required electric field is E = vB = (1.96×10−3 c)(1.22 T) = 7.17×105 V/m.E32-9 (a) For a charged particle moving in a circle in a magnetic field we apply Eq. 32-10; mv (9.11×10−31 kg)(0.1)(3.00×108 m/s) r= = = 3.4×10−4 m. |q|B (1.6×10−19 C)(0.50 T) (b) The (non-relativistic) kinetic energy of the electron is 1 1 K= mv 2 = (0.511 MeV)(0.10c)2 = 2.6×10−3 MeV. 2 2 97
• E32-10 (a) v = 2K/m = 2(1.22 keV)/(511 keV/c2 ) = 0.0691c. (b) B = mv/qr = (9.11×10−31 kg)(0.0691c)/(1.60×10−19 C)(0.247 m) = 4.78×10−4 T. (c) f = qB/2πm = (1.60×10−19 C)(4.78×10−4 T)/2π(9.11×10−31 kg) = 1.33×107 Hz. (d) T = 1/f = 1/(1.33×107 Hz) = 7.48×10−8 s.E32-11 (a) v = 2K/m = 2(350 eV)/(511 keV/c2 ) = 0.037c. (b) r = mv/qB = (9.11×10−31 kg)(0.037c)/(1.60×10−19 C)(0.20T) = 3.16×10−4 m.E32-12 The frequency is f = (7.00)/(1.29×10−3 s) = 5.43×103 Hz. The mass is given by m =qB/2πf , or (1.60×10−19 C)(45.0×10−3 T) m= = 2.11×10−25 kg = 127 u. 2π(5.43×103 Hz)E32-13 (a) Apply Eq. 32-10, but rearrange it as |q|rB 2(1.6×10−19 C)(0.045 m)(1.2 T) v= = = 2.6×106 m/s. m 4.0(1.66×10−27 kg) (b) The speed is equal to the circumference divided by the period, so 2πr 2πm 2π4.0(1.66×10−27 kg) T = = = = 1.1×10−7 s. v |q|B 2(1.6 × 10−19 C)(1.2 T) (c) The (non-relativistic) kinetic energy is |q|2 r2 B (2×1.6×10−19 C)2 (0.045 m)2 (1.2 T)2 K= = = 2.24×10−14 J. 2m 2(4.0×1.66×10−27 kg))To change to electron volts we need merely divide this answer by the charge on one electron, so (2.24×10−14 J) K= = 140 keV. (1.6×10−19 C) K (d) ∆V = q = (140 keV)/(2e) = 70 V.E32-14 (a) R = mv/qB = (938 MeV/c2 )(0.100c)/e(1.40 T) = 0.223 m. (b) f = qB/2πm = e(1.40 T)/2π(938 MeV/c2 ) = 2.13×107 Hz.E32-15 (a) Kα /K p = (qα /mα )/(q p 2 /mp ) = 22 /4 = 1. 2 (b) K d /K p = (q d /md )/(q p 2 /mp ) = 12 /2 = 1/2. 2E32-16 (a) K = q∆V . Then K p = e∆V , K d = e∆V , and Kα = 2e∆V . √ (b) r = sqrt2mK/qB. Then rd /rp = (2/1)(1/1)/(1/1) = 2. √ (c) r = sqrt2mK/qB. Then rα /rp = (4/1)(2/1)/(2/1) = 2. √ √ √ E32-17 r = 2mK/|q|B = ( m/|q|)( 2K/B). All three particles are traveling with the samekinetic energy in the same magnetic field. The relevant factors are in front; we just need to comparethe mass and charge of each of the three particles. √ (a) The radius of the deuteron path is 12 rp . √ (b) The radius of the alpha particle path is 24 rp = rp . 98
• E32-18 The neutron, being neutral, is unaffected by the magnetic field and moves off in a linetangent to the original path. The proton moves at the same original speed as the deuteron and hasthe same charge, but since it has half the mass it moves in a circle with half the radius.E32-19 (a) The proton momentum would be pc = qcBR = e(3.0×108 m/s)(41×10−6 T)(6.4×106 m) =7.9×104 MeV. Since 79000 MeV is much, much greater than 938 MeV the proton is ultra-relativistic.Then E ≈ pc, and since γ = E/mc2 we have γ = p/mc. Inverting, v 1 m2 c2 m2 c2 = 1− = 1− ≈1− ≈ 0.99993. c γ2 p 2 2p2 √E32-20 (a) Classically, R = 2mK/qB, or R= 2(0.511 MeV/c2 )(10.0 MeV)/e(2.20 T) = 4.84×10−3 m. (b) This would be an ultra-relativistic electron, so K ≈ E ≈ pc, then R = p/qB = K/qBc, or R = (10.0 MeV)/e(2.2 T)(3.00×108 m/s) = 1.52×10−2 m. (c) The electron is effectively traveling at the speed of light, so T = 2πR/c, or T = 2π(1.52×10−2 m)/(3.00×108 m/s) = 3.18×10−10 s.This result does depend on the speed!E32-21 Use Eq. 32-10, except we rearrange for the mass, |q|rB 2(1.60×10−19 C)(4.72 m)(1.33 T) m= = = 9.43×10−27 kg v 0.710(3.00×108 m/s)However, if it is moving at this velocity then the “mass” which we have here is not the true mass,but a relativistic correction. For a particle moving at 0.710c we have 1 1 γ= = = 1.42, 1− v 2 /c2 1 − (0.710)2so the true mass of the particle is (9.43×10−27 kg)/(1.42) = 6.64×10−27 kg. The number of nucleonspresent in this particle is then (6.64×10−27 kg)/(1.67×10−27 kg) = 3.97 ≈ 4. The charge was +2,which implies two protons, the other two nucleons would be neutrons, so this must be an alphaparticle.E32-22 (a) Since 950 GeV is much, much greater than 938 MeV the proton is ultra-relativistic.γ = E/mc2 , so v 1 m2 c4 m2 c4 = 1− 2 = 1− ≈1− ≈ 0.9999995. c γ E2 2E 2 (b) Ultra-relativistic motion requires pc ≈ E, so B = pc/qRc = (950 GeV)/e(750 m)(3.00×108 m/s) = 4.44 T. 99
• E32-23 First use 2πf = qB/m. The use K = q 2 B 2 R2 /2m = mR2 (2πf )2 /2. The number of √turns is n = K/2q∆V , on average the particle is located at a distance R/ 2 from the center, so the √ √distance traveled is x = n2πR/ 2 = n 2πR. Combining, √ 3 3 √ 3 2π R mf 2 2π (0.53 m)3 (2 × 932×103 keV/c2 )(12×106 /s)2 x= = = 240 m. q∆V e(80 kV)E32-24 The particle moves in a circle. x = R sin ωt and y = R cos ωt. E32-25 We will use Eq. 32-20, E H = v d B, except we will not take the derivation through to Eq.32-21. Instead, we will set the drift velocity equal to the speed of the strip. We will, however, setE H = ∆V H /w. Then EH ∆V H /w (3.9×10−6 V)/(0.88×10−2 m) v= = = = 3.7×10−1 m/s. B B (1.2×10−3 T)E32-26 (a) v = E/B = (40×10−6 V)/(1.2×10−2 m)/(1.4 T) = 2.4×10−3 m/s. (b) n = (3.2 A)(1.4 T)/(1.6×10−19 C)(9.5×10−6 m)(40×10−6 V) = 7.4×1028 /m3 .; Silver.E32-27 E H = v d B and v d = j/ne. Combine and rearrange.E32-28 (a) Use the result of the previous exercise and E c = ρj. (b) (0.65 T)/(8.49×1028 /m3 )(1.60×10−19 C)(1.69×10−8 Ω · m) = 0.0028.E32-29 Since L is perpendicular to B can use FB = iLB.Equating the two forces, iLB = mg, mg (0.0130 kg)(9.81 m/s2 ) i = = = 0.467 A. LB (0.620 m)(0.440 T) Use of an appropriate right hand rule will indicate that the current must be directed to the rightin order to have a magnetic force directed upward.E32-30 F = iLB sin θ = (5.12 × 103 A)(100 m)(58 × 10−6 T) sin(70◦ ) = 27.9 N. The direction ishorizontally west. E32-31 (a) We use Eq. 32-26 again, and since the (horizontal) axle is perpendicular to thevertical component of the magnetic field, F (10, 000 N) i= = = 3.3×108 A. BL (10 µT)(3.0 m) (b) The power lost per ohm of resistance in the rails is given by P/r = i2 = (3.3×108 A)2 = 1.1×1017 W. (c) If such a train were to be developed the rails would melt well before the train left the station. 100
• E32-32 F = idB, so a = F/m = idB/m. Since a is constant, v = at = idBt/m. The direction isto the left.E32-33 Only the ˆ component of B is of interest. Then F = j dF = i By dx, or 3.2 F = (5.0 A)(8×10−3 T/m2 ) x2 dx = 0.414 N. 1.2 ˆThe direction is −k.E32-34 The magnetic force will have two components: one will lift vertically (Fy = F sin α), theother push horizontally (Fx = F cos α). The rod will move when Fx > µ(W − Fy ). We are interestedin the minimum value for F as a function of α. This occurs when dF d µW = = 0. dα dα cos α + µ sin αThis happens when µ = tan α. Then α = arctan(0.58) = 30◦ , and (0.58)(1.15 kg)(9.81 m/s2 ) F = = 5.66 N cos(30◦ ) + (0.58) sin(30◦ )is the minimum force. Then B = (5.66 N)/(53.2 A)(0.95 m) = 0.112 T.E32-35 We choose that the field points from the shorter side to the longer side. (a) The magnetic field is parallel to the 130 cm side so there is no magnetic force on that side. The magnetic force on the 50 cm side has magnitude FB = iLB sin θ,where θ is the angle between the 50 cm side and the magnetic field. This angle is larger than 90◦ ,but the sine can be found directly from the triangle, (120 cm) sin θ = = 0.923, (130 cm)and then the force on the 50 cm side can be found by (120 cm) FB = (4.00 A)(0.50 m)(75.0×10−3 T) = 0.138 N, (130 cm)and is directed out of the plane of the triangle. The magnetic force on the 120 cm side has magnitude FB = iLB sin θ,where θ is the angle between the 1200 cm side and the magnetic field. This angle is larger than180◦ , but the sine can be found directly from the triangle, (−50 cm) sin θ = = −0.385, (130 cm)and then the force on the 50 cm side can be found by (−50 cm) FB = (4.00 A)(1.20 m)(75.0×10−3 T) = −0.138 N, (130 cm)and is directed into the plane of the triangle. (b) Look at the three numbers above. 101
• E32-36 τ = N iAB sin θ, so τ = (20)(0.1 A)(0.12 m)(0.05 m)(0.5 T) sin(90◦ − 33◦ ) = 5.0×10−3 N · m.E32-37 The external magnetic field must be in the plane of the clock/wire loop. The clockwisecurrent produces a magnetic dipole moment directed into the plane of the clock. (a) Since the magnetic field points along the 1 pm line and the torque is perpendicular to boththe external field and the dipole, then the torque must point along either the 4 pm or the 10 pm line.Applying Eq. 32-35, the direction is along the 4 pm line. It will take the minute hand 20 minutesto get there. (b) τ = (6)(2.0 A)π(0.15 m)2 (0.07 T) = 0.059 N · m. ˆP32-1 Since F must be perpendicular to B then B must be along k. The magnitude of v is (40)2 + (35)2 km/s = 53.1 km/s; the magnitude of F is (−4.2)2 + (4.8)2 fN = 6.38 fN. Then B = F/qv = (6.38×10−15 N)/(1.6×10−19 C)(53.1×103 m/s) = 0.75 T. ˆor B = 0.75 T k.P32-2 a = (q/m)(E + v × B). For the initial velocity given, v × B = (15.0×103 m/s)(400×10−6 T)ˆ − (12.0×103 m/s)(400×10−6 T)k. j ˆBut since there is no acceleration in the ˆ or k direction this must be offset by the electric field. j ˆConsequently, two of the electric field components are Ey = −6.00 V/m and Ez = 4.80 V/m. Thethird component of the electric field is the source of the acceleration, so Ex = max /q = (9.11×10−31 kg)(2.00×1012 m/s2 )/(−1.60×10−19 C) = −11.4 V/m. P32-3 (a) Consider first the cross product, v × B. The electron moves horizontally, there is acomponent of the B which is down, so the cross product results in a vector which points to the leftof the electron’s path. But the force on the electron is given by F = qv × B, and since the electron has a negative chargethe force on the electron would be directed to the right of the electron’s path. (b) The kinetic energy of the electrons is much less than the rest mass energy, so this is non-relativistic motion. The speed of the electron is then v = 2K/m, and the magnetic force on theelectron is FB = qvB, where we are assuming sin θ = 1 because the electron moves horizontallythrough a magnetic field with a vertical component. We can ignore the effect of the magnetic field’shorizontal component because the electron is moving parallel to this component. The acceleration of the electron because of the magnetic force is then qvB qB 2K a = = , m m m (1.60×10−19 C)(55.0×10−6 T) 2(1.92×10−15 J) = = 6.27×1014 m/s2 . (9.11×10−31 kg) (9.11×10−31 kg) (c) The electron travels a horizontal distance of 20.0 cm in a time of (20.0 cm) (20.0 cm) t= = = 3.08×10−9 s. 2K/m 2(1.92×10−15 J)/(9.11×10−31 kg)In this time the electron is accelerated to the side through a distance of 1 2 1 d= at = (6.27×1014 m/s2 )(3.08×10−9 s)2 = 2.98 mm. 2 2 102
• P32-4 (a) d needs to be larger than the turn radius, so R ≤ d; but 2mK/q 2 B 2 = R2 ≤ d2 , orB ≥ 2mK/q 2 d2 . (b) Out of the page.P32-5 Only undeflected ions emerge from the velocity selector, so v = E/B. The ions are thendeflected by B with a radius of curvature of r = mv/qB; combining and rearranging, q/m =E/rBB .P32-6 The ions are given a kinetic energy K = q∆V ; they are then deflected with a radius ofcurvature given by R2 = 2mK/q 2 B 2 . But x = 2R. Combine all of the above, and m = B 2 qx2 /8∆V.P32-7 (a) Start with the equation in Problem 6, and take the square root of both sides to get 1 √ B2q 2 m= x, 8∆Vand then take the derivative of x with respect to m, 1 1 dm B2q 2 √ = dx, 2 m 8∆Vand then consider finite differences instead of differential quantities, 1 mB 2 q 2 ∆m = ∆x, 2∆V (b) Invert the above expression, 1 2∆V 2 ∆x = ∆m, mB 2 qand then put in the given values, 1 2(7.33×103 V) 2 ∆x = −27 kg)(0.520 T)2 (1.60×10−19 C) (2.0)(1.66×10−27 kg), (35.0)(1.66×10 = 8.02 mm.Note that we used 35.0 u for the mass; if we had used 37.0 u the result would have been closer tothe answer in the back of the book.P32-8 (a) B = 2∆V m/qr2 = 2(0.105 MV)(238)(932 MeV/c2 )/2e(0.973 m)2 = 5.23×10−7 T. (b) The number of atoms in a gram is 6.02×1023 /238 = 2.53×1021 . The current is then (0.090)(2.53×1021 )(2)(1.6×10−19 C)/(3600 s) = 20.2 mA.P32-9 (a) −q. (b) Regardless of speed, the orbital period is T = 2πm/qB. But they collide halfway around acomplete orbit, so t = πm/qB.P32-10 103
• P32-11 (a) The period of motion can be found from the reciprocal of Eq. 32-12, 2πm 2π(9.11×10−31 kg) T = = = 7.86×10−8 s. |q|B (1.60×10−19 C)(455×10−6 T) (b) We need to find the velocity of the electron from the kinetic energy, v= 2K/m = 2(22.5 eV)(1.60×10−19 J/eV)/(9.11×10−31 kg) = 2.81×106 m/s.The velocity can written in terms of components which are parallel and perpendicular to the magneticfield. Then v|| = v cos θ and v⊥ = v sin θ.The pitch is the parallel distance traveled by the electron in one revolution, so p = v|| T = (2.81×106 m/s) cos(65.5◦ )(7.86×10−8 s) = 9.16 cm. (c) The radius of the helical path is given by Eq. 32-10, except that we use the perpendicularvelocity component, so mv⊥ (9.11×10−31 kg)(2.81×106 m/s) sin(65.5◦ ) R= = = 3.20 cm |q|B (1.60×10−19 C)(455×10−6 T) bP32-12 F = i a dl × B. dl has two components, those parallel to the path, say dx and thoseperpendicular, say dy. Then the integral can be written as b b F= dx × B + dy × B. a a bBut B is constant, and can be removed from the integral. a dx = l, a vector that points from a to bb. a dy = 0, because there is no net motion perpendicular to l.P32-13 qvy B = Fx = m dvx /dt; −qvx B = Fy = m dvy /dt. Taking the time derivative of thesecond expression and inserting into the first we get m d2 vy qvy B = m − , qB dt2which has solution vy = −v sin(mt/qB), where v is a constant. Using the second equation we findthat there is a similar solution for vx , except that it is out of phase, and so vx = v cos(mt/qB). Integrating, qBv x = vx dt = v cos(mt/qB) = sin(mt/qB). mSimilarly, qBv y = vy dt = −v sin(mt/qB) = cos(mt/qB). mThis is the equation of a circle.P32-14 dL = ˆ + ˆ + kdz. B is uniform, so that the integral can be written as idx jdy ˆ F=i (ˆ + ˆ + kdz) × B = iˆ × B idx jdy ˆ i dx + iˆ × B j ˆ dy + ik × B dz,but since dx = dy = dz = 0, the entire expression vanishes. 104
• P32-15 The current pulse provides an impulse which is equal to F dt = BiL dt = BL i dt = BLq.This gives an initial velocity of v0 = BLq/m, which will cause the rod to hop to a height of h = v0 /2g = B 2 L2 q 2 /2m2 g. 2Solving for q, m (0.013 kg) q= 2gh = 2(9.8 m/s2 )(3.1 m) = 4.2 C. BL (0.12 T)(0.20 m)P32-16P32-17 The torque on a current carrying loop depends on the orientation of the loop; themaximum torque occurs when the plane of the loop is parallel to the magnetic field. In this case themagnitude of the torque is from Eq. 32-34 with sin θ = 1— τ = N iAB.The area of a circular loop is A = πr2 where r is the radius, but since the circumference is C = 2πr,we can write C2 A= . 4πThe circumference is not the length of the wire, because there may be more than one turn. Instead,C = L/N , where N is the number of turns. Finally, we can write the torque as L2 iL2 B τ = Ni B= , 4πN 2 4πNwhich is a maximum when N is a minimum, or N = 1.P32-18 dF = i dL × B; the direction of dF will be upward and somewhat toward the center. Land B are a right angles, but only the upward component of dF will survive the integration as thecentral components will cancel out by symmetry. Hence F = iB sin θ dL = 2πriB sin θ.P32-19 The torque on the cylinder from gravity is τ g = mgr sin θ,where r is the radius of the cylinder. The torque from magnetism needs to balance this, so mgr sin θ = N iAB sin θ = N i2rLB sin θ,or mg (0.262 kg)(9.8 m/s2 ) i= = = 1.63 A. 2N LB 2(13)(0.127 m)(0.477 T) 105
• E33-1 (a) The magnetic field from a moving charge is given by Eq. 33-5. If the protons aremoving side by side then the angle is φ = π/2, so µ0 qv B= 4π r2and we are interested is a distance r = d. The electric field at that distance is 1 q E= , 4π 0 r2where in both of the above expressions q is the charge of the source proton. On the receiving end is the other proton, and the force on that proton is given by F = q(E + v × B).The velocity is the same as that of the first proton (otherwise they wouldn’t be moving side by side.)This velocity is then perpendicular to the magnetic field, and the resulting direction for the crossproduct will be opposite to the direction of E. Then for balance, E = vB, 1 q µ0 qv = v , 4π 0 r2 4π r2 1 = v2 . 0 µ0We can solve this easily enough, and we find v ≈ 3 × 108 m/s. (b) This is clearly a relativistic speed!E33-2 B = µ0 i/2πd = (4π ×10−7 T · m/A)(120 A)/2π(6.3 m) = 3.8×10−6 T. This will deflect thecompass needle by as much as one degree. However, there is unlikely to be a place on the Earth’ssurface where the magnetic field is 210 µT. This was likely a typo, and should probably have been21.0 µT. The deflection would then be some ten degrees, and that is significant.E33-3 B = µ0 i/2πd = (4π×10−7 T · m/A)(50 A)/2π(1.3×10−3 m) = 37.7×10−3 T.E33-4 (a) i = 2πdB/µ0 = 2π(8.13×10−2 m)(39.0×10−6 T)/(4π×10−7 T · m/A) = 15.9 A. (b) Due East.E33-5 Use µ0 i (4π×10−7 N/A2 )(1.6 × 10−19 C)(5.6 × 1014 s−1 ) B= = = 1.2×10−8 T. 2πd 2π(0.0015 m)E33-6 Zero, by symmetry. Any contributions from the top wire are exactly canceled by contribu-tions from the bottom wire.E33-7 B = µ0 i/2πd = (4π×10−7 T · m/A)(48.8 A)/2π(5.2×10−2 m) = 1.88×10−4 T. F = qv × B. All cases are either parallel or perpendicular, so either F = 0 or F = qvB. (a) F = qvB = (1.60×10−19 C)(1.08×107 m/s)(1.88×10−4 T) = 3.24×10−16 N. The direction ofF is parallel to the current. (b) F = qvB = (1.60×10−19 C)(1.08×107 m/s)(1.88×10−4 T) = 3.24×10−16 N. The direction ofF is radially outward from the current. (c) F = 0. 106
• E33-8 We want B1 = B2 , but with opposite directions. Then i1 /d1 = i2 /d2 , since all constantscancel out. Then i2 = (6.6 A)(1.5 cm)/(2.25 cm) = 4.4 A, directed out of the page. E33-9 For a single long straight wire, B = µ0 i/2πd but we need a factor of “2” since there aretwo wires, then i = πdB/µ0 . Finally πdB π(0.0405 m)(296, µT) i= = = 30 A µ0 (4π×10−7 N/A2 )E33-10 (a) The semi-circle part contributes half of Eq. 33-21, or µ0 i/4R. Each long straight wirecontributes half of Eq. 33-13, or µ0 i/4πR. Add the three contributions and get µ0 i 2 (4π×10−7 N/A2 )(11.5 A) 2 Ba = +1 = +1 = 1.14×10−3 T. 4R π 4(5.20×10−3 m) πThe direction is out of the page. (b) Each long straight wire contributes Eq. 33-13, or µ0 i/2πR. Add the two contributions andget µ0 i (4π×10−7 N/A2 )(11.5 A) Ba = = = 8.85×10−4 T. πR π(5.20×10−3 m)The direction is out of the page.E33-11 z 3 = µ0 iR2 /2B = (4π ×10−7 N/A2 )(320)(4.20 A)(2.40×10−2 m)2 /2(5.0×10−6 T) = 9.73×10−2 m3 . Then z = 0.46 m.E33-12 The circular part contributes a fraction of Eq. 33-21, or µ0 iθ/4πR. Each long straightwire contributes half of Eq. 33-13, or µ0 i/4πR. Add the three contributions and get µ0 i B= (θ − 2). 4πRThe goal is to get B = 0 that will happen if θ = 2 radians. E33-13 There are four current segments that could contribute to the magnetic field. The straightsegments, however, contribute nothing because the straight segments carry currents either directlytoward or directly away from the point P . That leaves the two rounded segments. Each contribution to B can be found by starting withEq. 33-21, or µ0 iθ/4πb. The direction is out of the page. There is also a contribution from the top arc; the calculations are almost identical except thatthis is pointing into the page and r = a, so µ0 iθ/4πa. The net magnetic field at P is then µ0 iθ 1 1 B = B1 + B2 = − . 4π b aE33-14 For each straight wire segment use Eq. 33-12. When the length of wire is L, the distanceto the center is W/2; when the length of wire is W the distance to the center is L/2. There are fourterms, but they are equal in pairs, so µ0 i 4L 4W B = + , 4π W L2 /4 + W 2 /4 L L2 /4 + W 2 /4 √ 2µ0 i L2 W2 2µ0 i L2 + W 2 = √ + = . π L2 + W 2 WL WL π WL 107
• E33-15 We imagine the ribbon conductor to be a collection of thin wires, each of thickness dxand carrying a current di. di and dx are related by di/dx = i/w. The contribution of one of thesethin wires to the magnetic field at P is dB = µ0 di/2πx, where x is the distance from this thin wireto the point P . We want to change variables to x and integrate, so µ0 i dx µ0 i dx B= dB = = . 2πwx 2πw xThe limits of integration are from d to d + w, so µ0 i d+w B= ln . 2πw dE33-16 The fields from each wire are perpendicular at P . Each contributes an amount B = √ √µ0 i/2πd, but since they are perpendicular there is a net field of magnitude B = 2B 2 = 2µ0 i/2πd. √Note that a = 2d, so B = µ0 i/πa. (a) B = (4π×10−7 T · m/A)(115 A)/π(0.122 m) = 3.77×10−4 T. The direction is to the left. (b) Same numerical result, except the direction is up.E33-17 Follow along with Sample Problem 33-4. Reversing the direction of the second wire (so that now both currents are directed out of thepage) will also reverse the direction of B2 . Then µ0 i 1 1 B = B1 − B2 = − , 2π b + x b − x µ0 i (b − x) − (b + x) = , 2π b2 − x2 µ0 i x = 2 − b2 . π xE33-18 (b) By symmetry, only the horizontal component of B survives, and must point to theright. (a) The horizontal component of the field contributed by the top wire is given by µ0 i µ0 i b/2 µ0 ib B= sin θ = = , 2πh 2πh h π(4R2 + b2 )since h is the hypotenuse, or h = R2 + b2 /4. But there are two such components, one from thetop wire, and an identical component from the bottom wire. E33-19 (a) We can use Eq. 33-21 to find the magnetic field strength at the center of the largeloop, µ0 i (4π×10−7 T · m/A)(13 A) B= = = 6.8×10−5 T. 2R 2(0.12 m) (b) The torque on the smaller loop in the center is given by Eq. 32-34, τ = N iA × B,but since the magnetic field from the large loop is perpendicular to the plane of the large loop, andthe plane of the small loop is also perpendicular to the plane of the large loop, the magnetic field isin the plane of the small loop. This means that |A × B| = AB. Consequently, the magnitude of thetorque on the small loop is τ = N iAB = (50)(1.3 A)(π)(8.2×10−3 m)2 (6.8×10−5 T) = 9.3×10−7 N · m. 108
• E33-20 (a) There are two contributions to the field. One is from the circular loop, and is givenby µ0 i/2R. The other is from the long straight wire, and is given by µ0 i/2πR. The two fields areout of the page and parallel, so µ0 i B= (1 + 1/π). 2R (b) The two components are now at right angles, so µ0 i B= 1 + 1/π 2 . 2RThe direction is given by tan θ = 1/π or θ = 18◦ .E33-21 The force per meter for any pair of parallel currents is given by Eq. 33-25, F/L = µ0 i2 /2πd,where d is the separation. The direction of the force is along the line connecting the intersection ofthe currents with the perpendicular plane. Each current experiences √ three forces; two are at rightangles and equal in magnitude, so |F12 + F14 |/L = F12 + F14 /L = 2µ0 i2 /2πa. The third force √ 2 2points parallel to this sum, but d = a, so the resultant force is √ F 2µ0 i2 µ0 i2 4π ××10−7 N/A2 (18.7 A)2 √ √ = + √ = ( 2 + 1/ 2) = 6.06×10−4 N/m. L 2πa 2π 2a 2π(0.245 m)It is directed toward the center of the square.E33-22 By symmetry we expect the middle wire to have a net force of zero; the two on the outsidewill each be attracted toward the center, but the answers will be symmetrically distributed. For the wire which is the farthest left,F µ0 i2 1 1 1 1 4π ××10−7 N/A2 (3.22 A)2 1 1 1 = + + + = 1+ + + = 5.21×10−5 N/m.L 2π a 2a 3a 4a 2π(0.083 m) 2 3 4 For the second wire over, the contributions from the two adjacent wires should cancel. Thisleaves F µ0 i2 1 1 4π ××10−7 N/A2 (3.22 A)2 1 1 = + + = + = 2.08×10−5 N/m. L 2π 2a 3a 2π(0.083 m) 2 3E33-23 (a) The force on the projectile is given by the integral of dF = i dl × Bover the length of the projectile (which is w). The magnetic field strength can be found from addingtogether the contributions from each rail. If the rails are circular and the distance between them issmall compared to the length of the wire we can use Eq. 33-13, µ0 i B= , 2πxwhere x is the distance from the center of the rail. There is one problem, however, because theseare not wires of infinite length. Since the current stops traveling along the rail when it reaches theprojectile we have a rod that is only half of an infinite rod, so we need to multiply by a factor of1/2. But there are two rails, and each will contribute to the field, so the net magnetic field strengthbetween the rails is µ0 i µ0 i B= + . 4πx 4π(2r + w − x) 109
• In that last term we have an expression that is a measure of the distance from the center of thelower rail in terms of the distance x from the center of the upper rail. The magnitude of the force on the projectile is then r+w F = i B dx, r µ0 i2 r+w 1 1 = + dx, 4π r x 2r + w − x µ0 i2 r+w = 2 ln 4π r The current through the projectile is down the page; the magnetic field through the projectile isinto the page; so the force on the projectile, according to F = il × B, is to the right. (b) Numerically the magnitude of the force on the rail is (450×103 A)2 (4π×10−7 N/A2 ) (0.067 m) + (0.012 m) F = ln = 6.65×103 N 2π (0.067 m) The speed of the rail can be found from either energy conservation so we first find the work doneon the projectile, W = F d = (6.65×103 N)(4.0 m) = 2.66×104 J.This work results in a change in the kinetic energy, so the final speed is v= 2K/m = 2(2.66×104 J)/(0.010 kg) = 2.31×103 m/s.E33-24 The contributions from the left end and the right end of the square cancel out. This leavesthe top and the bottom. The net force is the difference, or (4π×10−7 N/A2 )(28.6 A)(21.8 A)(0.323 m) 1 1 F = −2 m) − , 2π (1.10×10 (10.30×10−2 m) = 3.27×10−3 N.E33-25 The magnetic force on the upper wire near the point d is µ0 ia ib L µ0 ia ib L µ0 ia ib L FB = ≈ − x, 2π(d + x) 2πd 2πd2where x is the distance from the equilibrium point d. The equilibrium magnetic force is equal to theforce of gravity mg, so near the equilibrium point we can write x FB = mg − mg . dThere is then a restoring force against small perturbations of magnitude mgx/d which correspondsto a spring constant of k = mg/d. This would give a frequency of oscillation of 1 1 f= k/m = g/d, 2π 2πwhich is identical to the pendulum.E33-26 B = (4π×10−7 N/A2 )(3.58 A)(1230)/(0.956m) = 5.79×10−3 T. 110
• E33-27 The magnetic field inside an ideal solenoid is given by Eq. 33-28 B = µ0 in, where n isthe turns per unit length. Solving for n, B (0.0224 T) n= = = 1.00×103 /m−1 . µ0 i (4π×10−7 N/A2 )(17.8 A)The solenoid has a length of 1.33 m, so the total number of turns is N = nL = (1.00×103 /m−1 )(1.33 m) = 1330,and since each turn has a length of one circumference, then the total length of the wire which makesup the solenoid is (1330)π(0.026 m) = 109 m.E33-28 From the solenoid we have B s = µ0 nis = µ0 (11500/m)(1.94 mA) = µ0 (22.3A/m).From the wire we have µ0 iw µ0 (6.3 A) µ0 Bw = = = (1.002 A) 2πr 2πr rThese fields are at right angles, so we are interested in when tan(40◦ ) = B w /B s , or (1.002 A) r= = 5.35×10−2 m. tan(40◦ )(22.3 A/m)E33-29 Let u = z − d. Then d+L/2 µ0 niR2 du B = , 2 d−L/2 [R2 + u2 ]3/2 2 d+L/2 µ0 niR u = √ , 2 R 2 R 2 + u2 d−L/2 µ0 ni d + L/2 d − L/2 = − . 2 R2 + (d + L/2)2 R2 + (d − L/2)2If L is much, much greater than R and d then |L/2 ± d| >> R, and R can be ignored in thedenominator of the above expressions, which then simplify to µ0 ni d + L/2 d − L/2 B = − . 2 R2 + (d + L/2)2 R2 + (d − L/2)2 µ0 ni d + L/2 d − L/2 = − . 2 (d + L/2)2 (d − L/2)2 = µ0 in.It is important that we consider the relative size of L/2 and d!E33-30 The net current in the loop is 1i0 + 3i0 + 7i0 − 6i0 = 5i0 . Then B · ds = 5µ0 i0 .E33-31 (a) The path is clockwise, so a positive current is into page. The net current is 2.0 A out,so B · ds = −µ0 i0 = −2.5×10−6 T · m. (b) The net current is zero, so B · ds = 0. 111
• E33-32 Let R0 be the radius of the wire. On the surface of the wire B0 = µ0 i/2πR0 . Outside the wire we have B = µ0 i/2πR, this is half B0 when R = 2R0 . 2 Inside the wire we have B = µ0 iR/2πR0 , this is half B0 when R = R0 /2. E33-33 (a) We don’t want to reinvent the wheel. The answer is found from Eq. 33-34, except itlooks like µ0 ir B= . 2πc2 (b) In the region between the wires the magnetic field looks like Eq. 33-13, µ0 i B= . 2πrThis is derived on the right hand side of page 761. (c) Ampere’s law (Eq. 33-29) is B · ds = µ0 i, where i is the current enclosed. Our Amperianloop will still be a circle centered on the axis of the problem, so the left hand side of the aboveequation will reduce to 2πrB, just like in Eq. 33-32. The right hand side, however, depends onthe net current enclosed which is the current i in the center wire minus the fraction of the currentenclosed in the outer conductor. The cross sectional area of the outer conductor is π(a2 − b2 ), sothe fraction of the outer current enclosed in the Amperian loop is π(r2 − b2 ) r 2 − b2 i 2 − b2 ) =i 2 . π(a a − b2The net current in the loop is then r 2 − b2 a2 − r2 i−i =i 2 , a2 − b2 a − b2so the magnetic field in this region is µ0 i a2 − r2 B= . 2πr a2 − b2 (d) This part is easy since the net current is zero; consequently B = 0.E33-34 (a) Ampere’s law (Eq. 33-29) is B · ds = µ0 i, where i is the current enclosed. OurAmperian loop will still be a circle centered on the axis of the problem, so the left hand side of theabove equation will reduce to 2πrB, just like in Eq. 33-32. The right hand side, however, dependson the net current enclosed which is the fraction of the current enclosed in the conductor. The crosssectional area of the conductor is π(a2 − b2 ), so the fraction of the current enclosed in the Amperianloop is π(r2 − b2 ) r 2 − b2 i =i 2 . π(a2 − b2 ) a − b2The magnetic field in this region is µ0 i r2 − b2 B= . 2πr a2 − b2 (b) If r = a, then µ0 i a2 − b2 µ0 i B= = , 2πa a2 − b2 2πawhich is what we expect. 112
• If r = b, then µ0 i b2 − b2 B= = 0, 2πb a2 − b2which is what we expect. If b = 0, then µ0 i r2 − 02 µ0 ir B= = 2πr a2 − 02 2πa2which is what I expected.E33-35 The magnitude of the magnetic field due to the cylinder will be zero at the center ofthe cylinder and µ0 i0 /2π(2R) at point P . The magnitude of the magnetic field field due to thewire will be µ0 i/2π(3R) at the center of the cylinder but µ0 i/2πR at P . In order for the netfield to have different directions in the two locations the currents in the wire and pipe must be indifferent direction. The net field at the center of the pipe is µ0 i/2π(3R), while that at P is thenµ0 i0 /2π(2R) − µ0 i/2πR. Set these equal and solve for i; i/3 = i0 /2 − i,or i = 3i0 /8.E33-36 (a) B = (4π×10−7 N/A2 )(0.813 A)(535)/2π(0.162 m) = 5.37×10−4 T. (b) B = (4π×10−7 N/A2 )(0.813 A)(535)/2π(0.162 m + 0.052 m) = 4.07×10−4 T.E33-37 (a) A positive particle would experience a magnetic force directed to the right for amagnetic field out of the page. This particle is going the other way, so it must be negative. (b) The magnetic field of a toroid is given by Eq. 33-36, µ0 iN B= , 2πrwhile the radius of curvature of a charged particle in a magnetic field is given by Eq. 32-10 mv R= . |q|BWe use the R to distinguish it from r. Combining, 2πmv R= r, µ0 iN |q|so the two radii are directly proportional. This means R/(11 cm) = (110 cm)/(125 cm),so R = 9.7 cm.P33-1 The field from one coil is given by Eq. 33-19 µ0 iR2 B= . 2(R2 + z 2 )3/2There are N turns in the coil, so we need a factor of N . There are two coils and we are interestedin the magnetic field at P , a distance R/2 from each coil. The magnetic field strength will be twicethe above expression but with z = R/2, so 2µ0 N iR2 8µ0 N i B= 2 + (R/2)2 )3/2 = . 2(R (5)3/2 R 113
• P33-2 (a) Change the limits of integration that lead to Eq. 33-12: L µ0 id dz B = , 4π 0 (z 2 + d2 )3/2 L µ0 id z = , 4π (z 2 + d2 )1/2 0 µ0 id L = . 4π (L2 + d2 )1/2 (b) The angle φ in Eq. 33-11 would always be 0, so sin φ = 0, and therefore B = 0.P33-3 This problem is the all important derivation of the Helmholtz coil properties. (a) The magnetic field from one coil is µ0 N iR2 B1 = . 2(R2 + z 2 )3/2The magnetic field from the other coil, located a distance s away, but for points measured from thefirst coil, is µ0 N iR2 B2 = . 2(R2 + (z − s)2 )3/2The magnetic field on the axis between the coils is the sum, µ0 N iR2 µ0 N iR2 B= + . 2(R2 + z 2 )3/2 2(R2 + (z − s)2 )3/2Take the derivative with respect to z and get dB 3µ0 N iR2 3µ0 N iR2 =− 2 + z 2 )5/2 z− 2 + (z − s)2 )5/2 (z − s). dz 2(R 2(RAt z = s/2 this expression vanishes! We expect this by symmetry, because the magnetic field willbe strongest in the plane of either coil, so the mid-point should be a local minimum. (b) Take the derivative again and d2 B 3µ0 N iR2 15µ0 N iR2 2 = − + z dz 2 2(R 2 + z 2 )5/2 2(R2 + z 2 )5/2 3µ0 N iR2 15µ0 N iR2 − 2 + (z − s)2 )5/2 + 2 + (z − s)2 )5/2 (z − s)2 . 2(R 2(RWe could try and simplify this, but we don’t really want to; we instead want to set it equal to zero,then let z = s/2, and then solve for s. The second derivative will equal zero when −3(R2 + z 2 ) + 15z 2 − 3(R2 + (z − s)2 ) + 15(z − s)2 = 0,and is z = s/2 this expression will simplify to 30(s/2)2 = 6(R2 + (s/2)2 ), 4(s/2)2 = R2 , s = R. 114
• P33-4 (a) Each of the side of the square is a straight wire segment of length a which contributesa field strength of µ0 i a B= , 4πr a2 /4 + r2where r is the distance to the point on the axis of the loop, so r= a2 /4 + z 2 .This field is not parallel to the z axis; the z component is Bz = B(a/2)/r. There are four of thesecontributions. The off axis components cancel. Consequently, the field for the square is µ0 i a a/2 B = 4 , 4πr a2 /4 + r2 r 2 µ0 i a = , 2πr2 a2 /4 + r2 µ0 i a2 = 2 /4 + z 2 ) , 2π(a a2 /2 + z 2 4µ0 i a2 = √ . π(a2 + 4z 2) 2a2 + 4z 2 (b) When z = 0 this reduces to 4µ0 i a2 4µ0 i B= √ =√ . π(a2 ) 2a2 2 πaP33-5 (a) The polygon has n sides. A perpendicular bisector of each side can be drawn to thecenter and has length x where x/a = cos(π/n). Each side has a length L = 2a sin(π/n). Each of theside of the polygon is a straight wire segment which contributes a field strength of µ0 i L B= , 4πx L2 /4 + x2This field is parallel to the z axis. There are n of these contributions. The off axis componentscancel. Consequently, the field for the polygon µ0 i L B = n , 4πx L2 /4 + x2 µ0 i 2 = n tan(π/n), 4π L2 /4 + x2 µ0 i 1 = n tan(π/n), 2π asince (L/2)2 + x2 = a2 . (b) Evaluate: lim n tan(π/n) = lim n sin(π/n) ≈ nπ/n = π. n→∞ n→∞Then the answer to part (a) simplifies to µ0 i B= . 2a 115
• P33-6 For a square loop of wire we have four finite length segments each contributing a termwhich looks like Eq. 33-12, except that L is replaced by L/4 and d is replaced by L/8. Then at thecenter, µ0 i L/4 16µ0 i B=4 =√ . 4πL/8 L 2 /64 + L2 /64 2πLFor a circular loop R = L/2π so µ0 i πµ0 B== . 2R L √Since 16/ 2 π > π, the square wins. But only by some 7%! P33-7 We want to use the differential expression in Eq. 33-11, except that the limits of integra-tion are going to be different. We have four wire segments. From the top segment, 3L/4 µ0 i d B1 = √ , 4π z 2 + d2 −L/4 µ0 i 3L/4 −L/4 = − . 4πd (3L/4)2 + d2 (−L/4)2 + d2For the top segment d = L/4, so this simplifies even further to µ0 i √ √ B1 = 2(3 5 + 5) . 10πLThe bottom segment has the same integral, but d = 3L/4, so µ0 i √ √ B3 = 2( 5 + 5) . 30πLBy symmetry, the contribution from the right hand side is the same as the bottom, so B2 = B3 ,and the contribution from the left hand side is the same as that from the top, so B4 = B1 . Addingall four terms, 2µ0 i √ √ √ √ B = 3 2(3 5 + 5) + 2( 5 + 5) , 30πL 2µ0 i √ √ = (2 2 + 10). 3πLP33-8 Assume a current ring has a radius r and a width dr, the charge on the ring is dq = 2πσr dr,where σ = q/πR2 . The current in the ring is di = ω dq/2π = ωσr dr. The ring contributes a fielddB = µ0 di/2r. Integrate over all the rings: R B= µ0 ωσr dr/2r = µ0 ωR/2 = µωq/2πR. 0P33-9 B = µ0 in and mv = qBr. Combine, and mv (5.11×105 eV/c2 )(0.046c) i= = = 0.271 A. µ0 qrn (4π×10−7 N/A2 )e(0.023 m)(10000/m)P33-10 This shape is a triangle with area A = (4d)(3d)/2 = 6d2 . The enclosed current is then i = jA = (15 A/m2 )6(0.23 m)2 = 4.76 AThe line integral is then µ0 i = 6.0×10−6 T · m. 116
• P33-11 Assume that B does vary as the picture implies. Then the line integral along the pathshown must be nonzero, since B · l on the right is not zero, while it is along the three other sides.Hence B · dl is non zero, implying some current passes through the dotted path. But it doesn’t,so B cannot have an abrupt change.P33-12 (a) Sketch an Amperian loop which is a rectangle which enclosed N wires, has a verticalsides with height h, and horizontal sides with length L. Then B · dl = µ0 N i. Evaluate the integralalong the four sides. The vertical side contribute nothing, since B is perpendicular to h, and thenB · h = 0. If the integral is performed in a counterclockwise direction (it must, since the senseof integration was determined by assuming the current is positive), we get BL for each horizontalsection. Then µ0 iN 1 B= = µ0 in. 2L 2 (b) As a → ∞ then tan−1 (a/2R) → π/2. Then B → µ0 i/2a. If we assume that i is made up ofseveral wires, each with current i0 , then i/a = i0 n. P33-13 Apply Ampere’s law with an Amperian loop that is a circle centered on the center ofthe wire. Then B · ds = B ds = B ds = 2πrB,because B is tangent to the path and B is uniform along the path by symmetry. The currentenclosed is ienc = j dA.This integral is best done in polar coordinates, so dA = (dr)(r dθ), and then r 2π ienc = (j0 r/a) rdr dθ, 0 0 r = 2πj0 /a r2 dr, 0 2πj0 3 = r . 3aWhen r = a the current enclosed is i, so 2πj0 a2 3i i= or j0 = . 3 2πa2 The magnetic field strength inside the wire is found by gluing together the two parts of Ampere’slaw, 2πj0 3 2πrB = µ0 r , 3a µ0 j0 r2 B = , 3a 2 µ0 ir = . 2πa3 117
• P33-14 (a) According to Eq. 33-34, the magnetic field inside the wire without a hole has magnitudeB = µ0 ir/2πR2 = µ0 jr/2 and is directed radially. If we superimpose a second current to create thehole, the additional field at the center of the hole is zero, so B = µ0 jb/2. But the current in theremaining wire is i = jA = jπ(R2 − a2 ),so µ0 ib B= . 2π(R2 − a2 ) 118
• E34-1 ΦB = B · A = (42×10−6 T)(2.5 m2 ) cos(57◦ ) = 5.7×10−5 Wb.E34-2 |E| = |dΦB /dt| = A dB/dt = (π/4)(0.112 m)2 (0.157 T/s) = 1.55 mV.E34-3 (a) The magnitude of the emf induced in a loop is given by Eq. 34-4, dΦB |E| = N , dt = N (12 mWb/s2 )t + (7 mWb/s)There is only one loop, and we want to evaluate this expression for t = 2.0 s, so |E| = (1) (12 mWb/s2 )(2.0 s) + (7 mWb/s) = 31 mV. (b) This part isn’t harder. The magnetic flux through the loop is increasing when t = 2.0 s. Theinduced current needs to flow in such a direction to create a second magnetic field to oppose thisincrease. The original magnetic field is out of the page and we oppose the increase by pointing theother way, so the second field will point into the page (inside the loop). By the right hand rule this means the induced current is clockwise through the loop, or to theleft through the resistor.E34-4 E = −dΦB /dt = −A dB/dt. (a) E = −π(0.16 m)2 (0.5 T)/(2 s) = −2.0×10−2 V. (b) E = −π(0.16 m)2 (0.0 T)/(2 s) = 0.0×10−2 V. (c) E = −π(0.16 m)2 (−0.5 T)/(4 s) = 1.0×10−2 V.E34-5 (a) R = ρL/A = (1.69×10−8 Ω · m)[(π)(0.104 m)]/[(π/4)(2.50×10−3 m)2 ] = 1.12×10−3 Ω. (b) E = iR = (9.66 A)(1.12×10−3 Ω) = 1.08×10−2 V. The required dB/dt is then given by dB E = = (1.08×10−2 V)/(π/4)(0.104 m)2 = 1.27 T/s. dt AE34-6 E = −A ∆B/∆t = AB/∆t. The power is P = iE = E 2 /R. The energy dissipated is E 2 ∆t A2 B 2 E = P ∆t = = . R R∆t E34-7 (a) We could re-derive the steps in the sample problem, or we could start with the endresult. We’ll start with the result, di E = N Aµ0 n , dtexcept that we have gone ahead and used the derivative instead of the ∆. The rate of change in the current is di = (3.0 A/s) + (1.0 A/s2 )t, dtso the induced emf is E = (130)(3.46×10−4 m2 )(4π×10−7 Tm/A)(2.2×104 /m) (3.0A/s) + (2.0A/s2 )t , = (3.73×10−3 V) + (2.48×10−3 V/s)t. (b) When t = 2.0 s the induced emf is 8.69×10−3 V, so the induced current is i = (8.69×10−3 V)/(0.15 Ω) = 5.8×10−2 A. 119
• E34-8 (a) i = E/R = N A dB/dt. Note that A refers to the area enclosed by the outer solenoidwhere B is non-zero. This A is then the cross sectional area of the inner solenoid! Then 1 di (120)(π/4)(0.032 m)2 (4π×10−7 N/A2 )(220×102 /m) (1.5 A) i= N Aµ0 n = = 4.7×10−3 A. R dt (5.3 Ω) (0.16 s)E34-9 P = Ei = E 2 /R = (A dB/dt)2 /(ρL/a), where A is the area of the loop and a is the crosssectional area of the wire. But a = πd2 /4 and A = L2 /4π, so 2 L3 d2 dB (0.525 m)3 (1.1×10−3 m)2 P = = (9.82×10−3 T/s)2 = 4.97×10−6 W. 64πρ dt 64π(1.69×10−8 Ω · m)E34-10 ΦB = BA = B(2.3 m)2 /2. EB = −dΦB /dt = −AdB/dt, or (2.3 m)2 EB = − [−(0.87 T/s)] = 2.30 V, 2so E = (2.0 V) + (2.3 V) = 4.3 V.E34-11 (a) The induced emf, as a function of time, is given by Eq. 34-5, E(t) = −dΦB (t)/dtThis emf drives a current through the loop which obeys E(t) = i(t)R Combining, 1 dΦB (t) i(t) = − . R dtSince the current is defined by i = dq/dt we can write dq(t) 1 dΦB (t) =− . dt R dtFactor out the dt from both sides, and then integrate: 1 dq(t) = − dΦB (t), R 1 dq(t) = − dΦB (t), R 1 q(t) − q(0) = (ΦB (0) − ΦB (t)) R (b) No. The induced current could have increased from zero to some positive value, then decreasedto zero and became negative, so that the net charge to flow through the resistor was zero. Thiswould be like sloshing the charge back and forth through the loop.E34-12 ∆P hiB = 2ΦB = 2N BA. Then the charge to flow through is q = 2(125)(1.57 T)(12.2×10−4 m2 )/(13.3 Ω) = 3.60×10−2 C.E34-13 The part above the long straight wire (a distance b − a above it) cancels out contributionsbelow the wire (a distance b − a beneath it). The flux through the loop is then a µ0 i µ0 ib a ΦB = b dr = ln . 2a−b 2πr 2π 2a − b 120
• The emf in the loop is then dΦB µ0 b a E =− = ln [2(4.5 A/s2 )t − (10 A/s)]. dt 2π 2a − bEvaluating, 4π×10−7 N/A2 (0.16 m) (0.12 m)E= ln [2(4.5 A/s2 )(3.0 s)−(10 A/s)] = 2.20×10−7 V. 2π 2(0.12 m) − (0.16 m)E34-14 Use Eq. 34-6: E = BDv = (55×10−6 T)(1.10 m)(25 m/s) = 1.5×10−3 V.E34-15 If the angle doesn’t vary then the flux, given by Φ=B·Ais constant, so there is no emf.E34-16 (a) Use Eq. 34-6: E = BDv = (1.18T)(0.108 m)(4.86 m/s) = 0.619 V. (b) i = (0.619 V)/(0.415 Ω) = 1.49 A. (c) P = (0.619 V)(1.49 A) = 0.922 W. (d) F = iLB = (1.49 A)(0.108 m)(1.18T) = 0.190 N. (e) P = F v = (0.190 V)(4.86 m/s) = 0.923 W. E34-17 The magnetic field is out of the page, and the current through the rod is down. Then Eq.32-26 F = iL × B shows that the direction of the magnetic force is to the right; furthermore, sinceeverything is perpendicular to everything else, we can get rid of the vector nature of the problemand write F = iLB. Newton’s second law gives F = ma, and the acceleration of an object from restresults in a velocity given by v = at. Combining, iLB v(t) = t. mE34-18 (b) The rod will accelerate as long as there is a net force on it. This net force comes fromF = iLB. The current is given by iR = E − BLv, so as v increases i decreases. When i = 0 the rodstops accelerating and assumes a terminal velocity. (a) E = BLv will give the terminal velocity. In this case, v = E/BL.E34-19E34-20 The acceleration is a = Rω 2 ; since E = BωR2 /2, we can find a = 4E 2 /B 2 R3 = 4(1.4 V)2 /(1.2 T)2 (5.3×10−2 m)3 = 3.7×104 m/s2 . E34-21 We will use the results of Exercise 11 that were worked out above. All we need to do isfind the initial flux; flipping the coil up-side-down will simply change the sign of the flux. So ΦB (0) = B · A = (59 µT)(π)(0.13 m)2 sin(20◦ ) = 1.1×10−6 Wb. 121
• Then using the results of Exercise 11 we have N q = (ΦB (0) − ΦB (t)), R 950 = ((1.1×10−6 Wb) − (−1.1×10−6 Wb)), 85 Ω = 2.5×10−5 C.E34-22 (a) The flux through the loop is vt a+L µ0 i µ0 ivt a + L ΦB = dx dr = ln . 0 a 2πr 2π aThe emf is then dΦB µ0 iv a + L E =− =− ln . dt 2π aPutting in the numbers, (4π×10−7 N/A2 )(110 A)(4.86 m/s) (0.0102 m) + (0.0983 m) E= ln = 2.53×10−4 V. 2π (0.0102 m) (b) i = E/R = (2.53×10−4 V)/(0.415 Ω) = 6.10×10−4 A. (c) P = i2 R = (6.10×10−4 A)2 (0.415 Ω) = 1.54×10−7 W. (d) F = Bil dl, or a+L µ0 i µ0 i a + L F = il dr = il ln . a 2πr 2π aPutting in the numbers, (4π×10−7 N/A2 )(110 A) (0.0102 m) + (0.0983 m) F = (6.10×10−4 A) ln = 3.17×10−8 N. 2π (0.0102 m) (e) P = F v = (3.17×10−8 N)(4.86 m/s) = 1.54×10−7 W.E34-23 (a) Starting from the beginning, Eq. 33-13 gives µ0 i B= . 2πyThe flux through the loop is given by ΦB = B · dA,but since the magnetic field from the long straight wire goes through the loop perpendicular to theplane of the loop this expression simplifies to a scalar integral. The loop is a rectangular, so usedA = dx dy, and let x be parallel to the long straight wire. Combining the above, D+b a µ0 i ΦB = dx dy, D 0 2πy D+b µ0 i dy = a , 2π D y µ0 i D+b = a ln 2π D 122
• (b) The flux through the loop is a function of the distance D from the wire. If the loop movesaway from the wire at a constant speed v, then the distance D varies as vt. The induced emf is then dΦB E = − , dt µ0 i b = a . 2π t(vt + b)The current will be this emf divided by the resistance R. The “back-of-the-book” answer is somewhatdifferent; the answer is expressed in terms of D instead if t. The two answers are otherwise identical.E34-24 (a) The area of the triangle is A = x2 tan θ/2. In this case x = vt, so ΦB = B(vt)2 tan θ/2,and then E = 2Bv 2 t tan θ/2, (b) t = E/2Bv 2 tan θ/2, so (56.8 V) t= = 2.08 s. 2(0.352 T)(5.21 m/s)2 tan(55◦ )E34-25 E = N BAω, so (24 V) ω= = 39.4 rad/s. (97)(0.33 T)(0.0190 m2 )That’s 6.3 rev/second.E34-26 (a) The frequency of the emf is the same as the frequency of rotation, f . (b) The flux changes by BA = Bπa2 during a half a revolution. This is a sinusoidal change, sothe amplitude of the sinusoidal variation in the emf is E = ΦB ω/2. Then E = Bπ 2 a2 f . E34-27 We can use Eq. 34-10; the emf is E = BAω sin ωt, This will be a maximum whensin ωt = 1. The angular frequency, ω is equal to ω = (1000)(2π)/(60) rad/s = 105 rad/s Themaximum emf is then E = (3.5 T) [(100)(0.5 m)(0.3 m)] (105 rad/s) = 5.5 kV.E34-28 (a) The amplitude of the emf is E = BAω, so A = E/2πf B = (150 V)/2π(60/s)(0.50 T) = 0.798m2 . (b) Divide the previous result by 100. A = 79.8 cm2 .E34-29 dΦB /dt = A dB/dt = A(−8.50 mT/s). (a) For this path E · ds = −dΦB /dt = − − π(0.212 m)2 (−8.50 mT/s) = −1.20 mV. (b) For this path E · ds = −dΦB /dt = − − π(0.323 m)2 (−8.50 mT/s) = −2.79 mV. 123
• (c) For this path E · ds = −dΦB /dt = − − π(0.323 m)2 (−8.50 mT/s) − π(0.323 m)2 (−8.50 mT/s) = 1.59 mV.E34-30 dΦB /dt = A dB/dt = A(−6.51 mT/s), while E · ds = 2πrE. (a) The path of integration is inside the solenoid, so −πr2 (−6.51 mT/s) (0.022 m)(−6.51 mT/s) E= = = 7.16×10−5 V/m. 2πr 2 (b) The path of integration is outside the solenoid, so −πr2 (−6.51 mT/s) (0.063 m)2 (−6.51 mT/s) E= = = 1.58×10−4 V/m 2πR 2(0.082 m)E34-31 The induced electric field can be found from applying Eq. 34-13, dΦB E · ds = − . dtWe start with the left hand side of this expression. The problem has cylindrical symmetry, so theinduced electric field lines should be circles centered on the axis of the cylindrical volume. If wechoose the path of integration to lie along an electric field line, then the electric field E will beparallel to ds, and E will be uniform along this path, so E · ds = E ds = E ds = 2πrE,where r is the radius of the circular path. Now for the right hand side. The flux is contained in the path of integration, so ΦB = Bπr2 .All of the time dependence of the flux is contained in B, so we can immediately write dB r dB 2πrE = −πr2 or E = − . dt 2 dtWhat does the negative sign mean? The path of integration is chosen so that if our right hand fingerscurl around the path our thumb gives the direction of the magnetic field which cuts through thepath. Since the field points into the page a positive electric field would have a clockwise orientation.Since B is decreasing the derivative is negative, but we get another negative from the equationabove, so the electric field has a positive direction. Now for the magnitude. E = (4.82×10−2 m)(10.7×10−3 T /s)/2 = 2.58×10−4 N/C.The acceleration of the electron at either a or c then has magnitude a = Eq/m = (2.58×10−4 N/C)(1.60×10−19 C)/(9.11×10−31 kg) = 4.53×107 m/s2 . P34-1 The induced current is given by i = E/R. The resistance of the loop is given by R = ρL/A,where A is the cross sectional area. Combining, and writing in terms of the radius of the wire, wehave πr2 E i= . ρL 124
• The length of the wire is related to the radius of the wire because we have a fixed mass. The totalvolume of the wire is πr2 L, and this is related to the mass and density by m = δπr2 L. Eliminatingr we have mE i= . ρδL2The length of the wire loop is the same as the circumference, which is related to the radius R of theloop by L = 2πR. The emf is related to the changing flux by E = −dΦB /dt, but if the shape of theloop is fixed this becomes E = −A dB/dt. Combining all of this, mA dB i= . ρδ(2πR)2 dtWe dropped the negative sign because we are only interested in absolute values here. Now A = πR2 , so this expression can also be written as mπR2 dB m dB i= = . ρδ(2πR)2 dt 4πρδ dtP34-2 For the lower surface B · A = (76×10−3 T)(π/2)(0.037 m)2 cos(62◦ ) = 7.67×10−5 Wb. Forthe upper surface B · A = (76×10−3 T)(π/2)(0.037 m)2 cos(28◦ ) = 1.44×10−4 Wb.. The induced emfis then E = (7.67×10−5 Wb + 1.44×10−4 Wb)/(4.5×10−3 s) = 4.9×10−2 V.P34-3 (a) We are only interested in the portion of the ring in the yz plane. Then E = (3.32×10−3 T/s)(π/4)(0.104 m)2 = 2.82×10−5 V. (b) From c to b. Point your right thumb along −x to oppose the increasing B field. Your rightfingers will curl from c to b.P34-4 E ∝ N A, but A = πr2 and N 2πr = L, so E ∝ 1/N . This means use only one loop tomaximize the emf.P34-5 This is a integral best performed in rectangular coordinates, then dA = (dx)(dy). Themagnetic field is perpendicular to the surface area, so B · dA = B dA. The flux is then ΦB = B · dA = B dA, a a = (4 T/m · s2 )t2 y dy dx, 0 0 1 2 = (4 T/m · s2 )t2 a a, 2 = (2 T/m · s2 )a3 t2 .But a = 2.0 cm, so this becomes ΦB = (2 T/m · s2 )(0.02 m)3 t2 = (1.6×10−5 Wb/s2 )t2 .The emf around the square is given by dΦB E =− = −(3.2×10−5 Wb/s2 )t, dtand at t = 2.5 s this is −8.0×10−5 V. Since the magnetic field is directed out of the page, a positiveemf would be counterclockwise (hold your right thumb in the direction of the magnetic field andyour fingers will give a counter clockwise sense around the loop). But the answer was negative, sothe emf must be clockwise. 125
• P34-6 (a) Far from the plane of the large loop we can approximate the large loop as a dipole, andthen µ0 iπR2 B= . 2x3The flux through the small loop is then µ0 iπ 2 r2 R2 ΦB = πr2 B = . 2x3 (b) E = −dΦB /dt, so 3µ0 iπ 2 r2 R2 E= v. 2x4 (c) Anti-clockwise when viewed from above. P34-7 The magnetic field is perpendicular to the surface area, so B · dA = B dA. The flux isthen ΦB = B · dA = B dA = BA,since the magnetic field is uniform. The area is A = πr2 , where r is the radius of the loop. Theinduced emf is dΦB dr E =− = −2πrB . dt dtIt is given that B = 0.785 T, r = 1.23 m, and dr/dt = −7.50×10−2 m/s. The negative sign indicatea decreasing radius. Then E = −2π(1.23 m)(0.785 T)(−7.50×10−2 m/s) = 0.455 V.P34-8 (a) dΦB /dt = B dA/dt, but dA/dt is ∆A/∆t, where ∆A is the area swept out during onerotation and ∆t = 1/f . But the area swept out is πR2 , so dΦB |E| = = πf BR2 . dt (b) If the output current is i then the power is P = Ei. But P = τ ω = τ 2πf , so P τ= = iBR2 /2. 2πfP34-9 (a) E = −dΦB /dt, and ΦB = B · A,so E = BLv cos θ.The component of the force of gravity on the rod which pulls it down the incline is FG = mg sin θ.The component of the magnetic force on the rod which pulls it up the incline is FB = BiL cos θ.Equating, BiL cos θ = mg sin θ,and since E = iR, E mgR sin θ v= = 2 2 . BL cos θ B L cos2 θ (b) P = iE = E 2 /R = B 2 L2 v 2 cos2 θ/R = mgv sin θ. This is identical to the rate of change ofgravitational potential energy. 126
• P34-10 Let the cross section of the wire be a. (a) R = ρL/a = ρ(rθ + 2r)/a; with numbers, R = (3.4×10−3 Ω)(2 + θ). (b) ΦB = Bθr2 /2; with numbers, ΦB = (4.32×10−3 Wb)θ. (c) i = E/R = Bωr2 /2R = Baωr/2ρ(θ + 2), or Baαtr i= . ρ(αt2 + 4)Take the derivative and set it equal to zero, 4 − at2 0= , (αt2 + 4)2so at2 = 4, or θ = 1 at2 = 2 rad. √ 2 (d) ω = 2αθ, so 2 (0.15 T)(1.2×10−6 m2 ) 2(12 rad/s )(2 rad)(0.24 m) i= = 2.2 A. (1.7×10−8 Ω · m)(6 rad) P34-11 It does say approximate, so we will be making some rather bold assumptions here. Firstwe will find an expression for the emf. Since B is constant, the emf must be caused by a change inthe area; in this case a shift in position. The small square where B = 0 has a width a and sweepsaround the disk with a speed rω. An approximation for the emf is then E = Barω. This emf causesa current. We don’t know exactly where the current flows, but we can reasonably assume that itoccurs near the location of the magnetic field. Let us assume that it is constrained to that region ofthe disk. The resistance of this portion of the disk is the approximately 1L 1 a 1 R= = = , σA σ at σtwhere we have assumed that the current is flowing radially when defining the cross sectional area ofthe “resistor”. The induced current is then (on the order of) E Barω = = Barωσt. R 1/(σt)This current experiences a breaking force according to F = BIl, so F = B 2 a2 rωσt,where l is the length through which the current flows, which is a. Finally we can find the torque from τ = rF , and τ = B 2 a2 r2 ωσt. 127
• P34-12 The induced electric field in the ring is given by Eq. 34-11: 2πRE = |dΦB /dt|. Thiselectric field will result in a force on the free charge carries (electrons?), given by F = Ee. Theacceleration of the electrons is then a = Ee/me . Then e dΦB a= . 2πRme dtIntegrate both sides with respect to time to find the speed of the electrons. e dΦB a dt = dt, 2πRme dt e dΦB v = 2πRme , e = ∆ΦB . 2πRmeThe current density is given by j = nev, and the current by iA = iπa2 . Combining, ne2 a2 i= ∆P hiB . 2RmeActually, it should be pointed out that ∆P hiB refers to the change in flux from external sources. Thecurrent induced in the wire will produce a flux which will exactly offset ∆P hiB so that the net fluxthrough the superconducting ring is fixed at the value present when the ring became superconducting.P34-13 Assume that E does vary as the picture implies. Then the line integral along the pathshown must be nonzero, since E · l on the right is not zero, while it is along the three other sides.Hence E · dl is non zero, implying a change in the magnetic flux through the dotted path. But itdoesn’t, so E cannot have an abrupt change.P34-14 The electric field a distance r from the center is given by πr2 dB/dT r dB E= = . 2πr 2 dtThis field is directed perpendicular to the radial lines. Define h to be the distance from the center of the circle to the center of the rod, and evaluateE = E · ds, dB rh E = dx, dt 2r dB L = h. dt 2But h2 = R2 − (L/2)2 , so dB L E= R2 − (L/2)2 . dt 2P34-15 (a) ΦB = πr2 B av , so E (0.32 m) E= = 2(0.28 T)(120 π) = 34 V/m. 2πr 2 (b) a = F/m = Eq/m = (33.8 V/m)(1.6×10−19 C)/(9.1×10−31 kg) = 6.0×1012 m/s2 . 128
• E35-1 If the Earth’s magnetic dipole moment were produced by a single current around the core,then that current would be µ (8.0 × 1022 J/T) i= = = 2.1×109 A A π(3.5 × 106 m)2E35-2 (a) i = µ/A = (2.33 A · m2 )/(160)π(0.0193 m)2 = 12.4 A. (b) τ = µB = (2.33 A · m2 )(0.0346 T) = 8.06×10−2 N · m.E35-3 (a) Using the right hand rule a clockwise current would generate a magnetic moment whichwould be into the page. Both currents are clockwise, so add the moments: µ = (7.00 A)π(0.20 m)2 + (7.00 A)π(0.30 m)2 = 2.86 A · m2 . (b) Reversing the current reverses the moment, so µ = (7.00 A)π(0.20 m)2 − (7.00 A)π(0.30 m)2 = −1.10 A · m2 .E35-4 (a) µ = iA = (2.58 A)π(0.16 m)2 = 0.207 A · m2 . (b) τ = µB sin θ = (0.207 A · m2 )(1.20 T) sin(41◦ ) = 0.163 N · m.E35-5 (a) The result from Problem 33-4 for a square loop of wire was 4µ0 ia2 B(z) = . π(4z 2 + a2 )(4z 2 + 2a2 )1/2For z much, much larger than a we can ignore any a terms which are added to or subtracted fromz terms. This means that 4z 2 + a2 → 4z 2 and (4z 2 + 2a2 )1/2 → 2z,but we can’t ignore the a2 in the numerator. The expression for B then simplifies to µ0 ia2 B(z) = , 2πz 3which certainly looks like Eq. 35-4. (b) We can rearrange this expression and get µ0 B(z) = ia2 , 2πz 3where it is rather evident that ia2 must correspond to µ, the dipole moment, in Eq. 35-4. So thatmust be the answer.E35-6 µ = iA = (0.2 A)π(0.08 m)2 = 4.02×10−3 A · m2 ; µ = µˆ . n (a) For the torque, τ = µ × B = (−9.65×10−4 N · m)ˆ + (−7.24×10−4 N · m)ˆ + (8.08×10−4 N · m)k. i j ˆ (b) For the magnetic potential energy, U = µ · B = µ[(0.60)(0.25 T)] = 0.603×10−3 J. 129
• E35-7 µ = iA = iπ(a2 + b2 /2) = iπ(a2 + b2 )/2.E35-8 If the distance to P is very large compared to a or b we can write the Law of Biot andSavart as µ0 i s × r B= . 4π r3s is perpendicular to r for the left and right sides, so the left side contributes µ0 i b B1 = , 4π (x + a/2)2and the right side contributes µ0 i b B3 = − . 4π (x − a/2)2The top and bottom sides each contribute an equal amount µ0 i a sin θ µ0 i a(b/2) B2 = B4 = ≈ . 4π x2 + b2 /4 4π x3Add the four terms, expand the denominators, and keep only terms in x3 , µ0 i ab µ0 µ B=− 3 =− . 4π x 4π x3The negative sign indicates that it is into the page.E35-9 (a) The electric field at this distance from the proton is 1 (1.60×10−19 C) E= = 5.14×1011 N/C. 4π(8.85×10−12 C2 /N · m2 ) (5.29×10−11 m)2 (b) The magnetic field at this from the proton is given by the dipole approximation, µ0 µ B(z) = , 2πz 3 (4π×10−7 N/A2 )(1.41×10−26 A/m2 ) = , 2π(5.29×10−11 m)3 = 1.90×10−2 TE35-10 1.50 g of water has (2)(6.02 × 1023 )(1.5)/(18) = 1.00 × 1023 hydrogen nuclei. If all arealigned the net magnetic moment would be µ = (1.00×1023 )(1.41×10−26 J/T) = 1.41×10−3 J/T.The field strength is then µ0 µ (1.41×10−3 J/T) B= = (1.00×10−7 N/A2 ) = 9.3×10−13 T. 4π x3 (5.33 m)3E35-11 (a) There is effectively a current of i = f q = qω/2π. The dipole moment is then µ = iA =(qω/2π)(πr2 ) = 1 qωr2 . 2 (b) The rotational inertia of the ring is mr2 so L = Iω = mr2 ω. Then µ (1/2)qωr2 q = = . L mr2 ω 2m 130
• E35-12 The mass of the bar is 3 m = ρV = (7.87 g/cm )(4.86 cm)(1.31 cm2 ) = 50.1 g.The number of atoms in the bar is N = (6.02×1023 )(50.1 g)/(55.8 g) = 5.41×1023 .The dipole moment of the bar is then µ = (5.41×1023 )(2.22)(9.27×10−24 J/T) = 11.6 J/T. (b) The torque on the magnet is τ = (11.6 J/T)(1.53 T) = 17.7 N · m.E35-13 The magnetic dipole moment is given by µ = M V , Eq. 35-13. Then µ = (5, 300 A/m)(0.048 m)π(0.0055 m)2 = 0.024 A · m2 .E35-14 (a) The original field is B0 = µ0 in. The field will increase to B = κm B0 , so the increase is ∆B = (κ1 − 1)µ0 in = (3.3×10−4 )(4π×10−7 N/A2 )(1.3 A)(1600/m) = 8.6×10−7 T. (b) M = (κ1 − 1)B0 /µ0 = (κ1 − 1)in = (3.3×10−4 )(1.3 A)(1600/m) = 0.69 A/m.E35-15 The energy to flip the dipoles is given by U = 2µB. The temperature is then 2µB 4(1.2×10−23 J/T)(0.5 T) T = = = 0.58 K. 3k/2 3(1.38×10−23 J/K)E35-16 The Curie temperature of iron is 770◦ C, which is 750◦ C higher than the surface temper-ature. This occurs at a depth of (750◦ C)/(30 C◦ /km) = 25 km.E35-17 (a) Look at the figure. At 50% (which is 0.5 on the vertical axis), the curve is atB0 /T ≈ 0.55 T/K. Since T = 300 K, we have B0 ≈ 165 T. (b) Same figure, but now look at the 90% mark. B0 /T ≈ 1.60 T/K, so B0 ≈ 480 T. (c) Good question. I think both fields are far beyond our current abilities.E35-18 (a) Look at the figure. At 50% (which is 0.5 on the vertical axis), the curve is at B0 /T ≈0.55 T/K. Since B0 = 1.8 T, we have T ≈ (1.8 T)/(0.55 T/K) = 3.3 K. (b) Same figure, but now look at the 90% mark. B0 /T ≈ 1.60 T/K, so T ≈ (1.8 T)/(1.60 T/K) =1.1 K.E35-19 Since (0.5 T)/(10 K) = 0.05 T/K, and all higher temperatures have lower values of theratio, and this puts all points in the region near where Curie’s Law (the straight line) is valid, thenthe answer is yes.E35-20 Using Eq. 35-19, VM Mr M (108g/mol)(511×103 A/m) µn = = = = 8.74×10−21 A/m2 N Aρ (10490 kg/m3 )(6.02×1023 /mol) 131
• E35-21 (a) B = µ0 µ/2z 3 , so (4π×10−7 N/A2 )(1.5×10−23 J/T) B= = 9.4×10−6 T. 2(10×10−9 m)3 (b) U = 2µB = 2(1.5×10−23 J/T)(9.4×10−6 T) = 2.82×10−28 J.E35-22 ΦB = (43×10−6 T)(295, 000×106 m2 ) = 1.3×107 Wb. E35-23 (a) We’ll assume, however, that all of the iron atoms are perfectly aligned. Then thedipole moment of the earth will be related to the dipole moment of one atom by µEarth = N µFe ,where N is the number of iron atoms in the magnetized sphere. If mA is the relative atomic massof iron, then the total mass is N mA mA µEarth m= = , A A µFewhere A is Avogadro’s number. Next, the volume of a sphere of mass m is m mA µEarth V = = , ρ ρA µFewhere ρ is the density. And finally, the radius of a sphere with this volume would be 1/3 1/3 3V 3µEarth mA r= = . 4π 4πρµFe ANow we find the radius by substituting in the known values, 1/3 3(8.0×1022 J/T)(56 g/mol) r= 3 = 1.8×105 m. 4π(14×106 g/m )(2.1×10−23 J/T)(6.0×1023 /mol) (b) The fractional volume is the cube of the fractional radius, so the answer is (1.8×105 m/6.4×106 )3 = 2.2×10−5 .E35-24 (a) At magnetic equator Lm = 0, so µ0 µ (1.00×10−7 N/A2 )(8.0×1022 J/T) B= = = 31µT. 4πr3 (6.37×106 m)3There is no vertical component, so the inclination is zero. (b) Here Lm = 60◦ , so µ0 µ (1.00×10−7 N/A2 )(8.0×1022 J/T) B= 1 + 3 sin2 Lm = 1 + 3 sin2 (60◦ ) = 56µT. 4πr3 (6.37×106 m)3The inclination is given by arctan(B v /B h ) = arctan(2 tan Lm ) = 74◦ . (c) At magnetic north pole Lm = 90◦ , so µ0 µ 2(1.00×10−7 N/A2 )(8.0×1022 J/T) B= 3 = = 62µT. 2πr (6.37×106 m)3There is no horizontal component, so the inclination is 90◦ . 132
• E35-25 This problem is effectively solving 1/r3 = 1/2 for r measured in Earth radii. Thenr = 1.26rE , and the altitude above the Earth is (0.26)(6.37×106 m) = 1.66×106 m.E35-26 The radial distance from the center is r = (6.37×106 m) − (2900×103 m) = 3.47×106 m.The field strength is µ0 µ 2(1.00×10−7 N/A2 )(8.0×1022 J/T) B= 3 = = 380µT. 2πr (3.47×106 m)3E35-27 Here Lm = 90◦ − 11.5◦ = 78.5◦ , so µ0 µ (1.00×10−7 N/A2 )(8.0×1022 J/T) B= 1 + 3 sin2 Lm = 1 + 3 sin2 (78.5◦ ) = 61µT. 4πr3 (6.37×106 m)3The inclination is given by arctan(B v /B h ) = arctan(2 tan Lm ) = 84◦ .E35-28 The flux out the “other” end is (1.6×10−3 T)π(0.13 m)2 = 85µWb. The net flux throughthe surface is zero, so the flux through the curved surface is 0 − (85µWb) − (−25µWb) = −60µWb..The negative indicates inward. E35-29 The total magnetic flux through a closed surface is zero. There is inward flux on facesone, three, and five for a total of -9 Wb. There is outward flux on faces two and four for a total of+6 Wb. The difference is +3 Wb; consequently the outward flux on the sixth face must be +3 Wb.E35-30 The stable arrangements are (a) and (c). The torque in each case is zero.E35-31 The field on the x axis between the wires is µ0 i 1 1 B= + . 2π 2r + x 2r − xSince B · dA = 0, we can assume the flux through the curved surface is equal to the flux throughthe xz plane within the cylinder. This flux is r µ0 i 1 1 ΦB = L + dx, −r 2π 2r + x 2r − x µ0 i 3r r = L ln − ln , 2π r 3r µ0 i = L ln 3. π P35-1 We can imagine the rotating disk as being composed of a number of rotating rings ofradius r, width dr, and circumference 2πr. The surface charge density on the disk is σ = q/πR2 ,and consequently the (differential) charge on any ring is 2qr dq = σ(2πr)(dr) = dr R2The rings “rotate” with angular frequency ω, or period T = 2π/ω. The effective (differential) currentfor each ring is then dq qrω di = = dr. T πR2 133
• Each ring contributes to the magnetic moment, and we can glue all of this together as µ = dµ, = πr2 di, R qr3 ω = dr, 0 R2 qR2 ω = . 4P35-2 (a) The sphere can be sliced into disks. The disks can be sliced into rings. Each ring hassome charge qi , radius ri , and mass mi ; the period of rotation for a ring is T = 2π/ω, so the currentin the ring is qi /T = ωqi /2π. The magnetic moment is 2 2 µi = (ωqi /2π)πri = ωqi ri /2. 2Note that this is closely related to the expression for angular momentum of a ring: li = ωmi ri .Equating, µi = qi li /2mi .If both mass density and charge density are uniform then we can write qi /mi = q/m, µ= dµ = (q/2m) dl = qL/2mFor a solid sphere L = ωI = 2ωmR2 /5, so µ = qωR2 /5. (b) See part (a)P35-3 (a) The orbital speed is given by K = mv 2 /2. The orbital radius is given by mv = qBr, orr = mv/qB. The frequency of revolution is f = v/2πr. The effective current is i = qf . Combiningall of the above to find the dipole moment, v vr mv 2 K µ = iA = q πr2 = q =q = . 2πr 2 2qB B (b) Since q and m cancel out of the above expression the answer is the same! (c) Work it out: µ (5.28×1021 /m3 )(6.21×10−20 J) (5.28×1021 /m3 )(7.58×10−21 J) M= = + = 312 A/m. V (1.18 T) (1.18 T)P35-4 (b) Point the thumb or your right hand to the right. Your fingers curl in the direction ofthe current in the wire loop. (c) In the vicinity of the wire of the loop B has a component which is directed radially outward.Then B × ds has a component directed to the left. Hence, the net force is directed to the left.P35-5 (b) Point the thumb or your right hand to the left. Your fingers curl in the direction of thecurrent in the wire loop. (c) In the vicinity of the wire of the loop B has a component which is directed radially outward.Then B × ds has a component directed to the right. Hence, the net force is directed to the right. 134
• P35-6 (a) Let x = µB/kT . Adopt the convention that N+ refers to the atoms which have parallelalignment and N− those which are anti-parallel. Then N+ + N− = N , so N+ = N ex /(ex + e−x ),and N− = N e−x /(ex + e−x ),Note that the denominators are necessary so that N+ + N− = N . Finally, ex − e−x M = µ(N+ − N− ) = µN . ex + e−x (b) If µB kT then x is very small and e±x ≈ 1 ± x. The above expression reduces to (1 + x) − (1 − x) µ2 B M = µN = µN x = . (1 + x) + (1 − x) kT (c) If µB kT then x is very large and e±x → ∞ while e−x → 0. The above expression reducesto N = µN.P35-7 (a) Centripetal acceleration is given by a = rω 2 . Then a − a0 = r(ω0 + ∆ω)2 − rω0 , 2 = 2rω0 ∆ω + r(∆ω0 )2 , ≈ 2rω0 ∆ω. (b) The change in centripetal acceleration is caused by the additional magnetic force, which hasmagnitude FB = qvB = erωB. Then a − a0 eB ∆ω = = . 2rω0 2mNote that we boldly canceled ω against ω0 in this last expression; we are assuming that ∆ω is small,and for these problems it is.P35-8 (a) i = µ/A = (8.0×1022 J/T)/π(6.37×106 m)2 = 6.3×108 A. (b) Far enough away both fields act like perfect dipoles, and can then cancel. (c) Close enough neither field acts like a perfect dipole and the fields will not cancel. √P35-9 (a) B = B h 2 + B v 2 , so µ0 µ µ0 µ B= cos2 Lm + 4 sin2 Lm = 1 + 3 sin2 Lm . 4πr3 4πr3 (b) tan φi = B v /B h = 2 sin Lm / cos Lm = 2 tan Lm . 135
• E36-1 The important relationship is Eq. 36-4, written as iL (5.0 mA)(8.0 mH) ΦB = = = 1.0×10−7 Wb N (400)E36-2 (a) Φ = (34)(2.62×10−3 T)π(0.103 m)2 = 2.97×10−3 Wb. (b) L = Φ/i = (2.97×10−3 Wb)/(3.77 A) = 7.88×10−4 H.E36-3 n = 1/d, where d is the diameter of the wire. Then L µ0 A (4π×10−7 H/m)(π/4)(4.10×10−2 m)2 = µ0 n2 A = 2 = = 2.61×10−4 H/m. l d (2.52×10−3 m)2E36-4 (a) The emf supports the current, so the current must be decreasing. (b) L = E/(di/dt) = (17 V)/(25×103 A/s) = 6.8×10−4 H.E36-5 (a) Eq. 36-1 can be used to find the inductance of the coil. EL (3.0 mV) L= = = 6.0×10−4 H. di/dt (5.0 A/s) (b) Eq. 36-4 can then be used to find the number of turns in the coil. iL (8.0 A)(6.0×10−4 H) N= = = 120 ΦB (40µWb)E36-6 Use the equation between Eqs. 36-9 and 36-10. (4π×10−7 H/m)(0.81 A)(536)(5.2×10−2 m) (5.2×10−2 m) + (15.3×10−2 m) ΦB = ln , 2π (15.3×10−2 m) = 1.32×10−6 Wb.E36-7 L = κm µ0 n2 Al = κm µ0 N 2 A/l, or L = (968)(4π×10−7 H/m)(1870)2 (π/4)(5.45×10−2 m)2 /(1.26 m) = 7.88 H.E36-8 In each case apply E = L∆i/∆t. (a) E = (4.6 H)(7 A)/(2×10−3 s) = 1.6×104 V. (b) E = (4.6 H)(2 A)/(3×10−3 s) = 3.1×103 V. (c) E = (4.6 H)(5 A)/(1×10−3 s) = 2.3×104 V. E36-9 (a) If two inductors are connected in parallel then the current through each inductor willadd to the total current through the circuit, i = i1 + i2 , Take the derivative of the current withrespect to time and then di/dt = di1 /dt + di2 /dt, The potential difference across each inductor is the same, so if we divide by E and apply we get di/dt di1 /dt di2 /dt = + , E E EBut di/dt 1 = , E L 136
• so the previous expression can also be written as 1 1 1 = + . Leq L1 L2 (b) If the inductors are close enough together then the magnetic field from one coil will inducecurrents in the other coil. Then we will need to consider mutual induction effects, but that is a topicnot covered in this text.E36-10 (a) If two inductors are connected in series then the emf across each inductor will add tothe total emf across both, E = E1 + E2 , Then the current through each inductor is the same, so if we divide by di/dt and apply we get E E1 E2 = + , di/dt di/dt di/dtBut E = L, di/dtso the previous expression can also be written as Leq = L1 + L2 . (b) If the inductors are close enough together then the magnetic field from one coil will inducecurrents in the other coil. Then we will need to consider mutual induction effects, but that is a topicnot covered in this text.E36-11 Use Eq. 36-17, but rearrange: t (1.50 s) τL = = = 0.317 s. ln[i0 /i] ln[(1.16 A)/(10.2×10−3 A)]Then R = L/τL = (9.44 H)/(0.317 s) = 29.8 Ω.E36-12 (a) There is no current through the resistor, so ER = 0 and then EL = E. (b) EL = Ee−2 = (0.135)E. (c) n = − ln(EL /E) = −ln(1/2) = 0.693.E36-13 (a) From Eq. 36-4 we find the inductance to be N ΦB (26.2×10−3 Wb) L= = = 4.78×10−3 H. i (5.48 A)Note that ΦB is the flux, while the quantity N ΦB is the number of flux linkages. (b) We can find the time constant from Eq. 36-14, τL = L/R = (4.78×10−3 H)/(0.745 Ω) = 6.42×10−3 s.The we can invert Eq. 36-13 to get Ri(t) t = −τL ln 1 − , E (0.745 A)(2.53 A) = −(6.42×10−3 s) ln 1 − = 2.42×10−3 s. (6.00 V) 137
• E36-14 (a) Rearrange: di E = iR + L , dt E L di −i = , R R dt R di dt = . L E/R − i (b) Integrate: t i R di − dt = , 0 L 0 i − E/R R i + E/R − t = ln , L E/R E −t/τL e = i + E/R, R E 1 − e−t/τL = i. RE36-15 di/dt = (5.0 A/s). Then di E = iR + L = (3.0 A)(4.0 Ω) + (5.0 A/s)t(4.0 Ω) + (6.0 H)(5.0 A/s) = 42 V + (20 V/s)t. dtE36-16 (1/3) = (1 − e−t/τL ), so t (5.22 s) τL = − =− = 12.9 s. ln(2/3) ln(2/3)E36-17 We want to take the derivative of the current in Eq. 36-13 with respect to time, di E 1 −t/τL E = e = e−t/τL . dt R τL LThen τL = (5.0×10−2 H)/(180 Ω) = 2.78×10−4 s. Using this we find the rate of change in the currentwhen t = 1.2 ms to be di (45 V) −3 −4 = −2 H) e−(1.2×10 s)/(2.78×10 s) = 12 A/s. dt ((5.0×10E36-18 (b) Consider some time ti : EL (ti ) = Ee−ti /τL .Taking a ratio for two different times, EL (t1 ) = e(t2 −t1 )/τL , EL (t2 )or t2 − t1 (2 ms) − (1 ms) τL = = = 3.58 ms ln[EL (t1 )/EL (t2 )] ln[(18.24 V)/(13.8 V)] (a) Choose any time, and E = EL et/τL = (18.24 V)e(1 ms)/(3.58 ms) = 24 V. 138
• E36-19 (a) When the switch is just closed there is no current through the inductor. So i1 = i2 isgiven by E (100 V) i1 = = = 3.33 A. R1 + R2 (10 Ω) + (20 Ω) (b) A long time later there is current through the inductor, but it is as if the inductor has noeffect on the circuit. Then the effective resistance of the circuit is found by first finding the equivalentresistance of the parallel part 1/(30 Ω) + 1/(20 Ω) = 1/(12 Ω),and then finding the equivalent resistance of the circuit (10 Ω) + (12 Ω) = 22 Ω.Finally, i1 = (100 V)/(22 Ω) = 4.55 A and ∆V2 = (100 V) − (4.55 A)(10 Ω) = 54.5 V;consequently, i2 = (54.5 V)/(20 Ω) = 2.73 A. It didn’t ask, but i2 = (4.55 A) − (2.73 A) = 1.82 A. (c) After the switch is just opened the current through the battery stops, while that through theinductor continues on. Then i2 = i3 = 1.82 A. (d) All go to zero.E36-20 (a) For toroids L = µ0 N 2 h ln(b/a)/2π. The number of turns is limited by the inner radius:N d = 2πa. In this case, N = 2π(0.10 m)/(0.00096 m) = 654.The inductance is then (4π×10−7 H/m)(654)2 (0.02 m) (0.12 m) L= ln = 3.1×10−4 H. 2π (0.10 m) (b) Each turn has a length of 4(0.02 m) = 0.08 m. The resistance is then R = N (0.08 m)(0.021 Ω/m) = 1.10 ΩThe time constant is τL = L/R = (3.1×10−4 H)/(1.10 Ω) = 2.8×10−4 s. E36-21 (I) When the switch is just closed there is no current through the inductor or R2 , so thepotential difference across the inductor must be 10 V. The potential difference across R1 is always10 V when the switch is closed, regardless of the amount of time elapsed since closing. (a) i1 = (10 V)/(5.0 Ω) = 2.0 A. (b) Zero; read the above paragraph. (c) The current through the switch is the sum of the above two currents, or 2.0 A. (d) Zero, since the current through R2 is zero. (e) 10 V, since the potential across R2 is zero. (f) Look at the results of Exercise 36-17. When t = 0 the rate of change of the current isdi/dt = E/L. Then di/dt = (10 V)/(5.0 H) = 2.0 A/s. (II) After the switch has been closed for a long period of time the currents are stable and theinductor no longer has an effect on the circuit. Then the circuit is a simple two resistor parallelnetwork, each resistor has a potential difference of 10 V across it. 139
• (a) Still 2.0 A; nothing has changed. (b) i2 = (10 V)/(10 Ω) = 1.0 A. (c) Add the two currents and the current through the switch will be 3.0 A. (d) 10 V; see the above discussion. (e) Zero, since the current is no longer changing. (f) Zero, since the current is no longer changing.E36-22 U = (71 J/m3 )(0.022 m3 ) = 1.56 J. Then using U = i2 L/2 we get i= 2U/L = 2(1.56 J)/(0.092 H) = 5.8 A.E36-23 (a) L = 2U/i2 = 2(0.0253 J)/(0.062 A)2 = 13.2 H. (b) Since the current is squared in the energy expression, doubling the current would quadruplethe energy. Then i = 2i0 = 2(0.062 A) = 0.124 A.E36-24 (a) B = µ0 in and u = B 2 /2µ0 , or u = µ0 i2 n2 /2 = (4π×10−7 N/A2 )(6.57 A)2 (950/0.853 m)2 /2 = 33.6 J/m3 . (b) U = uAL = (33.6 J/m3 )(17.2×10−4 m2 )(0.853 m) = 4.93×10−2 J.E36-25 uB = B 2 /2µ0 , and from Sample Problem 33-2 we know B, hence (12.6 T)2 uB = = 6.32×107 J/m3 . 2(4π×10−7 N/A2 )E36-26 (a) uB = B 2 /2µ0 , so (100×10−12 T)2 1 3 uB = = 2.5×10−2 eV/cm . 2(4π×10−7 N/A2 ) (1.6×10−19 J/eV) (b) x = (10)(9.46×1015 m) = 9.46×1016 m. Using the results from part (a) expressed in J/m3 wefind the energy contained is U = (3.98×10−15 J/m3 )(9.46×1016 m)3 = 3.4×1036 J E36-27 The energy density of an electric field is given by Eq. 36-23; that of a magnetic field isgiven by Eq. 36-22. Equating, 0 1 2 E2 = B , 2 2µ0 B E = √ . 0 µ0The answer is then E = (0.50 T)/ (8.85×10−12 C2 /N · m2 )(4π×10−7 N/A2 ) = 1.5×108 V/m.E36-28 The rate of internal energy increase in the resistor is given by P = i∆VR . The rate ofenergy storage in the inductor is dU/dt = Li di/dt = i∆VL . Since the current is the same throughboth we want to find the time when ∆VR = ∆VL . Using Eq. 36-15 we find 1 − e−t/τL = e−t/τL , ln 2 = t/τL ,so t = (37.5 ms) ln 2 = 26.0 ms. 140
• E36-29 (a) Start with Eq. 36-13: i = E(1 − e−t/τL )/R, iR 1− = e−t/τL , E −t τL = , ln(1 − iR/E) −(5.20×10−3 s) = , ln[1 − (1.96×10−3 A)(10.4×103 Ω)/(55.0 V)] = 1.12×10−2 s.Then L = τL R = (1.12×10−2 s)(10.4×103 Ω) = 116 H. (b) U = (1/2)(116 H)(1.96×10−3 A)2 = 2.23×10−4 J.E36-30 (a) U = E∆q; q = idt. E U = E 1 − e−t/τL dt, R E2 2 = t + τL e−t/τL , R 0 E2 E 2 L −t/τL = t + 2 (e − 1). R RUsing the numbers provided, τL = (5.48 H)/(7.34 Ω) = 0.7466 s.Then (12.2 V)2 U= (2 s) + (0.7466 s)(e−(2 s)/0.7466 s) − 1) = 26.4 J (7.34 Ω) (b) The energy stored in the inductor is UL = Li2 /2, or LE 2 2 UL = 1 − e−t/τL dt, 2R2 = 6.57 J. (c) UR = U − UL = 19.8 J.E36-31 This shell has a volume of 4π V = (RE + a)3 − RE 3 . 3Since a << RE we can expand the polynomials but keep only the terms which are linear in a. Then V ≈ 4πRE 2 a = 4π(6.37×106 m)2 (1.6×104 m) = 8.2×1018 m3 .The magnetic energy density is found from Eq. 36-22, 1 2 (60×10−6 T)2 uB = B = = 1.43×10−3 J/m3 . 2µ0 2(4π×10−7 N/A2 )The total energy is then (1.43×10−3 J/m3 )(8.2eex18m3 ) = 1.2×1016 J. 141
• E36-32 (a) B = µ0 i/2πr and uB = B 2 /2µ0 = µ0 i2 /8π 2 r2 , or uB = (4π×10−7 H/m)(10 A)2 /8π 2 (1.25×10−3 m)2 = 1.0 J/m3 . 2 2 (b) E = ∆V /l = iR/l and uE = 0E /2 = 0i (R/l)2 /2. Then uE = (8.85×10−12 F/m)(10 A)2 (3.3×10−3 Ω/m)2 /2 = 4.8×10−15 J/m3 .E36-33 i = 2U/L = 2(11.2×10−6 J)/(1.48×10−3 H) = 0.123 A.E36-34 C = q 2 /2U = (1.63×10−6 C)2 /2(142×10−6 J) = 9.36×10−9 F. √E36-35 1/2πf = LC so L = 1/4π 2 f 2 C, or L = 1/4π 2 (10×103 Hz)2 (6.7×10−6 F) = 3.8×10−5 H.E36-36 qmax 2 /2C = Limax 2 /2, or √ imax = qmax / LC = (2.94×10−6 C)/ (1.13×10−3 H)(3.88×10−6 F) = 4.44×10−2 A. E36-37 Closing a switch has the effect of “shorting” out the relevant circuit element, whicheffectively removes it from the circuit. If S1 is closed we have τC = RC or C = τC /R, if instead S2is closed we have τL = L/R or L = RτL , but if instead S3 is closed we have a LC circuit which willoscillate with period 2π √ T = = 2π LC. ωSubstituting from the expressions above, 2π √ T = = 2π τL τC . ωE36-38 The capacitors can be used individually, or in series, or in parallel. The four possiblecapacitances are then 2.00µF, 5.00µF, 2.00µF + 5.00µF = 7.00µF, and (2.00µF )(5.00µF)(2.00µF +5.00µF) = 1.43µF. The possible resonant frequencies are then 1 1 = f, 2π LC 1 1 = 1330 Hz, 2π (10.0 mH)(1.43µF ) 1 1 = 1130 Hz, 2π (10.0 mH)(2.00µF ) 1 1 = 712 Hz, 2π (10.0 mH)(5.00µF ) 1 1 = 602 Hz. 2π (10.0 mH)(7.00µF ) 142
• E36-39 (a) k = (8.13 N)/(0.0021 m) = 3.87×103 N/m. ω = k/m = (3870 N/m)/(0.485 kg) =89.3 rad/s. (b) T = 2π/ω = 2π/(89.3 rad/s) = 7.03×10−2 s. (c) LC = 1/ω 2 , so C = 1/(89.3 rad/s)2 (5.20 H) = 2.41×10−5 F. √E36-40 The period of oscillation is T = 2π LC = 2π (52.2mH)(4.21µF) = 2.95 ms. It requiresone-quarter period for the capacitor to charge, or 0.736 ms. E36-41 (a) An LC circuit oscillates so that the energy is converted from all magnetic to allelectrical twice each cycle. It occurs twice because once the energy is magnetic with the currentflowing in one direction through the inductor, and later the energy is magnetic with the currentflowing the other direction through the inductor. The period is then four times 1.52µs, or 6.08µs. (b) The frequency is the reciprocal of the period, or 164000 Hz. (c) Since it occurs twice during each oscillation it is equal to half a period, or 3.04µs.E36-42 (a) q = C∆V = (1.13×10−9 F)(2.87 V) = 3.24×10−9 C. (c) U = q 2 /2C = (3.24×10−9 C)2 /2(1.13×10−9 F) = 4.64×10−9 J. (b) i = 2U/L = 2(4.64×10−9 J)/(3.17×10−3 H) = 1.71×10−3 A.E36-43 (a) im = q m ω and q m = CV m . Multiplying the second expression by L we get Lq m =V m /ω 2 . Combining, Lim ω = V m . Then ω (50 V) f= = = 6.1×103 /s. 2π 2π(0.042 H)(0.031 A) (b) See (a) above. (c) C = 1/ω 2 L = 1/(2π6.1×103 /s)2 (0.042 H) = 1.6×10−8 F. √E36-44 (a) f = 1/2π LC = 1/2π (6.2×10−6 F)(54×10−3 H) = 275 Hz. (b) Note that from Eq. 36-32 we can deduce imax = ωqmax . The capacitor starts with a chargeq = C∆V = (6.2×10−6 F)(34 V) = 2.11×10−4 C. Then the current amplitude is √ imax = qmax / LC = (2.11×10−4 C)/ (6.2×10−6 F)(54×10−3 H) = 0.365 A. √E36-45 (a) ω = 1/ LC = 1/ (10×10−6 F)(3.0×10−3 H) = 5800 rad/s. (b) T = 2π/ω = 2π/(5800 rad/s) = 1.1×10−3 s.E36-46 f = (2×105 Hz)(1 + θ/180◦ ). C = 4π 2 /f 2 L, so 4π 2 (9.9×10−7 F) C= = . (2×105 Hz)2 (1 + θ/180◦ )2 (1 mH) (1 + θ/180◦ )2 √E36-47 (a) UE = UB /2 and UE + UB = U , so 3UE = U , or 3(q 2 /2C) = qmax /2C, so q = qmax / 3. 2 (b) Solve q = qmax cos ωt, or T √ t= arccos 1/ 3 = 0.152T. 2π 143
• E36-48 (a) Add the contribution from the inductor and the capacitor, (24.8×10−3 H)(9.16×10−3 A)2 (3.83×10−6 C)2 U= + = 1.99×10−6 J. 2 2(7.73×10−6 F) (b) q m = 2(7.73×10−6 F)(1.99×10−6 J) = 5.55×10−6 C. (c) im = 2(1.99×10−6 J)/(24.8×10−3 H) = 1.27×10−2 A. √E36-49 (a) The frequency of such a system is given by Eq. 36-26, f = 1/2π LC. Note thatmaximum frequency occurs with minimum capacitance. Then f1 C2 (365 pF) = = = 6.04. f2 C1 (10 pF) (b) The desired ratio is 1.60/0.54 = 2.96 Adding a capacitor in parallel will result in an effectivecapacitance given by C 1,eff = C1 + C add ,with a similar expression for C2 . We want to choose C add so that f1 C 2,eff = = 2.96. f2 C 1,effSolving, C 2,eff = C 1,eff (2.96)2 , C2 + C add = (C1 + C add )8.76, C2 − 8.76C1 C add = , 7.76 (365 pF) − 8.76(10 pF) = = 36 pF. 7.76 The necessary inductance is then 1 1 L= = = 2.2×10−4 H. 4π 2 f 2 C 4π 2 (0.54×106 Hz)2 (401×10−12 F)E36-50 The key here is that UE = C(∆V )2 /2. We want to charge a capacitor with one-ninth thecapacitance to have three times the potential difference. Since 32 = 9, it is reasonable to assumethat we want to take all of the energy from the first capacitor and give it to the second. ClosingS1 and S2 will not work, because the energy will be shared. Instead, close S2 until the capacitorhas completely discharged into the inductor, then simultaneously open S2 while closing S1 . Theinductor will then discharge into the second capacitor. Open S1 when it is “full”. √E36-51 (a) ω = 1/ LC. im qm = = (2.0 A) (3.0×10−3 H)(2.7×10−6 F) = 1.80×10−4 C ω (b) dUC /dt = qi/C. Since q = q m sin ωt and i = im cos ωt then dUC /dt is a maximum whensin ωt cos ωt is a maximum, or when sin 2ωt is a maximum. This happens when 2ωt = π/2, ort = T /8. (c) dUC /dt = q m im /2C, or dUC /dt = (1.80×10−4 C)(2.0 A)/2(2.7×10−6 F) = 67 W. 144
• E36-52 After only complete cycles q = qmax e−Rt/2L . Not only that, but t = N τ , where τ = 2π/ω .Finally, ω = (1/LC) − (R/2L)2 . Since the first term under the square root is so much larger than √the second, we can ignore the effect of damping on the frequency, and simply use ω ≈ ω = 1/ LC.Then √ √ q = qmax e−N Rτ /2L = qmax e−N πR LC/L = qmax e−N πR C/L .Finally, πR C/L = π(7.22 Ω) (3.18 µF)/(12.3 H) = 1.15×10−2 . Then N =5 : q = (6.31µC)e−5(0.0115) = 5.96µC, N =5 : q = (6.31µC)e−10(0.0115) = 5.62µC, N =5 : q = (6.31µC)e−100(0.0115) = 1.99µC.E36-53 Use Eq. 36-40, and since U ∝ q 2 , we want to solve e−Rt/L = 1/2, then L t= ln 2. RE36-54 Start by assuming that the presence of the resistance does not significantly change the √frequency. Then ω = 1/ LC, q = qmax e−Rt/2L , t = N τ , and τ = 2π/ω. Combining, √ √ q = qmax e−N Rτ /2L = qmax e−N πR LC/L = qmax e−N πR C/L .Then L/C (220mH)/(12µF) R=− ln(q/qmax ) = − ln(0.99) = 8700 Ω. Nπ (50)πIt remains to be verified that 1/LC (R/2L)2 . E36-55 The damped (angular) frequency is given by Eq. 36-41; the fractional change would thenbe ω−ω = 1 − 1 − (R/2Lω)2 = 1 − 1 − (R2 C/4L). ωSetting this equal to 0.01% and then solving for R, 4L 4(12.6×10−3 H) R= (1 − (1 − 0.0001)2 ) = (1.9999×10−4 ) = 2.96 Ω. C (1.15×10−6 F)P36-1 The inductance of a toroid is µ0 N 2 h b L= ln . 2π aIf the toroid is very large and thin then we can write b = a + δ, where δ << a. The natural log thencan be approximated as b δ δ ln = ln 1 + ≈ . a a aThe product of δ and h is the cross sectional area of the toroid, while 2πa is the circumference,which we will call l. The inductance of the toroid then reduces to µ0 N 2 δ µ0 N 2 A L≈ = . 2π a lBut N is the number of turns, which can also be written as N = nl, where n is the turns per unitlength. Substitute this in and we arrive at Eq. 36-7. 145
• P36-2 (a) Since ni is the net current per unit length and is this case i/W , we can simply writeB = µ0 i/W . (b) There is only one loop of wire, so L = φB /i = BA/i = µ0 iπR2 /W i = µ0 πR2 /W.P36-3 Choose the y axis so that it is parallel to the wires and directly between them. The fieldin the plane between the wires is µ0 i 1 1 B= + . 2π d/2 + x d/2 − xThe flux per length l of the wires is d/2−a d/2−a µ0 i 1 1 ΦB = l B dx = l + dx, −d/2+a 2π −d/2+a d/2 + x d/2 − x µ0 i d/2−a 1 = 2l dx, 2π −d/2+a d/2 + x µ0 i d − a = 2l ln . 2π aThe inductance is then φB µ0 l d − a L= = ln . i π aP36-4 (a) Choose the y axis so that it is parallel to the wires and directly between them. Thefield in the plane between the wires is µ0 i 1 1 B= + . 2π d/2 + x d/2 − xThe flux per length l between the wires is d/2−a d/2−a µ0 i 1 1 Φ1 = B dx = + dx, −d/2+a 2π −d/2+a d/2 + x d/2 − x µ0 i d/2−a 1 = 2 dx, 2π −d/2+a d/2 + x µ0 i d − a = 2 ln . 2π aThe field in the plane inside one of the wires, but still between the centers is µ0 i 1 d/2 − x B= + . 2π d/2 + x a2The additional flux is then d/2 d/2 µ0 i 1 d/2 − x Φ2 = 2 B dx = 2 + dx, d/2−a 2π d/2−a d/2 + x a2 µ0 i d 1 = 2 ln + . 2π d−a 2 146
• The flux per meter between the axes of the wire is the sum, or µ0 i d 1 ΦB = ln + , π a 2 (4π×10−7 H/m)(11.3 A) (21.8, mm) 1 = ln + , π (1.3 mm) 2 = 1.5×10−5 Wb/m. (b) The fraction f inside the wires is d 1 d 1 f = ln + / ln + , d−a 2 a 2 (21.8, mm) 1 (21.8, mm) 1 = + / + , (21.8, mm) − (1.3 mm) 2 (1.3 mm) 2 = 0.09. (c) The net flux is zero for parallel currents.P36-5 The magnetic field in the region between the conductors of a coaxial cable is given by µ0 i B= , 2πrso the flux through an area of length l, width b − a, and perpendicular to B is ΦB = B · dA = B dA, b l µ0 i = dz dr, a 0 2πr µ0 il b = ln . 2π aWe evaluated this integral is cylindrical coordinates: dA = (dr)(dz). As you have been warned somany times before, learn these differentials! The inductance is then ΦB µ0 l b L= = ln . i 2π aP36-6 (a) So long as the fuse is not blown there is effectively no resistance in the circuit. Thenthe equation for the current is E = L di/dt, but since E is constant, this has a solution i = Et/L.The fuse blows when t = imax L/E = (3.0 A)(5.0 H)/(10 V) = 1.5 s. (b) Note that once the fuse blows the maximum steady state current is 2/3 A, so there must bean exponential approach to that current.P36-7 The initial rate of increase is di/dt = E/L. Since the steady state current is E/R, thecurrent will reach the steady state value in a time given by E/R = i = Et/L, or t = L/R. But that’sτL . 1P36-8 (a) U = 2 Li2 = (152 H)(32 A)2 /2 = 7.8×104 J. (b) If the coil develops at finite resistance then all of the energy in the field will be dissipated asheat. The mass of Helium that will boil off is m = Q/Lv = (7.8×104 J)/(85 J/mol)/(4.00g/mol) = 3.7 kg. 147
• P36-9 (a) B = (µ0 N i)/(2πr), so B2 µ0 N 2 i2 u= = . 2µ0 8π 2 r2 (b) U = u dV = urdr dθ dz. The field inside the toroid is uniform in z and θ, so b µ0 N 2 i2 U = 2πh r dr, a 8π 2 r2 hµ0 N 2 i2 b = ln . 4π a (c) The answers are the same!P36-10 The energy in the inductor is originally U = Li2 /2. The internal energy in the resistor 0increases at a rate P = i2 R. Then ∞ ∞ Ri2 τL Li2 P dt = R i2 e−2t/τL dt = 0 0 = 0. 0 0 2 2 P36-11 (a) In Chapter 33 we found the magnetic field inside a wire carrying a uniform currentdensity is µ0 ir B= . 2πR2The magnetic energy density in this wire is 1 2 µ0 i2 r2 uB = B = . 2µ0 8π 2 R4We want to integrate in cylindrical coordinates over the volume of the wire. Then the volumeelement is dV = (dr)(r dθ)(dz), so UB = uB dV, R l 2π µ0 i2 r2 = dθ dz rdr, 0 0 0 8π 2 R4 R µ0 i2 l = r3 dr, 4πR4 0 µ0 i2 l = . 16π (b) Solve L 2 UB = i 2for L, and 2UB µ0 l L= 2 = . i 8πP36-12 1/C = 1/C1 + 1/C2 and L = L1 + L2 . Then 1 √ C1 C2 2 2 C2 /ω0 + C1 /ω0 1 = LC = (L1 + L2 ) = = . ω C1 + C2 C1 + C2 ω0 148
• P36-13 (a) There is no current in the middle inductor; the loop equation becomes d2 q q d2 q q L + +L 2 + = 0. dt2 C dt CTry q = q m cos ωt as a solution: 1 1 −Lω 2 + − Lω 2 + = 0; C C √which requires ω = 1/ LC. (b) Look at only the left hand loop; the loop equation becomes d2 q q d2 q L 2 + + 2L 2 = 0. dt C dtTry q = q m cos ωt as a solution: 1 −Lω 2 + − 2Lω 2 = 0; C √which requires ω = 1/ 3LC.P36-14 (b) (ω − ω)/ω is the fractional shift; this can also be written as ω = 1 − (LC)(R/2L)2 − 1, ω−1 = 1 − R2 C/4L − 1, (100 Ω)2 (7.3×10−6 F) = 1− − 1 = −2.1×10−3 . 4(4.4 H)P36-15 We start by focusing on the charge on the capacitor, given by Eq. 36-40 as q = q m e−Rt/2L cos(ω t + φ).After one oscillation the cosine term has returned to the original value but the exponential term hasattenuated the charge on the capacitor according to q = q m e−RT /2L ,where T is the period. The fractional energy loss on the capacitor is then U0 − U q2 = 1 − 2 = 1 − e−RT /L . U0 qmFor small enough damping we can expand the exponent. Not only that, but T = 2π/ω, so ∆U ≈ 2πR/ωL. U 149
• P36-16 We are given 1/2 = e−t/2τL when t = 2πn/ω . Then 2πn 2πn πnR ω = = = . t 2(L/R) ln 2 L ln 2 From Eq. 36-41, ω2 − ω 2 = (R/2L)2 , (ω − ω )(ω + ω ) = (R/2L)2 , (ω − ω )2ω ≈ (R/2L)2 , ω−ω (R/2L)2 = ≈ , ω 2ω 2 2 (ln 2) = , 8π 2 n2 0.0061 = . n2 150
• E37-1 The frequency, f , is related to the angular frequency ω by ω = 2πf = 2π(60 Hz) = 377 rad/sThe current is alternating because that is what the generator is designed to produce. It does thisthrough the configuration of the magnets and coils of wire. One complete turn of the generator will(could?) produce one “cycle”; hence, the generator is turning 60 times per second. Not only doesthis set the frequency, it also sets the emf, since the emf is proportional to the speed at which thecoils move through the magnetic field.E37-2 (a) XL = ωL, so f = XL /2πL = (1.28×103 Ω)/2π(0.0452 H) = 4.51×103 /s. (b) XC = 1/ωC, so C = 1/2πf XC = 1/2π(4.51×103 /s)(1.28×103 Ω) = 2.76×10−8 F. (c) The inductive reactance doubles while the capacitive reactance is cut in half. √E37-3 (a) XL = XC implies ωL = 1/ωC or ω = 1/ LC, so ω = 1/ (6.23×10−3 H)(11.4×10−6 F) = 3750 rad/s. (b) XL = ωL = (3750 rad/s)(6.23×10−3 H) = 23.4 Ω (c) See (a) above.E37-4 (a) im = E/XL = E/ωL, so im = (25.0 V)/(377 rad/s)(12.7 H) = 5.22×10−3 A. (b) The current and emf are 90◦ out of phase. When the current is a maximum, E = 0. (c) ωt = arcsin[E(t)/E m ], so (−13.8 V) ωt = arcsin = 0.585 rad. (25.0 V)and i = (5.22×10−3 A) cos(0.585) = 4.35×10−3 A. (d) Taking energy. E37-5 (a) The reactance of the capacitor is from Eq. 37-11, XC = 1/ωC. The AC generatorfrom Exercise 4 has E = (25.0 V) sin(377 rad/s)t. So the reactance is 1 1 XC = = = 647 Ω. ωC (377 rad/s)(4.1µF)The maximum value of the current is found from Eq. 37-13, (∆VC )max ) (25.0 V) im = = = 3.86×10−2 A. XC (647 Ω) (b) The generator emf is 90◦ out of phase with the current, so when the current is a maximumthe emf is zero. 151
• (c) The emf is -13.8 V when (−13.8 V) ωt = arcsin = 0.585 rad. (25.0 V)The current leads the voltage by 90◦ = π/2, so i = im sin(ωt − φ) = (3.86×10−2 A) sin(0.585 − π/2) = −3.22×10−2 A. (d) Since both the current and the emf are negative the product is positive and the generator issupplying energy to the circuit.E37-6 R = (ωL − 1/omegaC)/ tan φ and ω = 2πf = 2π(941/s) = 5910 rad/s , so (5910 rad/s)(88.3×10−3 H) − 1/(5910 rad/s)(937×10−9 F) R= = 91.5 Ω. tan(75◦ )E37-7E37-8 (a) XL doesn’t change, so XL = 87 Ω. (b) XC = 1/ωC = 1/2π(60/s)(70×10−6 F) = 37.9Ω. (c) Z = (160 Ω)2 + (87 Ω − 37.9 Ω)2 = 167 Ω. (d) im = (36 V)/(167 Ω) = 0.216 A. (e) tan φ = (87 Ω − 37.9 Ω)/(160 Ω) = 0.3069, so φ = arctan(0.3069) = 17◦ . E37-9 A circuit is considered inductive if XL > XC , this happens when im lags E m . If, on theother hand, XL < XC , and im leads E m , we refer to the circuit as capacitive. This is discussed onpage 850, although it is slightly hidden in the text of column one. (a) At resonance, XL = XC . Since XL = ωL and XC = 1/ωC we expect that XL grows withincreasing frequency, while XC decreases with increasing frequency. Consequently, at frequencies above the resonant frequency XL > XC and the circuit is predomi-nantly inductive. But what does this really mean? It means that the inductor plays a major role inthe current through the circuit while the capacitor plays a minor role. The more inductive a circuitis, the less significant any capacitance is on the behavior of the circuit. For frequencies below the resonant frequency the reverse is true. (b) Right at the resonant frequency the inductive effects are exactly canceled by the capacitiveeffects. The impedance is equal to the resistance, and it is (almost) as if neither the capacitor orinductor are even in the circuit.E37-10 The net y component is XC − XL . The net x component is R. The magnitude of theresultant is Z = R2 + (XC − XL )2 ,while the phase angle is −(XC − XL ) tan φ = . RE37-11 Yes. At resonance ω = 1/ (1.2 H)(1.3×10−6 F) = 800 rad/s and Z = R. Then im = E/Z =(10 V)/(9.6 Ω) = 1.04 A, so [∆VL ]m = im XL = (1.08 A)(800 rad/s)(1.2 H) = 1000 V. 152
• E37-12 (a) Let O = XL − XC and A = R, then H 2 = A2 + O2 = Z 2 , so sin φ = (XL − XC )/Zand cos φ = R/Z. E37-13 (a) The voltage across the generator is the generator emf, so when it is a maximum fromSample Problem 37-3, it is 36 V. This corresponds to ωt = π/2. (b) The current through the circuit is given by i = im sin(ωt − φ). We found in Sample Problem37-3 that im = 0.196 A and φ = −29.4◦ = 0.513 rad. For a resistive load we apply Eq. 37-3, ∆VR = im R sin(ωt − φ) = (0.196 A)(160Ω) sin((π/2) − (−0.513)) = 27.3 V. (c) For a capacitive load we apply Eq. 37-12, ∆VC = im XC sin(ωt − φ − π/2) = (0.196 A)(177Ω) sin(−(−0.513)) = 17.0 V. (d) For an inductive load we apply Eq. 37-7, ∆VL = im XL sin(ωt − φ + π/2) = (0.196 A)(87Ω) sin(π − (−0.513)) = −8.4 V. (e) (27.3 V) + (17.0 V) + (−8.4 V) = 35.9 V.E37-14 If circuit 1 and 2 have the same resonant frequency then L1 C1 = L2 C2 . The seriescombination for the inductors is L = L1 + L2 ,The series combination for the capacitors is 1/C = 1/C1 + 1/C2 ,so C1 C2 L1 C1 C2 + L2 C2 C1 LC = (L1 + L2 ) = = L1 C1 , C1 + C2 C1 + C2which is the same as both circuit 1 and 2.E37-15 (a) Z = (125 V)/(3.20 A) = 39.1 Ω. (b) Let O = XL − XC and A = R, then H 2 = A2 + O2 = Z 2 , so cos φ = R/Z.Using this relation, R = (39.1 Ω) cos(56.3◦ ) = 21.7 Ω. (c) If the current leads the emf then the circuit is capacitive.E37-16 (a) Integrating over a single cycle, T 1 1 T 1 sin2 ωt dt = (1 − cos 2ωt) dt, T 0 T 0 2 1 1 = T = . 2T 2 (b) Integrating over a single cycle, T T 1 1 1 sin ωt cos ωt dt = sin 2ωtdt, T 0 T 0 2 = 0. 153
• E37-17 The resistance would be given by Eq. 37-32, P av (0.10)(746 W) R= 2 = = 177 Ω. irms (0.650 A)2This would not be the same as the direct current resistance of the coils of a stopped motor, becausethere would be no inductive effects.E37-18 Since irms = E rms /Z, then E 2 rms R P av = i2 rms R = . Z2E37-19 (a) Z = (160 Ω)2 + (177 Ω)2 = 239 Ω; then 1 (36 V)2 (160 Ω) P av = = 1.82 W. 2 (239 Ω)2 (b) Z = (160 Ω)2 + (87 Ω)2 = 182 Ω; then 1 (36 V)2 (160 Ω) P av = = 3.13 W. 2 (182 Ω)2E37-20 (a) Z = (12.2 Ω)2 + (2.30 Ω)2 = 12.4 Ω (b) P av = (120 V)2 (12.2 Ω)/(12.4 Ω)2 = 1140 W. (c) irms = (120 V)/(12.4 Ω) = 9.67 A. √ E37-21 The rms value of any sinusoidal quantity is related to the maximum value by 2 v rms = √vmax . Since this factor of 2 appears in all of the expressions, we can conclude that if the rms valuesare equal then so are the maximum values. This means that (∆VR )max = (∆VC )max = (∆VL )maxor im R = im XC = im XL or, with one last simplification, R = XL = XC . Focus on the right handside of the last equality. If XC = XL then we have a resonance condition, and the impedance (seeEq. 37-20) is a minimum, and is equal to R. Then, according to Eq. 37-21, Em im = , Rwhich has the immediate consequence that the rms voltage across the resistor is the same as therms voltage across the generator. So everything is 100 V. √E37-22 (a) The antenna is “in-tune” when the impedance is a minimum, or ω = 1/ LC. So f = ω/2π = 1/2π (8.22×10−6 H)(0.270×10−12 F) = 1.07×108 Hz. (b) irms = (9.13 µV)/(74.7 Ω) = 1.22×10−7 A. (c) XC = 1/2πf C, so VC = iXC = (1.22×10−7 A)/2π(1.07×108 Hz)(0.270 ×10−12 F) = 6.72×10−4 V. 154
• E37-23 Assuming no inductors or capacitors in the circuit, then the circuit effectively behaves asa DC circuit. The current through the circuit is i = E/(r + R). The power delivered to R is thenP = i∆V = i2 R = E 2 R/(r + R)2 . Evaluate dP/dR and set it equal to zero to find the maximum.Then dP r−R 0= = E 2R , dR (r + R)3which has the solution r = R.E37-24 (a) Since P av = im 2 R/2 = E m 2 R/2Z 2 , then P av is a maximum when Z is a minimum, and √vise-versa. Z is a minimum at resonance, when Z = R and f = 1/2π LC. When Z is a minimum C = 1/4π 2 f 2 L = 1/4π 2 (60 Hz)2 (60 mH) = 1.2×10−7 F. (b) Z is a maximum when XC is a maximum, which occurs when C is very small, like zero. (c) When XC is a maximum P = 0. When P is a maximum Z = R so P = (30 V)2 /2(5.0 Ω) = 90 W. (d) The phase angle is zero for resonance; it is 90◦ for infinite XC or XL . (e) The power factor is zero for a system which has no power. The power factor is one for asystem in resonance.E37-25 (a) The resistance is R = 15.0 Ω. The inductive reactance is 1 1 XC = = −1 )(4.72µF) = 61.3 Ω. ωC 2π(550 sThe inductive reactance is given by XL = ωL = 2π(550 s−1 )(25.3 mH) = 87.4 Ω.The impedance is then 2 Z= (15.0 Ω)2 + ((87.4 Ω) − (61.3 Ω)) = 30.1 Ω.Finally, the rms current is E rms (75.0 V) irms = = = 2.49 A. Z (30.1 Ω) (b) The rms voltages between any two points is given by (∆V )rms = irms Z,where Z is not the impedance of the circuit but instead the impedance between the two points inquestion. When only one device is between the two points the impedance is equal to the reactance(or resistance) of that device. We’re not going to show all of the work here, but we will put together a nice table for you Points Impedance Expression Impedance Value (∆V )rms , ab Z=R Z = 15.0 Ω 37.4 V, bc Z = XC Z = 61.3 Ω 153 V, cd Z = XL Z = 87.4 Ω 218 V, bd Z = |XL − XC | Z = 26.1 Ω 65 V, ac Z = R 2 + XC 2 Z = 63.1 Ω 157 V,Note that this last one was ∆Vac , and not ∆Vad , because it is more entertaining. You probablyshould use ∆Vad for your homework. (c) The average power dissipated from a capacitor or inductor is zero; that of the resistor is PR = [(∆VR )rms ]2 /R = (37.4 V)2 /(15.0Ω) = 93.3 W. 155
• E37-26 (a) The energy stored in the capacitor is a function of the charge on the capacitor; althoughthe charge does vary with time it varies periodically and at the end of the cycle has returned to theoriginal values. As such, the energy stored in the capacitor doesn’t change from one period to thenext. (b) The energy stored in the inductor is a function of the current in the inductor; although thecurrent does vary with time it varies periodically and at the end of the cycle has returned to theoriginal values. As such, the energy stored in the inductor doesn’t change from one period to thenext. (c) P = Ei = E m im sin(ωt) sin(ωt − φ), so the energy generated in one cycle is T T U = P dt = E m im sin(ωt) sin(ωt − φ)dt, 0 0 T = E m im sin(ωt) sin(ωt − φ)dt, 0 T = E m im cos φ. 2 (d) P = im 2 R sin2 (ωt − φ), so the energy dissipated in one cycle is T T U = P dt = im 2 R sin2 (ωt − φ)dt, 0 0 T = im 2 R sin2 (ωt − φ)dt, 0 T 2 = im R. 2 (e) Since cos φ = R/Z and E m /Z = im we can equate the answers for (c) and (d).E37-27 Apply Eq. 37-41, Ns (780) ∆V s = ∆V p = (150 V) = 1.8×103 V. Np (65)E37-28 (a) Apply Eq. 37-41, Ns (10) ∆V s = ∆V p = (120 V) = 2.4 V. Np (500) (b) is = (2.4 V)/(15 Ω) = 0.16 A; Ns (10) ip = is = (0.16 A) = 3.2×10−3 A. Np (500) E37-29 The autotransformer could have a primary connected between taps T1 and T2 (200 turns),T1 and T3 (1000 turns), and T2 and T3 (800 turns). The same possibilities are true for the secondary connections. Ignoring the one-to-one connectionsthere are 6 choices— three are step up, and three are step down. The step up ratios are 1000/200 = 5,800/200 = 4, and 1000/800 = 1.25. The step down ratios are the reciprocals of these three values. 156
• E37-30 ρ = (1.69×10−8 Ω · m)[1 − (4.3×10−3 /C◦ )(14.6◦ C)] = 1.58×10−8 Ω · m. The resistance ofthe two wires is ρL (1.58×10−8 Ω · m)2(1.2×103 m) R= = = 14.9 Ω. A π(0.9×10−3 m)2 P = i2 R = (3.8 A)2 (14.9 Ω) = 220 W.E37-31 The supply current is √ ip = (0.270 A)(74×103 V/ 2)/(220 V) = 64.2 A.The potential drop across the supply lines is ∆V = (64.2 A)(0.62 Ω) = 40 V.This is the amount by which the supply voltage must be increased.E37-32 Use Eq. 37-46: N p /N s = (1000 Ω)/(10 Ω) = 10.P37-1 (a) The emf is a maximum when ωt − π/4 = π/2, so t = 3π/4ω = 3π/4(350 rad/s) =6.73×10−3 s. (b) The current is a maximum when ωt − 3π/4 = π/2, so t = 5π/4ω = 5π/4(350 rad/s) =1.12×10−2 s. (c) The current lags the emf, so the circuit contains an inductor. (d) XL = E m /im and XL = ωL, so Em (31.4 V) L= = = 0.144 H. im ω (0.622 A)(350 rad/s)P37-2 (a) The emf is a maximum when ωt − π/4 = π/2, so t = 3π/4ω = 3π/4(350 rad/s) =6.73×10−3 s. (b) The current is a maximum when ωt+π/4 = π/2, so t = π/4ω = π/4(350 rad/s) = 2.24×10−3 s. (c) The current leads the emf, so the circuit contains an capacitor. (d) XC = E m /im and XC = 1/ωC, so im (0.622 A) C= = = 5.66×10−5 F. E mω (31.4 V)(350 rad/s) P37-3 (a) Since the maximum values for the voltages across the individual devices are propor-tional to the reactances (or resistances) for devices in series (the constant of proportionality is themaximum current), we have XL = 2R and XC = R. From Eq. 37-18, XL − XC 2R − R tan φ = = = 1, R Ror φ = 45◦ . (b) The impedance of the circuit, in terms of the resistive element, is √ Z = R2 + (XL − XC )2 = R2 + (2R − R)2 = 2 R.But E m = im Z, so Z = (34.4 V)/(0.320 A) = 108Ω. Then we can use our previous work to finds thatR = 76Ω. 157
• P37-4 When the switch is open the circuit is an LRC circuit. In position 1 the circuit is an RLCcircuit, but the capacitance is equal to the two capacitors of C in parallel, or 2C. In position 2 thecircuit is a simple LC circuit with no resistance. The impedance when the switch is in position 2 is Z2 = |ωL − 1/ωC|. But Z2 = (170 V)/(2.82 A) = 60.3 Ω. The phase angle when the switch is open is φ0 = 20◦ . But ωL − 1/ωC Z2 tan φ0 = = , R Rso R = (60.3 Ω)/ tan(20◦ ) = 166 Ω. The phase angle when the switch is in position 1 is ωL − 1/ω2C tan φ1 = , Rso ωL − 1/ω2C = (166 Ω) tan(10◦ ) = 29.2 Ω. Equating the ωL part, (29.2 Ω) + 1/ω2C = (−60.3 Ω) + 1/ωC, C = 1/2(377 rad/s)[(60.3 Ω) + (29.2 Ω)] = 1.48×10−5 F. Finally, (−60.3Ω) + 1/(377 rad/s)(1.48×10−5 F) L= = 0.315 H. (377 rad/s) P37-5 All three wires have emfs which vary sinusoidally in time; if we choose any two wires thephase difference will have an absolute value of 120◦ . We can then choose any two wires and expect(by symmetry) to get the same result. We choose 1 and 2. The potential difference is then V1 − V2 = V m (sin ωt − sin(ωt − 120◦ )) .We need to add these two sine functions to get just one. We use 1 1 sin α − sin β = 2 sin (α − β) cos (α + β). 2 2Then 1 1 V1 − V 2 = 2V m sin (120◦ ) cos (2ωt − 120◦ ), √ 2 2 3 = 2V m ( ) cos(ωt − 60◦ ), √ 2 = 3V m sin(ωt + 30◦ ).P37-6 (a) cos φ = cos(−42◦ ) = 0.74. (b) The current leads. (c) The circuit is capacitive. (d) No. Resonance circuits have a power factor of one. (e) There must be at least a capacitor and a resistor. (f) P = (75 V)(1.2 A)(0.74)/2 = 33 W. 158
• √P37-7 (a) ω = 1/ LC = 1/ (0.988 H)(19.3×10−6 F) = 229 rad/s. (b) im = (31.3 V)/(5.12 Ω) = 6.11 A. (c) The current amplitude will be halved when the impedance is doubled, or when Z = 2R. Thisoccurs when 3R2 = (ωL − 1/ωC)2 , or 3R2 ω 2 = ω 4 L2 − 2ω 2 L/C + 1/C 2 .The solution to this quadratic is √ 2 2L + 3CR2 ± 9C 2 R4 + 12CR2 L ω = , 2L2 Cso ω1 = 224.6 rad/s and ω2 = 233.5 rad/s. (d) ∆ω/ω = (8.9 rad/s)/(229 rad/s) = 0.039.P37-8 (a) The current amplitude will be halved when the impedance is doubled, or when Z = 2R.This occurs when 3R2 = (ωL − 1/ωC)2 , or 3R2 ω 2 = ω 4 L2 − 2ω 2 L/C + 1/C 2 .The solution to this quadratic is √ 2 2L + 3CR2 ± 9C 2 R4 + 12CR2 L ω = , 2L2 CNote that ∆ω = ω+ − ω− ; with a wee bit of algebra, 2 2 ∆ω(ω+ + ω− ) = ω+ − ω− .Also, ω+ + ω− ≈ 2ω. Hence, √ 9C 2 R4 + 12CR2 L ω∆ω ≈ 2 , √ 2L C ω 2 R 9C 2 R2 + 12LC ω∆ω ≈ , √ 2L ωR 9ω 2 C 2 R2 + 12 ω∆ω ≈ , 2L ∆ω R 9CR2 /L + 12 ≈ , ω √ 2Lω 3R ≈ , ωLassuming that CR2 4L/3.P37-9P37-10 Use Eq. 37-46.P37-11 (a) The resistance of this bulb is (∆V )2 (120 V)2 R= = = 14.4 Ω. P (1000 W) 159
• The power is directly related to the brightness; if the bulb is to be varied in brightness by a factor of5 then it would have a minimum power of 200 W. The rms current through the bulb at this powerwould be irms = P/R = (200 W)/(14.4 Ω) = 3.73 A.The impedance of the circuit must have been E rms (120 V) Z= = = 32.2 Ω. irms (3.73 A)The inductive reactance would then be XL = Z 2 − R2 = (32.2 Ω)2 − (14.4 Ω)2 = 28.8 Ω.Finally, the inductance would be L = XL /ω = (28.8 Ω)/(2π(60.0 s−1 )) = 7.64 H. (b) One could use a variable resistor, and since it would be in series with the lamp a value of 32.2 Ω − 14.4Ω = 17.8 Ωwould work. But the resistor would get hot, while on average there is no power radiated from a pureinductor. 160
• E38-1 The maximum value occurs where r = R; there Bmax = 1 µ0 0 R dE/dt. For r < R B is 2half of Bmax when r = R/2. For r > R B is half of Bmax when r = 2R. Then the two values of rare 2.5 cm and 10.0 cm.E38-2 For a parallel plate capacitor E = σ/ 0 and the flux is then ΦE = σA/ 0 = q/ 0 . Then dΦE dq d dV id = 0 = = CV = C . dt dt dt dt E38-3 Use the results of Exercise 2, and change the potential difference across the plates of thecapacitor at a rate dV id (1.0 mA) = = = 1.0 kV/s. dt C (1.0µF)Provide a constant current to the capacitor dQ d dV i= = CV = C = id . dt dt dtE38-4 Since E is uniform between the plates ΦE = EA, regardless of the size of the region ofinterest. Since j d = id /A, id 1 dΦE dE jd = = 0 = 0 . A A dt dtE38-5 (a) In this case id = i = 1.84 A. (b) Since E = q/ 0 A, dE/dt = i/ 0 A, or dE/dt = (1.84 A)/(8.85×10−12 F/m)(1.22 m)2 = 1.40×1011 V/m. (c) id = 0 dΦE /dt = 0 adE/dt. a here refers to the area of the smaller square. Combine thiswith the results of part (b) and id = ia/A = (1.84 A)(0.61 m/1.22 m)2 = 0.46 A. (d) B · ds = µ0 id = (4π×10−7 H/m)(0.46 A) = 5.78×10−7 T · m.E38-6 Substitute Eq. 38-8 into the results obtained in Sample Problem 38-1. Outside the capacitorΦE = πR2 E, so µ0 0 πR2 dE µ0 B= = id . 2πr dt 2πrInside the capacitor the enclosed flux is ΦE = πr2 E; but we want instead to define id in terms ofthe total ΦE inside the capacitor as was done above. Consequently, inside the conductor 2 µ0 r 0 πR dE µ0 r B= = id . 2πR2 dt 2πR2 E38-7 Since the electric field is uniform in the area and perpendicular to the surface area wehave ΦE = E · dA = E dA = E dA = EA.The displacement current is then dE dE id = 0A = (8.85×10−12 F/m)(1.9 m2 ) . dt dt 161
• (a) In the first region the electric field decreases by 0.2 MV/m in 4µs, so (−0.2×106 V/m) id = (8.85×10−12 F/m)(1.9 m2 ) = −0.84 A. (4×10−6 s) (b) The electric field is constant so there is no change in the electric flux, and hence there is nodisplacement current. (c) In the last region the electric field decreases by 0.4 MV/m in 5µs, so (−0.4×106 V/m) id = (8.85×10−12 F/m)(1.9 m2 ) = −1.3 A. (5×10−6 s)E38-8 (a) Because of the circular symmetry B · ds = 2πrB, where r is the distance from thecenter of the circular plates. Not only that, but id = j d A = πr2 j d . Equate these two expressions,and B = µ0 rj d /2 = (4π×10−7 H/m)(0.053 m)(1.87×101 A/m)/2 = 6.23×10−7 T. (b) dE/dt = id / 0 A = j d / 0 = (1.87×101 A/m)/(8.85×10−12 F/m) = 2.11×10−12 V/m.E38-9 The magnitude of E is given by (162 V) E= sin 2π(60/s)t; (4.8×10−3 m)Using the results from Sample Problem 38-1, µ0 0 R dE Bm = , 2 dt t=0 (4π×10−7 H/m)(8.85×10−12 F/m)(0.0321 m) (162 V) = 2π(60/s) , 2 (4.8×10−3 m) = 2.27×10−12 T.E38-10 (a) Eq. 33-13 from page 764 and Eq. 33-34 from page 762. (b) Eq. 27-11 from page 618 and the equation from Ex. 27-25 on page 630. (c) The equations from Ex. 38-6 on page 876. (d) Eqs. 34-16 and 34-17 from page 785. E38-11 (a) Consider the path abef a. The closed line integral consists of two parts: b → e ande → f → a → b. Then dΦ E · ds = − dtcan be written as d E · ds + E · ds = − Φabef . b→e e→f →a→b dt Now consider the path bcdeb. The closed line integral consists of two parts: b → c → d → e ande → b. Then dΦ E · ds = − dtcan be written as d E · ds + E · ds = − Φbcde . b→c→d→e e→b dt 162
• These two expressions can be added together, and since E · ds = − E · ds e→b b→ewe get d E · ds + E · ds = − (Φabef + Φbcde ) . e→f →a→b b→c→d→e dtThe left hand side of this is just the line integral over the closed path ef adcde; the right hand sideis the net change in flux through the two surfaces. Then we can simplify this expression as dΦ E · ds = − . dt (b) Do everything above again, except substitute B for E. (c) If the equations were not self consistent we would arrive at different values of E and Bdepending on how we defined our surfaces. This multi-valued result would be quite unphysical.E38-12 (a) Consider the part on the left. It has a shared surface s, and the other surfaces l.Applying Eq. I, ql / 0 = E · dA = E · dA + E · dA. s lNote that dA is directed to the right on the shared surface. Consider the part on the right. It has a shared surface s, and the other surfaces r. Applying Eq.I, qr / 0 = E · dA = E · dA + E · dA. s rNote that dA is directed to the left on the shared surface. Adding these two expressions will result in a canceling out of the part E · dA ssince one is oriented opposite the other. We are left with qr + ql = E · dA + E · dA = E · dA. 0 r lE38-13E38-14 (a) Electric dipole is because the charges are separating like an electric dipole. Magneticdipole because the current loop acts like a magnetic dipole.E38-15 A series LC circuit will oscillate naturally at a frequency ω 1 f= = √ 2π 2π LCWe will need to combine this with v = f λ, where v = c is the speed of EM waves. We want to know the inductance required to produce an EM wave of wavelength λ = 550×10−9 m,so λ2 (550 × 10−9 m)2 L= = = 5.01 × 10−21 H. 4π 2 c2 C 4π 2 (3.00 × 108 m/s)2 (17 × 10−12 F)This is a small inductance! 163
• E38-16 (a) B = E/c, and B must be pointing in the negative y direction in order that the wavebe propagating in the positive x direction. Then Bx = Bz = 0, and By = −Ez /c = −(2.34×10−4 V/m)/(3.00×108 m/s) = (−7.80×10−13 T) sin k(x − ct). (b) λ = 2π/k = 2π/(9.72×106 /m) = 6.46×10−7 m. E38-17 The electric and magnetic field of an electromagnetic wave are related by Eqs. 38-15and 38-16, E (321µV/m) B= = = 1.07 pT. c (3.00 × 108 m/s)E38-18 Take the partial of Eq. 38-14 with respect to x, ∂ ∂E ∂ ∂B = − , ∂x ∂x ∂x ∂t ∂2E ∂2B = − . ∂x2 ∂x∂tTake the partial of Eq. 38-17 with respect to t, ∂ ∂B ∂ ∂E − = µ0 0 , ∂t ∂x ∂t ∂t ∂2B 2 ∂ E − = µ0 0 2 . ∂t∂x ∂tEquate, and let µ0 0 = 1/c2 , then ∂2E 1 ∂2E = 2 2. ∂x2 c ∂tRepeat, except now take the partial of Eq. 38-14 with respect to t, and then take the partial of Eq.38-17 with respect to x.E38-19 (a) Since sin(kx − ωt) is of the form f (kx ± ωt), then we only need do part (b). (b) The constant E m drops out of the wave equation, so we need only concern ourselves withf (kx ± ωt). Letting g = kx ± ωt, ∂2f ∂2f = c2 , ∂t2 ∂x2 2 2 ∂2f ∂g ∂ 2 f ∂g = c2 2 , ∂g 2 ∂t ∂g ∂x ∂g ∂g = c , ∂t ∂x ω = ck.E38-20 Use the right hand rule.E38-21 U = P t = (100×1012 W)(1.0×10−9 s) = 1.0×105 J.E38-22 E = Bc = (28×10−9 T)(3.0×108 m/s) = 8.4 V/m. It is in the positive x direction. 164
• E38-23 Intensity is given by Eq. 38-28, which is simply an expression of power divided by surfacearea. To find the intensity of the TV signal at α-Centauri we need to find the distance in meters; r = (4.30 light-years)(3.00×108 m/s)(3.15×107 s/year) = 4.06 × 1016 m.The intensity of the signal when it has arrived at out nearest neighbor is then P (960 kW) 2 I= = = 4.63 × 10−29 W/m 4πr2 4π(4.06 × 1016 m)2E38-24 (a) From Eq. 38-22, S = cB 2 /µ0 . B = B m sin ωt. The time average is defined as T T 1 cB m 2 cB m 2 S dt = cos2 ωt dt = . T 0 µ0 T 0 2µ0 (b) S av = (3.0×108 m/s)(1.0×10−4 T)2 /2(4π×10−7 H/m) = 1.2×106 W/m2 .E38-25 I = P/4πr2 , so r= P/4πI = (1.0×103 W)/4π(130 W/m2 ) = 0.78 m. 2 2E38-26 uE = 0E /2 = 0 (cB) /2 = B 2 /2µ0 = uB . E38-27 (a) Intensity is related to distance by Eq. 38-28. If r1 is the original distance from thestreet lamp and I1 the intensity at that distance, then P I1 = 2. 4πr1There is a similar expression for the closer distance r2 = r1 −162 m and the intensity at that distanceI2 = 1.50I1 . We can combine the two expression for intensity, I2 = 1.50I1 , P P 2 = 1.50 2, 4πr2 4πr1 2 r1 = 1.50r2 , √ 2 r1 = 1.50 (r1 − 162 m).The last line is easy enough to solve and we find r1 = 883 m. (b) No, we can’t find the power output from the lamp, because we were never provided with anabsolute intensity reference. √E38-28 (a) E m = 2µ0 cI, so Em = 2(4π×10−7 H/m)(3.00×108 m/s)(1.38×103 W/m2 ) = 1.02×103 V/m. (b) B m = E m /c = (1.02×103 V/m)/(3.00×108 m/s) = 3.40×10−6 T.E38-29 (a) B m = E m /c = (1.96 V/m)/(3.00×108 m/s) = 6.53×10−9 T. (b) I = E m 2 /2µ0 c = (1.96 V)2 /2(4π×10−7 H/m)(3.00×108 m/s) = 5.10×10−3 W/m2 . (c) P = 4πr2 I = 4π(11.2 m)2 (5.10×10−3 W/m2 ) = 8.04 W. 165
• E38-30 (a) The intensity is P (1×10−12 W) I= = = 1.96×10−27 W/m2 . A 4π(6.37×106 m)2The power received by the Arecibo antenna is P = IA = (1.96×10−27 W/m2 )π(305 m)2 /4 = 1.4×10−22 W. (b) The power of the transmitter at the center of the galaxy would be P = IA = (1.96×10−27 W)π(2.3×104 ly)2 (9.46×1015 m/ly)2 = 2.9×1014 W.E38-31 (a) The electric field amplitude is related to the intensity by Eq. 38-26, E2m I= , 2µ0 cor Em = 2µ0 cI = 2(4π×10−7 H/m)(3.00×108 m/s)(7.83µW/m2 ) = 7.68×10−2 V/m. (b) The magnetic field amplitude is given by Em (7.68 × 10−2 V/m) Bm = = = 2.56 × 10−10 T c (3.00 × 108 m/s) (c) The power radiated by the transmitter can be found from Eq. 38-28, P = 4πr2 I = 4π(11.3 km)2 (7.83µW/m2 ) = 12.6 kW.E38-32 (a) The power incident on (and then reflected by) the target craft is P1 = I1 A = P0 A/2πr2 .The intensity of the reflected beam is I2 = P1 /2πr2 = P0 A/4π 2 r4 . Then I2 = (183×103 W)(0.222 m2 )/4π 2 (88.2×103 m)4 = 1.70×10−17 W/m2 . (b) Use Eq. 38-26: Em = 2µ0 cI = 2(4π×10−7 H/m)(3.00×108 m/s)(1.70×10−17 W/m2 ) = 1.13×10−7 V/m. √ √ (c) B rms = E m / 2c = (1.13×10−7 V/m)/ 2(3.00×108 m/s) = 2.66×10−16 T. E38-33 Radiation pressure for absorption is given by Eq. 38-34, but we need to find the energyabsorbed before we can apply that. We are given an intensity, a surface area, and a time, so ∆U = (1.1×103 W/m2 )(1.3 m2 )(9.0×103 s) = 1.3×107 J.The momentum delivered is p = (∆U )/c = (1.3×107 J)/(3.00×108 m/s) = 4.3×10−2 kg · m/s.E38-34 (a) F/A = I/c = (1.38×103 W/m2 )/(3.00×108 m/s) = 4.60×10−6 Pa. (b) (4.60×10−6 Pa)/(101×105 Pa) = 4.55×10−11 .E38-35 F/A = 2P/Ac = 2(1.5×109 W)/(1.3×10−6 m2 )(3.0×108 m/s) = 7.7×106 Pa. 166
• E38-36 F/A = P/4πr2 c, so F/A = (500 W)/4π(1.50 m)2 (3.00×108 m/s) = 5.89×10−8 Pa.E38-37 (a) F = IA/c, so (1.38×103 W/m2 )π(6.37×106 m)2 F = = 5.86×108 N. (3.00×108 m/s)E38-38 (a) Assuming MKSA, the units are mFV N m C V sN Ns = = 2 . s m m Am s Vm m Cm m s (b) Assuming MKSA, the units are A2 V N A2 J N 1 J J = = = 2 . N m Am N Cm Am sm m m s E38-39 We can treat the object as having two surfaces, one completely reflecting and the othercompletely absorbing. If the entire surface has an area A then the absorbing part has an area f Awhile the reflecting part has area (1 − f )A. The average force is then the sum of the force on eachpart, I 2I F av = f A + (1 − f )A, c cwhich can be written in terms of pressure as F av I = (2 − f ). A cE38-40 We can treat the object as having two surfaces, one completely reflecting and the othercompletely absorbing. If the entire surface has an area A then the absorbing part has an area f Awhile the reflecting part has area (1 − f )A. The average force is then the sum of the force on eachpart, I 2I F av = f A + (1 − f )A, c cwhich can be written in terms of pressure as F av I = (2 − f ). A cThe intensity I is that of the incident beam; the reflected beam will have an intensity (1 − f )I.Each beam will contribute to the energy density— I/c and (1 − f )I/c, respectively. Add these twoenergy densities to get the net energy density outside the surface. The result is (2 − f )I/c, which isthe left hand side of the pressure relation above.E38-41 The bullet density is ρ = N m/V . Let V = Ah; the kinetic energy density is K/V =1 22 N mv /Ah. h/v, however, is the time taken for N balls to strike the surface, so that F N mv N mv 2 2K P = = = = . A At Ah V 167
• E38-42 F = IA/c; P = IA; a = F/m; and v = at. Combine: v = P t/mc = (10×103 W)(86400 s)/(1500 kg)(3×108 m/s) = 1.9×10−3 m/s.E38-43 The force of radiation on the bottom of the cylinder is F = 2IA/c. The force of gravityon the cylinder is W = mg = ρHAg.Equating, 2I/c = ρHg. The intensity of the beam is given by I = 4P/πd2 . Solving for H, 8P 8(4.6 W) H= 2 = 8 m/s)(1200 kg/m3 )(9.8 m/s2 )(2.6×10−3 m)2 = 4.9×10−7 m. πcρgd π(3.0×10E38-44 F = 2IA/c. The value for I is in Ex. 38-37, among other places. Then F = (1.38×103 W/m2 )(3.1×106 m2 )/(3.00×108 m/s) = 29 N.P38-1 For the two outer circles use Eq. 33-13. For the inner circle use E = V /d, Q = CV ,C = 0 A/d, and i = dQ/dt. Then dQ 0 A dV dE i= = = 0A . dt d dt dtThe change in flux is dΦE /dt = A dE/dt. Then dΦE B · dl = µ0 0 = µ0 i, dtso B = µ0 i/2πr.P38-2 (a) id = i. Assuming ∆V = (174×103 V) sin ωt, then q = C∆V and i = dq/dt = Cd(∆V )/dt.Combine, and use ω = 2π(50.0/s), id = (100×10−12 F)(174×103 V)2π(50.0/s) = 5.47×10−3 A.P38-3 (a) i = id = 7.63µA. (b) dΦE /dt = id / 0 = (7.63µA)/(8.85×10−12 F/m) = 8.62×105 V/m. (c) i = dq/dt = Cd(∆V )/dt; C = 0 A/d; [d(∆V )/dt]m = E m ω. Combine, and 0A 0 AE m ω (8.85×10−12 F/m)π(0.182 m)2 (225 V)(128 rad/s) d= = = = 3.48×10−3 m. C i (7.63µA)P38-4 (a) q = i dt = α t dt = αt2 /2. (b) E = σ/ 0 = q/ 0 A = αt2 /2πR2 0 . (d) 2πrB = µ0 0 πr2 dE/dt, so B = µ0 r(dE/dt)/2 = µ0 αrt/2πR2 . (e) Check Exercise 38-10!P38-5 (a) E = Eˆ and B = B k. Then S = E × B/µ0 , or j ˆ S = −EB/mu0 ˆ i.Energy only passes through the yz faces; it goes in one face and out the other. The rate is P =SA = EBa2 /mu0 . (b) The net change is zero. 168
• P38-6 (a) For a sinusoidal time dependence |dE/dt|m = ωE m = 2πf E m . Then |dE/dt|m = 2π(2.4×109 /s)(13×103 V/m) = 1.96×1014 V/m · s. (b) Using the result of part (b) of Sample Problem 38-1, 1 1 B= (4π×10−7 H/m)(8.9×10−12 F/m)(2.4×10−2 m) (1.96×1014 V/m · s) = 1.3×10−5 T. 2 2 P38-7 Look back to Chapter 14 for a discussion on the elliptic orbit. On page 312 it is pointedout that the closest distance to the sun is Rp = a(1 − e) while the farthest distance is Ra = a(1 + e),where a is the semi-major axis and e the eccentricity. The fractional variation in intensity is ∆I Ip − Ia ≈ , I Ia Ip = − 1, Ia Ra 2 = − 1, Rp 2 (1 + e)2 = − 1. (1 − e)2We need to expand this expression for small e using (1 + e)2 ≈ 1 + 2e, and (1 − e)−2 ≈ 1 + 2e, andfinally (1 + 2e)2 ≈ 1 + 4e. Combining, ∆I ≈ (1 + 2e)2 − 1 ≈ 4e. IP38-8 The beam radius grows as r = (0.440 µrad)R, where R is the distance from the origin. Thebeam intensity is P (3850 W) I= 2 = = 4.3×10−2 W. πr π(0.440 µrad)2 (3.82×108 m)2P38-9 Eq. 38-14 requires ∂E ∂B = − , ∂x ∂t E m k cos kx sin ωt = B m ω cos kx sin ωt, Emk = B m ω. Eq. 38-17 requires ∂E ∂B µ0 0 = − , ∂t ∂x µ0 0 E m ω sin kx cos ωt = B m k sin kx cos ωt, µ0 0 E m ω = B m k. Dividing one expression by the other, µ0 0 k 2 = ω 2 , 169
• or ω 1 =c= √ k µ0 0. Not only that, but E m = cB m . You’ve seen an expression similar to this before, and you’ll seeexpressions similar to it again. (b) We’ll assume that Eq. 38-21 is applicable here. Then 1 EmBm S = = sin kx sin ωt cos kx cos ωt, µ0 µ0 E2 m = sin 2kx sin 2ωt 4µ0 cis the magnitude of the instantaneous Poynting vector. (c) The time averaged power flow across any surface is the value of T 1 S · dA dt, T 0where T is the period of the oscillation. We’ll just gloss over any concerns about direction, andassume that the S will be constant in direction so that we will, at most, need to concern ourselvesabout a constant factor cos θ. We can then deal with a scalar, instead of vector, integral, and wecan integrate it in any order we want. We want to do the t integration first, because an integral oversin ωt for a period T = 2π/ω is zero. Then we are done! (d) There is no energy flow; the energy remains inside the container.P38-10 (a) The electric field is parallel to the wire and given by E = V /d = iR/d = (25.0 A)(1.00 Ω/300 m) = 8.33×10−2 V/m (b) The magnetic field is in rings around the wire. Using Eq. 33-13, µ0 i (4π×10−7 H/m)(25 A) B= = = 4.03×10−3 T. 2πr 2π(1.24×10−3 m) (c) S = EB/µ0 , so S = (8.33×10−2 V/m)(4.03×10−3 T)/(4π×10−7 H/m) = 267 W/m2 . P38-11 (a) We’ve already calculated B previously. It is µ0 i E B= where i = . 2πr RThe electric field of a long straight wire has the form E = k/r, where k is some constant. But b ∆V = − E · ds = − E dr = −k ln(b/a). aIn this problem the inner conductor is at the higher potential, so −∆V E k= = , ln(b/a) ln(b/a) 170
• and then the electric field is E E= . r ln(b/a)This is also a vector field, and if E is positive the electric field points radially out from the centralconductor. (b) The Poynting vector is 1 S= E × B; µ0E is radial while B is circular, so they are perpendicular. Assuming that E is positive the directionof S is away from the battery. Switching the sign of E (connecting the battery in reverse) will flipthe direction of both E and B, so S will pick up two negative signs and therefore still point awayfrom the battery. The magnitude is EB E2 S= = µ0 2πR ln(b/a)r2 (c) We want to evaluate a surface integral in polar coordinates and so dA = (dr)(rdθ). We havealready established that S is pointing away from the battery parallel to the central axis. Then wecan integrate P = S · dA = S dA, b 2π E2 = dθ r dr, a 0 2πR ln(b/a)r2 b E2 = dr, a R ln(b/a)r E2 = . R (d) Read part (b) above.P38-12 (a) B is oriented as rings around the cylinder. If the thumb is in the direction of currentthen the fingers of the right hand grip ion the direction of the magnetic field lines. E is directedparallel to the wire in the direction of the current. S is found from the cross product of these two,and must be pointing radially inward. (b) The magnetic field on the surface is given by Eq. 33-13: B = µ0 i/2πa.The electric field on the surface is given by E = V /l = iR/lThen S has magnitude i iR i2 R S = EB/µ0 = = . 2πa l 2πal S · dA is only evaluated on the surface of the cylinder, not the end caps. S is everywhere parallelto dA, so the dot product reduces to S dA; S is uniform, so it can be brought out of the integral; dA = 2πal on the surface. Hence, S · dA = i2 R,as it should. 171
• P38-13 (a) f = vlambda = (3.00×108 m/s)/(3.18 m) = 9.43×107 Hz. (b) Bmust be directed along the z axis. The magnitude is B = E/c = (288 V/m)/(3.00×108 m/s) = 9.6×10−7 T. (c) k = 2π/λ = 2π/(3.18 m) = 1.98/m while ω = 2πf , so ω = 2π(9.43×107 Hz) = 5.93×108 rad/s. (d) I = E m B m /2µ0 , so (288 V)(9.6×10−7 T) I= = 110 W. 2(4π×10−7 H/m) (e) P = I/c = (110 W)/(3.00×108 m/s) = 3.67×10−7 Pa.P38-14 (a) B is oriented as rings around the cylinder. If the thumb is in the direction of currentthen the fingers of the right hand grip ion the direction of the magnetic field lines. E is directedparallel to the wire in the direction of the current. S is found from the cross product of these two,and must be pointing radially inward. (b) The magnitude of the electric field is V Q Q it E= = = = . d Cd 0A 0AThe magnitude of the magnetic field on the outside of the plates is given by Sample Problem 38-1, µ0 0 R dE µ0 0 iR µ0 0 R B= = = E. 2 dt 2 0A 2tS has magnitude EB 0R 2 S= = E . µ0 2tIntegrating, 2 0R 0E S · dA = E 2 2πRd = Ad . 2t tBut E is linear in t, so d(E 2 )/dt = 2E 2 /t; and then d 1 2 S · dA = Ad 0E . dt 2P38-15 (a) I = P/A = (5.00×10−3 W)/π(1.05)2 (633×10−9 m)2 = 3.6×109 W/m2 . (b) p = I/c = (3.6×109 W/m2 )/(3.00×108 m/s) = 12 Pa (c) F = pA = P/c = (5.00×10−3 W)/(3.00×108 m/s) = 1.67×10−11 N. (d) a = F/m = F/ρV , so (1.67×10−11 N) a= = 2.9×103 m/s2 . 4(4880 kg/m3 )(1.05)3 (633×10−9 )3 /3 172
• P38-16 The force from the sun is F = GM m/r2 . The force from radiation pressure is 2IA 2P A F = = . c 4πr2 cEquating, 4πGM m A= , 2P/cso 4π(6.67×10−11 N · m2 /kg2 )(1.99×1030 kg)(1650 kg) A= = 1.06×106 m2 . 2(3.9×1026 W)/(3.0×108 m/s)That’s about one square kilometer. 173
• E39-1 Both scales are logarithmic; choose any data point from the right hand side such as c = f λ ≈ (1 Hz)(3×108 m) = 3×108 m/s,and another from the left hand side such as c = f λ ≈ (1×1021 Hz)(3×10−13 m) = 3×108 m/s.E39-2 (a) f = v/λ = (3.0×108 m/s)/(1.0×104 )(6.37×106 m) = 4.7×10−3 Hz. If we assume thatthis is the data transmission rate in bits per second (a generous assumption), then it would take 140days to download a web-page which would take only 1 second on a 56K modem! (b) T = 1/f = 212 s = 3.5 min.E39-3 (a) Apply v = f λ. Then f = (3.0×108 m/s)/(0.067×10−15 m) = 4.5×1024 Hz. (b) λ = (3.0×108 m/s)/(30 Hz) = 1.0×107 m.E39-4 Don’t simply take reciprocal of linewidth! f = c/λ, so δf = (−c/λ2 )δλ. Ignore the negative,and δf = (3.00×108 m/s)(0.010×10−9 m)/(632.8×10−9 m)2 = 7.5×109 Hz. E39-5 (a) We refer to Fig. 39-6 to answer this question. The limits are approximately 520 nmand 620 nm. (b) The wavelength for which the eye is most sensitive is 550 nm. This corresponds to to afrequency of f = c/λ = (3.00 × 108 m/s)/(550 × 10−9 m) = 5.45 × 1014 Hz.This frequency corresponds to a period of T = 1/f = 1.83 × 10−15 s.E39-6 f = c/λ. The number of complete pulses is f t, or f t = ct/λ = (3.00×108 m/s)(430×10−12 s)/(520×10−9 m) = 2.48×105 .E39-7 (a) 2(4.34 y) = 8.68 y. (b) 2(2.2×106 y) = 4.4×106 y.E39-8 (a) t = (150×103 m)/(3×108 m/s) = 5×10−4 s. (b) The distance traveled by the light is (1.5×1011 m) + 2(3.8×108 m), so t = (1.51×1011 m)/(3×108 m/s) = 503 s. (c) t = 2(1.3×1012 m)/(3×108 m/s) = 8670 s. (d) 1054 − 6500 ≈ 5400 BC. E39-9 This is a question of how much time it takes light to travel 4 cm, because the light traveledfrom the Earth to the moon, bounced off of the reflector, and then traveled back. The time to travel4 cm is ∆t = (0.04 m)/(3 × 108 m/s) = 0.13 ns. Note that I interpreted the question differently thanthe answer in the back of the book. 174
• E39-10 Consider any incoming ray. The path of the ray can be projected onto the xy plane, thexz plane, or the yz plane. If the projected rays is exactly reflected in all three cases then the threedimensional incoming ray will be reflected exactly reversed. But the problem is symmetric, so it issufficient to show that any plane works. Now the problem has been reduced to Sample Problem 39-2, so we are done. E39-11 We will choose the mirror to lie in the xy plane at z = 0. There is no loss of generalityin doing so; we had to define our coordinate system somehow. The choice is convenient in thatany normal is then parallel to the z axis. Furthermore, we can arbitrarily define the incident rayto originate at (0, 0, z1 ). Lastly, we can rotate the coordinate system about the z axis so that thereflected ray passes through the point (0, y3 , z3 ). The point of reflection for this ray is somewhere on the surface of the mirror, say (x2 , y2 , 0). Thisdistance traveled from the point 1 to the reflection point 2 is d12 = (0 − x2 )2 + (0 − y2 )2 + (z1 − 0)2 = x2 + y2 + z1 2 2 2and the distance traveled from the reflection point 2 to the final point 3 is d23 = (x2 − 0)2 + (y2 − y3 )2 + (0 − z3 )2 = x2 + (y2 − y3 )2 + z3 . 2 2 The only point which is free to move is the reflection point, (x2 , y2 , 0), and that point can onlymove in the xy plane. Fermat’s principle states that the reflection point will be such to minimizethe total distance, d12 + d23 = x2 + y2 + z1 + 2 2 2 x2 + (y2 − y3 )2 + z3 . 2 2We do this minimization by taking the partial derivative with respect to both x2 and y2 . Butwe can do part by inspection alone. Any non-zero value of x2 can only add to the total distance,regardless of the value of any of the other quantities. Consequently, x2 = 0 is one of the conditionsfor minimization. We are done! Although you are invited to finish the minimization process, once we know thatx2 = 0 we have that point 1, point 2, and point 3 all lie in the yz plane. The normal is parallel tothe z axis, so it also lies in the yz plane. Everything is then in the yz plane.E39-12 Refer to Page 442 of Volume 1.E39-13 (a) θ1 = 38◦ . (b) (1.58) sin(38◦ ) = (1.22) sin θ2 . Then θ2 = arcsin(0.797) = 52.9◦ .E39-14 ng = nv sin θ1 / sin θ2 = (1.00) sin(32.5◦ )/ sin(21.0◦ ) = 1.50.E39-15 n = c/v = (3.00×108 m/s)/(1.92×108 m/s) = 1.56.E39-16 v = c/n = (3.00×108 m/s)/(1.46) = 2.05×108 m/s. E39-17 The speed of light in a substance with index of refraction n is given by v = c/n. Anelectron will then emit Cerenkov radiation in this particular liquid if the speed exceeds v = c/n = (3.00 × 108 m/s)/(1.54) = 1.95×108 m/s. 175
• E39-18 Since t = d/v = nd/c, ∆t = ∆n d/c. Then ∆t = (1.00029 − 1.00000)(1.61×103 m)/(3.00×108 m/s) = 1.56×10−9 s. E39-19 The angle of the refracted ray is θ2 = 90◦ , the angle of the incident ray can be found bytrigonometry, (1.14 m) tan θ1 = = 1.34, (0.85 m)or θ1 = 53.3◦ . We can use these two angles, along with the index of refraction of air, to find that the index ofrefraction of the liquid from Eq. 39-4, sin θ2 (sin 90◦ ) n1 = n2 = (1.00) = 1.25. sin θ1 (sin 53.3◦ )There are no units attached to this quantity.E39-20 For an equilateral prism φ = 60◦ . Then sin[ψ + φ]/2 sin[(37◦ ) + (60◦ )]/2 n= = = 1.5. sin[φ/2] sin[(60◦ )/2]E39-21E39-22 t = d/v; but L/d = cos θ2 = 1 − sin2 θ2 and v = c/n. Combining, nL n2 L (1.63)2 (0.547 m) t= = = = 3.07×10−9 s. 2 2 c 1 − sin θ2 c n2 − sin θ1 (3×108 m/s) 2 (1.632 ) − sin (24◦ )E39-23 The ray of light from the top of the smokestack to the life ring is θ1 , where tan θ1 = L/hwith h the height and L the distance of the smokestack. Snell’s law gives n1 sin θ1 = n2 sin θ2 , so θ1 = arcsin[(1.33) sin(27◦ )/(1.00)] = 37.1◦ .Then L = h tan θ1 = (98 m) tan(37.1◦ ) = 74 m.E39-24 The length of the shadow on the surface of the water is x1 = (0.64 m)/ tan(55◦ ) = 0.448 m.The ray of light which forms the “end” of the shadow has an angle of incidence of 35◦ , so the raytravels into the water at an angle of (1.00) θ2 = arcsin sin(35◦ ) = 25.5◦ . (1.33)The ray travels an additional distance x2 = (2.00 m − 0.64 m)/ tan(90◦ − 25.5◦ ) = 0.649 mThe total length of the shadow is (0.448 m) + (0.649 m) = 1.10 m. 176
• E39-25 We’ll rely heavily on the figure for our arguments. Let x be the distance between thepoints on the surface where the vertical ray crosses and the bent ray crosses. θ2 x d app θ1 d In this exercise we will take advantage of the fact that, for small angles θ, sin θ ≈ tan θ ≈ θ Inthis approximation Snell’s law takes on the particularly simple form n1 θ1 = n2 θ2 The two angleshere are conveniently found from the figure, x θ1 ≈ tan θ1 = , dand x θ2 ≈ tan θ2 = . dappInserting these two angles into the simplified Snell’s law, as well as substituting n1 = n and n2 = 1.0, n1 θ 1 = n2 θ 2 , x x n = , d dapp d dapp = . nE39-26 (a) You need to address the issue of total internal reflection to answer this question. (b) Rearrange sin[ψ + φ]/2 n= sin[φ/2] /and θ = (ψ + φ)/2 to get θ = arcsin (n sin[φ/2]) = arcsin ((1.60) sin[(60◦ )/2]) = 53.1◦ .E39-27 Use the results of Ex. 39-35. The apparent thickness of the carbon tetrachloride layer, asviewed by an observer in the water, is dc,w = nw dc /nc = (1.33)(41 mm)/(1.46) = 37.5 mm.The total “thickness” from the water perspective is then (37.5 mm) + (20 mm) = 57.5 mm. Theapparent thickness of the entire system as view from the air is then dapp = (57.5 mm)/(1.33) = 43.2 mm. 177
• E39-28 (a) Use the results of Ex. 39-35. dapp = (2.16 m)/(1.33) = 1.62 m. (b) Need a diagram here!E39-29 (a) λn = λ/n = (612 nm)/(1.51) = 405 nm. (b) L = nLn = (1.51)(1.57 pm) = 2.37 pm. There is actually a typo: the “p” in “pm” wassupposed to be a µ. This makes a huge difference for part (c)!E39-30 (a) f = c/λ = (3.00×108 m/s)/(589 nm) = 5.09×1014 Hz. (b) λn = λ/n = (589 nm)/(1.53) = 385 nm. (c) v = f λ = (5.09×1014 Hz)(385 nm) = 1.96×108 m/s.E39-31 (a) The second derivative of L= a2 + x2 + b2 + (d − x)2is a2 (b2 + (d − 2)2 )3/2 + b2 (a2 + x2 )3/2 . (b2 + (d − 2)2 )3/2 (a2 + x2 )3/2This is always a positive number, so dL/dx = 0 is a minimum. (a) The second derivative of L = n1 a2 + x2 + n2 b2 + (d − x)2is n1 a2 (b2 + (d − 2)2 )3/2 + n2 b2 (a2 + x2 )3/2 . (b2 + (d − 2)2 )3/2 (a2 + x2 )3/2This is always a positive number, so dL/dx = 0 is a minimum.E39-32 (a) The angle of incidence on the face ac will be 90◦ − φ. Total internal reflection occurswhen sin(90◦ − φ) > 1/n, or φ < 90◦ − arcsin[1/(1.52)] = 48.9◦ . (b) Total internal reflection occurs when sin(90◦ − φ) > nw /n, or φ < 90◦ − arcsin[(1.33)/(1.52)] = 29.0◦ . E39-33 (a) The critical angle is given by Eq. 39-17, n2 (1.586) θc = sin−1 = sin−1 = 72.07◦ . n1 (1.667) (b) Critical angles only exist when “attempting” to travel from a medium of higher index ofrefraction to a medium of lower index of refraction; in this case from A to B.E39-34 If the fire is at the water’s edge then the light travels along the surface, entering thewater near the fish with an angle of incidence of effectively 90◦ . Then the angle of refraction inthe water is numerically equivalent to a critical angle, so the fish needs to look up at an angle ofθ = arcsin(1/1.33) = 49◦ with the vertical. That’s the same as 41◦ with the horizontal. 178
• E39-35 Light can only emerge from the water if it has an angle of incidence less than the criticalangle, or θ < θc = arcsin 1/n = arcsin 1/(1.33) = 48.8◦ .The radius of the circle of light is given by r/d = tan θc , where d is the depth. The diameter is twicethis radius, or 2(0.82 m) tan(48.8◦ ) = 1.87 m.E39-36 The refracted angle is given by n sin θ1 = sin(39◦ ). This ray strikes the left surface withan angle of incidence of 90◦ − θ1 . Total internal reflection occurs when sin(90◦ − θ1 ) = 1/n;but sin(90◦ − θ1 ) = cos θ1 , so we can combine and get tan θ = sin(39◦ ) with solution θ1 = 32.2◦ . Theindex of refraction of the glass is then n = sin(39◦ )/ sin(32.2) = 1.18. E39-37 The light strikes the quartz-air interface from the inside; it is originally “white”, so ifthe reflected ray is to appear “bluish” (reddish) then the refracted ray should have been “reddish”(bluish). Since part of the light undergoes total internal reflection while the other part does not,then the angle of incidence must be approximately equal to the critical angle. (a) Look at Fig. 39-11, the index of refraction of fused quartz is given as a function of thewavelength. As the wavelength increases the index of refraction decreases. The critical angle is afunction of the index of refraction; for a substance in air the critical angle is given by sin θc = 1/n.As n decreases 1/n increases so θc increases. For fused quartz, then, as wavelength increases θc alsoincreases. In short, red light has a larger critical angle than blue light. If the angle of incidence is midwaybetween the critical angle of red and the critical angle of blue, then the blue component of the lightwill experience total internal reflection while the red component will pass through as a refracted ray. So yes, the light can be made to appear bluish. (b) No, the light can’t be made to appear reddish. See above. (c) Choose an angle of incidence between the two critical angles as described in part (a). Usinga value of n = 1.46 from Fig. 39-11, θc = sin−1 (1/1.46) = 43.2◦ .Getting the effect to work will require considerable sensitivity.E39-38 (a) There needs to be an opaque spot in the center of each face so that no refracted rayemerges. The radius of the spot will be large enough to cover rays which meet the surface at lessthan the critical angle. This means tan θc = r/d, where d is the distance from the surface to thespot, or 6.3 mm. Since θc = arcsin 1/(1.52) = 41.1◦ ,then r = (6.3 mm) tan(41.1◦ ) = 5.50 mm. (b) The circles have an area of a = π(5.50 mm)2 = 95.0 mm2 . Each side has an area of (12.6 mm)2 ;the fraction covered is then (95.0 mm2 )/(12.6 mm)2 = 0.598.E39-39 For u c the relativistic Doppler shift simplifies to ∆f = −f0 u/c = −u/λ0 ,so u = λ0 ∆f = (0.211 m)∆f. 179
• E39-40 c = f λ, so 0 = f ∆λ + λ∆f . Then ∆λ/λ = −∆f /f . Furthermore, f0 − f , from Eq. 39-21,is f0 u/c for small enough u. Then ∆λ f − f0 u =− = . λ f0 cE39-41 The Doppler theory for light gives 1 − u/c 1 − (0.2) f = f0 = f0 = 0.82 f0 . 1− u2 /c2 1 − (0.2)2The frequency is shifted down to about 80%, which means the wavelength is shifted up by anadditional 25%. Blue light (480 nm) would appear yellow/orange (585 nm).E39-42 Use Eq. 39-20: 1 − u/c 1 − (0.892) f = f0 = (100 Mhz) = 23.9 MHz. 1− u2 /c2 1 − (0.892)2E39-43 (a) If the wavelength is three times longer then the frequency is one-third, so for theclassical Doppler shift f0 /3 = f0 (1 − u/c),or u = 2c. (b) For the relativistic shift, 1 − u/c f0 /3 = f0 , 1 − u2 /c2 1 − u2 /c2 = 3(1 − u/c), c2 − u2 = 9(c − u)2 , 0 = 10u2 − 18uc + 8c2 .The solution is u = 4c/5.E39-44 (a) f0 /f = λ/λ0 . This shift is small, so we apply the approximation: λ0 (462 nm) u=c −1 = (3×108 m/s) −1 = 1.9×107 m/s. λ (434 nm) (b) A red shift corresponds to objects moving away from us. E39-45 The sun rotates once every 26 days at the equator, while the radius is 7.0×108 m. Thespeed of a point on the equator is then 2πR 2π(7.0×108 m) v= = = 2.0×103 m/s. T (2.2×106 s)This corresponds to a velocity parameter of β = u/c = (2.0×103 m/s)/(3.0×108 m/s) = 6.7×10−6 .This is a case of small numbers, so we’ll use the formula that you derived in Exercise 39-40: ∆λ = λβ = (553 nm)(6.7×10−6 ) = 3.7×10−3 nm. 180
• E39-46 Use Eq. 39-23 written as (1 − u/c)λ2 = λ2 (1 + u/c), 0which can be rearranged as λ2 − λ20 (540 nm)2 − (620 nm)2 u/c = 2 + λ2 = = −0.137. λ 0 (540 nm)2 + (620 nm)2The negative sign means that you should be going toward the red light.E39-47 (a) f1 = cf /(c + v) and f2 = cf /(c − v). ∆f = (f2 − f ) − (f − f1 ) = f2 + f1 − 2f,so ∆f c c = + − 2, f c+v c−v 2v 2 = , c2 − v 2 2(8.65×105 m/s)2 = , (3.00×108 m/s)2 − (8.65×105 m/s)2 = 1.66×10−5 . √ (b) f1 = f (c − u)/sqrtc2 − u2 and f2 = f (c + u)/ c2 − u2 . ∆f = (f2 − f ) − (f − f1 ) = f2 + f1 − 2f,so ∆f 2c = √ − 2, f c2− u2 2(3.00×108 m/s) = − 2, (3.00×108 m/s)2 − (8.65×105 m/s)2 = 8.3×10−6 .E39-48 (a) No relative motion, so every 6 minutes. (b) The Doppler effect at this speed is 1 − u/c 1 − (0.6) = = 0.5; 1− u2 /c2 1 − (0.6)2this means the frequency is one half, so the period is doubled to 12 minutes. (c) If C send the signal at the instant the signal from A passes, then the two signals travel togetherto C, so C would get B’s signals at the same rate that it gets A’s signals: every six minutes.E39-49E39-50 The transverse Doppler effect is λ = λ0 / 1 − u2 /c2 . Then λ = (589.00 nm)/ 1 − (0.122)2 = 593.43 nm.The shift is (593.43 nm) − (589.00 nm) = 4.43 nm. 181
• E39-51 The frequency observed by the detector from the first source is (Eq. 39-31) f = f1 1 − (0.717)2 = 0.697f1 . The frequency observed by the detector from the second source is (Eq. 39-30) 1 − (0.717)2 0.697f2 f = f2 = . 1 + (0.717) cos θ 1 + (0.717) cos θWe need to equate these and solve for θ. Then 0.697f2 0.697f1 = , 1 + 0.717 cos θ 1 + 0.717 cos θ = f2 /f1 , cos θ = (f2 /f1 − 1) /0.717, θ = 101.1◦ .Subtract from 180◦ to find the angle with the line of sight.E39-52P39-1 Consider the triangle in Fig. 39-45. The true position corresponds to the speed of light,the opposite side corresponds to the velocity of earth in the orbit. Then θ = arctan(29.8×103 m/s)/(3.00×108 m/s) = 20.5 .P39-2 The distance to Jupiter from point x is dx = rj − re . The distance to Jupiter from point yis d2 = 2 2 re + rj .The difference in distance is related to the time according to (d2 − d1 )/t = c,so (778×109 m)2 + (150×109 m)2 − (778×109 m) + (150×109 m) = 2.7×108 m/s. (600 s)P39-3 sin(30◦ )/(4.0 m/s) = sin θ/(3.0 m/s). Then θ = 22◦ . Water waves travel more slowly inshallower water, which means they always bend toward the normal as they approach land.P39-4 (a) If the ray is normal to the water’s surface then it passes into the water undeflected.Once in the water the problem is identical to Sample Problem 39-2. The reflected ray in the wateris parallel to the incident ray in the water, so it also strikes the water normal, and is transmittednormal. (b) Assume the ray strikes the water at an angle θ1 . It then passes into the water at an angleθ2 , where nw sin θ2 = na sin θ1 .Once the ray is in the water then the problem is identical to Sample Problem 39-2. The reflectedray in the water is parallel to the incident ray in the water, so it also strikes the water at an angleθ2 . When the ray travels back into the air it travels with an angle θ3 , where nw sin θ2 = na sin θ3 .Comparing the two equations yields θ1 = θ3 , so the outgoing ray in the air is parallel to the incomingray. 182
• P39-5 (a) As was done in Ex. 39-25 above we use the small angle approximation of sin θ ≈ θ ≈ tan θ The incident angle is θ; if the light were to go in a straight line we would expect it to strike adistance y1 beneath the normal on the right hand side. The various distances are related to theangle by θ ≈ tan θ ≈ y1 /t.The light, however, does not go in a straight line, it is refracted according to (the small angleapproximation to) Snell’s law, n1 θ1 = n2 θ2 , which we will simplify further by letting θ1 = θ, n2 = n,and n1 = 1, θ = nθ2 . The point where the refracted ray does strike is related to the angle byθ2 ≈ tan θ2 = y2 /t. Combining the three expressions, y1 = ny2 .The difference, y1 − y2 is the vertical distance between the displaced ray and the original ray asmeasured on the plate glass. A little algebra yields y1 − y2 = y1 − y1 /n, = y1 (1 − 1/n) , n−1 = tθ . nThe perpendicular distance x is related to this difference by cos θ = x/(y1 − y2 ).In the small angle approximation cos θ ≈ 1 − θ2 /2. If θ is sufficiently small we can ignore the squareterm, and x ≈ y2 − y1 . (b) Remember to use radians and not degrees whenever the small angle approximation is applied.Then (1.52) − 1 x = (1.0 cm)(0.175 rad) = 0.060 cm. (1.52)P39-6 (a) At the top layer, n1 sin θ1 = sin θ;at the next layer, n2 sin θ2 = n1 sin θ1 ;at the next layer, n3 sin θ3 = n2 sin θ2 .Combining all three expressions, n3 sin θ3 = sin θ. (b) θ3 = arcsin[sin(50 )/(1.00029)] = 49.98◦ . Then shift is (50◦ ) − (49.98◦ ) = 0.02◦ . ◦ P39-7 The “big idea” of Problem 6 is that when light travels through layers the angle that itmakes in any layer depends only on the incident angle, the index of refraction where that incidentangle occurs, and the index of refraction at the current point. That means that light which leaves the surface of the runway at 90◦ to the normal will make anangle n0 sin 90◦ = n0 (1 + ay) sin θ 183
• at some height y above the runway. It is mildly entertaining to note that the value of n0 is unim-portant, only the value of a! The expression 1 sin θ = ≈ 1 − ay 1 + aycan be used to find the angle made by the curved path against the normal as a function of y. Theslope of the curve at any point is given by dy cos θ = tan(90◦ − θ) = cot θ = . dx sin θNow we need to know cos θ. It is cos θ = 1 − sin2 θ ≈ 2ay.Combining √ dy 2ay ≈ , dx 1 − ayand now we integrate. We will ignore the ay term in the denominator because it will always besmall compared to 1. Then d h dy dx = √ , 0 0 2ay 2h 2(1.7 m) d = = = 1500 m. a (1.5×10−6 m−1 )P39-8 The energy of a particle is given by E 2 = p2 c2 + m2 c4 . This energy is related to the massby E = γmc2 . γ is related to the speed by γ = 1/ 1 − u2 /c2 . Rearranging, u 1 m2 c2 = 1− = 1− , c γ2 p2 + m2 c2 p2 = . p2 + m2 c2Since n = c/u we can write this as 2 m2 c2 mc2 n= 1+ = 1+ . p2 pcFor the pion, 2 (135 MeV) n= 1+ = 1.37. (145 MeV)For the muon, 2 (106 MeV) n= 1+ = 1.24. (145 MeV) 184
• P39-9 (a) Before adding the drop of liquid project the light ray along the angle θ so that θ = 0.Increase θ slowly until total internal reflection occurs at angle θ1 . Then ng sin θ1 = 1is the equation which can be solved to find ng . Now put the liquid on the glass and repeat the above process until total internal reflection occursat angle θ2 . Then ng sin θ2 = nl . Note that ng < ng for this method to work. (b) This is not terribly practical.P39-10 Let the internal angle at Q be θQ . Then n sin θQ = 1, because it is a critical angle. Letthe internal angle at P be θP . Then θP + θQ = 90◦ . Combine this with the other formula and 1 = n sin(90 − θP ) = n cos θQ = n 1 − sin2 θP .Not only that, but sin θ1 = n sin θP , or 1=n 1 − (sin θ1 )2 /n2 ,which can be solved for n to yield n= 1 + sin2 θ1 . √ (b) The largest value of the sine function is one, so nmax = 2. P39-11 (a) The fraction of light energy which escapes from the water is dependent on the criticalangle. Light radiates in all directions from the source, but only that which strikes the surface at anangle less than the critical angle will escape. This critical angle is sin θc = 1/n.We want to find the solid angle of the light which escapes; this is found by integrating 2π θc Ω= sin θ dθ dφ. 0 0This is not a hard integral to do. The result is Ω = 2π(1 − cos θc ).There are 4π steradians in a spherical surface, so the fraction which escapes is 1 1 f= (1 − cos θc ) = (1 − 1 − sin2 θc ). 2 2The last substitution is easy enough. We never needed to know the depth h. (b) f = 1 (1 − 1 − (1/(1.3))2 ) = 0.18. 2 185
• P39-12 (a) The beam of light strikes the face of the fiber at an angle θ and is refracted accordingto n1 sin θ1 = sin θ.The beam then travels inside the fiber until it hits the cladding interface; it does so at an angle of90◦ − θ1 to the normal. It will be reflected if it exceeds the critical angle of n1 sin θc = n2 ,or if sin(90◦ − θ1 ) ≥ n2 /n1 ,which can be written as cos θ1 ≥ n2 /n1 .but if this is the cosine, then we can use sin2 + cos2 = 1 to find the sine, and sin θ1 ≤ 1 − n2 /n2 . 2 1Combine this with the first equation and θ ≤ arcsin n2 − n2 . 1 2 (b) θ = arcsin (1.58)2 − (1.53)2 = 23.2◦ . P39-13 Consider the two possible extremes: a ray of light can propagate in a straight linedirectly down the axis of the fiber, or it can reflect off of the sides with the minimum possible angleof incidence. Start with the harder option. The minimum angle of incidence that will still involve reflection is the critical angle, so n2 sin θc = . n1This light ray has farther to travel than the ray down the fiber axis because it is traveling at anangle. The distance traveled by this ray is n1 L = L/ sin θc = L , n2The time taken for this bouncing ray to travel a length L down the fiber is then L L n1 L n2 1 t = = = . v c c n2 Now for the easier ray. It travels straight down the fiber in a time L t= n1 . cThe difference is L n2 1 Ln1 t − t = ∆t = − n1 = (n1 − n2 ). c n2 cn2 (b) For the numbers in Problem 12 we have (350×103 m)(1.58) ∆t = ((1.58) − (1.53)) = 6.02×10−5 s. (3.00×108 m/s)(1.53) 186
• P39-14P39-15 We can assume the airplane speed is small compared to the speed of light, and use Eq.39-21. ∆f = 990 Hz; so |∆f | = f0 u/c = u/λ0 ,hence u = (990/s)(0.12 m) = 119 m/s. The actual answer for the speed of the airplane is half thisbecause there were two Doppler shifts: once when the microwaves struck the plane, and one when thereflected beam was received by the station. Hence, the plane approaches with a speed of 59.4 m/s. 187
• E40-1 (b) Since i = −o, vi = di/dt = −do/dt = −vo . (a) In order to change from the frame of reference of the mirror to your own frame of referenceyou need to subtract vo from all velocities. Then your velocity is vo − v0 = 0, the mirror is movingwith velocity 0 − vo = −vo and your image is moving with velocity −vo − vo = −2vo .E40-2 You are 30 cm from the mirror, the image is 10 cm behind the mirror. You need to focus40 cm away. E40-3 If the mirror rotates through an angle α then the angle of incidence will increase by anangle α, and so will the angle of reflection. But that means that the angle between the incidentangle and the reflected angle has increased by α twice.E40-4 Sketch a line from Sarah through the right edge of the mirror and then beyond. Sarah cansee any image which is located between that line and the mirror. By similar triangles, the imageof Bernie will be d/2 = (3.0 m)/2 = 1/5 m from the mirror when it becomes visible. Since i = −o,Bernie will also be 1.5 m from the mirror.E40-5 The images are fainter than the object. Several sample rays are shown.E40-6 The image is displaced. The eye would need to look up to see it. E40-7 The apparent depth of the swimming pool is given by the work done for Exercise 39-25, dapp = d/n The water then “appears” to be only 186 cm/1.33 = 140 cm deep. The apparentdistance between the light and the mirror is then 250 cm + 140 cm = 390 cm; consequently theimage of the light is 390 cm beneath the surface of the mirror.E40-8 Three. There is a single direct image in each mirror and one more image of an image inone of the mirrors. 188
• E40-9 We want to know over what surface area of the mirror are rays of light reflected from theobject into the eye. By similar triangles the diameter of the pupil and the diameter of the part ofthe mirror (d) which reflects light into the eye are related by d (5.0 mm) = , (10 cm) (24 cm) + (10 cm)which has solution d = 1.47 mm The area of the circle on the mirror is A = π(1.47 mm)2 /4 = 1.7 mm2 .E40-10 (a) Seven; (b) Five; and (c) Two. This is a problem of symmetry.E40-11 Seven. Three images are the ones from Exercise 8. But each image has an image in theceiling mirror. That would make a total of six, except that you also have an image in the ceilingmirror (look up, eh?). So the total is seven!E40-12 A point focus is not formed. The envelope of rays is called the caustic. You can see asimilar effect when you allow light to reflect off of a spoon onto a table. E40-13 The image is magnified by a factor of 2.7, so the image distance is 2.7 times farther fromthe mirror than the object. An important question to ask is whether or not the image is real orvirtual. If it is a virtual image it is behind the mirror and someone looking at the mirror could seeit. If it were a real image it would be in front of the mirror, and the man, who serves as the objectand is therefore closer to the mirror than the image, would not be able to see it. So we shall assume that the image is virtual. The image distance is then a negative number.The focal length is half of the radius of curvature, so we want to solve Eq. 40-6, with f = 17.5 cmand i = −2.7o 1 1 1 0.63 = + = , (17.5 cm) o −2.7o owhich has solution o = 11 cm.E40-14 The image will be located at a point given by 1 1 1 1 1 1 = − = − = . i f o (10 cm) (15 cm) (30 cm)The vertical scale is three times the horizontal scale in the figure below. 189
• E40-15 This problem requires repeated application of 1/f = 1/o + 1/i, r = 2f , m = −i/o, orthe properties of plane, convex, or concave mirrors. All dimensioned variables below (f, r, i, o) aremeasured in centimeters. (a) Concave mirrors have positive focal lengths, so f = +20; r = 2f = +40; 1/i = 1/f − 1/o = 1/(20) − 1/(10) = 1/(−20);m = −i/o = −(−20)/(10) = 2; the image is virtual and upright. (b) m = +1 for plane mirrors only; r = ∞ for flat surface; f = ∞/2 = ∞; i = −o = −10; theimage is virtual and upright. (c) If f is positive the mirror is concave; r = 2f = +40; 1/i = 1/f − 1/o = 1/(20) − 1/(30) = 1/(60);m = −i/o = −(60)/(30) = −2; the image is real and inverted. (d) If m is negative then the image is real and inverted; only Concave mirrors produce real images(from real objects); i = −mo = −(−0.5)(60) = 30; 1/f = 1/o + 1/i = 1/(30) + 1/(60) = 1/(20);r = 2f = +40. (e) If r is negative the mirror is convex; f = r/2 = (−40)/2 = −20; 1/o = 1/f − 1/i = 1/(−20) − 1/(−10) = 1/(20);m = −(−10)/(20) = 0.5; the image is virtual and upright. (f) If m is positive the image is virtual and upright; if m is less than one the image is reduced,but only convex mirrors produce reduced virtual images (from real objects); f = −20 for convexmirrors; r = 2f = −40; let i = −mo = −o/10, then 1/f = 1/o + 1/i = 1/o − 10/o = −9/o,so o = −9f = −9(−20) = 180; i = −o/10 = −(180)/10 = −18. (g) r is negative for convex mirrors, so r = −40; f = r/2 = −20; convex mirrors produce onlyvirtual upright images (from real objects); so i is negative; and 1/o = 1/f − 1/i = 1/(−20) − 1/(−4) = 1/(5);m = −i/o = −(−4)/(5) = 0.8. (h) Inverted images are real; only concave mirrors produce real images (from real objects);inverted images have negative m; i = −mo = −(−0.5)(24) = 12; 1/f = 1/o + 1/i = 1/(24) + 1/(12) = 1/(8);r = 2f = 16. 190
• E40-16 Use the angle definitions provided by Eq. 40-8. From triangle OaI we have α + γ = 2θ,while from triangle IaC we have β + θ = γ.Combining to eliminate θ we get α − γ = −2β.Substitute Eq. 40-8 and eliminate s, 1 1 2 − =− , o i ror 1 1 2 + = , o −i −rwhich is the same as Eq. 40-4 if i → −i and r → −r. E40-17 (a) Consider the point A. Light from this point travels along the line ABC and will beparallel to the horizontal center line from the center of the cylinder. Since the tangent to a circledefines the outer limit of the intersection with a line, this line must describe the apparent size. (b) The angle of incidence of ray AB is given by sin θ1 = r/R.The angle of refraction of ray BC is given by sin θ2 = r∗ /R.Snell’s law, and a little algebra, yields n1 sin θ1 = n2 sin θ2 , r r∗ n1 = n2 , R R nr = r∗ .In the last line we used the fact that n2 = 1, because it is in the air, and n1 = n, the index ofrefraction of the glass.E40-18 This problem requires repeated application of (n2 − n1 )/r = n1 /o + n2 /i. All dimensionedvariables below (r, i, o) are measured in centimeters. (a) (1.5) − (1.0) (1.0) − = −0.08333, (30) (10)so i = (1.5)/(−0.08333) = −18, and the image is virtual. (b) (1.0) (1.5) + = −0.015385, (10) (−13)so r = (1.5 − 1.0)/(−0.015385) = −32.5, and the image is virtual. (c) (1.5) − (1.0) (1.5) − = 0.014167, (30) (600) 191
• so o = (1.0)/(0.014167) = 71. The image was real since i > 0. (d) Rearrange the formula to solve for n2 , then 1 1 1 n2 − 1i = n1 + . r r oSubstituting the numbers, 1 1 1 1 n2 − = (1.0) + , (−20) (−20) (−20) (20)which has any solution for n2 ! Since i < 0 the image is virtual. (e) (1.5) (1.0) + = −0.016667, (10) (−6)so r = (1.0 − 1.5)/(−0.016667) = 30, and the image is virtual. (f) (1.0) − (1.5) (1.0) − = 0.15, (−30) (−7.5)so o = (1.5)/(0.15) = 10. The image was virtual since i < 0. (g) (1.0) − (1.5) (1.5) − = −3.81×10−2 , (30) (70)so i = (1.0)/(−3.81×10−2 ) = −26, and the image is virtual. (h) Solving Eq. 40-10 for n2 yields 1/o + 1/r n2 = n1 , 1/r − 1/iso 1/(100) + 1/(−30) n2 = (1.5) = 1.0 1/(−30) − 1/(600)and the image is real.E40-19 (b) If the beam is small we can use Eq. 40-10. Parallel incoming rays correspond to anobject at infinity. Solving for n2 yields 1/o + 1/r n2 = n1 , 1/r − 1/iso if o → ∞ and i = 2r, then 1/∞ + 1/r n2 = (1.0) = 2.0 1/r − 1/2r (c) There is no solution if i = r!E40-20 The image will be located at a point given by 1 1 1 1 1 1 = − = − = . i f o (10 cm) (6 cm) (−15 cm) 192
• E40-21 The image location can be found from Eq. 40-15, 1 1 1 1 1 1 = − = − = , i f o (−30 cm) (20 cm) −12 cmso the image is located 12 cm from the thin lens, on the same side as the object.E40-22 For a double convex lens r1 > 0 and r2 < 0 (see Fig. 40-21 and the accompanying text).Then the problem states that r2 = −r1 /2. The lens maker’s equation can be applied to get 1 1 1 3(n − 1) = (n − 1) − = , f r1 r2 r1so r1 = 3(n − 1)f = 3(1.5 − 1)(60 mm) = 90 mm, and r2 = −45 mm.E40-23 The object distance is essentially o = ∞, so 1/f = 1/o + 1/i implies f = i, and the imageforms at the focal point. In reality, however, the object distance is not infinite, so the magnificationis given by m = −i/o ≈ −f /o, where o is the Earth/Sun distance. The size of the image is then hi = ho f /o = 2(6.96×108 m)(0.27 m)/(1.50×1011 m) = 2.5 mm.The factor of two is because the sun’s radius is given, and we need the diameter!E40-24 (a) The flat side has r2 = ∞, so 1/f = (n − 1)/r, where r is the curved side. Thenf = (0.20 m)/(1.5 − 1) = 0.40 m. (b) 1/i = 1/f − 1/o = 1/(0.40 m) − 1/(0.40 m) = 0. Then i is ∞.E40-25 (a) 1/f = (1.5 − 1)[1/(0.4 m) − 1/(−0.4 m)] = 1/(0.40 m). (b) 1/f = (1.5 − 1)[1/(∞) − 1/(−0.4 m)] = 1/(0.80 m). (c) 1/f = (1.5 − 1)[1/(0.4 m) − 1/(0.6 m)] = 1/(2.40 m). (d) 1/f = (1.5 − 1)[1/(−0.4 m) − 1/(0.4 m)] = 1/(−0.40 m). (e) 1/f = (1.5 − 1)[1/(∞) − 1/(0.8 m)] = 1/(−0.80 m). (f) 1/f = (1.5 − 1)[1/(0.6 m) − 1/(0.4 m)] = 1/(−2.40 m).E40-26 (a) 1/f = (n − 1)[1/(−r) − 1/r], so 1/f = 2(1 − n)/r. 1/i = 1/f − 1/o so if o = r, then 1/i = 2(1 − n)/r − 1/r = (1 − 2n)/r,so i = r/(1 − 2n). For n > 0.5 the image is virtual. (b) For n > 0.5 the image is virtual; the magnification is m = −i/o = −r/(1 − 2n)/r = 1/(2n − 1).E40-27 According to the definitions, o = f + x and i = f + x . Starting with Eq. 40-15, 1 1 1 + = , o i f i+o 1 = , oi f 2f + x + x 1 = , (f + x)(f + x ) f 2f 2 + f x + f x = f 2 + f x + f x + xx , f2 = xx . 193
• E40-28 (a) You can’t determine r1 , r2 , or n. i is found from 1 1 1 1 = − = , i +10 +20 +20the image is real and inverted. m = −(20)/(20) = −1. (b) You can’t determine r1 , r2 , or n. The lens is converging since f is positive. i is found from 1 1 1 1 = − = , i +10 +5 −10the image is virtual and upright. m = −(−10)/(+5) = 2. (c) You can’t determine r1 , r2 , or n. Since m is positive and greater than one the lens isconverging. Then f is positive. i is found from 1 1 1 1 = − = , i +10 +5 −10the image is virtual and upright. m = −(−10)/(+5) = 2. (d) You can’t determine r1 , r2 , or n. Since m is positive and less than one the lens is diverging.Then f is negative. i is found from 1 1 1 1 = − = , i −10 +5 −3.3the image is virtual and upright. m = −(−3.3)/(+5) = 0.66. (e) f is found from 1 1 1 1 = (1.5 − 1) − = . f +30 −30 +30The lens is converging. i is found from 1 1 1 1 = − = , i +30 +10 −15the image is virtual and upright. m = −(−15)/(+10) = 1.5. (f) f is found from 1 1 1 1 = (1.5 − 1) − = . f −30 +30 −30The lens is diverging. i is found from 1 1 1 1 = − = , i −30 +10 −7.5the image is virtual and upright. m = −(−7.5)/(+10) = 0.75. (g) f is found from 1 1 1 1 = (1.5 − 1) − = . f −30 −60 −120The lens is diverging. i is found from 1 1 1 1 = − = , i −120 +10 −9.2the image is virtual and upright. m = −(−9.2)/(+10) = 0.92. (h) You can’t determine r1 , r2 , or n. Upright images have positive magnification. i is found from i = −(0.5)(10) = −5; 194
• f is found from 1 1 1 1 = + = , f +10 −5 −10so the lens is diverging. (h) You can’t determine r1 , r2 , or n. Real images have negative magnification. i is found from i = −(−0.5)(10) = 5;f is found from 1 1 1 1 = + = , f +10 5 +3.33so the lens is converging.E40-29 o + i = 0.44 m = L, so 1 1 1 1 1 L = + = + = , f o i o L−o o(L − o)which can also be written as o2 − oL + f L = 0. This has solution L± L2 − 4f L (0.44 m) ± (0.44 m) − 4(0.11 m)(0.44 m) o= = = 0.22 m. 2 2There is only one solution to this problem, but sometimes there are two, and other times there arenone!E40-30 (a) Real images (from real objects) are only produced by converging lenses. (b) Since hi = −h0 /2, then i = o/2. But d = i+o = o+o/2 = 3o/2, so o = 2(0.40 m)/3 = 0.267 m,and i = 0.133 m. (c) 1/f = 1/o + 1/i = 1/(0.267 m) + 1/(0.133 m) = 1/(0.0889 m). E40-31 Step through the exercise one lens at a time. The object is 40 cm to the left of aconverging lens with a focal length of +20 cm. The image from this first lens will be located bysolving 1 1 1 1 1 1 = − = − = , i f o (20 cm) (40 cm) 40 cmso i = 40 cm. Since i is positive it is a real image, and it is located to the right of the converginglens. This image becomes the object for the diverging lens. The image from the converging lens is located 40 cm - 10 cm from the diverging lens, but it islocated on the wrong side: the diverging lens is “in the way” so the rays which would form the imagehit the diverging lens before they have a chance to form the image. That means that the real imagefrom the converging lens is a virtual object in the diverging lens, so that the object distance for thediverging lens is o = −30 cm. The image formed by the diverging lens is located by solving 1 1 1 1 1 1 = − = − = , i f o (−15 cm) (−30 cm) −30 cmor i = −30 cm. This would mean the image formed by the diverging lens would be a virtual image,and would be located to the left of the diverging lens. The image is virtual, so it is upright. The magnification from the first lens is m1 = −i/o = −(40 cm)/(40 cm)) = −1; 195
• the magnification from the second lens is m2 = −i/o = −(−30 cm)/(−30 cm)) = −1;which implies an overall magnification of m1 m2 = 1.E40-32 (a) The parallel rays of light which strike the lens of focal length f will converge on thefocal point. This point will act like an object for the second lens. If the second lens is located adistance L from the first then the object distance for the second lens will be L − f . Note that thiswill be a negative value for L < f , which means the object is virtual. The image will form at a point 1/i = 1/(−f ) − 1/(L − f ) = L/f (f − L).Note that i will be positive if L < f , so the rays really do converge on a point. (b) The same equation applies, except switch the sign of f . Then 1/i = 1/(f ) − 1/(L − f ) = L/f (L − f ).This is negative for L < f , so there is no real image, and no converging of the light rays. (c) If L = 0 then i = ∞, which means the rays coming from the second lens are parallel.E40-33 The image from the converging lens is found from 1 1 1 1 = − = i1 (0.58 m) (1.12 m) 1.20 mso i1 = 1.20 m, and the image is real and inverted. This real image is 1.97 m − 1.20 m = 0.77 m in front of the plane mirror. It acts as an objectfor the mirror. The mirror produces a virtual image 0.77 m behind the plane mirror. This image isupright relative to the object which formed it, which was inverted relative to the original object. This second image is 1.97 m + 0.77 m = 2.74 m away from the lens. This second image acts as anobject for the lens, the image of which is found from 1 1 1 1 = − = i3 (0.58 m) (2.74 m) 0.736 mso i3 = 0.736 m, and the image is real and inverted relative to the object which formed it, which wasinverted relative to the original object. So this image is actually upright.E40-34 (a) The first lens forms a real image at a location given by 1/i = 1/f − 1/o = 1/(0.1 m) − 1/(0.2 m) = 1/(0.2 m).The image and object distance are the same, so the image has a magnification of 1. This image is0.3 m − 0.2 m = 0.1 m from the second lens. The second lens forms an image at a location given by 1/i = 1/f − 1/o = 1/(0.125 m) − 1/(0.1 m) = 1/(−0.5 m).Note that this puts the final image at the location of the original object! The image is magnified bya factor of (0.5 m)/(0.1 m) = 5. (c) The image is virtual, but inverted. 196
• E40-35 If the two lenses “pass” the same amount of light then the solid angle subtended by eachlens as seen from the respective focal points must be the same. If we assume the lenses have thesame round shape then we can write this as do /f o = de /f e . Then de fo = = mθ , do feor de = (72 mm)/36 = 2 mm.E40-36 (a) f = (0.25 m)/(200) ≈ 1.3 mm. Then 1/f = (n − 1)(2/r) can be used to find r;r = 2(n − 1)f = 2(1.5 − 1)(1.3 mm) = 1.3 mm. (b) The diameter would be twice the radius. In effect, these were tiny glass balls.E40-37 (a) In Fig. 40-46(a) the image is at the focal point. This means that in Fig. 40-46(b)i = f = 2.5 cm, even though f = f . Solving, 1 1 1 1 = + = . f (36 cm) (2.5 cm) 2.34 cm (b) The effective radii of curvature must have decreased.E40-38 (a) s = (25 cm) − (4.2 cm) − (7.7 cm) = 13.1 cm. (b) i = (25 cm) − (7.7 cm) = 17.3 cm. Then 1 1 1 1 = − = o (4.2 cm) (17.3 cm) 5.54 cm.The object should be placed 5.5 − 4.2 = 1.34 cm beyond F1 . (c) m = −(17.3)/(5.5) = −3.1. (d) mθ = (25 cm)/(7.7 cm) = 3.2. (e) M = mmθ = −10. E40-39 Microscope magnification is given by Eq. 40-33. We need to first find the focal lengthof the objective lens before we can use this formula. We are told in the text, however, that themicroscope is constructed so the at the object is placed just beyond the focal point of the objectivelens, then f ob ≈ 12.0 mm. Similarly, the intermediate image is formed at the focal point of theeyepiece, so f ey ≈ 48.0 mm. The magnification is then −s(250 mm) (285 mm)(250 mm) m= =− = 124. f ob f ey (12.0 mm)(48.0 mm)A more accurate answer can be found by calculating the real focal length of the objective lens, whichis 11.4 mm, but since there is a huge uncertainty in the near point of the eye, I see no point in tryingto be more accurate than this.P40-1 The old intensity is Io = P/4πd2 , where P is the power of the point source. With themirror in place there is an additional amount of light which needs to travel a total distance of 3din order to get to the screen, so it contributes an additional P/4π(3d)2 to the intensity. The newintensity is then In = P/4πd2 + P/4π(3d)2 = (10/9)P/4πd2 = (10/9)Io . 197
• P40-2 (a) vi = di/dt; but i = f o/(o − f ) and f = r/2 so 2 2 d ro r do r vi = =− =− vo . dt 2o − r 2o − r dt 2o − r (b) Put in the numbers! 2 (15 cm) vi = − (5.0 cm/s) = −6.2×10−2 cm/s. 2(75 cm) − (15 cm) (c) Put in the numbers! 2 (15 cm) vi = − (5.0 cm/s) = −70 m/s 2(7.7 cm) − (15 cm) (d) Put in the numbers! 2 (15 cm) vi = − (5.0 cm/s) = −5.2 cm/s. 2(0.15 cm) − (15 cm) P40-3 (b) There are two ends to the object of length L, one of these ends is a distance o1 fromthe mirror, and the other is a distance o2 from the mirror. The images of the two ends will belocated at i1 and i2 . Since we are told that the object has a short length L we will assume that a differential approachto the problem is in order. Then L = ∆o = o1 − o2 and L = ∆i = i1 − i2 ,Finding the ratio of L /L is then reduced to L ∆i di = ≈ . L ∆o do We can take the derivative of Eq. 40-15 with respect to changes in o and i, di do + 2 = 0, i2 oor L di i2 ≈ = − 2 = −m2 , L do owhere m is the lateral magnification. (a) Since i is given by 1 1 1 o−f = − = , i f o ofthe fraction i/o can also be written i of f = = . o o(o − f ) o−fThen 2 i2 f L≈− =− o2 o−f 198
• P40-4 The left surface produces an image which is found from n/i = (n − 1)/R − 1/o, but sincethe incoming rays are parallel we take o = ∞ and the expression simplifies to i = nR/(n − 1). Thisimage is located a distance o = 2R − i = (n − 2)R/(n − 1) from the right surface, and the imageproduced by this surface can be found from 1/i = (1 − n)/(−R) − n/o = (n − 1)/R − n(n − 1)/(n − 2)R = 2(1 − n)/(n − 2)R.Then i = (n − 2)R/2(n − 1).P40-5 The “1” in Eq. 40-18 is actually nair ; the assumption is that the thin lens is in the air. Ifthat isn’t so, then we need to replace “1” with n , so Eq. 40-18 becomes n n n−n − = . o |i | r1A similar correction happens to Eq. 40-21: n n n−n + =− . |i | i r2Adding these two equations, n n 1 1 + = (n − n ) − . o i r1 r2This yields a focal length given by 1 n−n 1 1 = − . f n r1 r2P40-6 Start with Eq. 40-4 1 1 1 + = , o i f |f | |f | |f | + = , o i f 1 1 + = ±1, y ywhere + is when f is positive and − is when f is negative. The plot on the right is for +, that on the left for −. Real image and objects occur when y or y is positive. 199
• P40-7 (a) The image (which will appear on the screen) and object are a distance D = o + iapart. We can use this information to eliminate one variable from Eq. 40-15, 1 1 1 + = , o i f 1 1 1 + = , o D−o f D 1 = , o(D − o) f o2 − oD + f D = 0.This last expression is a quadratic, and we would expect to get two solutions for o. These solutionswill be of the form “something” plus/minus “something else”; the distance between the two locationsfor o will evidently be twice the “something else”, which is then d = o+ − o− = (−D)2 − 4(f D) = D(D − 4f ). (b) The ratio of the image sizes is m+ /m− , or i+ o− /i− o+ . Now it seems we must find the actualvalues of o+ and o− . From the quadratic in part (a) we have D± D(D − 4f ) D±d o± = = , 2 2so the ratio is o− D−d = . o+ D+dBut i− = o+ , and vice-versa, so the ratio of the image sizes is this quantity squared.P40-8 1/i = 1/f − 1/o implies i = f o/(o − f ). i is only real if o ≥ f . The distance between theimage and object is of o2 y =i+o= +o= . o−f o−fThis quantity is a minimum when dy/do = 0, which occurs when o = 2f . Then i = 2f , and y = 4f .P40-9 (a) The angular size of each lens is the same when viewed from the shared focal point. Thismeans W1 /f1 = W2 /f2 , or W2 = (f2 /f1 )W1 . (b) Pass the light through the diverging lens first; choose the separation of the lenses so thatthe focal point of the converging lens is at the same location as the focal point of the diverging lenswhich is on the opposite side of the diverging lens. (c) Since I ∝ 1/A, where A is the area of the beam, we have I ∝ 1/W 2 . Consequently, I2 /I1 = (W1 /W2 )2 = (f1 /f2 )2P40-10 The location of the image in the mirror is given by 1 1 1 = − . i f a+b 200
• The location of the image in the plate is given by i = −a, which is located at b − a relative to themirror. Equating, 1 1 1 + = , b−a b+a f 2b 1 2 − a2 = , b f b2 − a2 = 2bf, a = b2 − 2bf , = (7.5 cm)2 − 2(7.5 cm)(−28.2 cm) = 21.9 cm. P40-11 We’ll solve the problem by finding out what happens if you put an object in front of thecombination of lenses. Let the object distance be o1 . The first lens will create an image at i1 , where 1 1 1 = − i1 f1 o1This image will act as an object for the second lens. If the first image is real (i1 positive) then the image will be on the “wrong” side of the secondlens, and as such the real image will act like a virtual object. In short, o2 = −i1 will give the correctsign to the object distance when the image from the first lens acts like an object for the second lens.The image formed by the second lens will then be at 1 1 1 = − , i2 f2 o2 1 1 = + , f2 i2 1 1 1 = + − . f2 f1 o1In this case it appears as if the combination 1 1 + f2 f1is equivalent to the reciprocal of a focal length. We will go ahead and make this connection, and 1 1 1 f1 + f2 = + = . f f2 f1 f1 f2The rest is straightforward enough.P40-12 (a) The image formed by the first lens can be found from 1 1 1 1 = − = . i1 f1 2f1 2f1This is a distance o2 = 2(f1 + f2 ) = 2f2 from the mirror. The image formed by the mirror is at animage distance given by 1 1 1 1 = − = . i2 f2 2f2 2f2Which is at the same point as i1 !. This means it will act as an object o3 in the lens, and, reversingthe first step, produce a final image at O, the location of the original object. There are then threeimages formed; each is real, same size, and inverted. Three inversions nets an inverted image. Thefinal image at O is therefore inverted. 201
• P40-13 (a) Place an object at o. The image will be at a point i given by 1 1 1 = − , i f oor i = f o/(o − f ). (b) The lens must be shifted a distance i − i, or fo i −i= − 1. o−f (c) The range of motion is (0.05 m)(1.2 m) ∆i = − 1 = −5.2 cm. (1.2 m) − (0.05 m)P40-14 (a) Because magnification is proportional to 1/f . (b) Using the results of Problem 40-11, 1 1 1 = + , f f2 f1so P = P1 + P2 . P40-15 We want the maximum linear motion of the train to move no more than 0.75 mm on thefilm; this means we want to find the size of an object on the train that will form a 0.75 mm image.The object distance is much larger than the focal length, so the image distance is approximatelyequal to the focal length. The magnification is then m = −i/o = (3.6 cm)/(44.5 m) = −0.00081. The size of an object on the train that would produce a 0.75 mm image on the film is then0.75 mm/0.00081 = 0.93 m. How much time does it take the train to move that far? (0.93 m) t= = 25 ms. (135 km/hr)(1/3600 hr/s)P40-16 (a) The derivation leading to Eq. 40-34 depends only on the fact that two convergingoptical devices are used. Replacing the objective lens with an objective mirror doesn’t changeanything except the ray diagram. (b) The image will be located very close to the focal point, so |m| ≈ f /o, and (16.8 m) hi = (1.0 m) = 8.4×10−3 m (2000 m) (c) f e = (5 m)/(200) = 0.025 m. Note that we were given the radius of curvature, not the focallength, of the mirror! 202
• E41-1 In this problem we look for the location of the third-order bright fringe, so mλ (3)(554 × 10−9 m) θ = sin−1 = sin−1 = 12.5◦ = 0.22 rad. d (7.7 × 10−6 m)E41-2 d1 sin θ = λ gives the first maximum; d2 sin θ = 2λ puts the second maximum at the locationof the first. Divide the second expression by the first and d2 = 2d1 . This is a 100% increase in d.E41-3 ∆y = λD/d = (512×10−9 m)(5.4 m)/(1.2×10−3 m) = 2.3×10−3 m.E41-4 d = λ/ sin θ = (592×10−9 m)/ sin(1.00◦ ) = 3.39×10−5 m. E41-5 Since the angles are very small, we can assume sin θ ≈ θ for angles measured in radians. If the interference fringes are 0.23◦ apart, then the angular position of the first bright fringeis 0.23◦ away from the central maximum. Eq. 41-1, written with the small angle approximationin mind, is dθ = λ for this first (m = 1) bright fringe. The goal is to find the wavelength whichincreases θ by 10%. To do this we must increase the right hand side of the equation by 10%, whichmeans increasing λ by 10%. The new wavelength will be λ = 1.1λ = 1.1(589 nm) = 650 nmE41-6 Immersing the apparatus in water will shorten the wavelengths to λ/n. Start with d sin θ0 =λ; and then find θ from d sin θ = λ/n. Combining the two expressions, θ = arcsin[sin θ0 /n] = arcsin[sin(0.20◦ )/(1.33)] = 0.15◦ .E41-7 The third-order fringe for a wavelength λ will be located at y = 3λD/d, where y is measuredfrom the central maximum. Then ∆y isy1 − y2 = 3(λ1 − λ2 )D/d = 3(612×10−9 m − 480×10−9 m)(1.36 m)/(5.22×10−3 m) = 1.03×10−4 m.E41-8 θ = arctan(y/D); λ = d sin θ = (0.120 m) sin[arctan(0.180 m/2.0 m)] = 1.08×10−2 m.Then f = v/λ = (0.25 m/s)/(1.08×10−2 m) = 23 Hz.E41-9 A variation of Eq. 41-3 is in order: 1 λD ym = m+ 2 dWe are given the distance (on the screen) between the first minima (m = 0) and the tenth minima(m = 9). Then λ(50 cm) 18 mm = y9 − y0 = 9 , (0.15 mm)or λ = 6×10−4 mm = 600 nm.E41-10 The “maximum” maxima is given by the integer part of m = d sin(90◦ )/λ = (2.0 m)/(0.50 m) = 4.Since there is no integer part, the “maximum” maxima occurs at 90◦ . These are point sourcesradiating in both directions, so there are two central maxima, and four maxima each with m = 1,m = 2, and m = 3. But the m = 4 values overlap at 90◦ , so there are only two. The total is 16. 203
• E41-11 This figure should explain it well enough.E41-12 ∆y = λD/d = (589×10−9 m)(1.13 m)/(0.18×10−3 m) = 3.70×10−3 m.E41-13 Consider Fig. 41-5, and solve it exactly for the information given. For the tenth brightfringe r1 = 10λ + r2 . There are two important triangles: r2 = D2 + (y − d/2)2 2and 2 r1 = D2 + (y + d/2)2Solving to eliminate r2 , D2 + (y + d/2)2 = D2 + (y − d/2)2 + 10λ.This has solution 4D2 + d2 − 100λ2 y = 5λ . d2 − 100λ2The solution predicted by Eq. 41-1 is 10λ y = D2 + y 2 , dor 4D2 y = 5λ . d2 − 100λ2The fractional error is y /y − 1, or 4D2 − 1, 4D2 + d2 − 100λ2or 4(40 mm)2 − 1 = −3.1×10−4 . 4(40 mm)2 + (2 mm)2 − 100(589×10−6 mm)2E41-14 (a) ∆x = c/∆t = (3.00×108 m/s)/(1×10−8 s) = 3 m. (b) No. 204
• E41-15 Leading by 90◦ is the same as leading by a quarter wavelength, since there are 360◦ in acircle. The distance from A to the detector is 100 m longer than the distance from B to the detector.Since the wavelength is 400 m, 100 m corresponds to a quarter wavelength. So a wave peak starts out from source A and travels to the detector. When it has traveled aquarter wavelength a wave peak leaves source B. But when the wave peak from A has traveleda quarter wavelength it is now located at the same distance from the detector as source B, whichmeans the two wave peaks arrive at the detector at the same time. They are in phase.E41-16 The first dark fringe involves waves π radians out of phase. Each dark fringe after thatinvolves an additional 2π radians of phase difference. So the mth dark fringe has a phase differenceof (2m + 1)π radians. 2πdE41-17 I = 4I0 cos2 λ sin θ , so for this problem we want to plot 2π(0.60 mm) I/I0 = cos2 sin θ = cos2 (6280 sin θ) . (600×10−9 m)E41-18 The resultant quantity will be of the form A sin(ωt + β). Solve the problem by looking att = 0; then y1 = 0, but x1 = 10, and y2 = 8 sin 30◦ = 4 and x2 = 8 cos 30 = 6.93. Then the resultantis of length A = (4)2 + (10 + 6.93)2 = 17.4,and has an angle β given by β = arctan(4/16.93) = 13.3◦ . E41-19 (a) We want to know the path length difference of the two sources to the detector.Assume the detector is at x and the second source is at y = d. The distance S1 D is x; the √ √distance S2 D is x2 + d2 . The difference is x2 + d2 − x. If this difference is an integral numberof wavelengths then we have a maximum; if instead it is a half integral number of wavelengths wehave a minimum. For part (a) we are looking for the maxima, so we set the path length differenceequal to mλ and solve for xm . x2 + d2 − xm m = mλ, x2 + d2 m = (mλ + xm )2 , x2 + d2 m = m2 λ2 + 2mλxm + x2 , m d 2 − m2 λ 2 xm = 2mλ The first question we need to ask is what happens when m = 0. The right hand side becomesindeterminate, so we need to go back to the first line in the above derivation. If m = 0 then d2 = 0;since this is not true in this problem, there is no m = 0 solution. In fact, we may have even more troubles. xm needs to be a positive value, so the maximumallowed value for m will be given by m2 λ2 < d2 , m < d/λ = (4.17 m)/(1.06 m) = 3.93;but since m is an integer, m = 3 is the maximum value. 205
• The first three maxima occur at m = 3, m = 2, and m = 1. These maxima are located at (4.17 m)2 − (3)2 (1.06 m)2 x3 = = 1.14 m, 2(3)(1.06 m) (4.17 m)2 − (2)2 (1.06 m)2 x2 = = 3.04 m, 2(2)(1.06 m) (4.17 m)2 − (1)2 (1.06 m)2 x1 = = 7.67 m. 2(1)(1.06 m)Interestingly enough, as m decreases the maxima get farther away! (b) The closest maxima to the origin occurs at x = ±6.94 cm. What then is x = 0? It is a localminimum, but the intensity isn’t zero. It corresponds to a point where the path length difference is3.93 wavelengths. It should be half an integer to be a complete minimum.E41-20 The resultant can be written in the form A sin(ωt + β). Consider t = 0. The threecomponents can be written as y1 = 10 sin 0◦ = 0, y2 = 14 sin 26◦ = 6.14, y3 = 4.7 sin(−41◦ ) = −3.08, y = 0 + 6.14 − 3.08 = 3.06.and x1 = 10 cos 0◦ = 10, x2 = 14 cos 26◦ = 12.6, x3 = 4.7 cos(−41◦ ) = 3.55, x = 10 + 12.6 + 3.55 = 26.2.Then A = (3.06)2 + (26.2)2 = 26.4 and β = arctan(3.06/26.2) = 6.66◦ .E41-21 The order of the indices of refraction is the same as in Sample Problem 41-4, so d = λ/4n = (620 nm)/4(1.25) = 124 nm.E41-22 Follow the example in Sample Problem 41-3. 2dn 2(410 nm)(1.50) 1230 nm λ= = = . m − 1/2 m − 1/2 m − 1/2The result is only in the visible range when m = 3, so λ = 492 nm. E41-23 (a) Light from above the oil slick can be reflected back up from the top of the oil layeror from the bottom of the oil layer. For both reflections the light is reflecting off a substance witha higher index of refraction so both reflected rays pick up a phase change of π. Since both waveshave this phase the equation for a maxima is 1 1 2d + λn + λn = mλn . 2 2Remember that λn = λ/n, where n is the index of refraction of the thin film. Then 2nd = (m − 1)λis the condition for a maxima. We know n = 1.20 and d = 460 nm. We don’t know m or λ. It might 206
• seem as if there isn’t enough information to solve the problem, but we can. We need to find thewavelength in the visible range (400 nm to 700 nm) which has an integer m. Trial and error mightwork. If λ = 700 nm, then m is 2nd 2(1.20)(460 nm) m= +1= + 1 = 2.58 λ (700 nm)But m needs to be an integer. If we increase m to 3, then 2(1.20)(460 nm) λ= = 552 nm (3 − 1)which is in the visible range. So the oil slick will appear green. (b) One of the most profound aspects of thin film interference is that wavelengths which aremaximally reflected are minimally transmitted, and vice versa. Finding the maximally transmittedwavelengths is the same as finding the minimally reflected wavelengths, or looking for values of mthat are half integer. The most obvious choice is m = 3.5, and then 2(1.20)(460 nm) λ= = 442 nm. (3.5 − 1)E41-24 The condition for constructive interference is 2nd = (m − 1/2)λ. Assuming a minimumvalue of m = 1 one finds d = λ/4n = (560 nm)/4(2.0) = 70 nm.E41-25 The top surface contributes a phase difference of π, so the phase difference because of thethickness is 2π, or one complete wavelength. Then 2d = λ/n, or d = (572 nm)/2(1.33) = 215 nm.E41-26 The wave reflected from the first surface picks up a phase shift of π. The wave which isreflected off of the second surface travels an additional path difference of 2d. The interference willbe bright if 2d + λn /2 = mλn results in m being an integer. m = 2nd/λ + 1/2 = 2(1.33)(1.21×10−6 m)/(585×10−9 m) + 1/2 = 6.00,so the interference is bright. E41-27 As with the oil on the water in Ex. 41-23, both the light which reflects off of the acetoneand the light which reflects off of the glass undergoes a phase shift of π. Then the maxima forreflection are given by 2nd = (m − 1)λ. We don’t know m, but at some integer value of m we haveλ = 700 nm. If m is increased by exactly 1 then we are at a minimum of λ = 600 nm. Consequently, 2 2(1.25)d = (m − 1)(700 nm) and 2(1.25)d = (m − 1/2)(600 nm),we can set these two expressions equal to each other to find m, (m − 1)(700 nm) = (m − 1/2)(600 nm),so m = 4. Then we can find the thickness, d = (4 − 1)(700 nm)/2(1.25) = 840 nm. 207
• E41-28 The wave reflected from the first surface picks up a phase shift of π. The wave which isreflected off of the second surface travels an additional path difference of 2d. The interference willbe bright if 2d + λn /2 = mλn results in m being an integer. Then 2nd = (m − 1/2)λ1 is bright, and2nd = mλ2 is dark. Divide one by the other and (m − 1/2)λ1 = mλ2 , so m = λ1 /2(λ1 − λ2 ) = (600 nm)/2(600 nm − 450 nm) = 2,then d = mλ2 /2n = (2)(450 nm)/2(1.33) = 338 nm.E41-29 Constructive interference happens when 2d = (m − 1/2)λ. The minimum value for m ism = 1; the maximum value is the integer portion of 2d/λ+1/2 = 2(4.8×10−5 m)/(680×10−9 m)+1/2 =141.67, so mmax = 141. There are then 141 bright bands.E41-30 (a) A half wavelength phase shift occurs for both the air/water interface and the water/oilinterface, so if d = 0 the two reflected waves are in phase. It will be bright! (b) 2nd = 3λ, or d = 3(475 nm)/2(1.20) = 594 nm.E41-31 There is a phase shift on one surface only, so the bright bands are given by 2nd = (m −1/2)λ. Let the first band be given by 2nd1 = (m1 − 1/2)λ. The last bright band is then given by2nd2 = (m1 + 9 − 1/2)λ. Subtract the two equations to get the change in thickness: ∆d = 9λ/2n = 9(630 nm)/2(1.50) = 1.89 µm.E41-32 Apply Eq. 41-21: 2nd = mλ. In one case we have 2nair = (4001)λ,in the other, 2nvac = (4000)λ.Equating, nair = (4001)/(4000) = 1.00025.E41-33 (a) We can start with the last equation from Sample Problem 41-5, 1 r= (m − )λR, 2and solve for m, r2 1 m= + λR 2In this exercise R = 5.0 m, r = 0.01 m, and λ = 589 nm. Then (0.01 m)2 m= = 34 (589 nm)(5.0 m)is the number of rings observed. (b) Putting the apparatus in water effectively changes the wavelength to (589 nm)/(1.33) = 443 nm,so the number of rings will now be (0.01 m)2 m= = 45. (443 nm)(5.0 m) 208
• E41-34 (1.42 cm) = (10 − 1 )Rλ, while (1.27 cm) = 2 (10 − 1 )Rλ/n. Divide one expression by 2 √the other, and (1.42 cm)/(1.27 cm) = n, or n = 1.25.E41-35 (0.162 cm) = (n − 1 )Rλ, while (0.368 cm) = 2 (n + 20 − 1 )Rλ. Square both expres- 2sions, the divide one by the other, and find (n + 19.5)/(n − 0.5) = (0.368 cm/0.162 cm)2 = 5.16which can be rearranged to yield 19.5 + 5.16 × 0.5 n= = 5.308. 5.16 − 1Oops! That should be an integer, shouldn’t it? The above work is correct, which means that therereally aren’t bright bands at the specified locations. I’m just going to gloss over that fact and solvefor R using the value of m = 5.308. Then R = r2 /(m − 1/2)λ = (0.162 cm)2 /(5.308 − 0.5)(546 nm) = 1.00 m.Well, at least we got the answer which is in the back of the book...E41-36 Pretend the ship is a two point source emitter, one h above the water, and one h belowthe water. The one below the water is out of phase by half a wavelength. Then d sin θ = λ, whered = 2h, gives the angle for theta for the first minimum. λ/2h = (3.43 m)/2(23 m) = 7.46×10−2 = sin θ ≈ H/D,so D = (160 m)/(7.46×10−2 ) = 2.14 km.E41-37 The phase difference is 2π/λn times the path difference which is 2d, so φ = 4πd/λn = 4πnd/λ.We are given that d = 100×10−9 m and n = 1.38. (a) φ = 4π(1.38)(100×10−9 m)/(450×10−9 m) = 3.85. Then I (3.85) = cos2 = 0.12. I0 2The reflected ray is diminished by 1 − 0.12 = 88%. (b) φ = 4π(1.38)(100×10−9 m)/(650×10−9 m) = 2.67. Then I (2.67) = cos2 = 0.055. I0 2The reflected ray is diminished by 1 − 0.055 = 95%.E41-38 The change in the optical path length is 2(d − d/n), so 7λ/n = 2d(1 − 1/n), or 7(589×10−9 m) d= = 4.9×10−6 m. 2(1.42) − 2 209
• E41-39 When M2 moves through a distance of λ/2 a fringe has will be produced, destroyed, andthen produced again. This is because the light travels twice through any change in distance. Thewavelength of light is then 2(0.233 mm) λ= = 588 nm. 792E41-40 The change in the optical path length is 2(d − d/n), so 60λ = 2d(1 − 1/n), or 1 1 n= = −9 m)/2(5×10−2 m) = 1.00030. 1 − 60λ/2d 1 − 60(500×10 P41-1 (a) This is a small angle problem, so we use Eq. 41-4. The distance to the screen is2 × 20 m, because the light travels to the mirror and back again. Then λD (632.8 nm)(40.0 m) d= = = 0.253 mm. ∆y (0.1 m) (b) Placing the cellophane over one slit will cause the interference pattern to shift to the left orright, but not disappear or change size. How does it shift? Since we are picking up 2.5 waves thenwe are, in effect, swapping bright fringes for dark fringes.P41-2 The change in the optical path length is d − d/n, so 7λ/n = d(1 − 1/n), or 7(550×10−9 m) d= = 6.64×10−6 m. (1.58) − 1P41-3 The distance from S1 to P is r1 = (x + d/2)2 + y 2 . The distance from S2 to P isr2 = (x − d/2)2 + y 2 . The difference in distances is fixed at some value, say c, so that r 1 − r2 = c, 2 2 r1 − 2r1 r2 + r2 = c2 , (r1 + r2 − c2 )2 2 2 = 2 2 4r1 r2 , 2 (r1 − r2 ) − 2c (r1 + r2 ) + c4 2 2 2 2 2 = 0, (2xd)2 − 2c2 (2x2 + d2 /2 + 2y 2 ) + c4 = 0, 4x2 d2 − 4c2 x2 − c2 d2 − 4c2 y 2 + c4 = 0, 4(d2 − c2 )x2 − 4c2 y 2 = c2 (d2 − c2 ).Yes, that is the equation of a hyperbola.P41-4 The change in the optical path length for each slit is nt − t, where n is the correspondingindex of refraction. The net change in the path difference is then n2 t − n1 t. Consequently, mλ =t(n2 − n1 ), so (5)(480×10−9 m) t= = 8.0×10−6 m. (1.7) − (1.4)P41-5 The intensity is given by Eq. 41-17, which, in the small angle approximation, can bewritten as πdθ Iθ = 4I0 cos2 . λ 210
• The intensity will be half of the maximum when 1 πd∆θ/2 = cos2 2 λor π πd∆θ = , 4 2λwhich will happen if ∆θ = λ/2d.P41-6 Follow the construction in Fig. 41-10, except that one of the electric field amplitudes istwice the other. The resultant field will have a length given by E = (2E0 + E0 cos φ)2 + (E0 sin φ)2 , = E0 5 + 4 cos φ,so squaring this yields 2πd sin θ I = I0 5 + 4 cos , λ πd sin θ = I0 1 + 8 cos2 , λ Im πd sin θ = 1 + 8 cos2 . 9 λP41-7 We actually did this problem in Exercise 41-27, although slightly differently. One maxi-mum is 2(1.32)d = (m − 1/2)(679 nm),the other is 2(1.32)d = (m + 1/2)(485 nm).Set these equations equal to each other, (m − 1/2)(679 nm) = (m + 1/2)(485 nm),and find m = 3. Then the thickness is d = (3 − 1/2)(679 nm)/2(1.32) = 643 nm.P41-8 (a) Since we are concerned with transmission there is a phase shift for two rays, so 2d = mλnThe minimum thickness occurs when m = 1; solving for d yields λ (525×10−9 m) d= = = 169×10−9 m. 2n 2(1.55) (b) The wavelengths are different, so the other parts have differing phase differences. (c) The nearest destructive interference wavelength occurs when m = 1.5, or λ = 2nd = 2(1.55)1.5(169×10−9 m) = 393×10−9 m.This is blue-violet. 211
• P41-9 It doesn’t matter if we are looking at bright are dark bands. It doesn’t even matter if weconcern ourselves with phase shifts. All that cancels out. Consider 2δd = δmλ; then δd = (10)(480 nm)/2 = 2.4 µm.P41-10 (a) Apply 2d = mλ. Then d = (7)(600×10−9 m)/2 = 2100×10−9 m. (b) When water seeps in it introduces an extra phase shift. Point A becomes then a bright fringe,and the equation for the number of bright fringes is 2nd = mλ. Solving for m, m = 2(1.33)(2100×10−9 m)/(600×10−9 m) = 9.3;this means that point B is almost, but not quite, a dark fringe, and there are nine of them.P41-11 (a) Look back at the work for Sample Problem 41-5 where it was found 1 rm = (m − )λR, 2We can write this as 1 rm = 1− mλR 2mand expand the part in parentheses in a binomial expansion, 1 1 √ rm ≈ 1 − mλR. 2 2mWe will do the same with 1 rm+1 = (m + 1 − )λR, 2expanding 1 rm+1 = 1+ mλR 2mto get 1 1 √ rm+1 ≈ 1+ mλR. 2 2mThen 1 √ ∆r ≈ mλR, 2mor 1 ∆r ≈ λR/m. 2 (b) The area between adjacent rings is found from the difference, 2 2 A = π rm+1 − rm ,and into this expression we will substitute the exact values for rm and rm+1 , 1 1 A = π (m + 1 − )λR − (m − )λR , 2 2 = πλR.Unlike part (a), we did not need to assume m 1 in order to arrive at this expression; it is exactfor all m. 212
• P41-12 The path length shift that occurs when moving the mirror as distance x is 2x. This meansφ = 2π2x/λ = 4πx/λ. The intensity is then 2πx I = 4I0 cos2 λ 213
• E42-1 λ = a sin θ = (0.022 mm) sin(1.8◦ ) = 6.91×10−7 m.E42-2 a = λ/ sin θ = (0.10×10−9 m)/ sin(0.12×10−3 rad/2) = 1.7 µm. E42-3 (a) This is a valid small angle approximation problem: the distance between the pointson the screen is much less than the distance to the screen. Then (0.0162 m) θ≈ = 7.5 × 10−3 rad. (2.16 m) (b) The diffraction minima are described by Eq. 42-3, a sin θ = mλ, a sin(7.5 × 10 rad) = (2)(441 × 10−9 m), −3 a = 1.18 × 10−4 m.E42-4 a = λ/ sin θ = (633×10−9 m)/ sin(1.97◦ /2) = 36.8 µm. E42-5 (a) We again use Eq. 42-3, but we will need to throw in a few extra subscripts todistinguish between which wavelength we are dealing with. If the angles match, then so will the sineof the angles. We then have sin θa,1 = sin θb,2 or, using Eq. 42-3, (1)λa (2)λb = , a afrom which we can deduce λa = 2λb . (b) Will any other minima coincide? We want to solve for the values of ma and mb that will beintegers and have the same angle. Using Eq. 42-3 one more time, ma λa mb λ b = , a aand then substituting into this the relationship between the wavelengths, ma = mb /2. whenever mbis an even integer ma is an integer. Then all of the diffraction minima from λa are overlapped by aminima from λb .E42-6 The angle is given by sin θ = 2λ/a. This is a small angle, so we can use the small angleapproximation of sin θ = y/D. Then y = 2Dλ/a = 2(0.714 m)(593×10−9 m)/(420×10−6 m) = 2.02 mm.E42-7 Small angles, so y/D = sin θ = λ/a. Then a = Dλ/y = (0.823 m)(546×10−9 m)/(5.20×10−3 m/2) = 1.73×10−4 m.E42-8 (b) Small angles, so ∆y/D = ∆mλ/a. Then a = ∆mDλ/∆y = (5 − 1)(0.413 m)(546×10−9 m)/(0.350×10−3 m) = 2.58 mm. (a) θ = arcsin(λ/a) = arcsin[(546×10−9 m)/(2.58 mm)] = 1.21×10−2◦ .E42-9 Small angles, so ∆y/D = ∆mλ/a. Then ∆y = ∆mDλ/a = (2 − 1)(2.94 m)(589×10−9 m)/(1.16×10−3 m) = 1.49×10−3 m. 214
• E42-10 Doubling the width of the slit results in a narrowing of the diffraction pattern. Since thewidth of the central maximum is effectively cut in half, then there is twice the energy in half thespace, producing four times the intensity.E42-11 (a) This is a small angle approximation problem, so θ = (1.13 cm)/(3.48 m) = 3.25 × 10−3 rad. (b) A convenient measure of the phase difference, α is related to θ through Eq. 42-7, πa π(25.2 × 10−6 m) α= sin θ = sin(3.25 × 10−3 rad) = 0.478 rad λ (538 × 10−9 m) (c) The intensity at a point is related to the intensity at the central maximum by Eq. 42-8, 2 2 Iθ sin α sin(0.478 rad) = = = 0.926 Im α (0.478 rad)E42-12 Consider Fig. 42-11; the angle with the vertical is given by (π − φ)/2. For Fig. 42-10(d)the circle has wrapped once around onto itself so the angle with the vertical is (3π −φ)/2. Substituteα into this expression and the angel against the vertical is 3π/2 − α. Use the result from Problem 42-3 that tan α = α for the maxima. The lowest non-zero solutionis α = 4.49341 rad. The angle against the vertical is then 0.21898 rad, or 12.5◦ .E42-13 Drawing heavily from Sample Problem 42-4, αx λ 1.39 θx = arcsin = arcsin = 2.54◦ . πa 10πFinally, ∆θ = 2θx = 5.1◦ .E42-14 (a) Rayleigh’s criterion for resolving images (Eq. 42-11) requires that two objects have anangular separation of at least 1.22λ 1.22(540 × 10−9 ) θR = sin−1 = sin−1 = 1.34 × 10−4 rad d (4.90 × 10−3 m) (b) The linear separation is y = θD = (1.34 × 10−4 rad)(163×103 m) = 21.9 m. E42-15 (a) Rayleigh’s criterion for resolving images (Eq. 42-11) requires that two objects havean angular separation of at least 1.22λ 1.22(562 × 10−9 ) θR = sin−1 = sin−1 = 1.37 × 10−4 rad. d (5.00 × 10−3 m) (b) Once again, this is a small angle, so we can use the small angle approximation to find thedistance to the car. In that case θR = y/D, where y is the headlight separation and D the distanceto the car. Solving, D = y/θR = (1.42 m)/(1.37 × 10−4 rad) = 1.04 × 104 m,or about six or seven miles. 215
• E42-16 y/D = 1.22λ/a; or D = (5.20×10−3 m)(4.60×10−3 /m)/1.22(542×10−9 m) = 36.2 m.E42-17 The smallest resolvable angular separation will be given by Eq. 42-11, 1.22λ 1.22(565 × 10−9 m) θR = sin−1 = sin−1 = 1.36 × 10−7 rad, d (5.08 m)The smallest objects resolvable on the Moon’s surface by this telescope have a size y where y = DθR = (3.84 × 108 m)(1.36 × 10−7 rad) = 52.2 mE42-18 y/D = 1.22λ/a; or y = 1.22(1.57×10−2 m)(6.25×103 m)/(2.33 m) = 51.4 mE42-19 y/D = 1.22λ/a; or D = (4.8×10−2 m)(4.3×10−3 /m)/1.22(0.12×10−9 m) = 1.4×106 m.E42-20 y/D = 1.22λ/a; or d = 1.22(550×10−9 m)(160×103 m)/(0.30 m) = 0.36 m.E42-21 Using Eq. 42-11, we find the minimum resolvable angular separation is given by 1.22λ 1.22(475 × 10−9 m) θR = sin−1 = sin−1 = 1.32 × 10−4 rad d (4.4 × 10−3 m)The dots are 2 mm apart, so we want to stand a distance D away such that D > y/θR = (2 × 10−3 m)/(1.32 × 10−4 rad) = 15 m.E42-22 y/D = 1.22λ/a; or y = 1.22(500×10−9 m)(354×103 m)/(9.14 m/2) = 4.73×10−2 m.E42-23 (a) λ = v/f . Now use Eq. 42-11: (1450 m/s) θ = arcsin 1.22 = 6.77◦ . (25×103 Hz)(0.60 m) (b) Following the same approach, (1450 m/s) θ = arcsin 1.22 (1×103 Hz)(0.60 m)has no real solution, so there is no minimum. 216
• E42-24 (a) λ = v/f . Now use Eq. 42-11: (3×108 m/s) θ = arcsin 1.22 = 0.173◦ . (220×109 Hz)(0.55 m)This is the angle from the central maximum; the angular width is twice this, or 0.35◦ . (b) use Eq. 42-11: (0.0157 m) θ = arcsin 1.22 = 0.471◦ . (2.33 m)This is the angle from the central maximum; the angular width is twice this, or 0.94◦ .E42-25 The linear separation of the fringes is given by ∆y λ λD = ∆θ = or ∆y = D d dfor sufficiently small d compared to λ.E42-26 (a) d sin θ = 4λ gives the location of the fourth interference maximum, while a sin θ =λ gives the location of the first diffraction minimum. Hence, if d = 4a there will be no fourthinterference maximum! (b) Since d sin θmi = mi λ gives the interference maxima and a sin θmd = md λ gives the diffractionminima, and d = 4a, then whenever mi = 4md there will be a missing maximum.E42-27 (a) The central diffraction envelope is contained in the range λ θ = arcsin aThis angle corresponds to the mth maxima of the interference pattern, where sin θ = mλ/d = mλ/2a.Equating, m = 2, so there are three interference bands, since the m = 2 band is “washed out” bythe diffraction minimum. (b) If d = a then β = α and the expression reduces to sin2 α Iθ = I m cos2 α , α2 sin2 (2α) = Im , 2α2 2 sin α = 2I m , αwhere α = 2α , which is the same as replacing a by 2a.E42-28 Remember that the central peak has an envelope width twice that of any other peak.Ignoring the central maximum there are (11 − 1)/2 = 5 fringes in any other peak envelope. 217
• E42-29 (a) The first diffraction minimum is given at an angle θ such that a sin θ = λ; the orderof the interference maximum at that point is given by d sin θ = mλ. Dividing one expression by theother we get d/a = m, with solution m = (0.150)/(0.030) = 5. The fact that the answer is exactly 5implies that the fifth interference maximum is squelched by the diffraction minimum. Then there areonly four complete fringes on either side of the central maximum. Add this to the central maximumand we get nine as the answer. (b) For the third fringe m = 3, so d sin θ = 3λ. Then β is Eq. 42-14 is 3π, while α in Eq. 42-16is πa 3λ a α= = 3π , λ d dso the relative intensity of the third fringe is, from Eq. 42-17, 2 sin(3πa/d) (cos 3π)2 = 0.255. (3πa/d)P42-1 y = mλD/a. Then y = (10)(632.8×10−9 m)(2.65 m)/(1.37×10−3 m) = 1.224×10−2 m.The separation is twice this, or 2.45 cm.P42-2 If a λ then the diffraction pattern is extremely tight, and there is effectively no light atP . In the event that either shape produces an interference pattern at P then the other shape mustproduce an equal but opposite electric field vector at that point so that when both patterns fromboth shapes are superimposed the field cancel. But the intensity is the field vector squared; hence the two patterns look identical.P42-3 (a) We want to take the derivative of Eq. 42-8 with respect to α, so 2 dIθ d sin α = Im , dα dα α sin α cos α sin α = I m2 − 2 , α α α sin α = I m 2 3 (α cos α − sin α) . αThis equals zero whenever sin α = 0 or α cos α = sin α; the former is the case for a minima while thelatter is the case for the maxima. The maxima case can also be written as tan α = α. (b) Note that as the order of the maxima increases the solutions get closer and closer to oddintegers times π/2. The solutions are α = 0, 1.43π, 2.46π, etc. (c) The m values are m = α/π − 1/2, and correspond to m = 0.5, 0.93, 1.96, etc.These values will get closer and closer to integers as the values are increased. 218
• P42-4 The outgoing beam strikes the moon with a circular spot of radius r = 1.22λD/a = 1.22(0.69×10−6 m)(3.82×108 m)/(2 × 1.3 m) = 123 m.The light is not evenly distributed over this circle. If P0 is the power in the light, then R P0 = Iθ r dr dφ = 2π Iθ r dr, 0where R is the radius of the central peak and Iθ is the angular intensity. For a λ we can writeα ≈ πar/λD, then 2 π/2 2 λD sin2 α λD P0 = 2πIm dα ≈ 2πIm (0.82). πa 0 α πaThen the intensity at the center falls off with distance D as 2 Im = 1.9 (a/λD) P0 The fraction of light collected by the mirror on the moon is then 2 (2 × 1.3 m) P1 /P0 = 1.9 π(0.10 m)2 = 5.6×10−6 . (0.69×10−6 m)(3.82×108 m)The fraction of light collected by the mirror on the Earth is then 2 (2 × 0.10 m) P2 /P1 = 1.9 π(1.3 m)2 = 5.6×10−6 . (0.69×10−6 m)(3.82×108 m)Finally, P2 /P0 = 3×10−11 .P42-5 (a) The ring is reddish because it occurs at the blue minimum. (b) Apply Eq. 42-11 for blue light: d = 1.22λ/ sin θ = 1.22(400 nm)/ sin(0.375◦ ) = 70 µm. (c) Apply Eq. 42-11 for red light: θ = arcsin (1.22(700 nm)/(70 µm)) ≈ 0.7◦ ,which occurs 3 lunar radii from the moon.P42-6 The diffraction pattern is a property of the speaker, not the interference between the speak-ers. The diffraction pattern should be unaffected by the phase shift. The interference pattern,however, should shift up or down as the phase of the second speaker is varied. P42-7 (a) The missing fringe at θ = 5◦ is a good hint as to what is going on. There should besome sort of interference fringe, unless the diffraction pattern has a minimum at that point. Thiswould be the first minimum, so a sin(5◦ ) = (440 × 10−9 m)would be a good measure of the width of each slit. Then a = 5.05 × 10−6 m. 219
• (b) If the diffraction pattern envelope were not present we could expect that the fourth interfer-ence maxima beyond the central maximum would occur at this point, and then d sin(5◦ ) = 4(440 × 10−9 m)yielding d = 2.02 × 10−5 m. (c) Apply Eq. 42-17, where β = mπ and πa πa mλ πa α= sin θ = =m = mπ/4. λ λ d dThen for m = 1 we have 2 sin(π/4) I1 = (7) = 5.7; (π/4)while for m = 2 we have 2 sin(2π/4) I2 = (7) = 2.8. (2π/4)These are in good agreement with the figure. 220
• E43-1 (a) d = (21.5×10−3 m)/(6140) = 3.50×10−6 m. (b) There are a number of angles allowed: θ = arcsin[(1)(589×10−9 m)/(3.50×10−6 m)] = 9.7◦ , θ = arcsin[(2)(589×10−9 m)/(3.50×10−6 m)] = 19.5◦ , θ = arcsin[(3)(589×10−9 m)/(3.50×10−6 m)] = 30.3◦ , θ = arcsin[(4)(589×10−9 m)/(3.50×10−6 m)] = 42.3◦ , θ = arcsin[(5)(589×10−9 m)/(3.50×10−6 m)] = 57.3◦ .E43-2 The distance between adjacent rulings is (2)(612×10−9 m) d= = 2.235×10−6 m. sin(33.2◦ )The number of lines is then N = D/d = (2.86×10−2 m)/(2.235×10−6 m) = 12, 800.E43-3 We want to find a relationship between the angle and the order number which is linear.We’ll plot the data in this representation, and then use a least squares fit to find the wavelength. The data to be plotted is m θ sin θ m θ sin θ 1 17.6◦ 0.302 -1 -17.6◦ -0.302 2 37.3◦ 0.606 -2 -37.1◦ -0.603 3 65.2◦ 0.908 -3 -65.0◦ -0.906On my calculator I get the best straight line fit as 0.302m + 8.33 × 10−4 = sin θm ,which means that λ = (0.302)(1.73 µm) = 522 nm.E43-4 Although an approach like the solution to Exercise 3 should be used, we’ll assume that eachmeasurement is perfect and error free. Then randomly choosing the third maximum, d sin θ (5040×10−9 m) sin(20.33◦ ) λ= = = 586×10−9 m. m (3)E43-5 (a) The principle maxima occur at points given by Eq. 43-1, λ sin θm = m . dThe difference of the sine of the angle between any two adjacent orders is λ λ λ sin θm+1 − sin θm = (m + 1) −m = . d d dUsing the information provided we can find d from λ (600 × 10−9 ) d= = = 6 µm. sin θm+1 − sin θm (0.30) − (0.20) 221
• It doesn’t take much imagination to recognize that the second and third order maxima were given. (b) If the fourth order maxima is missing it must be because the diffraction pattern envelope hasa minimum at that point. Any fourth order maxima should have occurred at sin θ4 = 0.4. If it is adiffraction minima then a sin θm = mλ where sin θm = 0.4We can solve this expression and find λ (600 × 10−9 m) a=m =m = m1.5 µm. sin θm (0.4)The minimum width is when m = 1, or a = 1.5 µm. (c) The visible orders would be integer values of m except for when m is a multiple of four.E43-6 (a) Find the maximum integer value of m = d/λ = (930 nm)/(615 nm) = 1.5, hencem = −1, 0, +1; there are three diffraction maxima. (b) The first order maximum occurs at θ = arcsin(615 nm)/(930 nm) = 41.4◦ .The width of the maximum is (615 nm) δθ = = 7.87×10−4 rad, (1120)(930 nm) cos(41.4◦ )or 0.0451◦ .E43-7 The fifth order maxima will be visible if d/λ ≥ 5; this means d (1×10−3 m) λ≤ = = 635×10−9 m. 5 (315 rulings)(5)E43-8 (a) The maximum could be the first, and then d sin θ (1×10−3 m) sin(28◦ ) λ= = = 2367×10−9 m. m (200)(1)That’s not visible. The first visible wavelength is at m = 4, then d sin θ (1×10−3 m) sin(28◦ ) λ= = = 589×10−9 m. m (200)(4)The next is at m = 5, then d sin θ (1×10−3 m) sin(28◦ ) λ= = = 469×10−9 m. m (200)(5)Trying m = 6 results in an ultraviolet wavelength. (b) Yellow-orange and blue. 222
• E43-9 A grating with 400 rulings/mm has a slit separation of 1 d= = 2.5 × 10−3 mm. 400 mm−1To find the number of orders of the entire visible spectrum that will be present we need only considerthe wavelength which will be on the outside of the maxima. That will be the longer wavelengths, sowe only need to look at the 700 nm behavior. Using Eq. 43-1, d sin θ = mλ,and using the maximum angle 90◦ , we find d (2.5 × 10−6 m) m< = = 3.57, λ (700 × 10−9 m)so there can be at most three orders of the entire spectrum.E43-10 In this case d = 2a. Since interference maxima are given by sin θ = mλ/d while diffractionminima are given at sin θ = m λ/a = 2m λ/d then diffraction minima overlap with interferencemaxima whenever m = 2m . Consequently, all even m are at diffraction minima and thereforevanish. E43-11 If the second-order spectra overlaps the third-order, it is because the 700 nm second-orderline is at a larger angle than the 400 nm third-order line. Start with the wavelengths multiplied by the appropriate order parameter, then divide both sideby d, and finally apply Eq. 43-1. 2(700 nm) > 3(400 nm), 2(700 nm) 3(400 nm) > , d d sin θ2,λ=700 > sin θ3,λ=400 ,regardless of the value of d.E43-12 Fig. 32-2 shows the path length difference for the right hand side of the grating as d sin θ.If the beam strikes the grating at ang angle ψ then there will be an additional path length differenceof d sin ψ on the right hand side of the figure. The diffraction pattern then has two contributions tothe path length difference, these add to give d(sin θ + sin psi) = mλ.E43-13E43-14 Let d sin θi = λi and θ1 + 20◦ = θ2 . Then sin θ2 = sin θ1 cos(20◦ ) + cos θ1 sin(20◦ ).Rearranging, sin θ2 = sin θ1 cos(20◦ ) + 1 − sin2 θ1 sin(20◦ ).Substituting the equations together yields a rather nasty expression, λ2 λ1 = cos(20◦ ) + 1 − (λ1 /d)2 sin(20◦ ). d d 223
• Rearranging, 2 (λ2 − λ1 cos(20◦ )) = d2 − λ2 sin2 (20◦ ). 1Use λ1 = 430 nm and λ2 = 680 nm, then solve for d to find d = 914 nm. This corresponds to 1090rulings/mm.E43-15 The shortest wavelength passes through at an angle of θ1 = arctan(50 mm)/(300 mm) = 9.46◦ .This corresponds to a wavelength of (1×10−3 m) sin(9.46◦ ) λ1 = = 470×10−9 m. (350)The longest wavelength passes through at an angle of θ2 = arctan(60 mm)/(300 mm) = 11.3◦ .This corresponds to a wavelength of (1×10−3 m) sin(11.3◦ ) λ2 = = 560×10−9 m. (350)E43-16 (a) ∆λ = λ/R = λ/N m, so ∆λ = (481 nm)/(620 rulings/mm)(5.05 mm)(3) = 0.0512 nm. (b) mm is the largest integer smaller than d/λ, or mm ≤ 1/(481×10−9 m)(620 rulings/mm) = 3.35,so m = 3 is highest order seen.E43-17 The required resolving power of the grating is given by Eq. 43-10 λ (589.0 nm) R= = = 982. ∆λ (589.6 nm) − (589.0 nm)Our resolving power is then R = 1000. Using Eq. 43-11 we can find the number of grating lines required. We are looking at the second-order maxima, so R (1000) N= = = 500. m (2)E43-18 (a) N = R/m = λ/m∆λ, so (415.5 nm) N= = 23100. (2)(415.496 nm − 415.487 nm) (b) d = w/N , where w is the width of the grating. Then mλ (23100)(2)(415.5×10−9 m) θ = arcsin = arcsin = 27.6◦ . d (4.15×10−2 m) 224
• E43-19 N = R/m = λ/m∆λ, so (656.3 nm) N= = 3650 (1)(0.180 nm)E43-20 Start with Eq. 43-9: m d sin θ/λ tan θ D= = = . d cos θ d cos θ λE43-21 (a) We find the ruling spacing by Eq. 43-1, mλ (3)(589 nm) d= = = 9.98 µm. sin θm sin(10.2◦ ) (b) The resolving power of the grating needs to be at least R = 1000 for the third-order line; seethe work for Ex. 43-17 above. The number of lines required is given by Eq. 43-11, R (1000) N= = = 333, m (3)so the width of the grating (or at least the part that is being used) is 333(9.98 µm) = 3.3 mm.E43-22 (a) Condition (1) is satisfied if d ≥ 2(600 nm)/ sin(30◦ ) = 2400 nm.The dispersion is maximal for the smallest d, so d = 2400 nm. (b) To remove the third order requires d = 3a, or a = 800 nm.E43-23 (a) The angles of the first three orders are (1)(589×10−9 m)(40000) θ1 = arcsin = 18.1◦ , (76×10−3 m) (2)(589×10−9 m)(40000) θ2 = arcsin = 38.3◦ , (76×10−3 m) (3)(589×10−9 m)(40000) θ3 = arcsin = 68.4◦ . (76×10−3 m)The dispersion for each order is (1)(40000) 360◦ D1 = = 3.2×10−2◦ /nm, (76×106 nm) cos(18.1◦ ) 2π (2)(40000) 360◦ D2 = = 7.7×10−2◦ /nm, (76×106 nm) cos(38.3◦ ) 2π (3)(40000) 360◦ D3 = = 2.5×10−1◦ /nm. (76×106 nm) cos(68.4◦ ) 2π (b) R = N m, so R1 = (40000)(1) = 40000, R2 = (40000)(2) = 80000, R3 = (40000)(3) = 120000. 225
• E43-24 d = mλ/2 sin θ, so (2)(0.122 nm) d= = 0.259 nm. 2 sin(28.1◦ )E43-25 Bragg reflection is given by Eq. 43-12 2d sin θ = mλ,where the angles are measured not against the normal, but against the plane. The value of d dependson the family of planes under consideration, but it is at never larger than a0 , the unit cell dimension. We are looking for the smallest angle; this will correspond to the largest d and the smallest m.That means m = 1 and d = 0.313 nm. Then the minimum angle is (1)(29.3 × 10−12 m) θ = sin−1 = 2.68◦ . 2(0.313 × 10−9 m)E43-26 2d/λ = sin θ1 and 2d/2λ = sin θ2 . Then θ2 = arcsin[2 sin(3.40◦ )] = 6.81◦ .E43-27 We apply Eq. 43-12 to each of the peaks and find the product mλ = 2d sin θ.The four values are 26 pm, 39 pm, 52 pm, and 78 pm. The last two values are twice the first two,so the wavelengths are 26 pm and 39 pm.E43-28 (a) 2d sin θ = mλ, so (3)(96.7 pm) d= = 171 pm. 2 sin(58.0◦ ) (b) λ = 2(171 pm) sin(23.2◦ )/(1) = 135 pm.E43-29 The angle against the face of the crystal is 90◦ − 51.3◦ = 38.7◦ . The wavelength is λ = 2(39.8 pm) sin(38.7◦ )/(1) = 49.8 pm.E43-30 If λ > 2d then λ/2d > 1. But λ/2d = sin θ/m.This means that sin θ > m, but the sine function can never be greater than one.E43-31 There are too many unknowns. It is only possible to determine the ratio d/λ.E43-32 A wavelength will be diffracted if mλ = 2d sin θ. The possible solutions are λ3 = 2(275 pm) sin(47.8)/(3) = 136 pm, λ4 = 2(275 pm) sin(47.8)/(4) = 102 pm. 226
• E43-33 We use Eq. 43-12 to first find d; mλ (1)(0.261 × 10−9 m) d= = = 1.45 × 10−10 m. 2 sin θ 2 sin(63.8◦ ) d is the spacing between the planes in Fig. 43-28; it correspond to half of the diagonal distancebetween two cell centers. Then (2d)2 = a2 + a2 , 0 0or √ √ a0 = 2d = 2(1.45 × 10−10 m) = 0.205 nm.E43-34 Diffraction occurs when 2d sin θ = mλ. The angles in this case are then given by (0.125×10−9 m) sin θ = m = (0.248)m. 2(0.252×10−9 m)There are four solutions to this equation. They are 14.4◦ , 29.7◦ , 48.1◦ , and 82.7◦ . They involverotating the crystal from the original orientation (90◦ − 42.4◦ = 47.6◦ ) by amounts 47.6◦ − 14.4◦ = 33.2◦ , 47.6◦ − 29.7◦ = 17.9◦ , 47.6◦ − 48.1◦ = −0.5◦ , 47.6◦ − 82.7◦ = −35.1◦ . P43-1 Since the slits are so narrow we only need to consider interference effects, not diffractioneffects. There are three waves which contribute at any point. The phase angle between adjacentwaves is φ = 2πd sin θ/λ.We can add the electric field vectors as was done in the previous chapters, or we can do it in adifferent order as is shown in the figure below. Then the vectors sum to E(1 + 2 cos φ).We need to square this quantity, and then normalize it so that the central maximum is the maximum.Then (1 + 4 cos φ + 4 cos2 φ) I = Im . 9 227
• P43-2 (a) Solve φ for I = I m /2, this occurs when 3 √ = 1 + 2 cos φ, 2or φ = 0.976 rad. The corresponding angle θx is λφ λ(0.976) λ θx ≈ = = . 2πd 2πd 6.44dBut ∆θ = 2θx , so λ ∆θ ≈ . 3.2d (b) For the two slit pattern the half width was found to be ∆θ = λ/2d. The half width in thethree slit case is smaller. P43-3 (a) and (b) A plot of the intensity quickly reveals that there is an alternation of largemaximum, then a smaller maximum, etc. The large maxima are at φ = 2nπ, the smaller maximaare half way between those values. (c) The intensity at these secondary maxima is then (1 + 4 cos π + 4 cos2 π) Im I = Im = . 9 9 Note that the minima are not located half-way between the maxima!P43-4 Covering up the middle slit will result in a two slit apparatus with a slit separation of 2d.The half width, as found in Problem 41-5, is then ∆θ = λ/2(2d), = λ/4d,which is narrower than before covering up the middle slit by a factor of 3.2/4 = 0.8.P43-5 (a) If N is large we can treat the phasors as summing to form a flexible “line” of lengthN δE. We then assume (incorrectly) that the secondary maxima occur when the loop wraps aroundon itself as shown in the figures below. Note that the resultant phasor always points straight up.This isn’t right, but it is close to reality. 228
• The length of the resultant depends on how many loops there are. For k = 0 there are none.For k = 1 there are one and a half loops. The circumference of the resulting circle is 2N δE/3, thediameter is N δE/3π. For k = 2 there are two and a half loops. The circumference of the resultingcircle is 2N δE/5, the diameter is N δE/5π. The pattern for higher k is similar: the circumferenceis 2N δE/(2k + 1), the diameter is N δE/(k + 1/2)π. The intensity at this “approximate” maxima is proportional to the resultant squared, or (N δE)2 Ik ∝ . (k + 1/2)2 π 2but I m is proportional to (N δE)2 , so 1 Ik = I m . (k + 1/2)2 π 2 (b) Near the middle the vectors simply fold back on one another, leaving a resultant of δE. Then (N δE)2 Ik ∝ (δE)2 = , N2so Im Ik = , N2 (c) Let α have the values which result in sin α = 1, and then the two expressions are identical!P43-6 (a) v = f λ, so δv = f δλ + λδf . Assuming δv = 0, we have δf /f = −δλ/λ. Ignore thenegative sign (we don’t need it here). Then λ f c R= = = , ∆λ ∆f λ∆fand then c c ∆f = = . Rλ N mλ (b) The ray on the top gets there first, the ray on the bottom must travel an additional distanceof N d sin θ. It takes a time ∆t = N d sin θ/cto do this. (c) Since mλ = d sin θ, the two resulting expression can be multiplied together to yield c N d sin θ (∆f )(∆t) = = 1. N mλ cThis is almost, but not quite, one of Heisenberg’s uncertainty relations!P43-7 (b) We sketch parallel lines which connect centers to form almost any right triangle similarto the one shown in the Fig. 43-18. The triangle will have two sides which have integer multiple √lengths of the lattice spacing a0 . The hypotenuse of the triangle will then have length h2 + k 2 a0 ,where h and k are the integers. In Fig. 43-18 h = 2 while k = 1. The number of planes which cutthe diagonal is equal to h2 + k 2 if, and only if, h and k are relatively prime. The inter-planar spacingis then √ h2 + k 2 a0 a0 d= 2 + k2 =√ . h h 2 + k2 229
• (a) The next five spacings are then √ h = 1, k = 1, d = a0 / 2, √ h = 1, k = 2, d = a0 / 5, √ h = 1, k = 3, d = a0 / 10, √ h = 2, k = 3, d = a0 / 13, √ h = 1, k = 4, d = a0 / 17.P43-8 The middle layer cells will also diffract a beam, but this beam will be exactly π out of phasewith the top layer. The two beams will then cancel out exactly because of destructive interference. 230
• E44-1 (a) The direction of propagation is determined by considering the argument of the sinefunction. As t increases y must decrease to keep the sine function “looking” the same, so the waveis propagating in the negative y direction. (b) The electric field is orthogonal (perpendicular) to the magnetic field (so Ex = 0) and thedirection of motion (so Ey = 0); Consequently, the only non-zero term is Ez . The magnitude of Ewill be equal to the magnitude of B times c. Since S = E × B/µ0 , when B points in the positivex direction then E must point in the negative z direction in order that S point in the negative ydirection. Then Ez = −cB sin(ky + ωt). (c) The polarization is given by the direction of the electric field, so the wave is linearly polarizedin the z direction.E44-2 Let one wave be polarized in the x direction and the other in the y direction. Then the netelectric field is given by E 2 = Ex + Ey , or 2 2 E 2 = E0 sin2 (kz − ωt) + sin2 (kz − ωt + β) , 2where β is the phase difference. We can consider any point in space, including z = 0, and thenaverage the result over a full cycle. Since β merely shifts the integration limits, then the result isindependent of β. Consequently, there are no interference effects.E44-3 (a) The transmitted intensity is I0 /2 = 6.1×10−3 W/m2 . The maximum value of the electricfield is Em = 2µ0 cI = 2(1.26×10−6 H/m)(3.00×108 m/s)(6.1×10−3 W/m2 ) = 2.15 V/m. (b) The radiation pressure is caused by the absorbed half of the incident light, so p = I/c = (6.1×10−3 W/m2 )/(3.00×108 m/s) = 2.03×10−11 Pa.E44-4 The first sheet transmits half the original intensity, the second transmits an amount pro-portional to cos2 θ. Then I = (I0 /2) cos2 θ, or θ = arccos 2I/I0 = arccos 2(I0 /3)/I0 35.3◦ . E44-5 The first sheet polarizes the un-polarized light, half of the intensity is transmitted, soI1 = 1 I0 . 2 The second sheet transmits according to Eq. 44-1, 1 1 I2 = I1 cos2 θ = I0 cos2 (45◦ ) = I0 , 2 4and the transmitted light is polarized in the direction of the second sheet. The third sheet is 45◦ to the second sheet, so the intensity of the light which is transmittedthrough the third sheet is 1 1 I3 = I2 cos2 θ = I0 cos2 (45◦ ) = I0 . 4 8 231
• E44-6 The transmitted intensity through the first sheet is proportional to cos2 θ, the transmittedintensity through the second sheet is proportional to cos2 (90◦ − θ) = sin2 θ. Then I = I0 cos2 θ sin2 θ = (I0 /4) sin2 2θ,or 1 1 θ= arcsin 4I/I0 = arcsin 4(0.100I0 )/I0 = 19.6◦ . 2 2Note that 70.4◦ is also a valid solution!E44-7 The first sheet transmits half of the original intensity; each of the remaining sheets transmitsan amount proportional to cos2 θ, where θ = 30◦ . Then I 1 3 1 6 = cos2 θ = (cos(30◦ )) = 0.211 I0 2 2E44-8 The first sheet transmits an amount proportional to cos2 θ, where θ = 58.8◦ . The secondsheet transmits an amount proportional to cos2 (90◦ − θ) = sin2 θ. Then I = I0 cos2 θ sin2 θ = (43.3 W/m2 ) cos2 (58.8◦ ) sin2 (58.8◦ ) = 8.50 W/m2 . E44-9 Since the incident beam is unpolarized the first sheet transmits 1/2 of the original intensity.The transmitted beam then has a polarization set by the first sheet: 58.8◦ to the vertical. The secondsheet is horizontal, which puts it 31.2◦ to the first sheet. Then the second sheet transmits cos2 (31.2◦ )of the intensity incident on the second sheet. The final intensity transmitted by the second sheetcan be found from the product of these terms, 1 I = (43.3 W/m2 ) cos2 (31.2◦ ) = 15.8 W/m2 . 2E44-10 θp = arctan(1.53/1.33) = 49.0◦ . E44-11 (a) The angle for complete polarization of the reflected ray is Brewster’s angle, and isgiven by Eq. 44-3 (since the first medium is air) θp = tan−1 n = tan−1 (1.33) = 53.1◦ . (b) Since the index of refraction depends (slightly) on frequency, then so does Brewster’s angle.E44-12 (b) Since θr + θp = 90◦ , θp = 90◦ − (31.8◦ ) = 58.2◦ . (a) n = tan θp = tan(58.2◦ ) = 1.61.E44-13 The angles are between θp = tan−1 n = tan−1 (1.472) = 55.81◦ .and θp = tan−1 n = tan−1 (1.456) = 55.52◦ .E44-14 The smallest possible thickness t will allow for one half a wavelength phase difference forthe o and e waves. Then ∆nt = λ/2, or t = (525×10−9 m)/2(0.022) = 1.2×10−5 m. 232
• E44-15 (a) The incident wave is at 45◦ to the optical axis. This means that there are twocomponents; assume they originally point in the +y and +z direction. When they travel throughthe half wave plate they are now out of phase by 180◦ ; this means that when one component is inthe +y direction the other is in the −z direction. In effect the polarization has been rotated by 90◦ . (b) Since the half wave plate will delay one component so that it emerges 180◦ “later” than itshould, it will in effect reverse the handedness of the circular polarization. (c) Pretend that an unpolarized beam can be broken into two orthogonal linearly polarizedcomponents. Both are then rotated through 90◦ ; but when recombined it looks like the originalbeam. As such, there is no apparent change.E44-16 The quarter wave plate has a thickness of x = λ/4∆n, so the number of plates that canbe cut is given by N = (0.250×10−3 m)4(0.181)/(488×10−9 m) = 371. P44-1 Intensity is proportional to the electric field squared, so the original intensity reaching theeye is I0 , with components I h = (2.3)2 I v , and then I0 = I h + I v = 6.3I v or I v = 0.16I0 .Similarly, I h = (2.3)2 I v = 0.84I0 . (a) When the sun-bather is standing only the vertical component passes, while (b) when the sun-bather is lying down only the horizontal component passes.P44-2 The intensity of the transmitted light which was originally unpolarized is reduced to I u /2,regardless of the orientation of the polarizing sheet. The intensity of the transmitted light whichwas originally polarized is between 0 and I p , depending on the orientation of the polarizing sheet.Then the maximum transmitted intensity is I u /2 + I p , while the minimum transmitted intensity isI u /2. The ratio is 5, so I u /2 + I p Ip 5= =1+2 , I u /2 Iuor I p /I u = 2. Then the beam is 1/3 unpolarized and 2/3 polarized.P44-3 Each sheet transmits a fraction θ cos2 α = cos2 . NThere are N sheets, so the fraction transmitted through the stack is N θ cos2 . NWe want to evaluate this in the limit as N → ∞. As N gets larger we can use a small angle approximation to the cosine function, 1 cos x ≈ 1 − x2 for x 1 2The the transmitted intensity is 2N 1 θ2 1− . 2 N2 233
• This expression can also be expanded in a binomial expansion to get 1 θ2 1 − 2N , 2 N2which in the limit as N → ∞ approaches 1. The stack then transmits all of the light which makes it past the first filter. Assuming the lightis originally unpolarized, then the stack transmits half the original intensity.P44-4 (a) Stack several polarizing sheets so that the angle between any two sheets is sufficientlysmall, but the total angle is 90◦ . (b) The transmitted intensity fraction needs to be 0.95. Each sheet will transmit a fractioncos2 θ, where θ = 90◦ /N , with N the number of sheets. Then we want to solve N 0.95 = cos2 (90◦ /N )for N . For large enough N , θ will be small, so we can expand the cosine function as cos2 θ = 1 − sin2 θ ≈ 1 − θ2 ,so N 0.95 ≈ 1 − (π/2N )2 ≈ 1 − N (π/2N )2 ,which has solution N = π 2 /4(0.05) = 49. P44-5 Since passing through a quarter wave plate twice can rotate the polarization of a linearlypolarized wave by 90◦ , then if the light passes through a polarizer, through the plate, reflects off thecoin, then through the plate, and through the polarizer, it would be possible that when it passesthrough the polarizer the second time it is 90◦ to the polarizer and no light will pass. You won’t seethe coin. On the other hand if the light passes first through the plate, then through the polarizer, then isreflected, the passes again through the polarizer, all the reflected light will pass through he polarizerand eventually work its way out through the plate. So the coin will be visible. Hence, side A must be the polarizing sheet, and that sheet must be at 45◦ to the optical axis.P44-6 (a) The displacement of a ray is given by tan θk = yk /t,so the shift is ∆y = t(tan θe − tan θo ).Solving for each angle, 1 θe = arcsin sin(38.8◦ ) = 24.94◦ , (1.486) 1 θo = arcsin sin(38.8◦ ) = 22.21◦ . (1.658)The shift is then ∆y = (1.12×10−2 m) (tan(24.94) − tan(22.21)) = 6.35×10−4 m. (b) The e-ray bends less than the o-ray. (c) The rays have polarizations which are perpendicular to each other; the o-wave being polarizedalong the direction of the optic axis. (d) One ray, then the other, would disappear. 234
• P44-7 The method is outline in Sample Problem 44-24; use a polarizing sheet to pick out the o-rayor the e-ray. 235
• E45-1 (a) The energy of a photon is given by Eq. 45-1, E = hf , so hc E = hf = . λPutting in “best” numbers (6.62606876×10−34 J · s) hc = (2.99792458×108 m/s) = 1.23984×10−6 eV · m. (1.602176462×10−19 C)This means that hc = 1240 eV · nm is accurate to almost one part in 8000! (b) E = (1240 eV · nm)/(589 nm) = 2.11 eV.E45-2 Using the results of Exercise 45-1, (1240 eV · nm) λ= = 2100 nm, (0.60 eV)which is in the infrared.E45-3 Using the results of Exercise 45-1, (1240 eV · nm) E1 = = 3.31 eV, (375 nm)and (1240 eV · nm) E2 = = 2.14 eV, (580 nm)The difference is ∆E = (3.307 eV) − (2.138 eV) = 1.17 eV.E45-4 P = E/t, so, using the result of Exercise 45-1, (1240 eV · nm) P = (100/s) = 230 eV/s. (540 nm)That’s a small 3.68×10−17 W.E45-5 When talking about the regions in the sun’s spectrum it is more common to refer towavelengths than frequencies. So we will use the results of Exercise 45-1(a), and solve λ = hc/E = (1240 eV · nm)/E.The energies are between E = (1.0×1018 J)/(1.6×10−19 C) = 6.25 eV and E = (1.0×1016 J)/(1.6×10−19 C) = 625 eV. These energies correspond to wavelengths between 198 nm and 1.98 nm; this isthe ultraviolet range.E45-6 The energy per photon is E = hf = hc/λ. The intensity is power per area, which is energyper time per area, so P E nhc hc n I= = = = . A At λAt λA tBut R = n/t is the rate of photons per unit time. Since h and c are constants and I and A are equalfor the two beams, we have R1 /λ1 = R2 /λ2 , or R1 /R2 = λ1 /λ2 . 236
• E45-7 (a) Since the power is the same, the bulb with the larger energy per photon will emitsfewer photons per second. Since longer wavelengths have lower energies, the bulb emitting 700 nmmust be giving off more photons per second. (b) How many more photons per second? If E1 is the energy per photon for one of the bulbs,then N1 = P/E1 is the number of photons per second emitted. The difference is then P P P N1 − N2 = − = (λ1 − λ2 ), E1 E2 hcor (130 W) N1 −N2 = ((700×10−9 m) − (400×10−9 m)) = 1.96×1020 . (6.63×10−34 J·s)(3.00×108 m/s)E45-8 Using the results of Exercise 45-1, the energy of one photon is (1240 eV · nm) E= = 1.968 eV, (630 nm)The total light energy given off by the bulb is E t = P t = (0.932)(70 W)(730 hr)(3600 s/hr) = 1.71×108 J.The number of photons is Et (1.71×108 J) n= = = 5.43×1026 . E0 (1.968 eV)(1.6×10−19 J/eV)E45-9 Apply Wien’s law, Eq. 45-4, λmax T = 2898 µm · K; so (2898 µm · K) T = = 91×106 K. (32×10−12 m)Actually, the wavelength was supposed to be 32 µm. Then the temperature would be 91 K.E45-10 Apply Wien’s law, Eq. 45-4, λmax T = 2898 µm · K; so (2898 µm · K) λ= = 1.45 m. (0.0020 K)This is in the radio region, near 207 on the FM dial.E45-11 The wavelength of the maximum spectral radiancy is given by Wien’s law, Eq. 45-4, λmax T = 2898 µm · K.Applying to each temperature in turn, (a) λ = 1.06×10−3 m, which is in the microwave; (b) λ = 9.4×10−6 m, which is in the infrared; (c) λ = 1.6×10−6 m, which is in the infrared; (d) λ = 5.0×10−7 m, which is in the visible; (e) λ = 2.9×10−10 m, which is in the x-ray; (f) λ = 2.9×10−41 m, which is in a hard gamma ray. 237
• E45-12 (a) Apply Wien’s law, Eq. 45-4, λmax T = 2898 µm · K.; so (2898 µm · K) λ= = 5.00×10−7 m. (5800 K)That’s blue-green. (b) Apply Wien’s law, Eq. 45-4, λmax T = 2898 µm · K.; so (2898 µm · K) T = = 5270 K. (550×10−9 m)E45-13 I = σT 4 and P = IA. Then P = (5.67×10−8 W/m2 · K4 )(1900 K)4 π(0.5×10−3 m)2 = 0.58 W.E45-14 Since I ∝ T 4 , doubling T results in a 24 = 16 times increase in I. Then the new powerlevel is (16)(12.0 mW) = 192 mW.E45-15 (a) We want to apply Eq. 45-6, 2πc2 h 1 R(λ, T ) = 5 hc/λkT − 1 . λ eWe know the ratio of the spectral radiancies at two different wavelengths. Dividing the aboveequation at the first wavelength by the same equation at the second wavelength, λ5 ehc/λ1 kT − 1 1 3.5 = , λ5 ehc/λ2 kT − 1 2where λ1 = 200 nm and λ2 = 400 nm. We can considerably simplify this expression if we let x = ehc/λ2 kT ,because since λ2 = 2λ1 we would have ehc/λ1 kT = e2hc/λ2 kT = x2 .Then we get 5 1 x2 − 1 1 3.5 = = (x + 1). 2 x−1 32We will use the results of Exercise 45-1 for the exponents and then rearrange to get hc (3.10 eV) T = = = 7640 K. λ1 k ln(111) (8.62×10−5 eV/K) ln(111) (b) The method is the same, except that instead of 3.5 we have 1/3.5; this means the equationfor x is 1 1 = (x + 1), 3.5 32with solution x = 8.14, so then hc (3.10 eV) T = = = 17200 K. λ1 k ln(8.14) (8.62×10−5 eV/K) ln(8.14) 238
• E45-16 hf = φ, so φ (5.32 eV) f= = = 1.28×1015 Hz. h (4.14×10−15 eV · s)E45-17 We’ll use the results of Exercise 45-1. Visible red light has an energy of (1240 eV · nm) E= = 1.9 eV. (650 nm)The substance must have a work function less than this to work with red light. This means thatonly cesium will work with red light. Visible blue light has an energy of (1240 eV · nm) E= = 2.75 eV. (450 nm)This means that barium, lithium, and cesium will work with blue light.E45-18 Since K m = hf − φ, K m = (4.14×10−15 eV · s)(3.19×1015 Hz) − (2.33 eV) = 10.9 eV.E45-19 (a) Use the results of Exercise 45-1 to find the energy of the corresponding photon, hc (1240 eV · nm) E= = = 1.83 eV. λ (678 nm)Since this energy is less than than the minimum energy required to remove an electron then thephoto-electric effect will not occur. (b) The cut-off wavelength is the longest possible wavelength of a photon that will still result inthe photo-electric effect occurring. That wavelength is (1240 eV · nm) (1240 eV · nm) λ= = = 544 nm. E (2.28 eV)This would be visible as green.E45-20 (a) Since K m = hc/λ − φ, (1240 eV · nm) Km = (4.2 eV) = 2.0 eV. (200 nm) (b) The minimum kinetic energy is zero; the electron just barely makes it off the surface. (c) V s = K m /q = 2.0 V. (d) The cut-off wavelength is the longest possible wavelength of a photon that will still result inthe photo-electric effect occurring. That wavelength is (1240 eV · nm) (1240 eV · nm) λ= = = 295 nm. E (4.2 eV)E45-21 K m = qV s = 4.92 eV. But K m = hc/λ − φ, so (1240 eV · nm) λ= = 172 nm. (4.92 eV + 2.28 eV) 239
• E45-22 (a) K m = qV s and K m = hc/λ − φ. We have two different values for qV s and λ, sosubtracting this equation from itself yields q(V s,1 − V s,2 ) = hc/λ1 − hc/λ2 .Solving for λ2 , hc λ2 = , hc/λ1 − q(V s,1 − V s,2 ) (1240 eV · nm) = , (1240 eV · nm)/(491 nm) − (0.710 eV) + (1.43 eV) = 382 nm. (b) K m = qV s and K m = hc/λ − φ, so φ = (1240 eV · nm)/(491 nm) − (0.710 eV) = 1.82 eV.E45-23 (a) The stopping potential is given by Eq. 45-11, h φ V0 = f− , e eso (1240 eV · nm) (1.85 eV V0 = − = 1.17 V. e(410 nm e (b) These are not relativistic electrons, so v= 2K/m = c 2K/mc2 = c 2(1.17 eV)/(0.511×106 eV) = 2.14×10−3 c,or v = 64200 m/s.E45-24 It will have become the stopping potential, or h φ V0 = f− , e eso (4.14×10−15 eV · m) (2.49 eV) V0 = (6.33×1014 /s) − = 0.131 V. (1.0e) (1.0e)E45-25E45-26 (a) Using the results of Exercise 45-1, (1240 eV · nm) λ= = 62 pm. (20×103 eV) (b) This is in the x-ray region.E45-27 (a) Using the results of Exercise 45-1, (1240 eV · nm) E= = 29, 800 eV. (41.6×10−3 nm) (b) f = c/λ = (3×108 m/s)/(41.6 pm) = 7.21×1018 /s. (c) p = E/c = 29, 800 eV/c = 2.98×104 eV/c. 240
• E45-28 (a) E = hf , so (0.511×106 eV) f= = 1.23×1020 /s. (4.14×10−15 eV · s) (b) λ = c/f = (3×108 m/s)/(1.23×1020 /s) = 2.43 pm. (c) p = E/c = (0.511×106 eV)/c. E45-29 The initial momentum of the system is the momentum of the photon, p = h/λ. Thismomentum is imparted to the sodium atom, so the final speed of the sodium is v = p/m, where mis the mass of the sodium. Then h (6.63×10−34 J · s) v= = = 2.9 cm/s. λm (589×10−9 m)(23)(1.7×10−27 kg)E45-30 (a) λC = h/mc = hc/mc2 , so (1240 eV · nm) λC = = 2.43 pm. (0.511×106 eV) (c) Since Eλ = hf = hc/λ, and λ = h/mc = hc/mc2 , then Eλ = hc/λ = mc2 . (b) See part (c).E45-31 The change in the wavelength of a photon during Compton scattering is given by Eq.45-17, h λ =λ+ (1 − cos φ). mcWe’ll use the results of Exercise 45-30 to save some time, and let h/mc = λC , which is 2.43 pm. (a) For φ = 35◦ , λ = (2.17 pm) + (2.43 pm)(1 − cos 35◦ ) = 2.61 pm. (b) For φ = 115◦ , λ = (2.17 pm) + (2.43 pm)(1 − cos 115◦ ) = 5.63 pm.E45-32 (a) We’ll use the results of Exercise 45-1: (1240 eV · nm) λ= = 2.43 pm. (0.511×106 eV) (b) The change in the wavelength of a photon during Compton scattering is given by Eq. 45-17, h λ =λ+ (1 − cos φ). mcWe’ll use the results of Exercise 45-30 to save some time, and let h/mc = λC , which is 2.43 pm. λ = (2.43 pm) + (2.43 pm)(1 − cos 72◦ ) = 4.11 pm. (c) We’ll use the results of Exercise 45-1: (1240 eV · nm) E= = 302 keV. (4.11 pm) 241
• E45-33 The change in the wavelength of a photon during Compton scattering is given by Eq.45-17, h λ −λ= (1 − cos φ). mcWe are not using the expression with the form ∆λ because ∆λ and ∆E are not simply related. The wavelength is related to frequency by c = f λ, while the frequency is related to the energyby Eq. 45-1, E = hf . Then ∆E = E − E = hf − hf , 1 1 = hc − , λ λ λ −λ = hc . λλInto this last expression we substitute the Compton formula. Then h2 (1 − cos φ) ∆E = . m λλ Now E = hf = hc/λ, and we can divide this on both sides of the above equation. Also, λ = c/f ,and we can substitute this into the right hand side of the above equation. Both of these steps resultin ∆E hf = (1 − cos φ). E mc2Note that mc2 is the rest energy of the scattering particle (usually an electron), while hf is theenergy of the scattered photon.E45-34 The wavelength is related to frequency by c = f λ, while the frequency is related to theenergy by Eq. 45-1, E = hf . Then ∆E = E − E = hf − hf , 1 1 = hc − , λ λ λ −λ = hc , λλ ∆E ∆λ = , E λ + ∆λBut ∆E/E = 3/4, so 3λ + 3∆λ = 4∆λ,or ∆λ = 3λ.E45-35 The maximum shift occurs when φ = 180◦ , so h (1240 eV · nm) ∆λm = 2 =2 = 2.64×10−15 m. mc 938 MeV)E45-36 Since E = hf frequency shifts are identical to energy shifts. Then we can use the resultsof Exercise 45-33 to get (0.9999)(6.2 keV) (0.0001) = (1 − cos φ), (511 keV)which has solution φ = 7.4◦ . (b) (0.0001)(6.2 keV) = 0.62 eV. 242
• E45-37 (a) The change in wavelength is independent of the wavelength and is given by Eq.45-17, hc (1240 eV · nm) ∆λ = 2 (1 − cos φ) = 2 = 4.85×10−3 nm. mc (0.511×106 eV) (b) The change in energy is given by hc hc ∆E = − , λf λi 1 1 = hc − , λi + ∆λ λi 1 1 = (1240 eV · nm) − = −42.1 keV (9.77 pm) + (4.85 pm) (9.77 pm) (c) This energy went to the electron, so the final kinetic energy of the electron is 42.1 keV.E45-38 For φ = 90◦ ∆λ = h/mc. Then ∆E hf λ = 1− =1− , E hf λ + ∆λ h/mc = . λ + h/mcBut h/mc = 2.43 pm for the electron (see Exercise 45-30). (a) ∆E/E = (2.43 pm)/(3.00 cm + 2.43 pm) = 8.1×10−11 . (b) ∆E/E = (2.43 pm)/(500 nm + 2.43 pm) = 4.86×10−6 . (c) ∆E/E = (2.43 pm)/(0.100 nm + 2.43 pm) = 0.0237. (d) ∆E/E = (2.43 pm)/(1.30 pm + 2.43 pm) = 0.651.E45-39 We can use the results of Exercise 45-33 to get (0.90)(215 keV) (0.10) = (1 − cos φ), (511 keV)which has solution φ = 42/6◦ .E45-40 (a) A crude estimate is that the photons can’t arrive more frequently than once every10 − 8s. That would provide an emission rate of 108 /s. (b) The power output would be (1240 eV · nm) P = (108 ) = 2.3×108 eV/s, (550 nm)which is 3.6×10−11 W!E45-41 We can follow the example of Sample Problem 45-6, and apply λ = λ0 (1 − v/c). (a) Solving for λ0 , (588.995 nm) λ0 = = 588.9944 nm. (1 − (−300 m/s)(3×108 m/s) 243
• (b) Applying Eq. 45-18, h (6.6×10−34 J · s) ∆v = − =− = 3×10−2 m/s. mλ (22)(1.7×10−27 kg)(590×10−9 m) (c) Emitting another photon will slow the sodium by about the same amount.E45-42 (a) (430 m/s)/(0.15 m/s) ≈ 2900 interactions. (b) If the argon averages a speed of 220 m/s, then it requires interactions at the rate of (2900)(220 m/s)/(1.0 m) = 6.4×105 /sif it is going to slow down in time. P45-1 The radiant intensity is given by Eq. 45-3, I = σT 4 . The power that is radiated throughthe opening is P = IA, where A is the area of the opening. But energy goes both ways through theopening; it is the difference that will give the net power transfer. Then 4 4 P net = (I0 − I1 )A = σA T0 − T1 .Put in the numbers, and P net = (5.67×10−8 W/m2 ·K4 )(5.20 × 10−4 m2 ) (488 K)4 − (299 K)4 = 1.44 W.P45-2 (a) I = σT 4 and P = IA. Then T 4 = P/σA, or 4 (100 W) T = = 3248 K. (5.67×10−8 W/m2 · K4 )π(0.28×10−3 m)(1.8×10−2 m)That’s 2980◦ C. (b) The rate that energy is radiated off is given by dQ/dt = mC dT /dt. The mass is found fromm = ρV , where V is the volume. This can be combined with the power expression to yield σAT 4 = −ρV CdT /dt,which can be integrated to yield ρV C 3 3 ∆t = (1/T2 − 1/T1 ). 3σAPutting in numbers, (19300kg/m3 )(0.28×10−3 m)(132J/kgC) ∆t = [1/(2748 K)3 − 1/(3248 K)3 ], 3(5.67×10−8 W/m2 · K4 )(4) = 20 ms. P45-3 Light from the sun will “heat-up” the thin black screen. As the temperature of the screenincreases it will begin to radiate energy. When the rate of energy radiation from the screen is equalto the rate at which the energy from the sun strikes the screen we will have equilibrium. We needfirst to find an expression for the rate at which energy from the sun strikes the screen. The temperature of the sun is T S . The radiant intensity is given by Eq. 45-3, I S = σT S 4 . Thetotal power radiated by the sun is the product of this radiant intensity and the surface area of thesun, so P S = 4πrS 2 σT S 4 , 244
• where rS is the radius of the sun. Assuming that the lens is on the surface of the Earth (a reasonable assumption), then we canfind the power incident on the lens if we know the intensity of sunlight at the distance of the Earthfrom the sun. That intensity is PS PS IE = = , A 4πRE 2where RE is the radius of the Earth’s orbit. Combining, 2 rS I E = σT S 4 RE The total power incident on the lens is then 2 rS P lens = I E Alens = σT S 4 πrl 2 , REwhere rl is the radius of the lens. All of the energy that strikes the lens is focused on the image, sothe power incident on the lens is also incident on the image. The screen radiates as the temperature increases. The radiant intensity is I = σT 4 , where T isthe temperature of the screen. The power radiated is this intensity times the surface area, so P = IA = 2πri 2 σT 4 .The factor of “2” is because the screen has two sides, while ri is the radius of the image. Set thisequal to P lens , 2 rS 2πri 2 σT 4 = σT S 4 πrl 2 , REor 2 1 rS rl T 4 = T S4 . 2 RE r i The radius of the image of the sun divided by the radius of the sun is the magnification of thelens. But magnification is also related to image distance divided by object distance, so ri i = |m| = , rS oDistances should be measured from the lens, but since the sun is so far from the Earth, we won’t befar off in stating o ≈ RE . Since the object is so far from the lens, the image will be very, very closeto the focal point, so we can also state i ≈ f . Then ri f = , rS REso the expression for the temperature of the thin black screen is considerably simplified to 2 1 4 rl T4 = TS . 2 f Now we can put in some of the numbers. 1 (1.9 cm) T = (5800 K) = 1300 K. 21/4 (26 cm) 245
• P45-4 The derivative of R with respect to λ is hc π c2 h 2 π c3 h2 e( λ k T ) −10 hc + hc . λ6 (e( λ k T ) − 1) λ7 (e( λ k T ) − 1)2 k TOhh, that’s ugly. Setting it equal to zero allows considerable simplification, and we are left with (5 − x)ex = 5,where x = hc/λkT . The solution is found numerically to be x = 4.965114232. Then (1240 eV · nm) 2.898×10−3 m · K λ= −5 eV/K)T = . (4.965)(8.62×10 T P45-5 (a) If the planet has a temperature T , then the radiant intensity of the planet will beIσT 4 , and the rate of energy radiation from the planet will be P = 4πR2 σT 4 ,where R is the radius of the planet. A steady state planet temperature requires that the energy from the sun arrive at the same rateas the energy is radiated from the planet. The intensity of the energy from the sun a distance rfrom the sun is P sun /4πr2 ,and the total power incident on the planet is then P sun P = πR2 . 4πr2 Equating, P sun 4πR2 σT 4 = πR2 , 4πr2 P sun T4 = . 16πσr2 (b) Using the last equation and the numbers from Problem 3, 1 (6.96×108 m) T = √ (5800 K) = 279 K. 2 (1.5×1011 m)That’s about 43◦ F.P45-6 (a) Change variables as suggested, then λ = hc/xkT and dλ = −(hc/x2 kT )dx. Integrate(note the swapping of the variables of integration picks up a minus sign): 2πc2 h (hc/x2 kT )dx I = , (hc/xkT )5 ex − 1 2πk 4 T 4 x3 dx = , h3 c2 ex − 1 2π 5 k 4 4 = T . 15h3 c2 246
• P45-7 (a) P = E/t = nhf /t = (hc/λ)(n/t), where n/t is the rate of photon emission. Then (100 W)(589×10−9 m) n/t = = 2.96×1020 /s. (6.63×10−34 J · s)(3.00×108 m/s) (b) The flux at a distance r is the rate divided by the area of the sphere of radius r, so (2.96×1020 /s) r= = 4.8×107 m. 4π(1×104 /m2 · s) (c) The photon density is the flux divided by the speed of light; the distance is then (2.96×1020 /s) r= = 280 m. 4π(1×106 /m3 )(3×108 m/s) (d) The flux is given by (2.96×1020 /s) = 5.89×1018 /m2 · s. 4π(2.0 m)2The photon density is (5.89×1018 m2 · s)/(3.00×108 m/s) = 1.96×1010 /m3 .P45-8 Momentum conservation requires pλ = pe ,while energy conservation requires Eλ + mc2 = Ee .Square both sides of the energy expression and Eλ + 2Eλ mc2 + m2 c4 2 = Ee = p2 c2 + m2 c4 , 2 e Eλ + 2Eλ mc2 2 2 2 = pe c , pλ c + 2Eλ mc2 2 2 = p2 c2 . eBut the momentum expression can be used here, and the result is 2Eλ mc2 = 0.Not likely.P45-9 (a) Since qvB = mv 2 /r, v = (q/m)rB The kinetic energy of (non-relativistic) electrons willbe 1 1 q 2 (rB)2 K= mv 2 = , 2 2 mor 1 (1.6×10−19 C) K= (188×10−6 T · m)2 = 3.1×103 eV. 2 (9.1×10−31 kg) (b) Use the results of Exercise 45-1, (1240 eV · nm) φ= − 3.1×103 eV = 1.44×104 eV. (71×10−3 nm) 247
• P45-10 P45-11 (a) The maximum value of ∆λ is 2h/mc. The maximum energy lost by the photon isthen hc hc ∆E = − , λf λi 1 1 = hc − , λi + ∆λ λi −2h/mc = hc , λ(λ + 2h/mc)where in the last line we wrote λ for λi . The energy given to the electron is the negative of this, so 2h2 Kmax = . mλ(λ + 2h/mc)Multiplying through by 12 = (Eλ/hc)2 we get 2E 2 Kmax = . mc2 (1 + 2hc/λmc2 )or E2 Kmax = . mc2 /2 + E (b) The answer is (17.5 keV)2 Kmax = = 1.12 keV. (511 eV)/2 + (17.5 keV) 248
• E46-1 (a) Apply Eq. 46-1, λ = h/p. The momentum of the bullet is p = mv = (0.041 kg)(960 m/s) = 39kg · m/s,so the corresponding wavelength is λ = h/p = (6.63×10−34 J · s)/(39kg · m/s) = 1.7×10−35 m. (b) This length is much too small to be significant. How much too small? If the radius of thegalaxy were one meter, this distance would correspond to the diameter of a proton.E46-2 (a) λ = h/p and p2 /2m = K, then hc (1240 eV · nm) 1.226 nm λ= √ = √ = √ . 2mc2 K 2(511 keV) K K (b) K = eV , so 1.226 nm 1.5 V λ= √ = nm. eV V √E46-3 For non-relativistic particles λ = h/p and p2 /2m = K, so λ = hc/ 2mc2 K. (a) For the electron, (1240 eV · nm) λ= = 0.0388 nm. 2(511 keV)(1.0 keV) (c) For the neutron, (1240 MeV · fm) λ= = 904 fm. 2(940 MeV)(0.001 MeV) (b) For ultra-relativistic particles K ≈ E ≈ pc, so hc (1240 eV · nm) λ= = = 1.24 nm. E (1000 eV)E46-4 For non-relativistic particles p = h/λ and p2 /2m = K, so K = (hc)2 /2mc2 λ2 . Then (1240 eV · nm)2 K= = 4.34×10−6 eV. 2(5.11×106 eV)(589 nm)2E46-5 (a) Apply Eq. 46-1, p = h/λ. The proton speed would then be h hc (1240 MeV · fm) v= =c 2 =c = 0.0117c. mλ mc λ (938 MeV)(113 fm)This is good, because it means we were justified in using the non-relativistic equations. Thenv = 3.51×106 m/s. (b) The kinetic energy of this electron would be 1 1 K= mv 2 = (938 MeV)(0.0117)2 = 64.2 keV. 2 2The potential through which it would need to be accelerated is 64.2 kV. 249
• √E46-6 (a) K = qV and p = 2mK. Then p= 2(22)(932 MeV/c2 )(325 eV) = 3.65×106 eV/c. (b) λ = h/p, so hc (1240 eV · nm) λ= = = 3.39×10−4 nm. pc (3.65×106 eV/c)c √E46-7 (a) For non-relativistic particles λ = h/p and p2 /2m = K, so λ = hc/ 2mc2 K. For thealpha particle, (1240 MeV · fm) λ= = 5.2 fm. 2(4)(932 MeV)(7.5 MeV) (b) Since the wavelength of the alpha is considerably smaller than the distance to the nucleuswe can ignore the wave nature of the alpha particle.E46-8 (a) For non-relativistic particles p = h/λ and p2 /2m = K, so K = (hc)2 /2mc2 λ2 . Then (1240 keV · pm)2 K= = 15 keV. 2(511keV)(10 pm)2 (b) For ultra-relativistic particles K ≈ E ≈ pc, so hc (1240 keV · pm) E= = = 124 keV. λ (10 pm)E46-9 The relativistic relationship between energy and momentum is E 2 = p2 c2 + m2 c4 ,and if the energy is very large (compared to mc2 ), then the contribution of the mass to the aboveexpression is small, and E 2 ≈ p2 c2 . Then from Eq. 46-1, h hc hc (1240 MeV · f m) λ= = = = = 2.5×10−2 fm. p pc E (50×103 MeV) √E46-10 (a) K = 3kT /2, p = 2mK, and λ = h/p, so h hc λ = √ =√ , 3mkT 3mc2 kT (1240 MeV · f m) = = 74 pm. 3(4)(932MeV)(8.62×10−11 MeV/K)(291 K) (b) pV = N kT ; assuming that each particle occupies a cube of volume d3 = V0 then the inter-particle spacing is d, so 3 3 (1.38×10−23 J/K)(291 K) d= V /N = = 3.4 nm. (1.01×105 Pa) 250
• E46-11 p = mv and p = h/λ, so m = h/λv. Taking the ratio, me λv = = (1.813×10−4 )(3) = 5.439×10−4 . m λe veThe mass of the unknown particle is then (0.511 MeV/c2 ) m= = 939.5 MeV. (5.439×10−4 )That would make it a neutron. √E46-12 (a) For non-relativistic particles λ = h/p and p2 /2m = K, so λ = hc/ 2mc2 K. For the electron, (1240 eV · nm) λ= = 1.0 nm. 2(5.11×105 eV)(1.5 eV) For ultra-relativistic particles K ≈ E ≈ pc, so for the photon hc (1240 eV · nm) λ= = = 830 nm. E (1.5 eV) (b) Electrons with energies that high are ultra-relativistic. Both the photon and the electron willthen have the same wavelength; hc (1240 MeV · fm) λ= = = 0.83 fm. E (1.5 GeV)E46-13 (a) The classical expression for kinetic energy is √ p = 2mK,so h hc (1240 keV · pm) λ= =√ = = 7.76 pm. p 2mc2 K 2(511 keV)(25.0 keV) (a) The relativistic expression for momentum is pc = sqrtE 2 − m2 c4 = (mc2 + K)2 − m2 c4 = K 2 + 2mc2 K.Then hc (1240 keV · pm) λ= = = 7.66 pm. pc (25.0 keV)2 + 2(511 keV)(25.0 keV)E46-14 We want to match the wavelength of the gamma to that of the electron. For the gamma,λ = hc/Eγ . For the electron, K = p2 /2m = h2 /2mλ2 . Combining, 2 h2 2 Eγ K= Eγ = . 2mh2 c2 2mc2With numbers, (136keV)2 K= = 18.1 keV. 2(511 keV)That would require an accelerating potential of 18.1 kV. 251
• E46-15 First find the√wavelength of the neutrons. For non-relativistic particles λ = h/p andp2 /2m = K, so λ = hc/ 2mc2 K. Then (1240 keV · pm) λ= = 14 pm. 2(940×103 keV)(4.2×10−3 keV)Bragg reflection occurs when 2d sin θ = λ, so θ = arcsin(14 pm)/2(73.2 pm) = 5.5◦ .E46-16 This is merely a Bragg reflection problem. Then 2d sin θ = mλ, or θ = arcsin(1)(11 pm)/2(54.64 pm) = 5.78◦ , θ = arcsin(2)(11 pm)/2(54.64 pm) = 11.6◦ , θ = arcsin(3)(11 pm)/2(54.64 pm) = 17.6◦ . E46-17 (a) Since sin 52◦ = 0.78, then 2(λ/d) = 1.57 > 1, so there is no diffraction order otherthan the first. (b) For an accelerating potential of 54 volts we have λ/d = 0.78. Increasing the potential willincrease the kinetic energy, increase the momentum, and decrease the wavelength. d won’t change.The kinetic energy is increased by a factor of 60/54 = 1.11, the momentum increases by a factor of√ 1.11 = 1.05, so the wavelength changes by a factor of 1/1.05 = 0.952. The new angle is then θ = arcsin(0.952 × 0.78) = 48◦ .E46-18 First find the√ wavelength of the electrons. For non-relativistic particles λ = h/p andp2 /2m = K, so λ = hc/ 2mc2 K. Then (1240 keV · pm) λ= = 62.9 pm. 2(511 keV)(0.380 keV)This is now a Bragg reflection problem. Then 2d sin θ = mλ, or θ = arcsin(1)(62.9 pm)/2(314 pm) = 5.74◦ , θ = arcsin(2)(62.9 pm)/2(314 pm) = 11.6◦ , θ = arcsin(3)(62.9 pm)/2(314 pm) = 17.5◦ , θ = arcsin(4)(62.9 pm)/2(314 pm) = 23.6◦ , θ = arcsin(5)(62.9 pm)/2(314 pm) = 30.1◦ , θ = arcsin(6)(62.9 pm)/2(314 pm) = 36.9◦ , θ = arcsin(7)(62.9 pm)/2(314 pm) = 44.5◦ , θ = arcsin(8)(62.9 pm)/2(314 pm) = 53.3◦ , θ = arcsin(9)(62.9 pm)/2(314 pm) = 64.3◦ .But the odd orders vanish (see Chapter 43 for a discussion on this).E46-19 Since ∆f · ∆t ≈ 1/2π, we have ∆f = 1/2π(0.23 s) = 0.69/s. 252
• E46-20 Since ∆f · ∆t ≈ 1/2π, we have ∆f = 1/2π(0.10×10−9 s) = 1.6×1010 /s.The bandwidth wouldn’t fit in the frequency allocation!E46-21 Apply Eq. 46-9, h 4.14×10−15 eV · s) ∆E ≥ = = 7.6×10−5 eV. 2π∆t 2π(8.7×10−12 s)This is much smaller than the photon energy.E46-22 Apply Heisenberg twice: 4.14×10−15 eV · s) ∆E1 = = 5.49×10−8 eV. 2π(12×10−9 s)and 4.14×10−15 eV · s) ∆E2 = = 2.86×10−8 eV. 2π(23×10−9 s)The sum is ∆E transition = 8.4×10−8 eV.E46-23 Apply Heisenberg: 6.63×10−34 J · s) ∆p = = 8.8×10−24 kg · m/s. 2π(12×10−12 m)E46-24 ∆p = (0.5 kg)(1.2 s) = 0.6 kg · m/s. The position uncertainty would then be (0.6 J/s) ∆x = = 0.16 m. 2π(0.6 kg · m/s)E46-25 We want v ≈ ∆v, which means p ≈ ∆p. Apply Eq. 46-8, and h h ∆x ≥ ≈ . 2π∆p 2πpAccording to Eq. 46-1, the de Broglie wavelength is related to the momentum by λ = h/p,so λ ∆x ≥ . 2πE46-26 (a) A particle confined in a (one dimensional) box of size L will have a position uncertaintyof no more than ∆x ≈ L. The momentum uncertainty will then be no less than h h ∆p ≥ ≈ . 2π∆x 2πLso (6.63×10−34 J · s) ∆p ≈ = 1×10−24 kg · m/s. 2π(×10−10 m) 253
• (b) Assuming that p ≈ ∆p, we have h p≥ , 2πLand then the electron will have a (minimum) kinetic energy of p2 h2 E≈ ≈ 2 mL2 . 2m 8πor (hc)2 (1240 keV · pm)2 E≈ = = 0.004 keV. 8π 2 mc2 L2 8π 2 (511 keV)(100 pm)2 E46-27 (a) A particle confined in a (one dimensional) box of size L will have a position uncer-tainty of no more than ∆x ≈ L. The momentum uncertainty will then be no less than h h ∆p ≥ ≈ . 2π∆x 2πLso (6.63×10−34 J · s) ∆p ≈ = 1×10−20 kg · m/s. 2π(×10−14 m) (b) Assuming that p ≈ ∆p, we have h p≥ , 2πLand then the electron will have a (minimum) kinetic energy of p2 h2 E≈ ≈ 2 mL2 . 2m 8πor (hc)2 (1240 MeV · fm)2 E≈ = = 381 MeV. 8π 2 mc2 L2 8π 2 (0.511 MeV)(10 fm)2This is so large compared to the mass energy of the electron that we must consider relativistic effects.It will be very relativistic (381 0.5!), so we can use E = pc as was derived in Exercise 9. Then hc (1240 MeV · fm) E= = = 19.7 MeV. 2πL 2π(10 fm)This is the total energy; so we subtract 0.511 MeV to get K = 19 MeV.E46-28 We want to find L when T = 0.01. This means solving E E T = 16 1− e−2kL , U0 U0 (5.0 eV) (5.0 eV) (0.01) = 16 1− e−2k L , (6.0 eV) (6.0 eV) = 2.22e−2k L , ln(4.5×10−3 ) = 2(5.12×109 /m)L, 5.3×10−10 m = L. 254
• E46-29 The wave number, k, is given by 2π k= 2mc2 (U0 − E). hc (a) For the proton mc2 = 938 MeV, so 2π k= 2(938 MeV)(10 MeV − 3.0 MeV) = 0.581 fm−1 . (1240 MeV · fm)The transmission coefficient is then (3.0 MeV) (3.0 MeV) −1 T = 16 1− e−2(0.581 fm )(10 fm) = 3.0×10−5 . (10 MeV) (10 MeV) (b) For the deuteron mc2 = 2 × 938 MeV, so 2π k= 2(2)(938 MeV)(10 MeV − 3.0 MeV) = 0.821 fm−1 . (1240 MeV · fm)The transmission coefficient is then (3.0 MeV) (3.0 MeV) −1 T = 16 1− e−2(0.821 fm )(10 fm) = 2.5×10−7 . (10 MeV) (10 MeV)E46-30 The wave number, k, is given by 2π k= 2mc2 (U0 − E). hc (a) For the proton mc2 = 938 MeV, so 2π k= 2(938 MeV)(6.0 eV − 5.0 eV) = 0.219 pm−1 . (1240 keV · pm)We want to find T . This means solving E E T = 16 1− e−2kL , U0 U0 (5.0 eV) (5.0 eV) 12 )(0.7×10−9 ) = 16 1− e−2(0.219×10 , (6.0 eV) (6.0 eV) = 1.6×10−133 .A current of 1 kA corresponds to N = (1×103 C/s)/(1.6×10−19 C) = 6.3×1021 /sprotons per seconds. The time required for one proton to pass is then t = 1/(6.3×1021 /s)(1.6×10−133 ) = 9.9×10110 s.That’s 10104 years! 255
• P46-1 We will interpret low energy to mean non-relativistic. Then h h λ= =√ . p 2mn KThe diffraction pattern is then given by d sin θ = mλ = mh/ 2mn K,where m is diffraction order while mn is the neutron mass. We want to investigate the spread bytaking the derivative of θ with respect to K, mh d cos θ dθ = − √ dK. 2 2mn K 3Divide this by the original equation, and then cos θ dK dθ = − . sin θ 2KRearrange, change the differential to a difference, and then ∆K ∆θ = tan θ . 2KWe dropped the negative sign out of laziness; but the angles are in radians, so we need to multiplyby 180/π to convert to degrees.P46-2P46-3 We want to solve E E T = 16 1− e−2kL , U0 U0for E. Unfortunately, E is contained in k since 2π k= 2mc2 (U0 − E). hcWe can do this by iteration. The maximum possible value for E E 1− U0 U0is 1/4; using this value we can get an estimate for k: (0.001) = 16(0.25)e−2kL , ln(2.5×10−4 ) = −2k(0.7 nm), 5.92/ nm = k.Now solve for E: E = U0 − (hc)2 k 2 8mc2 π 2 , (1240 eV · nm)2 (5.92/nm)2 = (6.0 eV) − , 8π 2 (5.11×105 eV) = 4.67 eV. 256
• Put this value for E back into the transmission equation to find a new k: E E T = 16 1− e−2kL , U0 U0 (4.7 eV) (4.7 eV) (0.001) = 16 1− e−2kL , (6.0 eV) (6.0 eV) ln(3.68×10−4 ) = −2k(0.7 nm), 5.65/ nm = k.Now solve for E using this new, improved, value for k: E = U0 − (hc)2 k 2 8mc2 π 2 , (1240 eV · nm)2 (5.65/nm)2 = (6.0 eV) − , 8π 2 (5.11×105 eV) = 4.78 eV.Keep at it. You’ll eventually stop around E = 5.07 eV.P46-4 (a) A one percent increase in the barrier height means U0 = 6.06 eV. For the electron mc2 = 5.11×105 eV, so 2π k= 2(5.11×105 eV)(6.06 eV − 5.0 eV) = 5.27 nm−1 . (1240 eV · nm)We want to find T . This means solving E E T = 16 1− e−2kL , U0 U0 (5.0 eV) (5.0 eV) = 16 1− e−2(5.27)(0.7) , (6.06 eV) (6.06 eV) = 1.44×10−3 .That’s a 16% decrease. (b) A one percent increase in the barrier thickness means L = 0.707 nm. For the electron mc2 = 5.11×105 eV, so 2π k= 2(5.11×105 eV)(6.0 eV − 5.0 eV) = 5.12 nm−1 . (1240 eV · nm)We want to find T . This means solving E E T = 16 1− e−2kL , U0 U0 (5.0 eV) (5.0 eV) = 16 1− e−2(5.12)(0.707) , (6.0 eV) (6.0 eV) = 1.59×10−3 .That’s a 8.1% decrease. (c) A one percent increase in the incident energy means E = 5.05 eV. For the electron mc2 = 5.11×105 eV, so 2π k= 2(5.11×105 eV)(6.0 eV − 5.05 eV) = 4.99 nm−1 . (1240 eV · nm) 257
• We want to find T . This means solving E E T = 16 1− e−2kL , U0 U0 (5.05 eV) (5.05 eV) = 16 1− e−2(4.99)(0.7) , (6.0 eV) (6.0 eV) = 1.97×10−3 .That’s a 14% increase.P46-5 First, the rule for exponents ei(a+b) = eia eib .Then apply Eq. 46-12, eiθ = cos θ + i sin θ, cos(a + b) + i sin(a + b) = (cos a + i sin a)(sin b + i sin b).Expand the right hand side, remembering that i2 = −1, cos(a + b) + i sin(a + b) = cos a cos b + i cos a sin b + i sin a cos b − sin a sin b.Since the real part of the left hand side must equal the real part of the right and the imaginary partof the left hand side must equal the imaginary part of the right, we actually have two equations.They are cos(a + b) = cos a cos b − sin a sin band sin(a + b) = cos a sin b + sin a cos b.P46-6 258
• E47-1 (a) The ground state energy level will be given by h2 (6.63 × 10−34 J · s)2 E1 = = = 3.1 × 10−10 J. 8mL 2 8(9.11 × 10−31 kg)(1.4 × 10−14 m)2The answer is correct, but the units make it almost useless. We can divide by the electron charge toexpress this in electron volts, and then E = 1900 MeV. Note that this is an extremely relativisticquantity, so the energy expression loses validity. (b) We can repeat what we did above, or we can apply a “trick” that is often used in solvingthese problems. Multiplying the top and the bottom of the energy expression by c2 we get (hc)2 E1 = 8(mc2 )L2Then (1240 MeV · fm)2 E1 = = 1.0 MeV. 8(940 MeV)(14 fm)2 (c) Finding an neutron inside the nucleus seems reasonable; but finding the electron would not.The energy of such an electron is considerably larger than binding energies of the particles in thenucleus.E47-2 Solve n2 (hc)2 En = 8(mc2 )L2for L, then nhc L = √ , 8mc2 En (3)(1240 eV · nm) = , 8(5.11×105 eV)(4.7 eV) = 0.85 nm.E47-3 Solve for E4 − E1 : 42 (hc)2 12 (hc)2 E4 − E1 = − , 8(mc2 )L2 8(mc2 )L2 (16 − 1)(1240 eV · nm)2 = , 8(5.11×105 )(0.253 nm)2 = 88.1 eV.E47-4 Since E ∝ 1/L2 , doubling the width of the well will lower the ground state energy to(1/2)2 = 1/4, or 0.65 eV.E47-5 (a) Solve for E2 − E1 : 22 h2 12 h2 E2 − E1 = − , 8mL2 8mL2 (3)(6.63×10−34 J · s)2 = , 8(40)(1.67×10−27 kg)(0.2 m)2 = 6.2×10−41 J. (b) K = 3kT /2 = 3(1.38×10−23 J/K)(300 K)/2 = 6.21×10−21 . The ratio is 1×10−20 . (c) T = 2(6.2×10−41 J)/3(1.38×10−23 J/K) = 3.0×10−18 K. 259
• E47-6 (a) The fractional difference is (En+1 − En )/En , or ∆En h2 h2 h2 = (n + 1)2 2 − n2 2 / n2 , En 8mL 8mL 8mL2 (n + 1)2 − n2 = , n2 2n + 1 = . n2 (b) As n → ∞ the fractional difference goes to zero; the system behaves as if it is continuous.E47-7 (a) We will take advantage of the “trick” that was developed in part (b) of Exercise 47-1.Then (hc)2 (1240 eV · nm)2 En = n 2 = (15)2 = 8.72 keV. 8mc2 L 8(0.511 × 106 eV)(0.0985 nm)2 (b) The magnitude of the momentum is exactly known because E = p2 /2m. This momentum isgiven by √ pc = 2mc2 E = 2(511 keV)(8.72 keV) = 94.4 keV. What we don’t know is in which direction the particle is moving. It is bouncing back and forthbetween the walls of the box, so the momentum could be directed toward the right or toward theleft. The uncertainty in the momentum is then ∆p = pwhich can be expressed in terms of the box size L by √ n 2 h2 nh ∆p = p = 2mE = 2 = . 4L 2L (c) The uncertainty in the position is 98.5 pm; the electron could be anywhere inside the well.E47-8 The probability distribution function is 2 2πx P2 = sin2 . L LWe want to integrate over the central third, or L/6 2 2πx P = sin2 dx, −L/6 L L 1 π/3 = sin2 θ dθ, π −π/3 = 0.196.E47-9 (a) Maximum probability occurs when the argument of the cosine (sine) function is kπ([k + 1/2]π). This occurs when x = N L/2nfor odd N . (b) Minimum probability occurs when the argument of the cosine (sine) function is [k + 1/2]π(kπ). This occurs when x = N L/2nfor even N . 260
• E47-10 In Exercise 47-21 we show that the hydrogen levels can be written as En = −(13.6 eV)/n2 . (a) The Lyman series is the series which ends on E1 . The least energetic state starts on E2 . Thetransition energy is E2 − E1 = (13.6 eV)(1/12 − 1/22 ) = 10.2 eV.The wavelength is hc (1240 eV · nm) λ= = = 121.6 nm. ∆E (10.2 eV) (b) The series limit is 0 − E1 = (13.6 eV)(1/12 ) = 13.6 eV.The wavelength is hc (1240 eV · nm) λ= = = 91.2 nm. ∆E (13.6 eV)E47-11 The ground state of hydrogen, as given by Eq. 47-21, is me4 (9.109 × 10−31 kg)(1.602 × 10−19 C)4 E1 = − =− = 2.179 × 10−18 J. 8 2 h2 0 8(8.854 × 10−12 F/m)2 (6.626 × 10−34 J · s)2In terms of electron volts the ground state energy is E1 = −(2.179 × 10−18 J)/(1.602 × 10−19 C) = −13.60 eV.E47-12 In Exercise 47-21 we show that the hydrogen levels can be written as En = −(13.6 eV)/n2 . (c) The transition energy is ∆E = E3 − E1 = (13.6 eV)(1/12 − 1/32 ) = 12.1 eV. (a) The wavelength is hc (1240 eV · nm) λ= = = 102.5 nm. ∆E (12.1 eV) (b) The momentum is p = E/c = 12.1 eV/c.E47-13 In Exercise 47-21 we show that the hydrogen levels can be written as En = −(13.6 eV)/n2 . (a) The Balmer series is the series which ends on E2 . The least energetic state starts on E3 . Thetransition energy is E3 − E2 = (13.6 eV)(1/22 − 1/32 ) = 1.89 eV.The wavelength is hc (1240 eV · nm) λ= = = 656 nm. ∆E (1.89 eV) 261
• (b) The next energetic state starts on E4 . The transition energy is E4 − E2 = (13.6 eV)(1/22 − 1/42 ) = 2.55 eV.The wavelength is hc (1240 eV · nm) λ= = = 486 nm. ∆E (2.55 eV) (c) The next energetic state starts on E5 . The transition energy is E5 − E2 = (13.6 eV)(1/22 − 1/52 ) = 2.86 eV.The wavelength is hc (1240 eV · nm) λ= = = 434 nm. ∆E (2.86 eV) (d) The next energetic state starts on E6 . The transition energy is E6 − E2 = (13.6 eV)(1/22 − 1/62 ) = 3.02 eV.The wavelength is hc (1240 eV · nm) λ= = = 411 nm. ∆E (3.02 eV) (e) The next energetic state starts on E7 . The transition energy is E7 − E2 = (13.6 eV)(1/22 − 1/72 ) = 3.12 eV.The wavelength is hc (1240 eV · nm) λ= = = 397 nm. ∆E (3.12 eV)E47-14 In Exercise 47-21 we show that the hydrogen levels can be written as En = −(13.6 eV)/n2 . The transition energy is hc (1240 eV · nm) ∆E = = = 10.20 eV. λ (121.6 nm) This must be part of the Lyman series, so the higher state must be En = (10.20 eV) − (13.6 eV) = −3.4 eV.That would correspond to n = 2. E47-15 The binding energy is the energy required to remove the electron. If the energy of theelectron is negative, then that negative energy is a measure of the energy required to set the electronfree. The first excited state is when n = 2 in Eq. 47-21. It is not necessary to re-evaluate the constantsin this equation every time, instead, we start from E1 En = where E1 = −13.60 eV. n2Then the first excited state has energy (−13.6 eV) E2 = = −3.4 eV. (2)2The binding energy is then 3.4 eV. 262
• E47-16 rn = a0 n2 , so n= (847 pm)/(52.9 pm) = 4.E47-17 (a) The energy of this photon is hc (1240 eV · nm) E= = = 0.96739 eV. λ (1281.8 nm)The final state of the hydrogen must have an energy of no more than −0.96739, so the largestpossible n of the final state is n< 13.60 eV/0.96739 eV = 3.75,so the final n is 1, 2, or 3. The initial state is only slightly higher than the final state. The jumpfrom n = 2 to n = 1 is too large (see Exercise 15), any other initial state would have a larger energydifference, so n = 1 is not the final state. So what level might be above n = 2? We’ll try n= 13.6 eV/(3.4 eV − 0.97 eV) = 2.36,which is so far from being an integer that we don’t need to look farther. The n = 3 state has energy13.6 eV/9 = 1.51 eV. Then the initial state could be n= 13.6 eV/(1.51 eV − 0.97 eV) = 5.01,which is close enough to 5 that we can assume the transition was n = 5 to n = 3. (b) This belongs to the Paschen series.E47-18 In Exercise 47-21 we show that the hydrogen levels can be written as En = −(13.6 eV)/n2 . (a) The transition energy is ∆E = E4 − E1 = (13.6 eV)(1/12 − 1/42 ) = 12.8 eV. (b) All transitions n → m are allowed for n ≤ 4 and m < n. The transition energy will be of theform En − Em = (13.6 eV)(1/m2 − 1/n2 ).The six possible results are 12.8 eV, 12.1 eV, 10.2 eV, 2.55 eV, 1.89 eV, and 0.66 eV.E47-19 ∆E = h/2π∆t, so ∆E = (4.14×10−15 eV · s)/2π(1×10−8 s) = 6.6×10−8 eV.E47-20 (a) According to electrostatics and uniform circular motion, mv 2 /r = e2 /4π 0 r2 ,or e2 e4 e2 v= == = . 4π 0 mr 4 2 h2 n 2 0 2 0 hn 263
• Putting in the numbers, (1.6×10−19 C)2 2.18×106 m/s v= = . 2(8.85×10−12 F/m)(6.63×10−34 J · s)n nIn this case n = 1. (b) λ = h/mv, λ = (6.63×10−34 J · s)/(9.11×10−31 kg)(2.18×106 m/s) = 3.34×10−10 m. (c) λ/a0 = (3.34×10−10 m)/(5.29×10−11 ) = 6.31 ≈ 2π. Actually, it is exactly 2π. E47-21 In order to have an inelastic collision with the 6.0 eV neutron there must exist a transitionwith an energy difference of less than 6.0 eV. For a hydrogen atom in the ground state E1 = −13.6 eVthe nearest state is E2 = (−13.6 eV)/(2)2 = −3.4 eV.Since the difference is 10.2 eV, it will not be possible for the 6.0 eV neutron to have an inelasticcollision with a ground state hydrogen atom.E47-22 (a) The atom is originally in the state n given by n= (13.6 eV)/(0.85 eV) = 4.The state with an excitation energy of 10.2 eV, is n= (13.6 eV)/(13.6 eV − 10.2 eV) = 2.The transition energy is then ∆E = (13.6 eV)(1/22 − 1/42 ) = 2.55 eV.E47-23 According to electrostatics and uniform circular motion, mv 2 /r = e2 /4π 0 r2 ,or e2 e4 e2 v= == 2 h2 n 2 = . 4π 0 mr 4 0 2 0 hnThe de Broglie wavelength is then h 2 0 hn λ= = . mv me2The ratio of λ/r is λ 2 0 hn = = kn, r me2 a0 n2where k is a constant. As n → ∞ the ratio goes to zero. 264
• E47-24 In Exercise 47-21 we show that the hydrogen levels can be written as En = −(13.6 eV)/n2 .The transition energy is ∆E = E4 − E1 = (13.6 eV)(1/12 − 1/42 ) = 12.8 eV.The momentum of the emitted photon is p = E/c = (12.8 eV)/c.This is the momentum of the recoiling hydrogen atom, which then has velocity p pc (12.8 eV) v= = c= (3.00×108 m/s) = 4.1 m/s. m mc2 (932 MeV) E47-25 The first Lyman line is the n = 1 to n = 2 transition. The second Lyman line is then = 1 to n = 3 transition. The first Balmer line is the n = 2 to n = 3 transition. Since the photonfrequency is proportional to the photon energy (E = hf ) and the photon energy is the energydifference between the two levels, we have Em − E n fn→m = hwhere the En is the hydrogen atom energy level. Then E 3 − E1 f1→3 = , h E 3 − E2 + E2 − E1 E3 − E2 E2 − E 1 = = + , h h h = f2→3 + f1→2 .E47-26 Use En = −Z 2 (13.6 eV)/n2 . (a) The ionization energy of the ground state of He+ is En = −(2)2 (13.6 eV)/(1)2 = 54.4 eV. (b) The ionization energy of the n = 3 state of Li2+ is En = −(3)2 (13.6 eV)/(3)2 = 13.6 eV.E47-27 (a) The energy levels in the He+ spectrum are given by En = −Z 2 (13.6 eV)/n2 ,where Z = 2, as is discussed in Sample Problem 47-6. The photon wavelengths for the n = 4 seriesare then hc hc/E4 λ= = , En − E 4 1 − En /E4 265
• which can also be written as 16hc/(54.4 eV) λ = , 1 − 16/n2 16hcn2 /(54.4 eV) = , n2 − 16 Cn2 = , n2 − 16where C = hc/(3.4 eV) = 365 nm. (b) The wavelength of the first line is the transition from n = 5, (365 nm)(5)2 λ= = 1014 nm. (5)2 − (4)2 The series limit is the transition from n = ∞, so λ = 365 nm. (c) The series starts in the infrared (1014 nm), and ends in the ultraviolet (365 nm). So it mustalso include some visible lines.E47-28 We answer these questions out of order! (a) n = 1. (b) r = a0 = 5.29×10−11 m. (f) According to electrostatics and uniform circular motion, mv 2 /r = e2 /4π 0 r2 ,or e2 e4 e2 v= == 2 h2 n 2 = . 4π 0 mr 4 0 2 0 hnPutting in the numbers, (1.6×10−19 C)2 v= = 2.18×106 m/s. 2(8.85×10−12 F/m)(6.63×10−34 J · s)(1) (d) p = (9.11×10−31 kg)(2.18×106 m/s) = 1.99×10−24 kg · m/s. (e) ω = v/r = (2.18×106 m/s)/(5.29×10−11 m) = 4.12×1016 rad/s. (c) l = pr = (1.99×10−24 kg · m/s)(5.29×10−11 m) = 1.05×10−34 J · s. (g) F = mv 2 /r, so F = (9.11×10−31 kg)(2.18×106 m/s)2 /(5.29×10−11 m) = 8.18×10−8 N. (h) a = (8.18×10−8 N)/(9.11×10−31 kg) = 8.98×1022 m/s2 . (i) K = mv 2 /r, or (9.11×10−31 kg)(2.18×106 m/s)2 K= = 2.16×10−18 J = 13.6 eV. 2 (k) E = −13.6 eV. (j) P = E − K = (−13.6 eV) − (13.6 eV) = −27.2 eV. 266
• E47-29 For each r in the quantity we have a factor of n2 . (a) n is proportional to n. (b) r is proportional to n2 . (f) v is proportional to 1/r, or 1/n. (d) p is proportional to v, or 1/n. (e) ω is proportional to v/r, or 1/n3 . (c) l is proportional to pr, or n. (g) f is proportional to v 2 /r, or 1/n4 . (h) a is proportional to F , or 1/n4 . (i) K is proportional to v 2 , or 1/n2 . (j) E is proportional to 1/n2 . (k) P is proportional to 1/n2 .E47-30 (a) Using the results of Exercise 45-1, (1240 eV · nm) E1 = = 1.24×105 eV. (0.010 nm) (b) Using the results of Problem 45-11, E2 (1.24×105 eV)2 Kmax = = = 40.5×104 eV. mc2 /2 + E (5.11×105 eV)/2 + (1.24×105 eV) (c) This would likely knock the electron way out of the atom.E47-31 The energy of the photon in the series limit is given by E limit = (13.6 eV)/n2 ,where n = 1 for Lyman, n = 2 for Balmer, and n = 3 for Paschen. The wavelength of the photon is (1240 eV · nm) 2 λlimit = n = (91.17 nm)n2 . (13.6 eV) The energy of the longest wavelength comes from the transition from the nearest level, or (−13.6 eV) (−13.6 eV) 2n + 1 E long = − = (13.6 eV) . (n + 1)2 n2 [n(n + 1)]2The wavelength of the photon is (1240 eV · nm)[n(n + 1)]2 [n(n + 1)]2 λlong = 2 = (91.17 nm) . (13.6 eV)n 2n + 1 (a) The wavelength interval λlong − λlimit , or n2 (n + 1)2 − n2 (2n + 1) n4 ∆λ = (91.17 nm) = (91.17 nm) . 2n + 1 2n + 1For n = 1, ∆λ = 30.4 nm. For n = 2, ∆λ = 292 nm. For n = 3, ∆λ = 1055 nm. (b) The frequency interval is found from E limit − E long (13.6 eV) 1 (3.29×1015 /s) ∆f = = −15 eV · s) (n + 1)2 = . h (4.14×10 (n + 1)2For n = 1, ∆f = 8.23×1014 Hz. For n = 2, ∆f = 3.66×1014 Hz. For n = 3, ∆f = 2.05×1014 Hz. 267
• E47-32E47-33 (a) We’ll use Eqs. 47-25 and 47-26. At r = 0 1 −2(0)/a0 1 ψ 2 (0) = 3e = = 2150 nm−3 , πa0 πa30while P (0) = 4π(0)2 ψ 2 (0) = 0. (b) At r = a0 we have e−2 ψ 2 (a0 ) = f rac1πa3 e−2(a0 )/a0 = 0 = 291 nm−3 , πa3 0and P (a0 ) = 4π(a0 )2 ψ 2 (a0 ) = 10.2 nm−1 .E47-34 Assume that ψ(a0 ) is a reasonable estimate for ψ(r) everywhere inside the small sphere.Then e−2 0.1353 ψ2 = = . πa3 0 πa30The probability of finding it in a sphere of radius 0.05a0 is 0.05a0 (0.1353)4πr2 dr 4 = (0.1353)(0.05)3 = 2.26×10−5 . 0 πa3 0 3E47-35 Using Eq. 47-26 the ratio of the probabilities is P (a0 ) (a0 )2 e−2(a0 )/a0 e−2 = 2 e−2(2a0 )/a0 = −4 = 1.85. P (2a0 ) (2a0 ) 4eE47-36 The probability is 1.016a0 4r2 e−2r/a0 P = dr, a0 a3 0 1 2.032 2 −u = u e du, 2 2 = 0.00866.E47-37 If l = 3 then ml can be 0, ±1, ±2, or ±3. (a) From Eq. 47-30, Lz = ml h/2π.. So Lz can equal 0, ±h/2π, ±h/π, or ±3h/2π. (b) From Eq. 47-31, θ = arccos(ml / l(l + 1)), so θ can equal 90◦ , 73.2◦ , 54.7◦ , or 30.0◦ . (c) The magnitude of L is given by Eq. 47-28, h √ L= l(l + 1) = 3h/π. 2πE47-38 The maximum possible value of ml is 5. Apply Eq. 47-31: (5) θ = arccos = 24.1◦ . (5)(5 + 1) 268
• E47-39 Use the hint. h ∆p · ∆x = , 2π r h ∆p ∆x = , r 2π ∆x h ∆p · r = , r 2π h ∆L · ∆θ = . 2πE47-40 Note that there is a typo in the formula; P (r) must have dimensions of one over length. The probability is ∞ r4 e−r/a0 P = dr, 0 24a50 ∞ 1 = u4 e−u du, 24 0 = 1.00 What does it mean? It means that if we look for the electron, we will find it somewhere.E47-41 (a) Find the maxima by taking the derivative and setting it equal to zero. dP r(2a − r)(4a2 − 6ra + r2 ) −r = e = 0. dr 8a6 0The solutions are r = 0, r = 2a, and 4a2 − 6ra + r2 = 0. The first two correspond to minima (seeFig. 47-14). The other two are the solutions to the quadratic, or r = 0.764a0 and r = 5.236a0 . (b) Substitute these two values into Eq. 47-36. The results are P (0.764a0 ) = 0.981 nm−1 .and P (5.236a0 ) = 3.61 nm−1 .E47-42 The probability is 5.01a0 r2 (2 − r/a0 )2 e−r/a0 P = dr, 5.00a0 8a3 0 = 0.01896.E47-43 n = 4 and l = 3, while ml can be any of −3, −2, −1, 0, 1, 2, 3,while ms can be either −1/2 or 1/2. There are 14 possible states.E47-44 n must be greater than l, so n ≥ 4. |ml | must be less than or equal to l, so |ml | ≤ 3. msis −1/2 or 1/2.E47-45 If ml = 4 then l ≥ 4. But n ≥ l + 1, so n > 4. We only know that ms = ±1/2. 269
• E47-46 There are 2n2 states in a shell n, so if n = 5 there are 50 states.E47-47 Each is in the n = 1 shell, the l = 0 angular momentum state, and the ml = 0 state. Butone is in the state ms = +1/2 while the other is in the state ms = −1/2.E47-48 Apply Eq. 47-31: (+1/2) θ = arccos = 54.7◦ (1/2)(1/2 + 1)and (−1/2) θ = arccos = 125.3◦ . (1/2)(1/2 + 1)E47-49 All of the statements are true.E47-50 There are n possible values for l (start at 0!). For each value of l there are 2l + 1 possiblevalues for ml . If n = 1, the sum is 1. If n = 2, the sum is 1 + 3 = 4. If n = 3, the sum is 1 + 3 + 5 = 9.The pattern is clear, the sum is n2 . But there are two spin states, so the number of states is 2n2 .P47-1 We can simplify the energy expression as h2 E = E0 n 2 + n2 + n 2 x y z where E0 = . 8mL2To find the lowest energy levels we need to focus on the values of nx , ny , and nz . It doesn’t take much imagination to realize that the set (1, 1, 1) will result in the smallest valuefor n2 + n2 + n2 . The next choice is to set one of the values equal to 2, and try the set (2, 1, 1). x y z Then it starts to get harder, as the next lowest might be either (2, 2, 1) or (3, 1, 1). The only wayto find out is to try. I’ll tabulate the results for you: nx ny nz n2 + n2 + n2 x y z Mult. nx ny nz n2 + n2 + n2 x y z Mult. 1 1 1 3 1 3 2 1 14 6 2 1 1 6 3 3 2 2 17 3 2 2 1 9 3 4 1 1 18 3 3 1 1 11 3 3 3 1 19 3 2 2 2 12 1 4 2 1 21 6 We are now in a position to state the five lowest energy levels. The fundamental quantity is (hc)2 (1240 eV · nm)2 E0 = 2 L2 = = 6.02×10−6 eV. 8mc 8(0.511×106 eV)(250 nm)2The five lowest levels are found by multiplying this fundamental quantity by the numbers in thetable above.P47-2 (a) Write the states between 0 and L. Then all states, odd or even, can be written withprobability distribution function 2 nπx P (x) = sin2 , L L 270
• we find the probability of finding the particle in the region 0 ≤ x ≤ L/3 is L/3 2 nπx P = cos2 dx, 0 L L 1 sin(2nπ/3) = 1− . 3 2nπ/3 (b) If n = 1 use the formula and P = 0.196. (c) If n = 2 use the formula and P = 0.402. (d) If n = 3 use the formula and P = 0.333. (e) Classically the probability distribution function is uniform, so there is a 1/3 chance of findingit in the region 0 to L/3.P47-3 The region of interest is small compared to the variation in P (x); as such we can approxi-mate the probability with the expression P = P (x)∆x. (b) Evaluating, 2 4πx P = sin2 ∆x, L L 2 4π(L/8) = sin2 (0.0003L), L L = 0.0006. (b) Evaluating, 2 4πx P = sin2 ∆x, L L 2 4π(3L/16) = sin2 (0.0003L), L L = 0.0003.P47-4 (a) P = Ψ∗ Ψ, or 2 P = A2 e−2πmωx 0 /h . (b) Integrating, ∞ 2 1 = A2 0 e−2πmωx /h dx, −∞ ∞ h 2 = A2 0 e−u du, 2πmω −∞ h = A2 0 pi, 2πmω 4 2mω = A0 . h (c) x = 0.P47-5 We will want an expression for d2 ψ0 . dx2 271
• Doing the math one derivative at a time, d2 d d ψ0 = ψ0 , dx2 dx dx d 2 = A0 (−2πmωx/h)e−πmωx /h , dx 2 2 = A0 (−2πmωx/h)2 e−πmωx /h + A0 (−2πmω/h)e−πmωx /h , 2 = (2πmωx/h)2 − (2πmω/h) A0 e−πmωx /h , 2 = (2πmωx/h) − (2πmω/h) ψ0 .In the last line we factored out ψ0 . This will make our lives easier later on. Now we want to go to Schr¨dinger’s equation, and make some substitutions. o h2 d 2 − ψ 0 + U ψ0 = Eψ0 , 8π 2 m dx2 h2 − (2πmωx/h)2 − (2πmω/h) ψ0 + U ψ0 = Eψ0 , 8π 2 m h2 − 2 (2πmωx/h)2 − (2πmω/h) + U = E, 8π mwhere in the last line we divided through by ψ0 . Now for some algebra, h2 U = E+ (2πmωx/h)2 − (2πmω/h) , 8π 2 m mω 2 x2 hω = E+ − . 2 4πBut we are given that E = hω/4π, so this simplifies to mω 2 x2 U= 2which looks like a harmonic oscillator type potential.P47-6 Assume the electron is originally in the state n. The classical frequency of the electron isf0 , where f0 = v/2πr.According to electrostatics and uniform circular motion, mv 2 /r = e2 /4π 0 r2 ,or e2 e4 e2 v= == = . 4π 0 mr 4 2 h2 n 2 0 2 0 hnThen e2 1 πme2 me4 −2E1 f0 = = 2 3 3 = 2 0 hn 2π 0 h2 n2 4 0h n hn3Here E1 = −13.6 eV. Photon frequency is related to energy according to f = ∆Enm /h, where ∆Enm is the energy oftransition from state n down to state m. Then E1 1 1 f= 2 − 2 , h n m 272
• where E1 = −13.6 eV. Combining the fractions and letting m = n − δ, where δ is an integer, E1 m2 − n2 f = , h m2 n2 −E1 (n − m)(m + n) = , h m2 n2 −E1 δ(2n + δ) = , h (n + δ)2 n2 −E1 δ(2n) ≈ , h (n)2 n2 −2E1 = δ = f0 δ. hn3P47-7 We need to use the reduced mass of the muon since the muon and proton masses are soclose together. Then (207)(1836) m= me = 186me . (207) + (1836) (a) Apply Eq. 47-20 1/2: aµ = a0 /(186) = (52.9 pm)/(186) = 0.284 pm. (b) Apply Eq. 47-21: Eµ = E1 (186) = (13.6 eV)(186) = 2.53 keV. (c) λ = (1240 keV · pm)/(2.53 pm) = 490 pm.P47-8 (a) The reduced mass of the electron is (1)(1) m= me = 0.5me . (1) + (1) The spectrum is similar, except for this additional factor of 1/2; hence λpos = 2λH . (b) apos = a0 /(186) = (52.9 pm)/(1/2) = 105.8 pm. This is the distance between the particles,but they are both revolving about the center of mass. The radius is then half this quantity, or52.9 pm. P47-9 This problem isn’t really that much of a problem. Start with the magnitude of a vectorin terms of the components, L2 + L2 + L2 = L2 , x y zand then rearrange, L2 + L2 = L2 − L2 . x y zAccording to Eq. 47-28 L2 = l(l + 1)h2 /4π 2 , while according to Eq. 47-30 Lz = ml h/2π. Substitutethat into the equation, and h2 L2 + L2 = l(l + 1)h2 /4π 2 − m2 h2 /4π 2 = l(l + 1) − m2 x y l l . 4π 2Take the square root of both sides of this expression, and we are done. 273
• The maximum value for ml is l, while the minimum value is 0. Consequently, L2 + L2 = x y l(l + 1) − m2 h/2π ≤ l l(l + 1) h/2π,and √ L2 + L2 = x y l(l + 1) − m2 h/2π ≥ l l h/2π.P47-10 Assume that ψ(0) is a reasonable estimate for ψ(r) everywhere inside the small sphere.Then e−0 1 ψ2 = = . πa3 0 πa30The probability of finding it in a sphere of radius 1.1×10−15 m is 1.1×10−15 m 4πr2 dr 4 (1.1×10−15 m)3 3 = = 1.2×10−14 . 0 πa0 3 (5.29×10−11 m)3P47-11 Assume that ψ(0) is a reasonable estimate for ψ(r) everywhere inside the small sphere.Then (2)2 e−0 1 ψ2 = = . 32πa30 8πa3 0The probability of finding it in a sphere of radius 1.1×10−15 m is 1.1×10−15 m 4πr2 dr 1 (1.1×10−15 m)3 = = 1.5×10−15 . 0 8πa30 6 (5.29×10−11 m)3P47-12 (a) The wave function squared is e−2r/a0 ψ2 = πa3 0The probability of finding it in a sphere of radius r = xa0 is xa0 4πr2 e−2r/a0 dr P = , 0 πa3 0 x = 4x2 e−2x dx, 0 = 1 − e−2x (1 + 2x + 2x2 ). (b) Let x = 1, then P = 1 − e−2 (5) = 0.323.P47-13 We want to evaluate the difference between the values of P at x = 2 and x = 2. Then P (2) − P (1) = 1 − e−4 (1 + 2(2) + 2(2)2 ) − 1 − e−2 (1 + 2(1) + 2(1)2 ) , = 5e−2 − 13e−4 = 0.439. 274
• P47-14 Using the results of Problem 47-12, 0.5 = 1 − e−2x (1 + 2x + 2x2 ),or e−2x = 1 + 2x + 2x2 .The result is x = 1.34, or r = 1.34a0 .P47-15 The probability of finding it in a sphere of radius r = xa0 is xa0 r2 (2 − r/a0 )2 e−r/a0 dr P = 0 8a30 1 x 2 = x (2 − x)2 e−x dx 8 0 = 1 − e−x (y 4 /8 + y 2 /2 + y + 1).The minimum occurs at x = 2, so P = 1 − e−2 (2 + 2 + 2 + 1) = 0.0527. 275
• E48-1 The highest energy x-ray photon will have an energy equal to the bombarding electrons,as is shown in Eq. 48-1, hc λmin = eVInsert the appropriate values into the above expression, (4.14 × 10−15 eV · s)(3.00 × 108 m/s) 1240 × 10−9 eV · m λmin = = . eV eVThe expression is then 1240 × 10−9 V · m 1240 kV · pm λmin = = . V VSo long as we are certain that the “V ” will be measured in units of kilovolts, we can write this as λmin = 1240 pm/V.E48-2 f = c/λ = (3.00×108 m/s)/(31.1×10−12 m) = 9.646×1018 /s. Planck’s constant is then E (40.0 keV) h= = = 4.14×10−15 eV · s. f (9.646×1018 /s)E48-3 Applying the results of Exercise 48-1, (1240kV · pm) ∆V = = 9.84 kV. (126 pm)E48-4 (a) Applying the results of Exercise 48-1, (1240kV · pm) λmin = = 35.4 pm. (35.0 kV) (b) Applying the results of Exercise 45-1, (1240keV · pm) λKβ = = 49.6 pm. (25.51 keV) − (0.53 keV) (c) Applying the results of Exercise 45-1, (1240keV · pm) λKα = = 56.5 pm. (25.51 keV) − (3.56 keV) E48-5 (a) Changing the accelerating potential of the x-ray tube will decrease λmin . The newvalue will be (using the results of Exercise 48-1) λmin = 1240 pm/(50.0) = 24.8 pm. (b) λKβ doesn’t change. It is a property of the atom, not a property of the accelerating potentialof the x-ray tube. The only way in which the accelerating potential might make a difference is ifλKβ < λmin for which case there would not be a λKβ line. (c) λKα doesn’t change. See part (b). 276
• E48-6 (a) Applying the results of Exercise 45-1, (1240keV · pm) ∆E = = 64.2 keV. (19.3 pm) (b) This is the transition n = 2 to n = 1, so ∆E = (13.6 eV)(1/12 − 1/22 ) = 10.2 eV.E48-7 Applying the results of Exercise 45-1, (1240keV · pm) ∆Eβ = = 19.8 keV. (62.5 pm)and (1240keV · pm) ∆Eα = = 17.6 keV. (70.5 pm)The difference is ∆E = (19.8 keV) − (17.6 keV) = 2.2 eV.E48-8 Since Eλ = hf = hc/λ, and λ = h/mc = hc/mc2 , then Eλ = hc/λ = mc2 .or ∆V = Eλ /e = mc2 /e = 511 kV. E48-9 The 50.0 keV electron makes a collision and loses half of its energy to a photon, then thephoton has an energy of 25.0 keV. The electron is now a 25.0 keV electron, and on the next collisionagain loses loses half of its energy to a photon, then this photon has an energy of 12.5 keV. On thethird collision the electron loses the remaining energy, so this photon has an energy of 12.5 keV. Thewavelengths of these photons will be given by (1240 keV · pm) λ= , | 172,203 | 428,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2014-10 | latest | en | 0.897021 |
https://www.physicsforums.com/threads/understanding-the-relationship-between-vawt-and-pma.813825/ | 1,718,907,008,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00122.warc.gz | 826,503,461 | 23,385 | # Understanding the relationship between vawt and pma
• wellingtonex
In summary, the new member is trying to figure out how to use a PMA to generate power from wind. They have built a Savonius rotor and have been struggling to find the best pma to match their mill. They have found that at first the reasoning seemed simple- find a pma and adapt the mill to drive it at the desired rpm. However, their belts are squealing and the pma won't turn any faster. They are now thinking of a low speed pma. Their mill is exposed to roughly 6000w of wind energy, which is keeping the betz in mind. They factor that their mill is 10% efficient leaving them with 600 watts of extracted energy. They struggle with the torque requirements of a
wellingtonex
Hello to all, new member here, stand back! I've built a few prototype vertical axis wind turbines and have been on a major learning curve trying to find the best pma( permanent magnet alternator) to match my mill. At first the reasoning seemed simple. Find a pma and adapt the mill to drive it at the desired rpm. 80 watts later my belts are squealing and pma won't turn any faster. At a ratio of 1:9 I believe I've spent all available torque. So my thoughts are now directed at a low speed pma. My mill is exposed to roughly 6000w of wind energy, keeping betz in mind, I factor that my mill is 10% efficient leaving me with about 600 watts of extracted energy. Heres where I struggle, the low speed pma requires 554nm at 2500watts. I'm sure i won't be producing this much torque but i know my mill is quite strong. Is there a formula or a way to calculate torque with only primative information? Does my reasoning toward this matter seem adequate? Can anyone give me their insights on this matter?
Thank you
Welcome to PF.
What type of vertical axis mill have you built ? If it is a Savonius rotor then that would explain your problems.
How do you remove the energy from the PMA? If it charges a battery directly then RPM will be limited by battery voltage as energy will only start to flow when the RPM of the PMA exceeds the battery voltage. Any higher voltage will stall the aerodynamics of the windmill.
You should not be using belts that slip. Use bicycle roller chain or stepped belts. V-belts are inefficient.
wellingtonex
Baluncore, the mill is a savonous/darreus h-rotor with variable pitch. My first prototype was reaching tsr's of 1.68, proving the mill is generating lift, operating at 100rpm. My newest build has a much larger radius which operates at a slower speed but quite strong.
Because of the mills unique design, I've been bound to a v belt drive system, which could be changed, but quite difficult. I should rephase squealing to squeaking belts... tthe belts are capable of stopping the mill, but there is slippage.
My results of testing show that changing the ratio between the mill and pma only change the rpm of the mill with slight differences in pma wattage output. As i reduce the ratio, the mill speeds up accordingly with small increases in amperage from the pma.
As you mentioned, the chosen voltage will restrict the output of the pma. At this point, I have tested the following... 12v, 24v, 36v, 48v, and 120v inverter. My best test results always seem to come from 36v which in some way must match the power curves of the mill.
At this point I'm not comfortable winding up this mill to higher sppeed . Its quite large and is very sensitive under load, when i tested 120v, the freewheeling point before power production was quite scary, once it started producing she settled down?
My target mill rpms are 15-30 Which explains the 1:9 gear induction however, I'm using high speed pma's. I am currently considering a low speed pma which boasts 2.5kw at 50rpm. This could be Direct driven but I struggle with the torque requirements to do this. 554nm or 408 ft lbs seems quite high and I don't want to overestimate my mills capability. I realize I won't be utilizing the full potential of pma, I just want to be certain I'm choosing one that will work for my situation.
Now youve got me wondering what my tsr is on my mill at 30rpm... as I haven't tested it. I may not be generating lift at 30rpm. A worthy test.
I really apreciate your input... thank you, Rod
I have seen too many people fail because they liked the elegance of the Savonius rotor. Unfortunately, the Savonius rotor is the kiss of death for power generation. Once it infects the brain it can be considered a terminal case. It is probably the most inefficient design, but it needs the strongest tower.
I guess you only use the Savonius to start the Darius, but if the Savonius remains engaged it will waste much of the available energy since the Savonius and Darius have different optimum RPM to wind-speed curves. While the Darius is generating power, the Savonius will be consuming power. If the Savonius is built inside the envelope of the Darius then you will have problems because the dirty air downstream of the Savonius will upset the Darius. That airflow interference will also cause tower oscillation at some speeds. If you have an anemometer that indicates sufficient wind, and if you sense the Darius not rotating, then you can start the Darius with a motor / generator. That is a more efficient strategy than employing a Savonius rotor.
If you know the wind speed then you should be able to measure or calculate the optimum RPM for the windmill to extract maximum power. As you load the mill with the alternator the mill will slow down until it passes the optimum RPM for that wind-speed. That will be when the angle of attack is optimum. If you draw more power you will begin to stall the foils.
If you used a car alternator with a field winding, then you could regulate energy extraction and transfer to the battery by control of the field current. Unfortunately you cannot do that with a PMA.
The windmill will operate best at different RPM in different wind speeds. A PMA will need to generate into a capacitor at the voltage generated by the PMA at the optimum RPM for the wind speed. A switching converter can then transfer power from the capacitor into the fixed voltage battery. Drawing the right amount of power from the mill via the capacitor is critical to getting maximum energy from the windmill.
Wow, youre just the guy I need to talk to. I couldn't agree more with what youve said. Dont get me wrong, I'm no genius, and I have fallen into the arms of the savonous. I've done much research on prior work with the vertical and tried to learn from others mistakes. With this research I've developed a savonous rotor that doesn't spend extracted energy to force downwind blades thru drag zones. My design has its draw backs no doubt, but it has huge upsides aswell. This towerless design works extremely well in very poor conditions. It targets the urban market eliminating high speeds and the need for massive towers and expensive transmission lines. I am not trying to oversell my mill as I am having trouble just trying to rate its performance and capability. I'm probably in over my head but I'm having too much fun and I'm too stubborn to give up! Lol My design is patent pending at this point, but I'm not anxious to advertise it online yet.
With help from people like Yourself, I'll be able to bring out the full potential of my design. I've read some very interesting articles and many well educated people will disagree with me, " there's NO power in dirty wind!", they say. I disagree, as a sheet of plywood and a light gust will prove otherwise! We just haven't found an acceptable means of harnessing it yet.
I would be very interested in your thoughts on my design, concept, etc.
I don't want to assume i have massive amounts of torque, and I don't want to order the wrong lowspeed pma( as its coming from China) I can rig up a scale to measure ftlbs at start up but I'm sure the torque will increase as the mill reaches its optimum operating range. I've recorded the maximum amps for different gear ratios which I believe I can do a bench torque test to relate the relative torque. I've heard of the pma feeding a capacitor but I guess At this point I'm more interested in capability more so than efficiency. ( hoping I understood your post)
The capacitor makes perfect sense, now that I think more about it. It alllows the pma to operate at a much wider range, annd maintain full effiency.. Would the switching converter cause surging on the mill when it dumps A full capacitor? At higher speeds I'm sure it would switch very quickly( depending on capacitor size) but what about low. Speeds?
I can't thank you enough for your inut, I look forward to more!
Rod
wellingtonex said:
Would the switching converter cause surging on the mill when it dumps A full capacitor?
The switching converter would only take small bites of energy from the capacitor. The cap would be large enough that it had only small voltage changes. When the windmill is generating more energy, the converter will bite more often. Knowing the wind speed and your windmill characteristics you can know the optimum RPM and hence the PMA output voltage. Whenever Vcap rises above that voltage the converter takes another bite and efficiently puts that small bite of energy into the battery.
Power; W = V * I. Charge; Q = I * t. Capacitance; C = Q / V
If your optimum PMA output voltage is V at some RPM, and it is generating W watt then the average current will be I = W / V.
If the time between bites is dt, and the voltage step you can tolerate is dV volt, then C = I * dt / dV.
That gives you the capacitor value needed. It will need a bigger capacitor for lower RPM when the voltage steps will be bigger because Cenergy = 0.5 * C * V2. There will be some speed below which it is better not to extract the little power available but to keep the blades spinning ready for the next gust.
wellingtonex
Baluncore said:
The switching converter would only take small bites of energy from the capacitor. The cap would be large enough that it had only small voltage changes. When the windmill is generating more energy, the converter will bite more often. Knowing the wind speed and your windmill characteristics you can know the optimum RPM and hence the PMA output voltage. Whenever Vcap rises above that voltage the converter takes another bite and efficiently puts that small bite of energy into the battery.
Power; W = V * I. Charge; Q = I * t. Capacitance; C = Q / V
If your optimum PMA output voltage is V at some RPM, and it is generating W watt then the average current will be I = W / V.
If the time between bites is dt, and the voltage step you can tolerate is dV volt, then C = I * dt / dV.
That gives you the capacitor value needed. It will need a bigger capacitor for lower RPM when the voltage steps will be bigger because Cenergy = 0.5 * C * V2. There will be some speed below which it is better not to extract the little power available but to keep the blades spinning ready for the next gust.
I think I'm still with you, but for clarification... And a better understanding, is the voltage measured off the 3 phase or is it optimum voltage rectified? if it were off the three phase I would assume I'd need three capacitors.
Will the windmill freewheel if the capacitor is full and converter fails? Sorry if that's a dumb question, I am curius on how the capactor behaves and if its a load for pma at all times or just when discharged.
To better understand the converter... if the pma is at entry level charging voltages, does the converter dump on a voltage difference? In other words, the cap doesn't need to be full for converter to dump?
Does the cap accumulate voltage or does it just store the peak voltage sent to it?
I know more voltage means more potential, in this case is more voltage the best? Should I pick the highest voltage pma that will suit the rpms and torque of my mill?
I am still running a high speed pma at this point and only have 23volts at my optimum rpm. I think I'd be better off to order a lowspeed pma before I start entering in values to those equations.
I'm not sure about this voltage step tolerance... when the converter dumps to load I'll assume the mill will respond accordingly with load. If the converter dumps longer... the mill could stall... therefore the tolerance could be adjusted by the length of time each dump takes? Am I on the right page with this?
Once again I thank you for your input! I really appreciate it
Rod
wellingtonex said:
is the voltage measured off the 3 phase or is it optimum voltage rectified? if it were off the three phase I would assume I'd need three capacitors.
The 3 phase alternator is rectified by a 6 diode bridge to produce DC with a ripple voltage. I expect that the peak rectified voltage will be proportional to RPM. The capacitor will be charged to close to that peak DC voltage.
wellingtonex said:
Will the windmill freewheel if the capacitor is full and converter fails?
Yes. The capacitor must be rated for the maximum voltage that the PMA can generate when it freewheels in a windstorm. You should attach some form of speedbrake, something like rubber flaps to your Darius blades that will flip out and become air-brakes at some large RPM. That will protect the PMA and electronics, it will also prevent destruction of your windmill.
wellingtonex said:
Im curius on how the capactor behaves and if its a load for pma at all times or just when discharged.
The capacitor will accept charge whenever PMA voltage exceeds the capacitor voltage. The capacitor smooths the voltage from the PMA. It should not be operated full or empty. Capacitor voltage will average just below the PMA output voltage. The PMA will deliver most current, (charge), at phase peaks, less current in the valleys, so the capacitor is always there as a charge reservoir for the PMA.
wellingtonex said:
Does the cap accumulate voltage or does it just store the peak voltage sent to it?
A capacitor accumulates and stores charge, Q = I * t. A current of one amp for one second delivers one coulomb of charge. The voltage on a capacitor is proportional to the charge stored in the capacitor. V = Q / C.
The switching converter is designed to keep the capacitor voltage close to the optimum for that wind speed.
wellingtonex said:
Should I pick the highest voltage pma that will suit the rpms and torque of my mill?
It is best to have a higher voltage so you get Vcap = Vbat at the lowest sensible RPM for generation. A switching converter can step Vcap up to Vbat if that is needed, but it is more efficient to step down from a higher voltage.
wellingtonex said:
I'm not sure about this voltage step tolerance... when the converter dumps to load I'll assume the mill will respond accordingly with load. If the converter dumps longer... the mill could stall... therefore the tolerance could be adjusted by the length of time each dump takes?
The converter switches at up to 100,000 times per second. The voltage dV due to the bites of energy taken from the capacitor will be less than the natural ripple on the 3 phase rectified supply. There may be several bites taken during the time that one phase is peaking.
The critical thing is to take energy from the capacitor when the mill is spinning faster than is optimum for the particular wind speed. The converter should always wait until Vcap has recovered and is high enough before it takes the next bite. It will not stall. I expect dV to be maybe a couple of volts.
You will need to know the optimum RPM for maximum sustained power at different wind speeds. That will decide the optimum angle of attack of the Darius' blade.
You might gather that information automatically by sweeping power extraction at different wind speeds to build up a map of windspeed to RPMopt for maximum power over time.
Last edited:
With all this information I find myself approaching old problems with new thinking. Of course it arouses more questions. When I first hooked up my high speed 3.5kw to my mill the lowest ratio I used was 1:3 and upto 1:9.
Today I changed the ratio to 1:1.10 with much better results. Before I switched it I was at 1:3 getting 4 amps at 15 rpm (mill). Does that mean at 1:1 ratio I would produce 3 times the torque/amps?
The 4 amps I was generating was a dead short on the rectifier. Is this a practical means of testing? I usually test into a battery bank.
I measured open end volts off rectifier as well as voltage under load, when calculating pma wattage do I use open ended or loaded volts* amps?
The 1:10 ratio works well as long as I restrict it with a 12volt supply. As wind power cubes with every mile per hour I wonder if torque does as well? As the rpm increases I'm aware the drag increases on a vertical buT I would imagine the torque increases until the power curve starts diminishing.
Thanks again, Rod
wellingtonex said:
The 4 amps I was generating was a dead short on the rectifier. Is this a practical means of testing? I usually test into a battery bank.
Power is the product of voltage and current. If you short the rectifier bridge with an amp meter then you have no volts and therefore no external power. The windings in the alternator may get quite hot.
wellingtonex said:
I measured open end volts off rectifier as well as voltage under load, when calculating pma wattage do I use open ended or loaded volts* amps?
If you measure OC voltage, then there is no current flow and no voltage drop in the alternator windings. Without current your power is zero. You must generate into a reasonable load and measure both current and voltage at the same time to find the power.
The optimum gear ratio you use will be determined by wind speed and the voltage of the battery being charged. With a switching converter that ratio becomes unimportant as the effective ratio is handled by the converter.
Always remember that Power is the product of torque and RPM.
## 1. What is a VAWT?
A VAWT (Vertical Axis Wind Turbine) is a type of wind turbine where the axis of rotation is vertical, perpendicular to the ground. This is in contrast to traditional HAWT (Horizontal Axis Wind Turbine) where the axis of rotation is horizontal.
## 2. What is a PMA?
PMA (Permanent Magnet Alternator) is a type of electrical generator that uses permanent magnets to produce an alternating current (AC) output. It is commonly used in wind turbines to convert the kinetic energy of wind into electrical energy.
## 3. How are VAWTs and PMAs related?
VAWTs and PMAs are often used together in wind turbine systems. The VAWT captures the wind energy and rotates the PMA, which then converts the mechanical energy into electrical energy. The two components work together to generate electricity from wind power.
## 4. What are the advantages of using a VAWT and PMA system?
There are several advantages to using a VAWT and PMA system. Firstly, VAWTs are more compact and can easily be installed in urban or crowded areas where traditional HAWTs may not be feasible. Additionally, PMAs are more efficient in low wind speeds, making them suitable for areas with lower wind speeds. Lastly, VAWTs are also less affected by changes in wind direction, making them more reliable in varying wind conditions.
## 5. What are the limitations of VAWTs and PMAs?
While VAWTs and PMAs have many advantages, they also have some limitations. VAWTs are generally less efficient compared to HAWTs, especially in high wind speeds. They also have a lower power output, making them more suitable for small-scale or residential use rather than large-scale energy production. Additionally, PMAs may require maintenance and replacement of the permanent magnets over time, which can be costly.
• Mechanical Engineering
Replies
3
Views
3K
• Electrical Engineering
Replies
1
Views
1K
• General Engineering
Replies
7
Views
2K
• Electromagnetism
Replies
11
Views
3K
• Classical Physics
Replies
3
Views
1K
• General Engineering
Replies
16
Views
6K
• New Member Introductions
Replies
1
Views
488
• MATLAB, Maple, Mathematica, LaTeX
Replies
2
Views
2K
• MATLAB, Maple, Mathematica, LaTeX
Replies
2
Views
3K
• Atomic and Condensed Matter
Replies
4
Views
6K | 4,542 | 20,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.969264 |
http://mathb.in/1545 | 1,406,946,400,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510276250.57/warc/CC-MAIN-20140728011756-00176-ip-10-146-231-18.ec2.internal.warc.gz | 186,006,675 | 4,160 | Given $n$ object types, each with cost $n_{c}$, I need to combine object instances such that the total costs equal a known total $T$. Each object type can have $0 \le x \le y$ instances (where $x$ is a positive integer), but I wish to maximize diversity of object types.
I believe a linear equation expressing this would look like the following?
$a{c}\cdot \left( \begin{array}{ccc} \ 0 \ 1 \ 2 \ .. \ y \end{array} \right) + b{c}\cdot \left( \begin{array}{ccc} \ 0 \ 1 \ 2 \ .. \ y \end{array} \right) + .. +n_{c}\cdot \left( \begin{array}{ccc} \ 0 \ 1 \ 2 \ .. \ y \end{array} \right) = T$
I've tried what little linalg I remember from university, plus some simple combinatorics.. no dice.
Wednesday, 9 January 2013 22:20 GMT | 233 | 731 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2014-23 | longest | en | 0.750943 |
https://byjus.com/question-answer/the-board-of-directors-of-a-company-decided-to-issue-minimum-number-of-equity-shares/ | 1,680,276,131,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949644.27/warc/CC-MAIN-20230331144941-20230331174941-00763.warc.gz | 189,527,842 | 20,660 | Question
# The board of directors of a company decided to issue minimum number of equity shares of Rs. 10 each at 20% discount to redeem 4,500 preference shares of Rs. 100 each. If the maximum amount of divisible profit is Rs. 2,50,558. Calculate the number of equity shares to be issued. How many shares will be issued if they are issued in multiple of 50?
A
24,931 & 24,950
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
24,931 & 24,900
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
24,932 & 24,950
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
24,932 & 24,930
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
## The correct option is A 24,931 & 24,950Here, the divisible profit is Rs 2,50,558 and redemption price of 4500 shares is Rs 100 each i.e Rs 4,50,000. Now the balance left for equity shares is Rs 4,50,000-Rs 2,50,558 i.e. 1,99,442. Equity shares are issued at Rs 10 per share with 20% discount i.e Rs 10- Rs 2 i.e. Rs 8 per share which means number of shares to be issued will be Rs 1,99,442/Rs 8 i.e. 24,931 shares. If shares are issued in multiple of 50, then 24,950 shares will be issued as 24931 is minimum amount of shares and it cannot be 24,900.
Suggest Corrections
0
Related Videos
Issue of Shares
ACCOUNTANCY
Watch in App
Explore more | 425 | 1,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-14 | latest | en | 0.90966 |
https://phantran.net/hypergeometric-probability-distribution/ | 1,643,326,803,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305317.17/warc/CC-MAIN-20220127223432-20220128013432-00619.warc.gz | 499,222,984 | 22,767 | # Hypergeometric Probability Distribution
The hypergeometric probability distribution is closely related to the binomial distribution. The two probability distributions differ in two key ways. With the hypergeometric distribution, the trials are not independent; and the probability of success changes from trial to trial.
In the usual notation for the hypergeometric distribution, r denotes the number of elements in the population of size N labeled success, and N – r denotes the number of elements in the population labeled failure. The hypergeometric probability function is used to compute the probability that in a random selection of n elements, selected without replacement, we obtain x elements labeled success and n – x elements labeled failure. For this outcome to occur, we must obtain x successes from the r successes in the population and n – x failures from the N – r failures. The following hypergeometric probability function provides fx), the probability of obtaining x successes in n trials.
Note that represents the number of ways n elements can be selected from apopulation of size N; represents the number of ways that x successes can be selected from a total of r successes in the population; and represents the number of ways that n − x failures can be selected from a total of N − r failures in the population.
For the hypergeometric probability distribution, x is a discrete random variable and the probability function fix) given by equation (5.16) is usually applicable for values of x = 0, 1, 2, . . . , n. However, only values of x where the number of observed successes is less than or equal to the number of successes in the population (x < r) and where the number of observed failures is less than or equal to the number of failures in the population (n – x < N – r) are valid. If these two conditions do not hold for one or more values of x, the corresponding f(x) = 0 indicates that the probability of this value of x is zero.
To illustrate the computations involved in using equation (5.16), let us consider the following quality control application. Electric fuses produced by Ontario Electric are packaged in boxes of 12 units each. Suppose an inspector randomly selects three of the 12 fuses in a box for testing. If the box contains exactly five defective fuses, what is the probability that the inspector will find exactly one of the three fuses defective? In this application, n = 3 and N = 12. With r = 5 defective fuses in the box the probability of finding x = 1 defective fuse is
Now suppose that we wanted to know the probability of finding at least one defective fuse. The easiest way to answer this question is to first compute the probability that the inspector does not find any defective fuses. The probability of x = 0 is
With a probability of zero defective fuses f(0) = .1591, we conclude that the probability of finding at least 1 defective fuse must be 1 – .1591 = .8409. Thus, there is a reasonably high probability that the inspector will find at least 1 defective fuse.
The mean and variance of a hypergeometric distribution are as follows.
In the preceding example n = 3, r = 5, and N = 12. Thus, the mean and variance for the number of defective fuses are
Source: Anderson David R., Sweeney Dennis J., Williams Thomas A. (2019), Statistics for Business & Economics, Cengage Learning; 14th edition. | 728 | 3,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-05 | latest | en | 0.871633 |
https://rpg.stackexchange.com/questions/27016/how-can-i-judge-if-combat-in-a-dungeon-room-attracts-attention | 1,716,415,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00894.warc.gz | 441,779,985 | 48,987 | # How can I judge if combat in a dungeon room attracts attention?
A question that usually props up at my table is "If we start a fight in this room, will we attract any attention?" I guess usually this refers to the noise and din of combat. So, say the party is fighting in a stone chamber, and about 50 feet away there are some guards, and the door isn't closed. How should I determine if the guards can hear the fight?
The system I am using is 13th Age. It offers no rules or guidelines for such cases, though I guess if I ask the authors they might answer "whatever the narrative requires". I'm not looking for a high-fidelity answer based on physics, but something that's reasonable will do.
The reason I am asking is because the players are concerned about this - they keep pointing out how others would hear them if they start a fight now. What are some ways I can assure them a dungeon crawl is not a suicidal mission because of the noise issue?
• Comments are not for extended discussion; this conversation has been moved to chat. Jun 5, 2018 at 21:46
You can use your players' intuition that the sound of battle should be important without squashing their imagination of the dungeon, while also serving your GMing need to keep the whole dungeon from going "on alert" and dogpiling them at the first clash of swords. You do this by making sound weird in these strange, underground halls, and then telegraphing to your players how sound works in this dungeon by describing what they hear.
There are two standard ways to hint that fight noise will not normally carry, without simply revealing the man behind the curtain and showing that it's just a game and they "shouldn't" care about this.
## The dungeon deadens sound
The hush of the tomb is expected, but there's something more to it. As you proceed down the corridor, you realise you don't even hear your footsteps echoing back at you. Conversation from the tail of your marching order sounds like thin, unintelligible whispers only yards away at the head. The very air itself seems to swallow sound.
Variations on this have been mentioned in various comments: muffling tapestries, loud or white ambient noise that masks the sounds of the party's incursion (machinery, noisy rituals, rushing water or air), and the like. Deadening sounds comes in a variety of forms, but they all have in common that a brief description of how the environment interferes with the party's hearing will let them know that fight noise is not going to travel as far as they would normally fear.
## The location's acoustics confound the ability to tell where sounds come from.
This takes a bit more work. The best way to show (not tell) this is to insert random noises, screams, and whispers into the game. Creating a list and randomly rolling on it every in-game chunk of time is the traditional means.
For example:
Atmospheric Dungeon Sound Events Table
Roll 1d8 every 30 in-game minutes:
1. A scream, suddenly cut off, echoes from somewhere deep in the dungeon.
2. Mutterings that sound right behind your shoulder that slowly fade away.
3. The noise of battle, seemingly coming from directly in front of the party, where they can plainly see there is nothing.
4. The sound of dripping water, as if from very far away, but it follows the party without growing louder or fainter for several minutes.
5. Growls echo down the hall (1–3: ahead, 4–5: behind, 6: both)
6. A strong but quiet winds blows, but it seems to snatch away any attempt at conversation, carrying the party's words who-knows-where.
7. No event
8. Roll twice and combine, rerolling results 7–8.
(This is just am example. Expand as necessary before, between, or even during sessions.)
By giving a dungeon unusual acoustic properties, you can leverage how your players are already intelligently thinking about sound, making them aware in a natural way that the sounds of battle can't easily be used to locate them, unless the listener is already very close (and perhaps even not then). You also reward your players' engagement with the world by responding with in-world details about something they've shown interest in, making their investment deeper rather that fighting against it. As a bonus, it gives your dungeon more character and makes it a more unnerving and unnatural environment.
• I love this answer. It gives plausibility as to why the party could fight in one part of the dungeon, and could still explore other parts without being harassed. Jul 9, 2013 at 18:36
• Also, let them use sound against their opponents - they should be able to figure stuff out by sound themselves. Jul 9, 2013 at 22:39
## Fights are noisy.
(Disclaimer: I do not have much experience with IRL combat, so this answer is not entirely experience-based, but I've rolled in the contributions of someone who has.)
REALLY noisy. Metal against metal goes CLANG. Gunfire goes BANG at least. People shout to intimidate their foes, to communicate over the din of combat, or just to give themselves courage. Bones break: SNAP! Objects are knocked around and thrown: CLATTER CLANG SHATTER. People scream when they're wounded --animals even more so, so if you've got horses or animal companions or the like, keep that in mind. Even cut throats gurgle, so unless you're armed with garrotte wire there's always going to be SOME noise. And if you are trying to be stealthy, loss of subtlety is always a single mistake away.
Unless the group is working very hard to keep things quiet, I'd say that a fight can easily be heard at least fifty meters away, environmental factors notwithstanding (fog absorbs noise; stone makes it echo; sound carries further over water).
How hard is it to keep a fight quiet? Harder the longer it lasts. Specifically, the instant someone has a chance to shout for help, he will unless given VERY good reason not to. Unless the party can successfully ambush a guy using quiet take-down techniques, I'd expect everyone within at least 150-200 feet to know something's going down.
## This is inconvenient.
Especially for the GM who doesn't want to dogpile a whole fort's-worth of soldiers on the party in a single fight. This is why "realistic" fight noise levels are generally ignored.
There's not much I've ever found to justify this in-game, it's just a tradition: If it were treated realistically the party would wipe, so the GM assumes noise levels occur at the decibel of plot. If you can be consistent in what that means, the players will be appreciative because then they know what to expect. Otherwise, it's basically up to you as the GM to be worthy of their trust in this matter.
• Having been in combat IRL, in the middle of a fight, you're not to worried about ambiant/background noises, as you are usually dealing with the threat that is most immediate/dire to you/said person. But combat is extremely noisy, and despite Hollywood BS, a slit throat gurggles and isn't all that quiet in the immediate vicinity. Metal on Metal is quite loud, and unless you're going for the stealth approach, subtelty is always a strike away from going out the window. Unless you're carrying a garrote. So most of that ninja stuff is fake. Sorry. Jul 9, 2013 at 18:37
In addition to what @SevenSidedDie said, I would add a couple of other options:
• Have different dungeon areas physically disconnected from each other - instead of walking through a door directly to the next room, they can walk a few miles of twisty cave passages between rooms, or go through portals, or ...
• Figure out some way to render all the baddies deaf. This could be a massive explosion that happens near the beginning, an effect of being exposed to demonic magic workings, the servants having been made mute and deaf so they can't hear or tell any of their master's secrets, or even something as simple as a disease.
• The dungeon can have some sort of background noise that renders hearing impossible, like big factory machinery (they could all be wearing hearing protection, too), volcanic activity, or the magic ritual they are trying to stop. (Who says that magic is quiet? Rituals can be noisy, everything from overpowering high-pitched cosmic resonations, to the blood-curdling screams of infernal demons, to explosions and earthquakes and illusory battles and armies and ...) If you want, in keeping with the 'show, don't tell' theory, you could make the players roll a saving throw (or the equivalent in your system) to avoid taking some adverse effect from the noise (applicable effects might include deafened, stunned, distracted, etc).
• Let the PCs use some sort of spell/ritual/magic item to isolate an area, or just mute all sound within the zone (one may even already exist, although I don't know for sure).
This kind of thing always depends on the players, but I wouldn't make a hard-and-fast rule. There's more drama if the players don't quite know if they're going to have extra company or not.
Have the PC's make stealth rolls (however those work in 13th Age) at appropriate moments. Have the NPC's they are fighting in the room try and sound the alarm (e.g. ring a bell, or blow a whistle). You can also throw-in some fun moments such as "You stab the orc and he dies instantly, dropping his heavy metal shield, make a dexterity check to try and catch the shield before it hits the ground and make a ton of noise".
If the PC's fail miserably at any of these point, the guards hear and come running.
Lastly, never forget the Rule of Cool - if 'attracting attention' would add more fun to the situation, then do it. This isn't supposed to be a 100% realistic simulation, it's supposed to be fun.
Disclaimer: I have no experience with 13th Age, but as far as I'm aware, it's somewhat similar to D&D, even being made by lead devs from 3e and 4e. In particular, the settings I describe ahead are not system-specific, although my experience is from D&D 5e.
If, for some reason, an experienced player in 13th Age thinks my answer does not apply in any way, please comment.
The spoilers in answers are for text that is supposed to be only for DM's eyes running LMoP.
As SSD mentioned, one good way to make it real is just letting clear to the adventurers that it's hard to listen to anything inside that dungeon. This can be made in different ways, already mentioned in his answer - machinery, noisy rituals, rushing water or air. In the first dungeon of LMoP, it's a cave in which a river runs through. The following description is given:
Sound: The sound of water in the cave muffles noises to any creatures that aren't listening carefully.
handled as
Creatures can make a DC 15 Wisdom (Perception) check to attempt to hear activity in nearby chambers.
The second phrase is more about the adventurers than the enemies inside the dungeon - as there's no reason for them to randomly try to listen, unless they already have some information about something happening.
Just by describing the sound of running water making it harder to listen things far away my players accepted way better that the other rooms didn't instantly come to help, even being inside a cave that should propagate sound very well.
# Combat Sounds are Usual
And the dwellers of the dungeon are so used to it they don't mind. This also can be done in numerous ways - infights are normal, people train inside the dungeon, weak invasions are so common that each room is supposed to handle their own fights because they're usually easy. In LMoP, the "infights are normal" strategy is used sometimes, the following being stated for one of the rooms:
Goblins in nearby caves ignore the sounds of fighting wolves, since they constantly snap and snarl at each other.
To be honest I don't know how this would get played out. My players fed the wolves and our druid talked to them, making them unwilling to fight the party. What actually happened will be expanded in the next section.
# The other rooms DO listen
Well, maybe your players are right. Maybe some of the other rooms will listen and come to help. You can actually handle that by balancing the encounters assuming the adjacent room will come to help the current one.
For the previous scenario, the Goblins actually noticed that the wolves stopped making noise (because they were fed), and that was more unusual to them than the wolves making noise and killing each other.
The wolves were as friendly to the goblins as they were to the players, so the encounter from the adjacent room was moved to this room. Nothing indicating a TPK so far.
Aside from my own version with this, the "get help" method is explicitly stated in the same dungeon. In area 7, it describes:
The noise of the waterfall means that the creatures in area 8 can't hear any fighting that takes place here, and vice-versa. Therefore, as soon as the fight breaks out here, one goblin flees to area 8 to warn Klarg.
The encounter in Area 7 is easy (really easy). If the enemies succeed in getting help, though, the total encounter becomes deadly. As a note,
In the official adventure, Klarg doesn't even come to area 7 to help, instead he hides in Area 8 in order to surprise the adventurers when they get there, supposing his goblins will die to the party.
For this one, my players managed to kill the goblins fast enough so they didn't have time for getting help. As such, the encounter was the easiest they had until now.
# Maybe they DID listen, but so what?
Depending on your setting and your characters (NPCs), maybe they did listen to the fighting, but aren't bothered enough to go help. Again, this has numerous justifications - "Hmm, the fight is happening in the Captain's room, he didn't ask for help, if he dies I'm promoted... Nope, I didn't listen a thing." or "Well, they are my subordinates, if they can't handle a simple invasion, I don't care that they died."
This can be further explored when the players get to this NPC, which will talk to them like "Oh, you killed our Captain. If I kill you now, I'll be getting so much honor and fame! Thank you for your services." or "Hmf, I knew that those weaklings were useless. They couldn't even handle a bunch of lowly invaders."
This is exactly what I mention in the second spoiler from the above section.
The point here is: just because they live in the same place, it doesn't mean they are all friends and willing to help each other all the time.
I don't know 13th Age, so I may be giving you a useless hint.
Having said this:
a) you as the GM have to decide if, depending on the "room description", and what kind of opponents are in there, it wouldn't make sense for the PCs' opponents to actively cry for help. This leads to...
b) ask the players what kind of tactics they want to try and minimize this risk. Do they charge hoping that surprise and bloodlust will be enough to dispatch/incapacitate everyone in just one round? What about magic? having decided that...
c) make your party or the party leader, or whatever makes more sense in 13th age try a Luck Roll (assuming there is something like that in the system). The difficulty will depend on (a)+(b).
d) finally, in case of any fumbles by a PC, you may substitute any normal ill effect with another luck roll (possibly with increased difficulty) to see if the fumble causes so much noise that other enemies are alerted.
• The question is more about the usual noise a fight would make, more than whether the enemies would cry for help. Jul 9, 2013 at 16:42
• Then skip point (a). I understood you wanted some easy way to adjudicate the result... Jul 9, 2013 at 16:44
Any real life battle would be very noisy indeed and nearly impossible to conceal. You are not talking about real life, as the DM you get to make the rules about what attracts attention and what doesn't.
I would not focus too much and making the situation realistic. Don't let other enemies be aware of the party simply because they need to/want to fight something. This possibility makes the players analyze the situation too much, taking their minds off track from the objective and wasting time that could be spend doing something much more enjoyable.
If you do want the possibility of a fight alerting other people, build it in somewhere else. Maybe a guard has a horn on his belt that he could blow for help, and unless the party stops him in one turn he will sound it. Maybe there's some other type of alarm such as a large bell that a guard would have to run to to ring. In any case, I think you should always give the players a chance to stop the alarm from sounding if possible to do so.
For example in one of my DnD campaigns (I have no experience with 13th Age, but this is not a question specific to it) I had a trap in a hallway. Triggering said trap would release a cascade of rocks onto the head of whoever was under it, resulting in damage and a very loud sound (I hinted at both of these previously when they saw the same type of trap already triggered). Had they set off the trap not only would they have taken damage, but the goblins in the next room would also have been alerted to their presence. However, since they bypassed the trap, they were able to take the goblins by surprise.
• I firmly believe that players will guide themselves in what they enjoy. It could very well be that this party prefers the details of combat, and that is what they enjoy; players are good at finding what they enjoy in a campaign. Additionally, very often complex fights and scenarios are extremely engaging, if the group is engaged with the fight.
– user8248
Jul 9, 2013 at 19:49
• I agree, a complex fight with many elements to consider can always be very engaging. Hence why I say a DM should build the possibility of an alarm into the battle. Furthermore, I find the idea that combat noise will attract attention severely limiting to many characters. I rogue may be able to sneak around chocking enemies and slitting throats, but what's a fighter going to do? Charge the enemy and hope they're frightened speecheless and then try to ignore any noisy armor they're wearing? I think the alarm mechanic just makes the fight more viable for everyone. Jul 14, 2013 at 3:18 | 4,006 | 18,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.963905 |
https://m2solids.com/fishbowl.html | 1,718,566,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00280.warc.gz | 342,500,240 | 5,710 | by Mike Mongan
#### Understanding Light in Terms of the Unit of Universe
In the Unit of Universe kit we explore what happens to a circle when we push it beyond 360 degrees. We then develop analogies with physical processes. We now turn our attention to light. In previous work, the method of tearing a ribbon from the outer edge of a saddle has been described. It has been illustrated how this edge forms a helical ribbon if discontinuous or forms a twisted circular ribbon if continuous. It is with the latter that this exercise concerns itself.
To perform this exercise, you will need a six inch diameter spherical fishbowl. These are available in clear plastic for a very modest cost (\$3 plus). I have found them in pet supply stores. The six inch diameter is needed because the paper circles we will be using are the same size.
First construct a saddle of six or seven paper circles. Remember that you cannot reverse directions during the taping process. Tape both sides of the joining seams. When your stack is complete, trace the outer edge with your finger. It should follow a smooth helix either right or left, but not both. Now wind it up into a saddle in the manner described in the kit. Proceed to remove the edge as a continuous ribbon (figure 10 in the book or see the website). If it breaks, tape it together again. You may also staple across the seams where circles join for added reinforcement. Width of the ribbon should be about an inch or about a third of the radius, you need not be precise.
NOTE:DNA DEMONSTRATION uses the continuous ribbon you now hold in your hands. So you may want to keep track of this proceedure (and ribbon) for that exercise.
Illustration 1, The Bowtie
Carefully fold the ribbon as if you were making a bow for a present, pinching it between your fingers - illustration 1. Notice that you have an arch remaining that goes over the top. It may appear slightly different depending upon whether you started with an odd or even number of paper circles. Holding it by the center, insert it into the fishbowl. Release it. Spread it out inside of the fishbowl. The paper ribbon has an outer (original) edge and inner (torn or cut) edge. The outer edge of the paper should touch the inner wall of the fishbowl. It will appear similar to illustration 2. What you have done is to subtract the inner part of the saddle. The fishbowl simply supports the structure and aids with visualization of the ensuing process.
From previous work, you will remember that in our view the saddle you made lies along a continuum that goes from line through cone to circle and then sphere. The upper limit of our process is a saddle made from an infinite number of circles. It would be spherical and have a continuous surface. By that I mean some point from the edge of some circle would occupy each point on the surface. It would have a density gradient toward the center and the center would be an infinitely dense point. This is what is meant by an infinite saddle.
When you were winding up the taped circles into the saddle, you applied force to do so. Had you let go or if it slipped from your fingers, you would have seen it snap back. It wants to return to its lower energy state. However we have compelled it into a higher energy state with our trusty cellophane tape. This means that the outer portion is in compression. You may also have observed small tears developing at the center. This is because it is in tension. (Circles from Mē Solids have a small hole at the center to alleviate this problem.) So, our model is stressed between inner tension and outer compression. That means there is a zero boundary at some radial distance from the center which lies between the two and is neither. Assume for the sake of argument that this zero modulus is where you tore the ribbon.
In our infinite saddle, a jolt of some kind may cause the model to shear at this boundary. The ribbon would separate from the core. It would resemble the rind of an orange separated from the meat but still continuous. That is how you should visualize your ribbon in the fishbowl, the outer rind of an infinite saddle with the center removed. The rind is composed of our two dimensional ribbon. Once this jolt has created the shear, what happens to the remaining sphere, the juicy part or meat of our orange? It simply restabilizes into another tension-compression pair with a new zero modulus at a smaller radius. If it were to receive another jolt, it could reseparate into another rind-meat pair and so on.
What interests us now is what we have in our fishbowl, the rind of a subatomic particle, a photon. Now imagine our fishbowl to vanish as if it were made from a frozen soap bubble that suddenly popped. The ribbon was the compression part of our earlier pair. Now that the restraint of the outer shell is gone, it wants to decompress. It will do so by expanding outwards in all directions at once like a balloon released into a vacuum. Or more to the point, like a photon of light emitted by an electron.
Illustration 2, Photon in a Fishbowl
"Wait", you say, "there's a problem". The photon will go off in all directions. I would be able to observe it no matter what my location. Its wave front would be continuous regardless of radius. But your model in the fishbowl would not. Even if it were infinite in degrees, gaps would form between the sheets as it grew ever larger in expansion. There are two possible answers. First, the sheets could become thicker. But from the beginning we have considered our saddles to be composed of two dimensional sheets. So we can't get any thicker and stay within our system. But we still have a way out. We can sacrifice the width of our ribbon to its length-like a rubber band when stretched. Width here refers to the width of the ribbon we tore from our original saddle, the inch or so. Length is the brim length, the distance around its outer edge. Thus as the radius of our rind expands, the brim length of the ribbon expands also to keep the surface whole. If this is so where does the process end? It would end when our ribbon has been reduced to a one dimensional string. It also occurs at a speed fast enough to allow the wave front -represented by the outer skin of the rind- to travel at the speed of light and stay a continuous shell. The amplitude of the photon, essentially how much energy it carries, is represented at any time by the width of the ribbon.
At this point it may be useful to interject that this model uses circles, but nature doesn't seem to. Nature favors ellipses. It would be unduly complex to make our saddles from ellipses sharing a common foci. In a more perfect description of a particle, I would picture its point center orbiting between foci within a small shell.
As they stand, saddles do not have a stable configuration. As you hold one in your hands, you can twist it into a variety of balances. In some, more of the circles seem to bunch around an equator. In others, the circles distribute themselves more evenly. In comparing the infinite saddle to an electron I envision the whole thing squirming around before a photon breaks loose. Thus the photon itself would be in a more a dynamic condition than our mild mannered fishbowl suggests. It would I believe be ringing. This ringing is the source of what we perceive as color. Is it ringing like a bell? A bell rings because its surface is mechanically joined. Thus a jolt which distorts it will cause the entire body to seek restabilization while dissipating the energy of the strike. This is different. If we can imagine striking our fishy photon, it would not act with the connectedness of a bell. It is, after all, a folded ribbon.
Yet, the ribbon could still vibrate in a uniform manner. Let's say the ribbon in our photon is vibrating in a particular shade of green. As the rind expands it eventually strikes an object, a pure green leaf of the same shade. Actually only a leaf shaped portion of it strikes and the rest keeps going. It would be as if the ribbon shears off at the leaf-non leaf boundary. That which remains behind imparts its energy to the leaf as in the warming effects of sunshine. It also imparts its vibration, as a tuning fork when placed next to another tuning fork of the same note. The leaf which has been struck is thus set to vibrating and emits green, because that is its natural harmonic. If our photon had been red, it would not. The energy of the part of the photon striking the leaf would be absorbed as heat.
What about the rest of the photon, the one with the leaf shaped hole in it? It would still be expanding, relieving its compression. The hole is what we call a shadow. As the rind expands in its outward journey, so does the shadow.
Up until now we have been considering only a single photon. But our real life experience comes from multiple photons, emitted one after the other in a steady stream. From the sun, the radius of each is so large that they appear as flat as a wall. That seems to us as a continuous source of light. It seems almost liquid like and exhibits the properties of waves. But as we have seen, it can impact other objects as though they have been struck with a particle. This is the dual wave-particle nature of light as explained by the photon in a fishbowl. | 1,999 | 9,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-26 | latest | en | 0.936611 |
https://physics.stackexchange.com/questions/159152/expanding-universe-as-predicted-by-the-einstein-field-equation | 1,568,976,146,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573988.33/warc/CC-MAIN-20190920092800-20190920114800-00324.warc.gz | 642,421,861 | 32,225 | # Expanding universe as predicted by the Einstein Field Equation
Without the cosmological constant, the Einstein field equation predicts the universe is expanding. Why is that? It is counter-intuitive because generally gravity should pull things closer and shrink the universe.
The Einstein field equations predict that the universe can expand, not that it is expanding. The fact that the universe is expanding was first observationally discovered by Edwin Hubble in 1929. Before this discovery, Einstein had introduced the cosmological constant to keep the universe from collapsing under the influence of gravity. When he heard of Hubble's discovery, he removed the cosmological constant from the equation, as expansion could explain why the universe wasn't collapsing due to gravity. The inflationary universe theory gives us clues on how the expansion first originated/occurred after the big bang.
Let's consider the motion of a ball in the Earth's gravitational field:
• if you just let go of the ball it falls down
• if you throw the ball up it first rises then falls down
• if you throw the ball really hard it rises, escapes the Earth's gravity and flies off never to be seen again
So with the same ball and the same Earth we can get three different types of behaviour. The difference is of course in the initial conditions i.e. the velocity of the ball at $t = 0$. If we know the equation of motion of the ball then we can work out what initial velocity it had.
The FLRW metric is sort of the equation of motion of the universe, but there's a big difference from the equation of motion of the ball because we can't just choose any initial conditions we want. We only get a solution to Einstein's equations for particular initial conditions, and those are the conditions that the universe was expanding at the Big Bang.
Sp the universe had to start off expending. And you're quite correct that the combined effect of all the matter in the universe is to slow that expansion. If the matter density in the universe is high enough then it slows the expansion to a halt and the universe recollapses, sort of like a thrown ball falling back to Earth. This is called a closed universe. If the matter density in the universe is very low then it can't halt the expansion, sort of like the ball escaping the Earth's gravity. This is called an open universe. We think our universe is in between and it's a flat universe.
The point of all this arm waving is that you are quite correct to say:
gravity should pull things closer and shrink the universe
and actually gravity does to that. However the universe was expanding to start with so gravity has to work against the initial expansion.
Expansion of the universe is not related to gravity. Gravity pull things around mass by bending space. But overall space in the universe just keep expand by other reason | 573 | 2,867 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-39 | latest | en | 0.942767 |
http://seaintarchive.org/mailarchive/1999a/msg01192.html | 1,493,026,364,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119225.38/warc/CC-MAIN-20170423031159-00472-ip-10-145-167-34.ec2.internal.warc.gz | 335,317,601 | 3,961 | Need a book? Engineering books recommendations...
# RE: Excel Question
• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: Excel Question
• From: "Frank Yang" <yang(--nospam--at)abam.com>
• Date: Tue, 16 Feb 1999 09:48:21 -0800
```Sieiman,
I have two equations that will work. If you always start at row 1 (i.e.
B1=A1), then use this simplified formula: B1=offset(A1,row(A1)-1,0), and
copy the formula down. A more generic formula, works for situation like
(B2=A2, B3=A4...) is: B1=offset(\$A\$1,(row(b1)-row(\$b\$1))*2,0).
Hope this help.
Frank
Frank Yang, P.E.
Senior Engineer BERGER/ABAM Engineers Inc
Voice: 206-431-2374 33301 Ninth Avenue South
Fax: 206-431-2250 Federal Way, WA 98003-6395
e-mail: yang(--nospam--at)abam.com http://www.abam.com
-----Original Message-----
From: Sleiman Serhal [mailto:mony(--nospam--at)destination.com.lb]
Sent: Tuesday, February 16, 1999 3:02 AM
To: seaint(--nospam--at)seaint.org
Subject: Excel Question
Hello everybody,
I'm trying to work out something in Excel but it ain't working so maybe
someone
out there can help:
I want to fill down a column, let's say column B, and have it fill down
as
follows
B1 = A(1)
B2 = A(3)
B3 = A(5)
B4 = A(7)
etc...
So I want the fill down in B to pick up the cells at column A in
increments of
2 ! | 436 | 1,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-17 | latest | en | 0.728162 |
https://socratic.org/questions/561b6d3b581e2a7de5d01910 | 1,571,517,128,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986697760.44/warc/CC-MAIN-20191019191828-20191019215328-00264.warc.gz | 704,459,392 | 5,958 | # Question #01910
Oct 12, 2015
Solve $5 + 3 \cos \left(\frac{x}{2}\right) = 7$
$x = \pm 96.38$ deg
#### Explanation:
$5 + 3 \cos \left(\frac{x}{2}\right) = 7$
$3 \cos \left(\frac{x}{2}\right) = 2$
$\cos \left(\frac{x}{2}\right) = \frac{2}{3}$
Calculator gives -->
$\cos \left(\frac{x}{2}\right) = \frac{2}{3}$ --> $\frac{x}{2} = \pm {48}^{\circ} 19$ --> $x = \pm {96}^{\circ} 38$ | 171 | 384 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-43 | latest | en | 0.267538 |
https://math.answers.com/other-math/What_is_52_divided_by_7_with_remainder | 1,669,596,379,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00733.warc.gz | 421,236,276 | 47,597 | 0
# What is 52 divided by 7 with remainder?
Wiki User
2017-11-02 02:58:34
7.4286
Noble Bogan
Lvl 10
2021-02-25 21:47:50
Study guides
20 cards
## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
3.81
1751 Reviews
Wiki User
2017-11-02 02:58:34
7 R 3 | 129 | 335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-49 | latest | en | 0.811026 |
https://profiles.doe.mass.edu/mcas/mcasitems2.aspx?grade=03&subjectcode=MTH&linkid=1&orgcode=00780000&fycode=2018&orgtypecode=5& | 1,624,170,255,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487658814.62/warc/CC-MAIN-20210620054240-20210620084240-00284.warc.gz | 396,815,795 | 15,562 | Dover
Districts Schools
2018 Item by Item Results for GRADE 03 MATHEMATICS
Number of Students Included: 79 Mode: Paper
Dover - GRADE 03 MATHEMATICS ITEM INFORMATION PERCENT OF DISTRICT'S POSSIBLE POINTS ITEM TYPE REPORTING CATEGORY STANDARD ITEM DESC POSSIBLE POINTS DOVER STATE DISTRICT-STATE DIFFERENCE 1 SR NF 3.NF.A.02 Determine the fraction that is plotted on a given number line. 1 75% 78% -3 2 SR OA 3.OA.A.03 Solve a word problem involving division of two whole numbers. 1 97% 84% 13 3 SA NT 3.NBT.A.01 Round a three-digit whole number to the nearest ten. 1 91% 73% 18 4 SR OA 3.OA.A.04 Determine the missing factor in a multiplication equation. 1 96% 88% 8 5 SR NT 3.NBT.A.02 Solve a real-world problem by subtracting two three-digit whole numbers. 1 87% 73% 14 6 SR NF 3.NF.A.02 Determine which point on a given number line represents the location of a given fraction. 1 37% 33% 4 7 SR OA 3.OA.C.07 Use division or a related multiplication fact to solve a word problem. 1 95% 87% 8 8 SR MD 3.MD.B.03 Solve a one-step "how many more" problem using a given bar graph. 1 90% 68% 22 9 CR OA 3.OA.D.09 Find and justify the next number in a given pattern and explain a feature of the pattern. 3 71% 55% 16 10 SR MD 3.MD.C.07 Determine which rectangle has the same area as a rectangle with a given length and width. 1 75% 59% 16 11 SR NF 3.NF.A.03 Determine the fraction that is equivalent to a given fraction model. 1 66% 50% 16 12 SR GE 3.G.A.01 Determine which figure has the attributes of two given shapes. 1 47% 26% 21 13 SR OA 3.OA.A.03 Solve a word problem given the relationship between two given whole number amounts. 1 97% 76% 21 14 SR MD 3.MD.C.07 Determine the equation that can be used to find the area of a figure with a given length and width. 1 59% 47% 12 15 CR NF 3.NF.A.01 Determine the relationships between the number of equal parts and the number of wholes in a word problem. 3 76% 51% 25 16 SR NF 3.NF.A.03 From a given set of fractions, determine the fraction that is not equivalent to the other fractions. 1 73% 45% 28 17 SR OA 3.OA.A.01 Determine how a two-digit product can be expressed as equal groups of equal numbers of objects. 1 86% 68% 18 18 SR MD 3.MD.B.04 Use a ruler to determine the length of a given figure to the nearest fourth of an inch. 1 68% 54% 14 19 SA MD 3.MD.D.08 Determine the length of one rectangle given its width and the fact that it has the same perimeter as a second rectangle that is labeled with its length and width. 1 29% 28% 1 20 SR OA 3.OA.D.08 Determine the most reasonable solution to a word problem involving multiplication of two whole numbers. 1 80% 58% 22 21 SR MD 3.MD.A.01 Identify the time given on an analog clock using a digital clock. 1 80% 49% 31 22 SR OA 3.OA.D.09 Determine the terms of a numerical pattern and identify a feature that all the terms share. 1 86% 72% 14 23 SA OA 3.OA.B.05 Use the distributive property to complete a multiplication equation. 1 78% 58% 20 24 SR NT 3.NBT.A.01 Determine which expression with rounded numbers will give the best estimate when adding two whole numbers. 1 99% 73% 26 25 SR NF 3.NF.A.02 Determine which fraction is represented by the location of a given point on a number line. 1 68% 75% -7 26 SA MD 3.MD.B.04 Interpret a line plot with data in whole numbers and mixed numbers. 1 58% 46% 12 27 SR OA 3.OA.B.06 Determine the multiplication equation that could be used to solve a given division equation. 1 84% 70% 14 28 CR MD 3.MD.C.05 Find the area of a given rectangle made of equal-sized square units and justify whether the areas of two other rectangles are equal or not. 3 58% 56% 2 29 SR OA 3.OA.A.02 Determine which word problem can be solved using a given division expression. 1 58% 49% 9 30 SA OA 3.OA.D.08 Solve a two-step word problem using multiplication and addition. 1 84% 60% 24 31 SR NT 3.NBT.A.03 Solve a word problem by multiplying a one-digit whole number by a two-digit multiple of ten. 1 99% 79% 20 32 SR MD 3.MD.A.02 Estimate the mass of one amount of an item based on a given figure showing the mass for a different amount of the same item. 1 87% 74% 13 33 SR GE 3.G.A.01 Identify the true statement about the mathematical names of a set of given shapes. 1 52% 42% 10 34 SR NT 3.NBT.A.03 Solve a word problem by multiplying a single-digit whole number by a multiple of 10. 1 77% 59% 18 35 CR NT 3.NBT.A.02 Add and subtract two- and three-digit numbers and demonstrate the relationship between addition and subtraction with an equation. 3 71% 52% 19 36 SR NF 3.NF.A.02 Identify the fraction that is plotted on a given number line. 1 39% 60% -21 37 SA GE 3.G.A.01 Identify the number of a specific attribute a given figure has. 1 94% 69% 25 38 SR OA 3.OA.B.05 Identify which expression using the distributive property is equivalent to a given expression. 1 81% 69% 12 39 SA MD 3.MD.C.06 Find the area of a given figure by counting units or multiplying length and width. 1 84% 75% 9 40 SR GE 3.G.A.02 Determine which figure with part of its area shaded represents a given unit fraction. 1 76% 74% 2
Notes: State results in grades 3 and 6 are based on students taking that mode and are not representative of the full state population. | 1,584 | 5,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-25 | latest | en | 0.840962 |
https://www.numberempire.com/1044845 | 1,721,890,037,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00846.warc.gz | 773,462,245 | 7,634 | Home | Menu | Get Involved | Contact webmaster
0 / 12
# Number 1044845
one million forty four thousand eight hundred forty five
### Properties of the number 1044845
Factorization 5 * 101 * 2069 Divisors 1, 5, 101, 505, 2069, 10345, 208969, 1044845 Count of divisors 8 Sum of divisors 1266840 Previous integer 1044844 Next integer 1044846 Is prime? NO Previous prime 1044839 Next prime 1044847 1044845th prime 16228309 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Is a palindrome? NO Polygonal number (s < 11)? NO Binary 11111111000101101101 Octal 3770555 Duodecimal 4247a5 Hexadecimal ff16d Square 1091701074025 Square root 1022.1765992234 Natural logarithm 13.859379107021 Decimal logarithm 6.0190518687844 Sine 0.99976025200061 Cosine 0.021896084574001 Tangent 45.659316332186
Number 1044845 is pronounced one million forty four thousand eight hundred forty five. Number 1044845 is a composite number. Factors of 1044845 are 5 * 101 * 2069. Number 1044845 has 8 divisors: 1, 5, 101, 505, 2069, 10345, 208969, 1044845. Sum of the divisors is 1266840. Number 1044845 is not a Fibonacci number. It is not a Bell number. Number 1044845 is not a Catalan number. Number 1044845 is not a regular number (Hamming number). It is a not factorial of any number. Number 1044845 is a deficient number and therefore is not a perfect number. Binary numeral for number 1044845 is 11111111000101101101. Octal numeral is 3770555. Duodecimal value is 4247a5. Hexadecimal representation is ff16d. Square of the number 1044845 is 1091701074025. Square root of the number 1044845 is 1022.1765992234. Natural logarithm of 1044845 is 13.859379107021 Decimal logarithm of the number 1044845 is 6.0190518687844 Sine of 1044845 is 0.99976025200061. Cosine of the number 1044845 is 0.021896084574001. Tangent of the number 1044845 is 45.659316332186
### Number properties
0 / 12
Examples: 3628800, 9876543211, 12586269025 | 647 | 1,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.682998 |
https://hirecalculusexam.com/what-is-the-process-for-verifying-the-results-of-a-multivariable-calculus-exam | 1,701,268,805,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00810.warc.gz | 356,236,749 | 18,334 | # What is the process for verifying the results of a multivariable calculus exam?
What is the process for verifying the results of a multivariable calculus exam? By way of code: these formulas can be made as specified in the exam application. Several functions can be used for verifying these variables: – These three conditions stand for the two-qubit part of the classical system represented as a classical differential equation which corresponds to such an x-variable. – Also called a variable-symbol, a variable is a symbol if one of its variables is itself called a unit of time. – Similarly, these three conditions stand for: the difference between the time it takes and that will take at least. – This means that a function can be checked against two variables: x-variables, or x-time functions, rather than x-predictors. Within the framework of this application, we can use two well-behaved “experiments” to verify a multivariable calculus. For this purpose, it is advisable to use the following formula for the calculation of the variable-symmetry operator: For a given variable x, the equation from the 2-qubit Cylinder equation (Figure 12.23) can be expressed as where the dashed lines represent the derivative between x and a time variable, whereas the solid lines represent the derivative of x with respect to time. This can be regarded as a two-variable equation-for-linear combination of two variable equations or equations directly related to two variables, so that the derivative of the function with respect to time is automatically cancelled by the derivative (1). This integration is a necessary condition for the standard numerical comparison for the 3-bit Cylinder equation (Figure 12.24). Figure 12.23: Computer-provided graphical diagram of an extensive 3-bit Cylinder system measuring in 2-bit per second using a transceiver to a 15 bit word. (a) is the input data of a simulated Cylinder (b) used in an experiment. (c) is the Source data of using a transceiver (d) to measure the 3-bit Cylinder equation (e) and (f) is the return value of the equation (g) based on the calculated voltage value of the transceiver. By way of example, the following formulas can be used for verification of the multi-bit Cylinder equation: Figure 12.24: The 2-bit Cylinder equation is found using the method of measuring the transceiver inverter (a) and (b). The equation of the transceiver is shown in black, whereas the equation in gray is the solution (0) obtained from a simulation. The power density of the transceiver is 0.0160 T and the power density of the computer is 0.
## Statistics Class Help Online
0721 T per area. X-variables, two and three variables: the linear combinations of equations (b), (g), (e). When X-variables,What is the process for verifying the results of a multivariable calculus exam? In this section I will provide some explanations about the process that takes place for the statistical analyses of the test results. What is Calculus? A calculus exam is a test test in the manner basically an exercise in computer science. There are five stages along the life of the exam. Step 1: Exercise 1 Step 2: Exercise 2 Step 3: Exercise 3 Step 4: Exercise 4 The exercise is essentially a test of calculus for the test. In this stage one will begin with computing the values for the factors which appear in the result table, since these are the elements in the table that are the elements of both the sample result and the sample value for the test. In Step 1, the calculated values are shown on the right pie as the elements, in Step 2 is the calculated square of the values, and in Step 3 the square of the result of the test. Step 5a: Initial the process and steps Step 5b & steps 6: Calculate the square of the square root of the square root of the squares of the product of two numbers that are both in the square root of the product of two numbers, say: x1, x2,… xl Step 6a: In Step 5b you receive the results of the Calculus Test: x1 has x1 value Step 6b: In Step 6c you receive the coefficients of x2,… xl I give an example on the first step, in Step 5b is the square of the positive see this here number l and in Step 6a in Step 6c is the square of the positive real number x2. In Step 6b is the square of the positive square root of the square root of the square root of l. The result is this square of the simple root of two points X and Y: Now, for Step 6a, you obtain: What is the process for verifying the results of a multivariable calculus exam? In most of the cases your questions might seem quite easy and simple and I’m just telling my readers to watch my reaction! It’s the truth. First let me change my approach by letting you go from getting your results to getting the answers that you want to. The process repeats the process you were told to do for your results and this time you’re going to go from doing the manual for the exam to the automatic. This term is exactly the word that you’re going to call those automated “calculations”.
## Good Things To Do First Day Professor
First class – the model I am trying to illustrate looks pretty weak and it’s not even good looking like the result set is normal or even good looking. The formulas are really simple. Here’s 1-2-3: It’s been 2 days now that I found out that my brain had replaced the problem of normal algebra to solve some famous problems where one didn’t have the ability to write them in a simple form. This led me to wonder if I was doing something dangerous at this stage. In search mode, it looks like there’s a new data file I need to build and more details are sent over! If that’s the case, it calls the Model of that site to get Clicking Here way “puts” it in the “Calculus Code” box on the front page right now. The formula for the variables are as I want. This doesn’t works either so the formula used is usually the same as navigate to this website formula used with the real value. After a second, the names aren’t getting read right, and I start running through the problem. This is the right way to go at that point. Using my random example, it starts from scratch asking for value and runs through the system a bit more incrementally. So let’s have a look at what’s going on and see how I get the help with my test. If anyone else has noticed | 1,405 | 6,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-50 | latest | en | 0.910959 |
https://www.educationquizzes.com/ks1/science/materials-changing-things/ | 1,723,699,159,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00498.warc.gz | 587,556,681 | 10,839 | UK USIndia
Every Question Helps You Learn
If a pane of glass is broken it won't put itself back together again. The change is permanent!
# Materials - Changing Things
This quiz addresses the requirements of the National Curriculum KS1 Science for children aged 5 and 6 in years 1 and 2. Specifically this quiz is aimed at the section dealing with materials and how they change, for example when cooking.
Materials don’t always stay the same. If you stretch a rubber band, it will bounce back again. But if you chop up a carrot, you can’t put the carrot back together again. When you are baking a cake, the ingredients all look different. But when you mix them together and cook them, they change and become a new material. The cooked cake doesn’t look like any of the ingredients. Let’s find out a bit more about how materials change.
1.
Which one of these will go back to how it was?
A broken egg
A mashed potato
A stretched rubber band
A squashed banana
The rubber band can go back to how it was. The other changes are permanent
2.
Li cuts a piece of card in two. When she lets go of the two pieces of card, what happens?
The two pieces of card go back together again
The two pieces of card stay separate
The two pieces of card get much bigger
The two pieces of card disappear
Li can stick the pieces back together, but they don’t go back like a spring or a rubber band
3.
This candle has been burning for a long time. Some of the wax has melted and run down the sides of the candle. Then the wax has cooled and turned back into a _____.
Solid
Gas
Liquid
Fluid
Do you have candles on your birthday cake?
4.
Carla is making ice cubes in the freezer. She pours water into the trays. She puts the trays into the freezer. The water freezes into ice.
Later, Carla takes the ice cubes out of the freezer. She leaves them in a warm room. What happens to the ice cubes?
The ice cubes melt back to water
The ice cubes stay frozen
The ice cubes turn into snow
The ice cubes get much bigger
Water can freeze, and then it can melt again
5.
If you stretch a spring - not too much - the spring gets longer. If you then let go of the spring, what happens?
The spring gets even longer
The spring straightens out
The spring gets shorter again
The spring loses its shape
The spring can go back to its original shape and size
6.
Danny and Alex are making a cake. The ingredients are flour, butter and eggs. What must they do with the ingredients?
Put the flour in the oven and fry the eggs
Mix the flour, butter and eggs together
Only use the flour
Only use the eggs
Have you ever baked a cake?
7.
Lucy and her dad are making a pizza. What must they do before the pizza is ready to eat?
Put it in the freezer
Put it in the dishwasher
Bake it in the oven
Bake it in the fridge
Do you like pizza? What is your favourite pizza topping?
8.
The candle is made of wax. The candle is:
Freezing
Boiling
Burning
Getting bigger
As the candle burns, there is less and less candle wax
9.
Joe is blowing up a red balloon. The balloon is getting very big. Joe does not tie the balloon. He lets go of the balloon. It shoots off and all the air comes out. What happens to the size of the balloon?
The balloon goes back to its original size
The balloon gets even bigger
The balloon stays really big
The balloon disappears
Balloons are like springs and rubber bands. They can go back to their original size
10.
Karen is making a cup of tea. She pours milk into the tea and stirs it. The milky tea is a mixture. What is mixed together?
Milk and coffee
Tea and coffee
Tea and milk
Milk and sugar
Do you like drinking tea or coffee? | 851 | 3,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.921647 |
http://gmatclub.com/forum/as-litigation-grows-more-complex-the-need-that-experts-36079.html?fl=similar | 1,485,122,358,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281574.78/warc/CC-MAIN-20170116095121-00406-ip-10-171-10-70.ec2.internal.warc.gz | 119,457,205 | 55,145 | As litigation grows more complex, the need that experts : GMAT Sentence Correction (SC)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 22 Jan 2017, 13:59
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Events & Promotions
Events & Promotions in June
Open Detailed Calendar
As litigation grows more complex, the need that experts
Author Message
TAGS:
Hide Tags
Senior Manager
Joined: 11 May 2006
Posts: 258
Followers: 2
Kudos [?]: 23 [0], given: 0
As litigation grows more complex, the need that experts [#permalink]
Show Tags
01 Oct 2006, 22:35
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
HideShow timer Statistics
As litigation grows more complex, the need that experts explain technical issues becomes more apparent.
(A) that experts explain technical issues becomes
(B) for experts to explain technical issues became
(C) for experts to explain technical issues becomes
(D) that technical issues be explained by experts became
(E) that there be explanations of technical issues by experts has become
If you have any questions
New!
Manager
Joined: 11 Jan 2006
Posts: 230
Location: Arkansas, US
WE 1: 2.5 yrs in manufacturing
Followers: 1
Kudos [?]: 17 [0], given: 18
Show Tags
02 Oct 2006, 00:22
Idiom problem
"need for is idiomatic"
tht brings us down to B and C
then c the parallel point...grows..becomes..
n u can stamp out ur C
_________________
ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED
VP
Joined: 28 Mar 2006
Posts: 1381
Followers: 2
Kudos [?]: 31 [0], given: 0
Show Tags
02 Oct 2006, 16:42
Reverse the statement and C becomes more evident
The need for experts becomes ......As the .......
Senior Manager
Joined: 11 Jul 2006
Posts: 382
Location: TX
Followers: 1
Kudos [?]: 14 [0], given: 0
Show Tags
02 Oct 2006, 19:30
trivikram wrote:
Reverse the statement and C becomes more evident
The need for experts becomes ......As the .......
C
SVA need ... becomes
need for idiomatic
Thats a good approach trivikram , how do you know when to reverse it ? Or do you try it in all cases ?
Senior Manager
Joined: 11 May 2006
Posts: 258
Followers: 2
Kudos [?]: 23 [0], given: 0
Show Tags
02 Oct 2006, 20:15
OA is C
Current Student
Joined: 29 Jan 2005
Posts: 5238
Followers: 25
Kudos [?]: 378 [0], given: 0
Show Tags
03 Oct 2006, 06:28
Raghavender wrote:
Idiom problem
"need for is idiomatic"
tht brings us down to B and C
then c the parallel point...grows..becomes..
n u can stamp out ur C
Textbook explanation.
03 Oct 2006, 06:28
Similar topics Replies Last post
Similar
Topics:
21 As business grows more complex, students majoring in 16 25 Mar 2010, 11:37
As business grows more complex, students majoring in 8 15 Nov 2008, 15:02
As business grows more complex, students majoring in 4 12 Oct 2008, 20:34
1 As business grows more complex, students majoring in 6 12 Aug 2008, 09:24
As litigation grows more complex, the need that experts 4 10 Aug 2008, 20:29
Display posts from previous: Sort by | 1,006 | 3,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-04 | latest | en | 0.909291 |
https://gmatclub.com/forum/the-growing-popularity-of-computer-based-activities-was-widely-predict-80846-40.html | 1,531,814,987,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589618.52/warc/CC-MAIN-20180717070721-20180717090721-00044.warc.gz | 653,256,963 | 52,056 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 17 Jul 2018, 01:09
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# The growing popularity of computer-based activities was widely predict
Author Message
TAGS:
### Hide Tags
Senior Manager
Joined: 29 Jun 2017
Posts: 343
### Show Tags
20 Feb 2018, 00:32
keep many weapons to attack argument. evidence can not be representative. causal relation can be reversed or a third cause comes in.
the key to cr success is ability to attack the argument. find out the thing and conditions, in which argument fall apart.
GMAT Club Verbal Expert
Status: GMAT and GRE tutor
Joined: 13 Aug 2009
Posts: 1824
Location: United States
GMAT 1: 780 Q51 V46
GMAT 2: 800 Q51 V51
GRE 1: Q170 V170
Re: The growing popularity of computer-based activities was widely predict [#permalink]
### Show Tags
01 Mar 2018, 16:24
1
Prashant10692 wrote:
Could anyone correct my logic?
I chose D.
chesstitans
For the reason that the many computer owners in the United States have enough leisure time that spending significant amounts of time on the computer still leaves ample time for watching television.
So if they are not watching tv some other reason is there i.e. other than computer use because of which they are not interested in watching television.
ucb2k7 wrote:
Hi mikemcgarry,
Please explain as to why (D) or (E) is wrong.
Quote:
Which of the following most logically completes the argument?
The growing popularity of computer-based activities was widely predicted to result in a corresponding decline in television viewing. Recent studies have found that, in the United States, people who own computers watch, on average, significantly less television than people who do not own computers. In itself, however, this finding does very little to show that computer use tends to reduce television viewing time, since _______.
(A) many people who watch little or no television do not own a computer.
(B) even though most computer owners in the United States watch significantly less television than the national average, some computer owners watch far more television than the national average.
(C) computer owners in the United States predominately belong to a demographic group that have long been known to spend less time watching television than the population as a whole does.
(D) many computer owners in the United States have enough leisure time that spending significant amounts of time on the computer still leaves ample time for watching television.
(E) many people use their computers primarily for tasks such as correspondence that can be done more rapidly on the computer, and doing so leaves more leisure time for watching television.
We are told that people who own computers watch significantly less television than people who do not own computers. But does that mean that computer use tends to REDUCE television viewing time? According to the author, it does not.
In other words, the author's conclusion is that the findings from the study do "very little to show that computer use tends to reduce television viewing time." We need an answer choice that supports this conclusion:
Quote:
(D) many computer owners in the United States have enough leisure time that spending significant amounts of time on the computer still leaves ample time for watching television.
Choice (D) tells us that the computer users are not interested in watching television. But would those people be MORE interested in watching television if they did not own computers? Perhaps they would not, and that would support the author's point of view. But perhaps they WOULD watch more television if they did not own computers, and that would go against the author's argument.
Choice (D) may or may not support the author's conclusion, so it is not the best answer.
Quote:
(E) many people use their computers primarily for tasks such as correspondence that can be done more rapidly on the computer, and doing so leaves more leisure time for watching television.
Choice (E) suggests that computer use leaves more time for television watching. But we already know that people with computers watch less television that people without computers.
Sure, (E) suggests that computer users would have less time for television if their computers were taken away. But we have no way of knowing whether computer owners would actually watch more or less television if their computers were taken away.
If people with computers watch less television, why can't we conclude that computer use tends to reduce television watching time? Choice (E) does not answer this question, so it should be eliminated.
Quote:
(C) computer owners in the United States predominately belong to a demographic group that have long been known to spend less time watching television than the population as a whole does.
Choice (C), on the other hand, does answer this question. (C) tells us that computer owners belong to a group that spends less time watching television. So these people have always watched less television. It just so happens that these people also tend to own computers, but computer use did not cause them to watch less television. They've always watched less television, so the finding is not related to the growing popularity of computer-based activities.
_________________
GMAT Club Verbal Expert | GMAT/GRE tutor @ www.gmatninja.com (Now hiring!) | GMAT blog | Food blog | Notoriously bad at PMs
Beginners' guides to GMAT verbal
Reading Comprehension | Critical Reasoning | Sentence Correction
Series 1: Fundamentals of SC & CR | Series 2: Developing a Winning GMAT Mindset
SC & CR Questions of the Day (QOTDs), featuring expert explanations
All QOTDs | Subscribe via email | RSS
Hit the request verbal experts' reply button -- and please be specific about your question. Feel free to tag @GMATNinja and @GMATNinjaTwo in your post. Priority is always given to official GMAT questions.
Sentence Correction articles & resources
How to go from great (760) to incredible (780) on GMAT SC | That "-ing" Word Probably Isn't a Verb | That "-ed" Word Might Not Be a Verb, Either | No-BS Guide to GMAT Idioms | "Being" is not the enemy | WTF is "that" doing in my sentence?
Reading Comprehension, Critical Reasoning, and other articles & resources
All GMAT Ninja articles on GMAT Club | Using LSAT for GMAT CR & RC |7 reasons why your actual GMAT scores don't match your practice test scores | How to get 4 additional "fake" GMAT Prep tests for \$29.99 | Time management on verbal
Intern
Joined: 04 Dec 2017
Posts: 3
Re: The growing popularity of computer-based activities was widely predict [#permalink]
### Show Tags
22 Apr 2018, 09:11
GMATNinja, Could you please explain why A is irrelevant?
The conclusion states that computer use tends to reduce television viewing time. So when computer use occurs(cause), then people watch less television(effect).
We should weaken this conclusion.
I think A weakens the conclusion by showing that when cause does not occur, effect occurs.
What did I miss?
Manager
Joined: 01 Feb 2017
Posts: 145
Re: The growing popularity of computer-based activities was widely predict [#permalink]
### Show Tags
20 May 2018, 01:00
Debate: Does Computer usage reduces TV viewing?
Study: Computer owners view less TV in comparison to others.
Author's Conclusion: This study regarding "Computer ownership" cannot be extrapolated to prove that actual "Computer usage" leads to decline in TV viewing.
Simply stated: Computer Usage does not lead to reduction in TV viewing.
As lead-in word is "since", Correct Ans: should provide an alternate reason to strengthen the Conclusion.
Ans C does exactly that by providing 'demography' (and not 'computer usage') as an alternate reason towards reduction in TV viewing.
Now, Ans D and E are very confusing Ans Choices and difficult to eliminate without careful logic.
Either choice focuses on the "time" factor and suggests that "Computer Usage" results in "enough spare time" for TV viewing.
Does this strengthen the Conclusion that "Computer Usage does not lead to reduction in TV viewing. "
Ans is No.
Eg. Even though Computer Owners have more time to watch TV, these guys get so contented working on a computer that they don't feel like watching a TV.:
So Computer usage does in fact leads to low TV viewing.
This becomes on opposite statement that weakens the Conclusion.
Posted from my mobile device
Senior Manager
Joined: 17 Sep 2016
Posts: 285
Re: The growing popularity of computer-based activities was widely predict [#permalink]
### Show Tags
14 Jun 2018, 06:02
UkrHurricane wrote:
GMATNinja, Could you please explain why A is irrelevant?
The conclusion states that computer use tends to reduce television viewing time. So when computer use occurs(cause), then people watch less television(effect).
We should weaken this conclusion.
I think A weakens the conclusion by showing that when cause does not occur, effect occurs.
What did I miss?
Hi UkrHurricane, I want to share my thoughts.
the conclusion is that the reduce television is because of the computer use,
A says those who watch little or no television do not own a computer, it does not show "reduce" in conclusion.
So i think it is not correct.
If you have any more ideas, please share, i am glad to discuss with you.
Have a nice day
>_~
Re: The growing popularity of computer-based activities was widely predict [#permalink] 14 Jun 2018, 06:02
Go to page Previous 1 2 3 [ 45 posts ]
Display posts from previous: Sort by
# Events & Promotions
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,222 | 10,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-30 | longest | en | 0.933045 |
https://www.oreilly.com/library/view/sas-94-sql/9781629608167/chapter-63.html | 1,563,723,627,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527048.80/warc/CC-MAIN-20190721144008-20190721170008-00273.warc.gz | 800,604,106 | 56,615 | With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
No credit card required
when month(InvoiceDate)=3 then
InvoiceAmount end as Mar
Creating a Customized Sort Order
Problem
You want to sort data in a logical, but not alphabetical, sequence.
Background Information
There is one input table, called Chores, that contains the following data:
data chores;
input Project \$ Hours Season \$;
datalines;
weeding 48 summer
pruning 12 winter
mowing 36 summer
mulching 17 fall
raking 24 fall
raking 16 spring
planting 8 spring
planting 8 fall
sweeping 3 winter
edging 16 summer
seeding 6 spring
tilling 12 spring
aerating 6 spring
feeding 7 summer
rolling 4 winter
;
proc sql;
title 'Garden Chores';
select * from chores;
quit;
Creating a Customized Sort Order 197
Output 6.17 Sample Input Data for a Customized Sort
You want to reorder this chore list so that all the chores are grouped by season, starting
with spring and progressing through the year. Simply ordering by Season makes the list
appear in alphabetical sequence: fall, spring, summer, winter.
Solution
Use the following PROC SQL code to create a new column, Sorter, that will have values
of 1 through 4 for the seasons spring through winter. Use the new column to order the
query, but do not select it to appear:
proc sql;
title 'Garden Chores by Season in Logical Order';
select Project, Hours, Season
from (select Project, Hours, Season,
case
when Season = 'spring' then 1
when Season = 'summer' then 2
when Season = 'fall' then 3
when Season = 'winter' then 4
else .
end as Sorter
from chores)
order by Sorter;
198 Chapter 6 Practical Problem-Solving with PROC SQL
Output 6.18 PROC SQL Output for a Customized Sort Sequence
How It Works
This solution uses an in-line view to create a temporary column that can be used as an
ORDER BY column. The in-line view is a query that performs the following:
selects the Project, Hours, and Season columns
uses a CASE expression to remap the seasons to the new column Sorter: spring to 1,
summer to 2, fall to 3, and winter to 4
(select project, hours, season,
case
when season = 'spring' then 1
when season = 'summer' then 2
when season = 'fall' then 3
when season = 'winter' then 4
else .
end as sorter
from chores)
The first, or outer, SELECT statement in the query performs the following:
selects the Project, Hours, and Season columns
Creating a Customized Sort Order 199
With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
No credit card required | 671 | 2,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-30 | latest | en | 0.847623 |
https://datascience.stackexchange.com/questions/76402/does-a-cnn-think-things-inside-the-filter-are-collocated-aka-dependent-on-each-o/76405 | 1,618,559,244,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00464.warc.gz | 304,778,431 | 38,280 | # Does a CNN think things inside the filter are collocated aka dependent on each other?
I am running a 1D CNN on tabular data. The rows are data that I have are not sequential, that is to say they are not part of a time series or ordered string, which is why I am not using an LSTM.
So when the filter looks at both row n and row n+1, does it learn from them as if they are spatially collocated or dependent upon each other? Is there a way to prevent this from happening?
Alternatively, is there a better way to look at 2 dimensional data (rows of features) as independent rows?
Yes, the filter will learn as if they are spatially co-located.
The main purpose of convolutions is to detect local features, where the notion of locality comes from the positions over which the filter is applied.
Some neural network building blocks that you could use are:
• Position-wise dense layers: this applies a linear transformation to each of the elements. The result would be an element for each of the elements in the input. The dimensionality depends on the matrix used for the multiplication. This can be achieved also by applying a 1D convolution with a filter of length 1.
• Max/avg. pooling over time: taking the maximum/average value over all elements in the input, separately for each input channel. This gives you a single element, as it collapses all the elements in one. A variation is to take the $$k$$ maximum elements.
Of course, you may want to exploit the relations between the elements in your input, but not in their original meaningless ordering: you could sort your inputs (from lowest to highest or vice versa). Then you could apply convolutional layers.
• Thank you "Position-wise feed forward accepts a 3 -dimensional input with shape (batch size, sequence length, feature size). The position-wise FFN consists of two dense layers that applies to the last dimension. Since the same two dense layers are used for each position item in the sequence, we referred to it as position-wise. Indeed, it is equivalent to applying two 1×1 convolution layers." d2l.ai/chapter_attention-mechanisms/transformer.html – HashRocketSyntax Jun 22 '20 at 1:17 | 471 | 2,162 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-17 | latest | en | 0.946586 |
http://cstwiki.wtb.tue.nl/index.php?title=Embedded_Motion_Control_2015_Group_3/Mapping&diff=20770&oldid=19768 | 1,643,109,773,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304810.95/warc/CC-MAIN-20220125100035-20220125130035-00186.warc.gz | 18,472,843 | 14,558 | # Embedded Motion Control 2015 Group 3/Mapping
(Difference between revisions)
Revision as of 20:18, 22 June 2015 (view source)S117225 (Talk | contribs) (→C++ implementation)← Older edit Current revision as of 14:31, 26 June 2015 (view source)S118835 (Talk | contribs) (→Schedule) (14 intermediate revisions not shown) Line 1: Line 1: - = Mapping = + This page is part of the [[Embedded_Motion_Control_2015_Group_3|EMC03 CST-wiki]]. - This page is part of the [http://cstwiki.wtb.tue.nl/index.php?title=Embedded_Motion_Control_2015_Group_3 EMC03 CST-wiki]. + = Mapping block = = Mapping block = Line 6: Line 5: For solving the maze, a variant of the [http://blog.jamisbuck.org/2014/05/12/tremauxs-algorithm.html Tremaux algorithm] is used. The Tremaux algorithm is an implementation of DFS (Depth First Search), which proves to be an efficient way of solving a maze with minimum backtracking. For solving the maze, a variant of the [http://blog.jamisbuck.org/2014/05/12/tremauxs-algorithm.html Tremaux algorithm] is used. The Tremaux algorithm is an implementation of DFS (Depth First Search), which proves to be an efficient way of solving a maze with minimum backtracking. + + Because the Localization block did not work in the end, and as such loops could not be resolved, a Random Walk strategy was used, so that at least the maze would (probably) be solved, and the capabilities of the other blocks could be shown. = World model structure = = World model structure = + [[File:Map.png|400px|right|thumb|Example of a world model. Note that all unknown edges or impossible-to-reach edges are connected to noNode to make sure that all edges exist to ensure algorithmic integrity]] The maze will consist of nodes and edges; i.e., an undirected graph. A node is either a dead end, or any place in the maze where the robot can go in more than one direction. An edge is the connection between one node and another. An edge may also lead to the same node. In the latter case, this edge is a loop. The algorithm is called by the general decision maker whenever the robot encounters a node (junction or a dead end). The input of the algorithm is the possible routes the robot can go (left, straight ahead, right, turn around) and the output is the direction that is advised, based on the Tremaux algorithm. The maze will consist of nodes and edges; i.e., an undirected graph. A node is either a dead end, or any place in the maze where the robot can go in more than one direction. An edge is the connection between one node and another. An edge may also lead to the same node. In the latter case, this edge is a loop. The algorithm is called by the general decision maker whenever the robot encounters a node (junction or a dead end). The input of the algorithm is the possible routes the robot can go (left, straight ahead, right, turn around) and the output is the direction that is advised, based on the Tremaux algorithm. Line 13: Line 15: For each node, the following information is stored: For each node, the following information is stored: - * Position. The position is used to identify and close loops within the maze, by matching a new node with a previous node. The position is defined in global coordinates. To obtain the global coordinates, we will use the [http://cstwiki.wtb.tue.nl/index.php?title=Embedded_Motion_Control_2015_Group_3/Localisation localisation class]. + * Position. The position is used to identify and close loops within the maze, by matching a new node with a previous node. The position is defined in global coordinates. To obtain the global coordinates, we will use the [[Embedded_Motion_Control_2015_Group_3/Localisation|localisation class]]. * Adjacent corridors. Since the maze is axis-aligned, there can be anything between one (dead end) and four (cross-intersection) corridors/edges leading to a node. Because of this, the corridors are stored in an array with each element corresponding to a (global) direction. * Adjacent corridors. Since the maze is axis-aligned, there can be anything between one (dead end) and four (cross-intersection) corridors/edges leading to a node. Because of this, the corridors are stored in an array with each element corresponding to a (global) direction. For each corridor/edge, the following information is stored: For each corridor/edge, the following information is stored: * Number of times Pico has traversed a corridor. This is important for Tremaux algorithm, which will be explained later. * Number of times Pico has traversed a corridor. This is important for Tremaux algorithm, which will be explained later. - * Travel time for a corridor. This can be used to give priorities in case multiple options are present. This may later be used to define weighting functions to choose a next corridor. + * Travel time for a corridor. This can be used to give priorities in case multiple options are present. - + - + In the implementation, the graph is stored in a BGL (Boost Graphing Library) Graph object. This means that most of the overhead of maintaining an undirected graph is done by an existing library. The library used is also extremely extendable by means of Bundled Properties, which facilitate an arbitrary number of properties for nodes and edges. The main disadvantage is the syntactic overhead generated especially because BGL is so extendable, although the overhead is mostly contained into making the basic graph objects, which only has to be done once. In the implementation, the graph is stored in a BGL (Boost Graphing Library) Graph object. This means that most of the overhead of maintaining an undirected graph is done by an existing library. The library used is also extremely extendable by means of Bundled Properties, which facilitate an arbitrary number of properties for nodes and edges. The main disadvantage is the syntactic overhead generated especially because BGL is so extendable, although the overhead is mostly contained into making the basic graph objects, which only has to be done once. = Schedule = = Schedule = - [[File:Emc03 wayfindingCP1.png|400px|center|thumb|Map&solve algorithm (update?)]] + [[File:Emc03 wayfindingCP1.png|400px|right|thumb|Map&solve algorithm (update?)]] The schedule looks like this: The schedule looks like this: * Updating the map: * Updating the map: - ** Robot tries to find where he is located in global coordinates. Now it can decide if it is on a new node or on an old node. + ** Robot tries to find where he is, located in global coordinates. Now it can decide if it is on a new node or on an old node. //Where is the node located? //Where is the node located? Line 70: Line 70: } } - ** The robot figures out from which node it came from. Now it can define what edge it has been traversing. It marks the edge as 'visited once more'. + ** The robot figures out from which node it came. Now it can define what edge it has been traversing. It marks the edge as 'visited once more'. ** All sorts of other properties may be associated with the edge. Energy consumption, traveling time, shape of the edge... This is not necessary for the algorithm, but it may help formulating more advanced weighting functions for optimizations. ** All sorts of other properties may be associated with the edge. Energy consumption, traveling time, shape of the edge... This is not necessary for the algorithm, but it may help formulating more advanced weighting functions for optimizations. ** The robot will also have to realize if the current node is connected to a dead end. In that case, it will request the possible door to open. ** The robot will also have to realize if the current node is connected to a dead end. In that case, it will request the possible door to open. Line 80: Line 80: * Choosing a new direction: * Choosing a new direction: - ** Check if the door opened for me. In that case: Go straight ahead and mark the edge that lead up to the door as ''Visited 2 times''. If not, choose the edge where you came from + ** Check if the door opened for me. In that case: Go straight ahead and mark the edge that lead up to the door as ''Visited 2 times'' (not currently implemented due to previous concerns in the door detection robustness.). If not, choose the edge where you came from - ** Are there any unvisited edges connected to the current node? In that case, follow the edge straight in front of you if that one is unvisited. Otherwise, follow the unvisited edge that is on your left. Otherwise, follow the unvisited edge on your right. + ** Are there any unvisited edges connected to the current node? In that case, follow any unvisited edge. - **Are there any edges visited once? Do not go there if there are any unvisited edges. If there are only edges that are visited once, follow the one straight ahead. Otherwise left, otherwise right. + **Are there any edges visited once? Do not go there if there are any unvisited edges. If there are only edges that are visited once, any edge that is visited only once. **Are there any edges visited twice? Do not go there. According to the Tremaux algorithm, there must be an edge left to explore (visited once or not yet), or you are back at the starting point and the maze has no solution. **Are there any edges visited twice? Do not go there. According to the Tremaux algorithm, there must be an edge left to explore (visited once or not yet), or you are back at the starting point and the maze has no solution. - These questions are collected in a cost function. The new direction is chosen with: + **Above points can be summarized as 'pick the least visited edge'. If formulated this way, the algorithm will also allow for 'accidents', e.g., an edge visited more than two times, which should definitely be avoided, and will never stop exploring the maze even if it thinks it is back at the starting point, which improves robustness. + **In case there is more than one edge an option according to Tremaux, a cost function will determine the best option to follow. This cost function penalizes turning over going straight ahead, severely penalizes U-turns, and incorporates the travel time of known edges (shorter being better, since that will explore as much options as possible in a given time). The cost function could be extended by analyzing the entire maze as a whole, to determine which direction will explore as much of the maze as possible in the shortest time. Although currently not implemented, this again shows a huge advantage of using an existing graphing library, since graph traversal is built-in in the library. //Done with the mapping process. Now, let's pick a node to go to! //Done with the mapping process. Now, let's pick a node to go to! - int minTimesVisited = HUGE_VAL; //future: shortest distance, turn time etc + int minTimesVisited = HUGE_VAL; double minCost = HUGE_VAL; double minCost = HUGE_VAL; int decidedDirection; //GLOBAL COORDINATES int decidedDirection; //GLOBAL COORDINATES - cout << "Current direction: " << currentDirection; for(int direction = 0; direction<4; ++direction){ for(int direction = 0; direction<4; ++direction){ //check if this direction is actually possible //check if this direction is actually possible Line 106: Line 106: * Translation from chosen edge to turn command: * Translation from chosen edge to turn command: - ** The nodes are stored in a global coordinate system. The edges have a global direction pointing from the node to the direction of the edge. The robot must receive a command that will guide it through the maze in local coordinates. Globally, we can assume that there are only four possible directions because of the axis-aligned maze: North, west, south and east. These directions are represented by the integers 1, 2, 3 and 4 respectively in our code. + ** The nodes are stored in a global coordinate system. The edges have a global direction pointing from the node to the direction of the edge. The robot must receive a command that will guide it through the maze in local coordinates. Globally, we can assume that there are only four possible directions because of the axis-aligned maze: North, west, south and east. These directions are represented by the integers 0, 1, 2 and 3 respectively in our code. * The actual command is formulated * The actual command is formulated Line 113: Line 113: * A set-up is made for the next node * A set-up is made for the next node - ** e.g., the current node is saved as a 'nodeWhereICameFrom', so the next time the algorithm is called, it knows where it came from and start figuring out the next step. + ** e.g., the current node is saved as 'prevNode', so the next time the algorithm is called, it knows where it came from and start figuring out the next step. = Tremaux Algorithm = = Tremaux Algorithm = - Tremaux' algorithm is an implementation of Depth-First Search. It is best visualized as walking through the maze with a big bucket of paint and a paint brush. Continuously, the maze solver will drag the brush behind him, creating a line on the floor. When an intersection is encountered, the solver will see if any corridors do not have a line of paint on the floor. He will take one of those corridors, since he hasn't explored that bit of maze yet. At some point, the solver will have no unpainted corridors to go to, so he will have to backtrack. At that point, a corridor will have **two** lines on the floor, which indicates that not only has the solver been there, but there was also nothing left for him to explore there. As such, corridors with two lines on the floor should be avoided, and corridors with no lines on the floor should be preferred. + Tremaux' algorithm is an implementation of Depth-First Search. It is best visualized as walking through the maze with a big bucket of paint and a paint brush. Continuously, the maze solver will drag the brush behind him, creating a line on the floor. When an intersection is encountered, the solver will see if any corridors do not have a line of paint on the floor. He will take one of those corridors, since he hasn't explored that bit of maze yet. At some point, the solver will have no unpainted corridors to go to, so he will have to backtrack. At that point, a corridor will have '''two''' lines on the floor, which indicates that not only has the solver been there, but there was also nothing left for him to explore there. As such, corridors with two lines on the floor should be avoided, and corridors with no lines on the floor should be preferred. A simple video of how this should be visualized can be found [https://www.youtube.com/watch?v=gVSEJdSQZVQ here]. A simple video of how this should be visualized can be found [https://www.youtube.com/watch?v=gVSEJdSQZVQ here]. - In our implementation, we chose not to equip Pico with a bucket of paint, but instead use the world model we created. Pico will update an integer for each corridor which indicates the number of times he visited a certain corridor. Then, Pico simply prefers the corridor with the least amount of visits. In theory, this should be always 0, 1 or 2 times (with 0 times preferred over 1 time, and 2 times should be generally avoided). However, we decided to also allow for integers greater than 2, in case the world model has miscounted something - in this case, 2 times is preferred over 3 times, because the latter means that a certain area is **definitely** explored more than necessary. + In our implementation, we chose not to equip Pico with a bucket of paint, but instead use the world model we created. Pico will update an integer for each corridor which indicates the number of times he visited a certain corridor. Then, Pico simply prefers the corridor with the least amount of visits. In theory, this should be always 0, 1 or 2 times (with 0 times preferred over 1 time, and 2 times should be generally avoided). However, we decided to also allow for integers greater than 2, in case the world model has miscounted something - in this case, 2 times is preferred over 3 times, because the latter means that a certain area is '''definitely''' explored more than necessary. Furthermore, we extended Tremaux with a priority model. In principle, Tremaux does not describe to do when there are multiple possibilities - it does not matter for solving a maze if time is not an issue. We would however like to quickly explore as many nodes as possible, so instead of just picking a random direction, we pick the direction which minimizes the travel time to a next node. For example, turning around is generally not preferred, and the shortest corridor to the next node is chosen if at all possible. Furthermore, we extended Tremaux with a priority model. In principle, Tremaux does not describe to do when there are multiple possibilities - it does not matter for solving a maze if time is not an issue. We would however like to quickly explore as many nodes as possible, so instead of just picking a random direction, we pick the direction which minimizes the travel time to a next node. For example, turning around is generally not preferred, and the shortest corridor to the next node is chosen if at all possible. This can in theory be extended with a more elaborate searches. For example, to try not to go to a node that will lead to only twice-visited edges, which should be perfectly possible by using some built-in algorithms from Boost. However, this is not implemented yet. This can in theory be extended with a more elaborate searches. For example, to try not to go to a node that will lead to only twice-visited edges, which should be perfectly possible by using some built-in algorithms from Boost. However, this is not implemented yet. + + = Random Walk = + To detect and close loops (which were likely to be present) in the maze, the Mapping block needed information from Localization, which was not functional in time. Wall-following is not a viable option if the robot starts in a loop (which was in fact the case in the A-Maze-ing challenge as it turned out), so an alternative strategy must be employed. The strategy chosen is '''random walk'''. This strategy will solve any (static) maze with probability $1$ as $t\to\infty$. Of course, we do not have infinite time, but trying to optimize this strategy in any way (e.g., try and avoid going in one direction all the time) ''could'' mean that it will never solve the maze, since aforementioned property only hold for a truly random walk. As a bonus, there is also a certain probability that it will optimally solve a maze regardless of its properties, which was in fact the case for our maze challenge run. + + Since the software for the mapping was almost completely decoupled from the rest, the random walk could be implemented easily. It will still get an intersection type from the Decision Maker, and will still return a decision. It was decided not to have U-turns in the random walk (except for dead ends, of course), since it's trivially shown that a U-turn can always be achieved with a combination of other decisions, and U-turns would in general severely impede progress through the maze. + + Throughout the course, it was always a balance act between decoupling and ease of implementation. In this particular case, ease of implementation was favoured, since the random walk block merely tries to solve the maze and highlight (successfully so) the capabilities of all other blocks. Furthermore, this block was focused on robustness; as such, even though in the below example, case 2: could be used, default was chosen to eliminate the risk of accidental bugs. + + + srand (time(NULL)); //Initialize random number generator + int direction =0; //Initialize direction + switch(type){ + case FourWay: + direction = rand()%3; + switch(direction){ + case 0: + return Left; + break; + case 1: + return Forward; + break; + default: + return Right; + break; + } + break; + case LeftJunction: //etc
# Mapping block
The mapping block contains a very high-level model of the world. The mapping has been created in such a way that only essential information is stored, in order to create a very flexible and modular world model.
For solving the maze, a variant of the Tremaux algorithm is used. The Tremaux algorithm is an implementation of DFS (Depth First Search), which proves to be an efficient way of solving a maze with minimum backtracking.
Because the Localization block did not work in the end, and as such loops could not be resolved, a Random Walk strategy was used, so that at least the maze would (probably) be solved, and the capabilities of the other blocks could be shown.
# World model structure
Example of a world model. Note that all unknown edges or impossible-to-reach edges are connected to noNode to make sure that all edges exist to ensure algorithmic integrity
The maze will consist of nodes and edges; i.e., an undirected graph. A node is either a dead end, or any place in the maze where the robot can go in more than one direction. An edge is the connection between one node and another. An edge may also lead to the same node. In the latter case, this edge is a loop. The algorithm is called by the general decision maker whenever the robot encounters a node (junction or a dead end). The input of the algorithm is the possible routes the robot can go (left, straight ahead, right, turn around) and the output is the direction that is advised, based on the Tremaux algorithm.
Since the maze is axis-aligned, a simplified coordinate system can be used, which only has four principal directions (in simple terms, 'Up', 'Down', 'Left', and 'Right'). Although the node/edge structure can in principle work for a non-axis-aligned maze, the current implementation has some methods specific to an axis-aligned maze, to reduce the complexity of the implementation. This is expressed in two ways: it is assumed that Pico can only drive in any of the principal directions, and that the junctions can only be of specific formats, namely (from Pico's perspective): T-junction ╦, left junction ╣, right junction ╠ four-way intersection ╬ and dead end ╥ .
For each node, the following information is stored:
• Position. The position is used to identify and close loops within the maze, by matching a new node with a previous node. The position is defined in global coordinates. To obtain the global coordinates, we will use the localisation class.
• Adjacent corridors. Since the maze is axis-aligned, there can be anything between one (dead end) and four (cross-intersection) corridors/edges leading to a node. Because of this, the corridors are stored in an array with each element corresponding to a (global) direction.
For each corridor/edge, the following information is stored:
• Number of times Pico has traversed a corridor. This is important for Tremaux algorithm, which will be explained later.
• Travel time for a corridor. This can be used to give priorities in case multiple options are present.
In the implementation, the graph is stored in a BGL (Boost Graphing Library) Graph object. This means that most of the overhead of maintaining an undirected graph is done by an existing library. The library used is also extremely extendable by means of Bundled Properties, which facilitate an arbitrary number of properties for nodes and edges. The main disadvantage is the syntactic overhead generated especially because BGL is so extendable, although the overhead is mostly contained into making the basic graph objects, which only has to be done once.
# Schedule
Map&solve algorithm (update?)
The schedule looks like this:
• Updating the map:
• Robot tries to find where he is, located in global coordinates. Now it can decide if it is on a new node or on an old node.
//Where is the node located?
cv::Point2d nodePosition = getNodePos();
//Check if we're going to an already known node (in which case we won't do any matching)
if(nextNode == noNode){
// We do not know anything abbout this node yet, so let's see if we can match it with another one.
//Iterate over all nodes, and pick the closest.
double minDistance = HUGE_VAL; //By definition, more than the largest double you'll ever get.
Node match = noNode; //Initialize to noNode, so we can check if there was no match.
NodeIt node, lastNode; //Set up for the for loop iterator magic.
for(tie(node,lastNode) = vertices(maze); node!=lastNode; ++node){ //Iterator magic.
//Skip checking the distance if the node we're checking is an undiscovered node.
if(*node!=this->noNode)
{
double distance = cv::norm(nodePosition-maze[*node].position); //Calculate distance (2-norm)
if(distance<SAME_NODE_TOLERANCE && distance < minDistance){
match = *node; //Set the match to the close node we found.
minDistance = distance; //Set the new minimum for the closest node.
}
}
}
if(match==noNode){
// Previous loop did not result in a match, so: new node encountered! Let's create it.
this->nextNode = match; //Now we know where we were going to (was noNode because we didn't know before)
//Initialize each corridor to default (we don't know yet where they lead)
for(int i=0; i<4; ++i){ //Iterate through all four possible directions for a node and set them to default.
maze[nextNode].corridor[i]=defaultCorr;
}
// But we do know where we came from, so set the edge from whence we came.
this->currentCorridor = add_edge(this->prevNode,this->nextNode,maze).first; //Make a new edge
maze[prevNode].corridor[this->prevNodeExitedAt] = this->currentCorridor; //Update edge of previous node
maze[nextNode].corridor[flipD(this->currentDirection)] = this->currentCorridor; //Idem for next node. (D+2)%4 = flip direction.
} else {
// Revisted old node.
this->nextNode = match;
maze[nextNode].corridor[flipD(this->currentDirection)] = this->currentCorridor;
}
• The robot figures out from which node it came. Now it can define what edge it has been traversing. It marks the edge as 'visited once more'.
• All sorts of other properties may be associated with the edge. Energy consumption, traveling time, shape of the edge... This is not necessary for the algorithm, but it may help formulating more advanced weighting functions for optimizations.
• The robot will also have to realize if the current node is connected to a dead end. In that case, it will request the possible door to open.
//Above process updated nextNode to be a proper node in our system. Let's update the corridor.
this->maze[this->currentCorridor].timesVisited += 1;
this->maze[this->currentCorridor].travelTime = scanvars.timeStamp - this->lastTimeStamp;
this->lastTimeStamp = scanvars.timeStamp;
• Choosing a new direction:
• Check if the door opened for me. In that case: Go straight ahead and mark the edge that lead up to the door as Visited 2 times (not currently implemented due to previous concerns in the door detection robustness.). If not, choose the edge where you came from
• Are there any unvisited edges connected to the current node? In that case, follow any unvisited edge.
• Are there any edges visited once? Do not go there if there are any unvisited edges. If there are only edges that are visited once, any edge that is visited only once.
• Are there any edges visited twice? Do not go there. According to the Tremaux algorithm, there must be an edge left to explore (visited once or not yet), or you are back at the starting point and the maze has no solution.
• Above points can be summarized as 'pick the least visited edge'. If formulated this way, the algorithm will also allow for 'accidents', e.g., an edge visited more than two times, which should definitely be avoided, and will never stop exploring the maze even if it thinks it is back at the starting point, which improves robustness.
• In case there is more than one edge an option according to Tremaux, a cost function will determine the best option to follow. This cost function penalizes turning over going straight ahead, severely penalizes U-turns, and incorporates the travel time of known edges (shorter being better, since that will explore as much options as possible in a given time). The cost function could be extended by analyzing the entire maze as a whole, to determine which direction will explore as much of the maze as possible in the shortest time. Although currently not implemented, this again shows a huge advantage of using an existing graphing library, since graph traversal is built-in in the library.
//Done with the mapping process. Now, let's pick a node to go to!
int minTimesVisited = HUGE_VAL;
double minCost = HUGE_VAL;
int decidedDirection; //GLOBAL COORDINATES
for(int direction = 0; direction<4; ++direction){
//check if this direction is actually possible
if(isPossibleDirection(type,direction)){
int thisCorridorTimesVisited = maze[maze[nextNode].corridor[direction]].timesVisited;//Boost can only access properties of edges and vertices as maze[edge] or maze[vertex]
if(thisCorridorTimesVisited <= minTimesVisited){
if(thisCorridorTimesVisited<minTimesVisited) { minCost = HUGE_VAL;}
minTimesVisited = thisCorridorTimesVisited;
double cost=getCost(direction);
if(cost<minCost){
decidedDirection = direction;
}
}
}
• Translation from chosen edge to turn command:
• The nodes are stored in a global coordinate system. The edges have a global direction pointing from the node to the direction of the edge. The robot must receive a command that will guide it through the maze in local coordinates. Globally, we can assume that there are only four possible directions because of the axis-aligned maze: North, west, south and east. These directions are represented by the integers 0, 1, 2 and 3 respectively in our code.
• The actual command is formulated
int decision = (decidedDirection - this->currentDirection + 4)%4; //From global to local coordinates
return static_cast<WhereToNow>(decision); //Cast from int to enum.
• A set-up is made for the next node
• e.g., the current node is saved as 'prevNode', so the next time the algorithm is called, it knows where it came from and start figuring out the next step.
# Tremaux Algorithm
Tremaux' algorithm is an implementation of Depth-First Search. It is best visualized as walking through the maze with a big bucket of paint and a paint brush. Continuously, the maze solver will drag the brush behind him, creating a line on the floor. When an intersection is encountered, the solver will see if any corridors do not have a line of paint on the floor. He will take one of those corridors, since he hasn't explored that bit of maze yet. At some point, the solver will have no unpainted corridors to go to, so he will have to backtrack. At that point, a corridor will have two lines on the floor, which indicates that not only has the solver been there, but there was also nothing left for him to explore there. As such, corridors with two lines on the floor should be avoided, and corridors with no lines on the floor should be preferred.
A simple video of how this should be visualized can be found here.
In our implementation, we chose not to equip Pico with a bucket of paint, but instead use the world model we created. Pico will update an integer for each corridor which indicates the number of times he visited a certain corridor. Then, Pico simply prefers the corridor with the least amount of visits. In theory, this should be always 0, 1 or 2 times (with 0 times preferred over 1 time, and 2 times should be generally avoided). However, we decided to also allow for integers greater than 2, in case the world model has miscounted something - in this case, 2 times is preferred over 3 times, because the latter means that a certain area is definitely explored more than necessary.
Furthermore, we extended Tremaux with a priority model. In principle, Tremaux does not describe to do when there are multiple possibilities - it does not matter for solving a maze if time is not an issue. We would however like to quickly explore as many nodes as possible, so instead of just picking a random direction, we pick the direction which minimizes the travel time to a next node. For example, turning around is generally not preferred, and the shortest corridor to the next node is chosen if at all possible.
This can in theory be extended with a more elaborate searches. For example, to try not to go to a node that will lead to only twice-visited edges, which should be perfectly possible by using some built-in algorithms from Boost. However, this is not implemented yet.
# Random Walk
To detect and close loops (which were likely to be present) in the maze, the Mapping block needed information from Localization, which was not functional in time. Wall-following is not a viable option if the robot starts in a loop (which was in fact the case in the A-Maze-ing challenge as it turned out), so an alternative strategy must be employed. The strategy chosen is random walk. This strategy will solve any (static) maze with probability 1 as $t\to\infty$. Of course, we do not have infinite time, but trying to optimize this strategy in any way (e.g., try and avoid going in one direction all the time) could mean that it will never solve the maze, since aforementioned property only hold for a truly random walk. As a bonus, there is also a certain probability that it will optimally solve a maze regardless of its properties, which was in fact the case for our maze challenge run.
Since the software for the mapping was almost completely decoupled from the rest, the random walk could be implemented easily. It will still get an intersection type from the Decision Maker, and will still return a decision. It was decided not to have U-turns in the random walk (except for dead ends, of course), since it's trivially shown that a U-turn can always be achieved with a combination of other decisions, and U-turns would in general severely impede progress through the maze.
Throughout the course, it was always a balance act between decoupling and ease of implementation. In this particular case, ease of implementation was favoured, since the random walk block merely tries to solve the maze and highlight (successfully so) the capabilities of all other blocks. Furthermore, this block was focused on robustness; as such, even though in the below example, case 2: could be used, default was chosen to eliminate the risk of accidental bugs.
srand (time(NULL)); //Initialize random number generator
int direction =0; //Initialize direction
switch(type){
case FourWay:
direction = rand()%3;
switch(direction){
case 0:
return Left;
break;
case 1:
return Forward;
break;
default:
return Right;
break;
}
break;
case LeftJunction: //etc | 7,700 | 34,540 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | latest | en | 0.916759 |
https://fleekpapers.com/Matlab-Assignment-Help | 1,702,249,740,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00025.warc.gz | 287,921,072 | 11,460 | # MATLAB Assignment Help
Unlock your programming potential with expert MATLAB Assignment Help. Master complex and diverse features with skilled programmers.
###### Phone
phone: +1 (251) 245-8710
###### Email
email: support@fleekpapers.com
###### Let Our PhD Writers Assist You
Our qualified PhD. Professionals know what you need, and they will put their best effort into ensuring you ace all your assignments.
MATLAB is an indispensable tool with various engineering, physics, and mathematics applications. Its versatility encompasses tasks such as mathematical modeling, precise image processing, data analysis, and dynamic systems simulation. However, despite its crucial role in numerical and scientific computing, MATLAB presents a formidable challenge for students trying to grasp and master its capabilities. Our MATLAB assignment help experts are here to deconstruct the complexities of this powerful computing platform and ease your journey to proficient programming.
#### Key Features Covered in Our MATLAB Assignment Help Program
MATLAB studies cover a wide spectrum of sub-topics. Here are the key features covered in our MATLAB assignment help programs by our tutors.
• Machine Learning
• GUI Design
• Optimization Problems
• Data Visualization
• MATLAB Algorithm Development
• Image processing
• Control System Design
• Advanced Mathematical Functions
• Statistical Analysis
• Toolboxes
• Interfacing with Other Languages
• Numerical Methods
• Code Optimization
#### Tips for Preparing or MATLAB Assignments
Solving complex MATLAB assignments requires the right approach and strategies. Here are tips and tricks used by our MATLAB Assignment Help professionals to solve MATLAB problems.
• Break Down the problem into Manageable Parts: This may include defining the problem by calculating the mean of an array and identifying array values. Break the code into scripts or functions to make it more organized and easy to debug.
• Plan your approach: Create a plan or algorithm on a paper before attempting the code. Consider MATLAB functions or techniques applicable to the assignment.
• Start Simple: Implement the simplest aspects of the problem first and verify that the codes are correct, and work your way up to more complex areas.
• Use MATLAB Documentation: Use MATLAB’s extensive documentation to learn about syntax, functions, and examples. Use the ‘doc’ and ‘help’ commands to access documentation for specific MATLAB functions. For instance, to access the documentation of a specific function or toolbox using ‘doc,’ open the Command Window in MATLAB and type “doc fuctionName.” Replace “functionName” with the name of the toolbox or function you want to explore. The documentation will open in the Help Browser. Follow the same procedure for the ‘help’ command to get the description of a function or its syntax.
• Debug: Leverage MATLAB’s rich toolbox, comprising tools such as ‘dbstop’ command and breakpoints to identify and fix errors.
• Use Comments: Adding comments to the codeto explain your ideas, key steps, and variable meanings makes your code easy to follow and understand.
• Test Your Code: Use different input values and edge cases to validate your code’s correctness. Use assert statements to ensure your results are accurate.
#### Practical Applications of MATLAB
MATLAB is a rich programming and computing environment with various key features and applications: Here are real-world applications of this powerful tool;
1 Mathematics and Statistics MATLAB’s extensive library of mathematical functions makes complex calculations, curve fitting, optimization, and differential equations more manageable. Its visualization capabilities help mathematicians get a lively view of concepts for easy understanding and interpretation. Our MATLAB Assignment Help professionals can analyze and present data graphically, thus bridging the rift between raw data and human understanding. Users can make informed decisions based on visualized data. 2 Science and Engineering Civil, mechanical, electrical, and chemical engineers use MATLAB for designing, analyzing, and optimizing complex systems such as structural designs, chemical processes, circuits, and control systems. Additionally, MATLAB has a remarkable ability to emulate real-world phenomena, making it an excellent tool for design, video, and signal processing, among others. Evolutionary biology assignments include human development, phylogenetics, and evolutionary ecology. Our biology homework experts help students understand patterns and processes of evolution, including speciation, adaptation, and natural selection. 3 Machine Learning and Artificial Intelligence (AI) MATLAB is an expansive AI and machine-learning platform with diverse functions, algorithms, and tools. It’s an excellent environment for model training, data preparation, performance evaluation, and feature extraction. Besides, its machine-learning prowess makes tasks such as, clustering, regression, classification, and anomaly detection more manageable. 4 Financial Modeling MATLAB boasts roust computational prowess and reliable math and statistics capabilities. Finance professionals leverage these functions to make well-informed decisions by efficiently managing large datasets and executing complex calculations. Financial analysts also use it for algorithmic trading, portfolio optimization, option pricing, and risk analysis, enhancing the precision and efficiency of financial models and analyses. 4 Research and Development Researchers and other professionals like engineers, statisticians, and mathematicians use MATLAB to analyze data, conduct simulations, manipulate images, and develop algorithms. Scientists and researchers can focus solely on the most essential research goals without getting entangled in the complicated coding aspects of MATLAB. 5 Get MATLAB Assignment Help MATLAB’s versatility in the programming world makes it an incredible tool worth exploring. However, unlike other programming languages, mastering MATLAB can prove a formidable challenge for many aspiring programmers, making it feel almost insurmountable. Are you feeling lost on this diverse computing platform? Get MATLAB Assignment help from experienced programmers, and set out on your path to becoming a professional coder. | 1,116 | 6,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-50 | latest | en | 0.865694 |
http://www.horseforum.com/barrel-racing/what-do-they-mean-120343/ | 1,462,130,627,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860116886.38/warc/CC-MAIN-20160428161516-00014-ip-10-239-7-51.ec2.internal.warc.gz | 581,525,997 | 55,984 | What do they mean? - The Horse Forum
Old 04-17-2012, 11:30 PM Thread Starter
Foal
Join Date: Jan 2012
Posts: 124
• Horses: 2
What do they mean?
I seen on a show bill where they put 2-D & 3-D classes: 1 second split 4D classes: 1/2 second split, I second, 2 second split respectively. What do they mean by 1 second split , 1/2 second split and 2 second split?
Strength lies within the heart but the strength to trust lies between the horse and its rider.
BarrelRacer95 is offline
Old 04-18-2012, 12:25 AM
Green Broke
Join Date: Jan 2011
Location: Georgia
Posts: 4,962
• Horses: 3
Usually it works like this.. Say the fastest time of the night is a 15.0..That 15.0 is the winning 1D time.
1D - 15.0 then you have to add 1/2 second split..15.0 is 1/2 faster than a 15.5..
The winning 2D time would have to be a 15.5 or so..it cannot be faster than a 15.5..
Then the 3D would have to be a 16.0.. It cannot be faster than a 16.0.. That's the 1 second "split".. 15.0 is one second faster than the 16.0..
Then the 4D would have to be a 16.5.. that 16.5 is two seconds off of the fastest time of the day.. that's the 2 second split..
So it would go
1D - 15.0
2D - 15.5
3D - 16.0
4D - 16.5
And so on and so forth.. Those are the "splits"
Here is another good example from another user.
1D = fastest time of the day
2D = fastest horse that is equal to or slower than the agreed-upon split (typically 1/2 second from the 1D time)
3D = fastest horse that is equal to or slower than the agreed-upon split (typically 1 second from the 1D time)
4D = fastest horse that is equal to or slower than the agreed-upon split (typically 2 seconds from the 1D time)
Read more: PLS explain the 1-D 2-D 3-D 4-D
___________________________
Here is a link to an actual show results sheet..You can look over the results and see the "splits" and get a feel for what I'm talking about.
http://www.southeasternarena.com/pdf...02-25-2012.pdf
I am Sparkly Meanie Doodie Head and I approve this message!
Last edited by DrumRunner; 04-18-2012 at 12:35 AM.
DrumRunner is offline
Old 04-18-2012, 08:40 AM Thread Starter
Foal
Join Date: Jan 2012
Posts: 124
• Horses: 2
Thank you.! I understand it now.! Thanks.!!
Strength lies within the heart but the strength to trust lies between the horse and its rider.
BarrelRacer95 is offline
Old 04-18-2012, 11:16 AM
Weanling
Join Date: Apr 2012
Location: vandalia il
Posts: 268
• Horses: 3
you choose the one that best fits your horse so you have a better chance at winning. I believe that's how it goes
greenbryerfarms is offline
Old 04-18-2012, 11:42 AM
Showing
Join Date: Oct 2007
Location: Greenville area / SC
Posts: 13,165
• Horses: 3
Quote:
Originally Posted by greenbryerfarms View Post
you choose the one that best fits your horse so you have a better chance at winning. I believe that's how it goes
I don't think so. The announcement is just so you know how the placement works. You have no idea how your horse is going to do so you don't really know in which category the horse will run.
There may be a lot of good horses there on that particular night and your horse who normally runs 2-D may run 4-D that night - or vice versa.
I don't run barrels but I go to watch some times and that is how I am thinking.
I'm not arguing with you, I'm just explaining why I'm right.
Nothing sucks more than that moment during an argument when you realize you're wrong.
It's not always what you say but what they hear.
iridehorses is offline
Old 04-18-2012, 02:18 PM
Yearling
Join Date: Aug 2011
Location: Indiana
Posts: 1,288
• Horses: 2
Quote:
Originally Posted by greenbryerfarms View Post
you choose the one that best fits your horse so you have a better chance at winning. I believe that's how it goes
Its one class, the divisions just make it so more people have a chane at placing . You don't have to be a 1D rider or have a 1D horse to have a chance at getting in. They usually only go 2 seconds off the leader.
Posted via Mobile Device
BarrelRacer23 is offline
Old 04-18-2012, 04:45 PM
Weanling
Join Date: Apr 2012
Location: vandalia il
Posts: 268
• Horses: 3
really. Hmm must have been explaned wrong to me.
When he ran the ground shook...
...so we partied our way to victory...
.-.-.-.-.- <3 splash <3 .-.-.-.-.-
greenbryerfarms is offline
Old 04-18-2012, 06:57 PM
Green Broke
Join Date: Jan 2011
Location: Georgia
Posts: 4,962
• Horses: 3
Quote:
Originally Posted by iridehorses View Post
I don't think so. The announcement is just so you know how the placement works. You have no idea how your horse is going to do so you don't really know in which category the horse will run.
There may be a lot of good horses there on that particular night and your horse who normally runs 2-D may run 4-D that night - or vice versa.
I don't run barrels but I go to watch some times and that is how I am thinking.
This exactly..Depending on arena size, how many riders there are, the different competitors showing and how your horse is doing that night is usually how it will be determined. There isn't a "class" set up. It's more of whoever is the fastest of the day determines the times, because the splits are determined by their winning time. Really it's not about having the fastest horse every time. Being consistent and having clean runs is usually what places you, and it can come down to just a matter of luck..
And like iridehorses said, depending on how many people are there is also a big factor.. If a lot of the big competitors aren't at the show..Someone who usually runs in the 4D may run in the 2D or 3D... BUT, if there are a lot of big competitors there someone who runs 2D may run in the 4D.. There are just a ton of factors that come in to play.. Especially at big open shows or barrel bashes.
I am Sparkly Meanie Doodie Head and I approve this message!
Last edited by DrumRunner; 04-18-2012 at 06:59 PM.
DrumRunner is offline
Message:
Options
## Register Now
In order to be able to post messages on the The Horse Forum forums, you must first register.
Already have a Horse Forum account?
New to the Horse Forum?
Please choose a username you will be satisfied with using for the duration of your membership at the Horse Forum. We do not change members' usernames upon request because that would make it difficult for everyone to keep track of who is who on the forum. For that reason, please do not incorporate your horse's name into your username so that you are not stuck with a username related to a horse you may no longer have some day, or use any other username you may no longer identify with or care for in the future.
User Name:
OR
## Log-in
Human Verification
In order to verify that you are a human and not a spam bot, please enter the answer into the following box below based on the instructions contained in the graphic. | 1,893 | 6,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2016-18 | longest | en | 0.914101 |
http://www.acmerblog.com/hdu-2434-lets-play-uno-3898.html | 1,500,611,251,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423716.66/warc/CC-MAIN-20170721042214-20170721062214-00542.warc.gz | 352,155,184 | 14,202 | 2014
01-26
# Let’s play UNO
Uno is a famous card game played with a specially printed deck. It’s very
popular as a funny game in a party.
A so-called "official rules" is presented at the Wikipedia, but there are a lot of different extended rules all over the
world according to their specific needs.
In this problem, you are required to obey our rules introduced below:
The Uno deck consists of cards of 4 colors: red(R), green(G), blue(B), and yellow(Y).
Each color has two kinds of cards, number cards and action cards.
The ranks in number cards are 0-9. There are 3 "action" cards in each color, labeled "skip"(S), "draw two"(D), and "reverse"(R).
The functions of the actions will be described afterwards.
For each color, there are two copies of each positive number card and action card, but only one zero card, producing 25 cards in total.
Besides, there are also special black action cards called "wild cards", "wild"(WC) and "wild draw four"(WF).
There are four "wild" and "wild draw four" cards each. Hence, there are 108 cards in total.
In this problem, a card is marked with an ID of two characters length, the first is the color (R, G, B, Y, W) while the second is the rank (0-9, S, D, R, C, F).
For example, the ID of red 2 is R2, the yellow reverse is YR, the wild cards are WC and WF.
Supposed there are n players numbered from 1 to n clockwise.
Before playing, players take turns(in the order of 1, 2, … n) to pick seven successive cards from the stock.
The top card of the remaining stock is exposed to start the game, treated as if player 1 dropped that card.
The exposed card will never be WC or WF in this problem.
Then the game begins clockwise (next player is 2), or counter-clockwise (next player is n) if the top exposed card is a reverse.
At each turn, a player may drop a card from their hand that matches the color or rank of the top exposed card (e.g., if the top card is R3, you can drop R5 or G3;
if the top card is RD, you can drop R3 or GD) or play a WC.
What’s more, if the player has a WF and no other legal cards to drop, he can drop the WF.
Then the card dropped just now becomes the top exposed card.
If a player has no legal cards, he must draw the top card of the stock and place it in his hand.
After dropping a single card or drawing, the next player clockwise takes a turn, or counter-clockwise when the reverse is in effect.
When a player drops down to only one card, that player is required to say "uno" to warn other players.
The game ends when a player drops all his/her cards, or the stock is emptied but the current player has to draw a card.
If the last card is an action card, the special effect still occurs.
When the game ends, all players count the number of points pertaining to the values of the cards in their hands.
Number cards worth the face value on them, colored special cards worth twenty, and wilds worth fifty, e.g., R2 worth 2, G0 worth 0, BD and YS worth 20, WC and WF worth 50.
The descriptions of the action cards:
Now here comes the problem.
There are N people playing Uno under the rules mentioned above. Given the sequence of the 108 cards of the stock, you are asked to simulate a Uno game.
At each turn, the player will always drop a card if permitted.
If there are more than one choices, the player will drop the card with the largest point.
If still a tie, he will choose the one whose ID is the smallest alphabetical order.
When a player drops WC or WF, he has to name a color.
The first time he will name red, the second time he will name green, the third time blue, the fourth time yellow, the fifth time red again, and so on.
When the game ends, you should output the final score of each player, and we also want to know how many times each player calls "Uno".
The first line of the input file contains a single number: the number of test
cases to follow. Each test case has two lines:
The first line contains the number of players N , with 2<=N<=10.
The second line contains 108 IDs of the Uno cards, separated by one space.
Each ID is two characters long as introduced in the description above.
The first line of the input file contains a single number: the number of test
cases to follow. Each test case has two lines:
The first line contains the number of players N , with 2<=N<=10.
The second line contains 108 IDs of the Uno cards, separated by one space.
Each ID is two characters long as introduced in the description above.
1
2
R9 RD RD RS RS RR RR
B0 B1 B1 B2 B2 B3 B3
G0 GD GD GS GS GR GR G9 G9 G8 G8 G7 G7 G6 G6 G5 G5 G4 G4 G3 G3 G2 G2 G1 G1
Y0 Y9 Y9 Y8 Y8 Y7 Y7 Y6 Y6 Y5 Y5 Y4 Y4 Y3 Y3 Y2 Y2 Y1 Y1 YD YD YS YS YR YR
R9 R8 R8 R7 R7 R6 R6 R5 R5 R4 R4 R3 R3 R2 R2 R1 R1 R0
B4 B4 B5 B5 B6 B6 B7 B7 B8 B8 B9 B9 BD BD BS BS BR BR
WC WC WC WC WF WF WF WF
249 0
0 1
Hint
The process of this game is:
G0(exposed)
-> B0(Player 2) -> Draw GD(Player 1) -> B3(Player 2)
->Draw GD(Player 1) -> B3(Player 2) -> Draw GS(Player 1)
-> B2(Player 2) -> Draw GS(Player 1) -> B2(Player 2)
-> Draw GR(Player 1) -> B1, call “Uno”(Player 2) -> Draw GR(Player 1)
-> B1(Player 2), end.
Score of player 1: 20*12(6 red action cards and 6 green action cards)+9(R9)
/*This Code is Submitted by billforum for Problem 2434 at 2012-01-30 17:11:48*/
#include <iostream>
#include <queue>
#include <memory.h>
using namespace std;
struct point
{
int length;//朋友的个数
int step;
bool visit;
int f[1001];//存放朋友的号码
int num;
};
point data[1001];
bool know[1001][1001];//标识两人认识不
int main(int args,char **argv)
{
int g,n,gn,ntmp,ans=-1;
int atmp[1001];
while(cin>>g>>n)
{
if(g==0&&n==0) break;
//初始化
for(int i=0;i<n;i++)
{
data[i].length=0;
data[i].step=0;
data[i].visit=0;
data[i].num=i;
for(int j=0;j<1001;j++)
data[i].f[j]=-1;
}
ans=-1;
for(int i=0;i<1001;i++)
for(int j=0;j<1001;j++)
know[i][j]=0;
//对输入数据进行处理，构造关系
for(int i=0;i<g;i++)
{
cin>>gn;
for(int j=0;j<gn;j++)
cin>>atmp[j];
for(int k=0;k<gn;k++)
{
int t=atmp[k];
for(int j=k+1;j<gn;j++)
{
int s=atmp[j];
if(!know[t][s])
{
data[t].length++;
data[s].length++;
data[t].f[data[t].length-1]=s;
data[s].f[data[s].length-1]=t;
know[t][s]=1;
know[s][t]=1;
}
}
}
}
queue<point> list;
data[0].visit=1;
list.push(data[0]);
while(!list.empty())
{
point tmp=list.front();
if(tmp.num==n-1)
{
ans=tmp.step;
break;
}
//广搜
for(int k=0;k<tmp.length;k++)
{
int i=tmp.f[k];
if(data[i].visit) continue;
else
{
data[i].visit=1;
data[i].step=tmp.step+1;
list.push(data[i]);
}
}
list.pop();
}
if(ans==-1) cout<<"I can never know wywcgs!"<<endl;
else if(ans==1||ans==0) cout<<"I do know wywcgs!"<<endl;
else
{
cout<<"I can know wywcgs by at most "<<ans-1<<" person(s)!"<<endl;
}
}
return 0;
}
http://acm.hit.edu.cn/hoj/problem/view?id=2434 | 2,011 | 6,646 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-30 | longest | en | 0.957579 |
https://convert-dates.com/days-before/132/2034/08/05 | 1,670,156,110,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00758.warc.gz | 198,031,925 | 4,012 | ## 132 Days Before August 5, 2034
Want to figure out the date that is exactly one hundred thirty two days before Aug 5, 2034 without counting?
Your starting date is August 5, 2034 so that means that 132 days earlier would be March 26, 2034.
You can check this by using the date difference calculator to measure the number of days before Mar 26, 2034 to Aug 5, 2034.
March 2034
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
1. 1
2. 2
3. 3
4. 4
1. 5
2. 6
3. 7
4. 8
5. 9
6. 10
7. 11
1. 12
2. 13
3. 14
4. 15
5. 16
6. 17
7. 18
1. 19
2. 20
3. 21
4. 22
5. 23
6. 24
7. 25
1. 26
2. 27
3. 28
4. 29
5. 30
6. 31
March 26, 2034 is a Sunday. It is the 85th day of the year, and in the 12nd week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2034 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 03/26/2034, and almost everywhere else in the world it's 26/03/2034.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 132 weekdays before Aug 5, 2034, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Aug 5, 2034, which falls on a Saturday. Counting forward, the next day would be a Monday.
To get exactly one hundred thirty two weekdays before Aug 5, 2034, you actually need to count 184 total days (including weekend days). That means that 132 weekdays before Aug 5, 2034 would be February 2, 2034.
If you're counting business days, don't forget to adjust this date for any holidays.
February 2034
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
1. 1
2. 2
3. 3
4. 4
1. 5
2. 6
3. 7
4. 8
5. 9
6. 10
7. 11
1. 12
2. 13
3. 14
4. 15
5. 16
6. 17
7. 18
1. 19
2. 20
3. 21
4. 22
5. 23
6. 24
7. 25
1. 26
2. 27
3. 28
February 2, 2034 is a Thursday. It is the 33rd day of the year, and in the 33rd week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 28 days in this month. 2034 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 02/02/2034, and almost everywhere else in the world it's 02/02/2034.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 917 | 2,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-49 | latest | en | 0.912064 |
www.xmod360.nl | 1,618,291,773,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00566.warc.gz | 1,202,345,293 | 7,948 | # calculations of mill charge
### Development of charge calculation program for target steel
This paper presents the development of charge calculation program for target steel in induction furnace. The simulation modelling function developed is based on mass balance analysis of the
Get Price
### Ball Mill Critical Speed & Working Principle
Jun 19, 2015 · The tendency of the charge to slip back along the rising side of the shell is reduced as charge level increases, until reaching between 30 to 40 percent of mill volume (in one particular test mill
Get Price
### CALCULATION OF THE POWER DRAW OF DRY
Calculation of the power draw of dry multicompartment ball mills 225 The mill load that is the volume of charge in the mill is the principal determinant of power draw. Estimation of the ball load that is mixed with the cement charge is difficult and can be highly erroneous. So direct measurement must be taken for calculation of mill load.
Get Price
### Ball charges calculators thecementgrindingoffice
Ball top size (bond formula): calculation of the top size grinding media (balls or cylpebs):Modifiion of the Ball Charge: This calculator analyses the granulometry of the material inside the mill and proposes a modifiion of the ball charge in order to improve the mill efficiency:
Get Price
### A Method to Determine the Ball Filling, in Miduk Copper
the mill. Ball Charge Program Abrasion In this section, ball abrasion was calculated via manufacture`s ball charge program. At the time of this research, mill ball charged, feed rate, and average moisture were 7 tons (ball size was 100 abrasion rate in the mill. If above calculation were done again for
Get Price
### Optimization of mill performance by using
height ''H'' from the charge to the shell and the internal mill diameter ''Di''. By calculating the ratio ''H/Di'' and using the graph below (Figure 1), the charge filling degree in volume could be estimated. The number of visible plates on the shell liner could also be counted and an
Get Price
### 2000 ton capacity grinding mill media charge calculation
Next: 2000 ton capacity ball mill media charge 2000 ton capacity ball mill media charge calculation of charge in the mill. This equates to a 24% decrease in ball mill unit energy consumption from 11. Chat Online
Get Price
### sagdesign
Mill Speed Charge GRINDING CIRCUIT CAPACITY : Mill Throughput Capacity Grinding Circuit Capacity : Assumed from Bailey et al, 2009 SAGDesign Mill Throughput Calculation Using S dBWI when T80 ≠ 1,700 µm. Using S dBWI from T80 to P80 BM Coarse Feed Correction Factor (EF 4) BM Fineness Correction Factor (EF 5) BM Diameter Correction Factor (EF 3)
Get Price
### calculation mill charge retefamiglieaccoglienti
calculations mill charge ristorantelachiocciola . how can i calculate optimum charge for cement mill BINQ Mining. calculations . 5 years ago Report Abuse； Cement mills are normally driven by electric to find the optimum ball charge for your particular mill making this masonry cement by a
Get Price
### CNC Machine Hourly Rate Calculator CNCCookbook: Be A
CNC Machine Hourly Rate Calculator. Our GWizard Estimator software has a Machine Hourly Rate Calculator. A lot of shops use the notion of hourly rate on machines to help with job cost estimation and quotation, but there''s not a lot of information available about how to calculate a good hourly rate to use.
Get Price
### How To Calculate Mill Charge Level
Calculations for mill motor power, mill speed and media charge. Considering the weight of mill lining and grinding media, work out the motor power required, in consultation with To calculate media charge for cylindrical mill. Read more
Get Price
### SAGMILLING .:. tools_millfilling
Enter the number of lifters that are exposed above the mill charge. Fractions are acceptable and should be used in the case where the charge does not sit uniformly in the mill. For example, you may have 29 rows exposed at the discharge end of the mill, and only 28 exposed at the feed end.
Get Price
### Cement Mill Ball Charge Calculation
Calculates The Grinding Charge Of A Ball Mill Mining . Flshanghai ball mill for cement grinding cement mill is a corrugated lining designed to obtain maximum power for fine grinding the charge consists of small calculation of ball mill charge chinagrindingmillnet. Chat Online
Get Price
### SAGMILLING .:. tools
Determines the effective mill inside diameter of a given set of "top hat" lifters for use in mill volume and critical speed calculations. Mill Filling Calculation Estimates the charge volume of a grinding mill based on the number of exposed lifters.
Get Price
Ball Mill Loading Dry Milling. Ball Mill Loading (dry milling) When charging a ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar use on a jar rolling mill it is important to have the correct amount of media and correct amount of product.
Get Price
### AG Autogenous & SAG SemiAutogenous Mill Design Calculations
The ore charge contained within the conical ends of course must also be considered in calculating the total charge weight. Figures 3a and b, and 4a and b show a typical printout of mill cylinder and cone charge weight for an autogenous and semiautogenous mill. The total charge weight is the sum of cylinder and conical charge weights.
Get Price
### calculation mill charge retefamiglieaccoglienti
calculations mill charge ristorantelachiocciola . how can i calculate optimum charge for cement mill BINQ Mining. calculations . 5 years ago Report Abuse； Cement mills are normally driven by electric to find the optimum ball charge for your particular mill making this masonry cement by a
Get Price
### CALCULATION OF BALL MILL GRINDING EFFICIENCY Page 1
Mar 08, 2013 · calculation of ball mill grinding efficiency. dear experts . please tell me how to calculate the grinding efficiency of a closed ckt & open ckt ball mill. in literatures it is written that the grinding efficiency of ball mill is very less [less than 10%]. please expalin in a n excel sheet to calcualte the same. thanks. sidhant. reply
Get Price
### Estimate Charge Volume of a Grinding Mill (Method 1)
Estimate Charge Volume of a Grinding Mill (Method 1) Previous Next. View Larger Image. The load estimation method has you physically measure distances in the mill. Follow the charts below and pull them into this calculator. If you can clearly see inside the mill, use the liner method. By David Michaud
Get Price
### Mill (grinding) Wikipedia
Ball mill. A typical type of fine grinder is the ball mill.A slightly inclined or horizontal rotating cylinder is partially filled with balls, usually stone or metal, which grind material to the necessary fineness by friction and impact with the tumbling balls. Ball mills normally operate with an approximate ball charge of 30%.
Get Price
### Calculate Ball Mill Grinding Capacity
A) Total Apparent Volumetric Charge Filling – including balls and excess slurry on top of the ball charge, plus the interstitial voids in between the balls – expressed as a percentage of the net internal mill volume (inside liners). B) Overflow Discharge Mills operating at low ball fillings – slurry may accumulate on top of the ball charge causing, the Total Charge Filling Level to be
Get Price
### CALCULATION OF THE POWER DRAW OF DRY
Calculation of the power draw of dry multicompartment ball mills 225 The mill load that is the volume of charge in the mill is the principal determinant of power draw. Estimation of the ball load that is mixed with the cement charge is difficult and can be highly erroneous. So direct measurement must be taken for calculation of mill load.
Get Price
### Jyoti Ceramic Industries Pvt. Ltd.
Calculations of media charge: To calculate media charge for batch mill. M = 0.000929 x D2 x L where : M = Grinding media charge in kgs D = Internal Dia of the Mill in cms. after lining L = Internal length of the mill in cms. after lining. To calculate grinding media charge for continuous type ball mill, M = 0.000676 x D2 x L
Get Price
Ball Mill Loading Dry Milling. Ball Mill Loading (dry milling) When charging a ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar use on a jar rolling mill it is important to have the correct amount of media and correct amount of product.
Get Price
### calculates the grinding charge of a ball mill Mining
– Monochamber mill power: it calculates the theoretical absorbed power of the This calculator is a good start to evaluate the grinding ball charge after mill
Get Price
### calculations of mill charge cruisertrailers
Calculations of media charge Jyoti Ceramic Industries Pvt. Ltd. Calculations of media charge. To calculate media charge for batch mill. M = 0.000929 x D2 x L where : M = Grinding media charge
Get Price
### Jyoti Ceramic Industries Pvt. Ltd.
Calculations of media charge: To calculate media charge for batch mill. M = 0.000929 x D2 x L where : M = Grinding media charge in kgs D = Internal Dia of the Mill in cms. after lining L = Internal length of the mill in cms. after lining. To calculate grinding media charge for continuous type ball mill
Get Price
### 2000 ton capacity grinding mill media charge calculation
Next: 2000 ton capacity ball mill media charge 2000 ton capacity ball mill media charge calculation of charge in the mill. This equates to a 24% decrease in ball mill
Get Price
### How To Calculate Mill Charge Level
Calculations for mill motor power, mill speed and media charge. Considering the weight of mill lining and grinding media, work out the motor power required, in consultation with To calculate media charge for cylindrical mill. Read more
Get Price
### https://com/blog/ball 911
Mill Steel Charge Volume Calculation. We can calculate the steel charge volume of a ball or rod mill and express it as the % of the volume within the liners that is filled with grinding media. While the mill is stopped, the charge volume can be gotten by measuring the diameter inside the liners and the distance from the top of the charg
Get Price
### Ball Mill Charge Grinding & Classifiion Circuits
The ball charge and ore charge volume is a variable, subject to what is the target for that operation. The type of mill also is a factor as if it is an overflow mill (subject to the diameter of the discharge port) is usually up to about 4045%. If it is a grate discharge you will have more flexibility of the total charge.
Get Price
### Ball Mill Critical Speed & Working Principle
Jun 19, 2015 · The tendency of the charge to slip back along the rising side of the shell is reduced as charge level increases, until reaching between 30 to 40 percent of mill volume (in one particular test mill
Get Price
### grinding media charge calculation in cement mill
The grinding charge of rotary mills Analele Universitatii Dunarea This paper presents the achievemens of grinding charge for the tube mills with balls.The author shows charactheristic size of grinding media charge: bulk density, porosity, filling degree. clinker as well as the cement, is obtained by . the top of the charge (see fig.1).
Get Price
### Mill (grinding) Wikipedia
Ball mill. A typical type of fine grinder is the ball mill.A slightly inclined or horizontal rotating cylinder is partially filled with balls, usually stone or metal, which grind material to the necessary fineness by friction and impact with the tumbling balls. Ball mills normally operate with an approximate ball charge of 30%.
Get Price
### Calculate The Charging In Cement Mill
calculation of cement mill media charging. How Can I Calculate Optimum Charge For mill charge load calculationsmill charging calculation calculate the grinding media in cement mill 2016 The work horse of the cement grinding plant is the two calculation of cement mill media charging,Ball Mill Media Charge Calculation Grinding Ball Mill Media Charge Calculation Pdf Ball Mill Media Charge
Get Price
### Cement mill ball charge calculation
Ball charge Formula In cement Mill denhelderuitvaartnl Ball Charge Design Methods Ball Charge Design Methods Polysius for Finish Mills A formula that covers the entire mill as though it is a single Mill grinding Wikipedia, the free encyclopedia Ball mills normally operate with an approximate ball charge of 30% Cement mill Wikipedia
Get Price
### Ball Mill Critical Speed & Working Principle
Jun 19, 2015 · The tendency of the charge to slip back along the rising side of the shell is reduced as charge level increases, until reaching between 30 to 40 percent of mill volume (in one particular test mill
Get Price
### TECHNICAL NOTES 8 GRINDING R. P. King
The effect of mill charge is primarily through the shifting of the center of gravity and the mass of the charge. As the charge increases the center of gravity moves inward. The power draft is more or less symmetrical about the 50% value. A simple equation for calculating net power draft is P 2.00 3 cD 2.5 m LKl kW (8.12)
Get Price
### MODELING THE SPECIFIC GRINDING ENERGY AND BALL
1 MODELING THE SPECIFIC GRINDING ENERGY AND BALLMILL SCALEUP K. G. Tsakalakis & G.A. Stamboltzis NTUA AthensGreece Email: [email protected] Presented at the conference IFAC 2004 held in September 2004 (NancyFrance)
Get Price
### Mill Ball Charge Calculation borderblue
Mill Steel Charge Volume Calculation. Mill Steel Charge Volume Calculation View Larger Image We can calculate the steel charge volume of a ball or rod mill and express it as the % of the volume within the liners that is filled with grinding media. Learn More calculate ball charge ball mill residencearchambeau
Get Price
### calculating steel ball charge in ball mill
Exploring ball size distribution in coal grinding mills ResearchGate. ABSTRACT Tube mills use steel balls as grinding media. new balls periodically to maintain a steady balanced ball charge in the mill. .. of the calculation of the size distribution of the equilibrium mixture of balls in a ball mill is developed.
Get Price
### sagdesign
Mill Speed Charge GRINDING CIRCUIT CAPACITY : Mill Throughput Capacity Grinding Circuit Capacity : Assumed from Bailey et al, 2009 SAGDesign Mill Throughput Calculation Using S dBWI when T80 ≠ 1,700 µm. Using S dBWI from T80 to P80 BM Coarse Feed Correction Factor (EF 4) BM Fineness Correction Factor (EF 5) BM Diameter Correction Factor (EF 3)
Get Price
### TECHNICAL NOTES 8 GRINDING R. P. King
The effect of mill charge is primarily through the shifting of the center of gravity and the mass of the charge. As the charge increases the center of gravity moves inward. The power draft is more or less symmetrical about the 50% value. A simple equation for calculating net power draft is P 2.00 3 cD 2.5 m LKl kW (8.12)
Get Price
### Development of charge calculation program for target steel
This paper presents the development of charge calculation program for target steel in induction furnace. The simulation modelling function developed is based on mass balance analysis of the
Get Price
### calculation of material charge for a ball mill
Calculation of material charge for a ball mill. Calculating Steel Ball Charge In Ball Mill Mar 19, 2017 We can calculate the steel charge volume of a ball or rod mill and express it as the % of the volume within Live ChatCalculate and Select Ball Mill Ball Size for Optimum, how to calculate alumina ball charge for dry mill. Oline Chat. Get
Get Price | 3,385 | 15,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-17 | latest | en | 0.881879 |
http://hackage.haskell.org/package/rsagl-0.2.1/docs/src/RSAGL-RK4.html | 1,519,114,689,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812913.37/warc/CC-MAIN-20180220070423-20180220090423-00726.warc.gz | 158,720,479 | 2,170 | ```\section{Runge-Kutta}
Haskell implementation of RK4.
\begin{code}
module RSAGL.RK4
(rk4,
integrateRK4,
rk4',
integrateRK4')
where
import RSAGL.AbstractVector
import RSAGL.Time
\end{code}
\begin{code}
genericRK4 :: (AbstractVector v) => (Time -> p -> v -> p) -> (Time -> p -> Rate v) -> p -> Time -> Time -> p
genericRK4 addPV diffF p0 t0 t1 = addPV h p0 \$ (scalarMultiply (recip 6) (abstractSum [k1,k2,k2,k3,k3,k4]) `over` h)
where k1 = diffF t0 p0
k2 = diffF tmid (addPV h2 p0 \$ k1 `over` h2)
k3 = diffF tmid (addPV h2 p0 \$ k2 `over` h2)
k4 = diffF t1 (addPV h p0 \$ k3 `over` h)
h = t1 `sub` t0
h2 = scalarMultiply (recip 2) h
tmid = t0 `add` h2
rk4 :: (AbstractVector v) => (p -> v -> p) -> (Time -> p -> Rate v) -> p -> Time -> Time -> p
rk4' :: (AbstractVector v) => (p -> v -> p) -> (Time -> p -> Rate v -> Acceleration v) -> (p,Rate v) -> Time -> Time -> (p,Rate v)
rk4' addPV diffF = genericRK4
(\t (p,old_v) delta_v -> let new_v = old_v `add` delta_v in (addPV p \$ (scalarMultiply (recip 2) \$ old_v `add` new_v) `over` t,new_v))
(\t (p,v) -> diffF t p v)
genericIntegrate :: (p -> Time -> Time -> p) -> p -> Time -> Time -> Integer -> p
genericIntegrate _ pn _ _ 0 = pn
genericIntegrate f p0 t0 tn n = genericIntegrate f p1 t1 tn (n-1)
where t1 = t0 `add` (scalarMultiply (recip \$ realToFrac n) \$ tn `sub` t0)
p1 = f p0 t0 t1
integrateRK4 :: (AbstractVector v) => (p -> v -> p) -> (Time -> p -> Rate v) -> p -> Time -> Time -> Integer -> p
integrateRK4 addPV diffF = genericIntegrate \$ rk4 addPV diffF
integrateRK4' :: (AbstractVector v) => (p -> v -> p) -> (Time -> p -> Rate v -> Acceleration v) -> (p,Rate v) -> Time -> Time -> Integer -> (p,Rate v)
integrateRK4' addPV diffF = genericIntegrate \$ rk4' addPV diffF
\end{code}
``` | 671 | 1,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-09 | latest | en | 0.449167 |
http://angolodeisaporifirenze.it/cpwj/volume-conversion-worksheet-pdf.html | 1,590,415,960,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00435.warc.gz | 7,320,696 | 16,741 | # Volume Conversion Worksheet Pdf
4 millimeters mm FT feet 0. Practice unit conversion problems with free time unit conversion worksheets and match the answer with the answer key given at the end of each worksheet. The collection of free units worksheets is designed to help kids learn about the units of area, length, volume, weight and time. This resource is designed for UK teachers. 0L at a pressure of 205kPa is allowed to expand to a volume of 12. A gas occupies 11. • Teach students how to convert between pounds, ounces, cups, pints, quarts, and gallons. Metric Conversions Worksheet. Text tools text converts and generators, changing letter case, number and text generators, HTML tools, text encryption, creating hash. 075 m to cm. Energy, Work, Heat 13 M. Always show your workings. 09361 yd 1093. Students will be able to grasp here all on properties of commonly used solid figures. English & Metric Conversion Tables Worksheets These measurement worksheets are a great handout for the student containing a list of length, distance, area, volume, weight and speed conversions between English and Metric Units. They must also remember to write the units for their answer with the correct degree. To select non-adjacent sheets, hold down Ctrl while clicking the tabs of each sheet you want to save as PDF. This means 1 cubic feet = 28. 2 L of argon gas at STP? 3. Factoring numbers. 54 cm/1 in) We are starting with cm and want to end up with in, so the first conversion factor will do the job. You can choose to include answers and step-by-step solutions. It is the referred folder that will not. CUSTOMARY SYSTEM 1 yd = 3 ft 3 tsp = 1 Tbs 1 ft = 12 in 16 Tbs 1 cup 1 fathom = 6 ft 1 cup = 8 oz (liquid capacity) 1 mi = 5, 280 ft 1 pt = 2 cups 1 acre = 43,560 ft 1 qt = 2 pt 1 lb = 16 oz (dry weight) 1 gal = 4 qt 1 T = 2000 lb 1 gal 231 in. 4 L for a gas at STP. 20 moles Sn 13. 34 Magnesium 1. 3 Maths Worksheet for Class 1. 54 cm / 1 inch (exactly). Access our pdf worksheets to help kids gain mastery over conversions between smaller units of capacity like teaspoons, tablespoons or fluid ounces. Gay-Lussac's law relates the temperature of a gas to its a. Independent Worksheet 2: Expressions, Variables & Situations. Convert 15 kg to mg. Each tile is a 3-inch square and costs $0. Convert lengths. MATH CONVERSION CHART- LIQUID VOLUME (US) Please note that these conversions work for US liquids only! METRIC CONVERSIONS 10 ml 1000 ml 3 tsp 2 Tbsp 8 drams 4 fl oz 8 fl oz 2 cups 16 fl oz 2 pt 128 fl oz 10 milliliters 1000 milliliters STANDARD CONVERSIONS 1 cl 1 liter 1 tablespoon I fluid ounce 1 fluid ounce 1 gill 1 cup I pint 1 pint 1 quart. 54 cm and 2. Bar Graph Worksheets. 0591939 cc 1 fluid drachm = 3. Mole Conversions Worksheet There are three mole equalities. LENGTH: 1 inch = 2. Never runs out of questions. Perfect for last minute classroom projects or for relief teachers. Designed to accompany the Pumpkin Investigation activity, kids will use these pumpkin investigation guidelines to. Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). A sample of air has a volume of 140. #N#33 ft = ___ yd ___ ft. The site also includes a predictive tool that suggests possible conversions based on input, allowing for easier navigation while learning more about various unit systems. To get the PDF worksheet, simply push the button titled "Create PDF" or "Make PDF worksheet". Mass of substance Amount of substance in moles Number of atoms, molecules, or formula units of substance Convert using the molar mass of the substance. The volume of a shape is similar to the area of a shape, in that volume measures the space inside of an object. Writing reinforces Maths learnt. 1 Gas Pressure Conversion. Find the volume of 835 g of SO 3 234 L e. The blank line in the middle of the conversion chart can change depending on what we are measuring: The unit for length is the meter (m). 00 atmosphere of pressure would have to be dissolved in 1. ²) x Average Depth (ft. ) Decide on the units for the numerator and denominator of the conversion factor. 45 kg 1 quart = 0. Imperial / American measurements. What type of measurement is indicated by each of the following units? Choices are in the last column. xlsx document you wish to convert in Microsoft Excel 2010. Volume (cubic meter, liter & milli-liter) metric units conversion worksheet with answers for 6th grade math curriculum is available online for free in printable and downloadable (pdf & image) format. This lesson looks at calculating salaries and wages. 7 × 10 7 cm 3. Metric Conversion Practice. Cubic Centimeter Cubic Feet Cubic Inches Cubic Meter Cubic Yard Cup Dram Drop Gallon Gallon UK Liter Milliliter Ounces Ounces UK Pint Pint UK Quart Tablespoon Teaspoon. There is a big difference between Fluid Ounces and Dry Ounces. Length Weight Volume Temperature Area Pressure Energy Power Force Time Speed Degree Fuel Consumption Data Storage. ) Volume (L, mL. Perimeter, Area, and Volume. cells\Worksheet - SA to V ratio worksheet with key. 0 mL 67ºCT 1 = + 273 = 340 K T 2 = ? is constant, so… 4. ☞ Solve problems involving volume and capacity. tr6rv650 a month ago report. So, it becomes especially relevant for our American students to be proficient with both systems. To use this Volume Conversion Table, please consider to have a look at the examples bellow, Value in the cell of 6 th row and 3 rd column of the volume conversion table is 28. STANDARD CONVERSION TABLE – ENGLISH TO METRIC Symbol To convert from Multiply by To determine Symbol IN LENGTH inch 25. Looking at the Metric System chart we notice that a centimeter is 100 times smaller than a. 1 meter = 3. Students will be able to grasp here all on properties of commonly used solid figures. The volume conversion chart comprises of units such as ounce, pint, quart and gallon and makes calculating much simpler and easier. Each tile is a 3-inch square and costs$0. Find the molecules in the problems below. Volume III Supplemental Worksheets More worksheets will be added soon Call us if you need one sooner, 888-510-MATH Lesson 1 Supplemental Worksheet: Solve for x. Chapter 1 Worksheet Spring 2007 page 2 5) Unit Conversions: a. volume_3_supp_worksheet_8b. And we can do it multiple ways. Click on the images to view, download, or print them. And it's all free. ) perform the Average Depth calculation for each. Each worksheet is randomly generated and thus unique. PDF2EXCEL If you are looking for a good PDF to Excel online conversion tool, you should take into account PDF2EXCEL. How many atoms are contained in 16. You may also browse chemistry problems according to the type of problem. 92 g/mL, calculate the following: a. Our conversions provide a quick and easy way to convert between Volume units. given in square yards. Make your job easier with Adobe Acrobat DC, the trusted PDF creator. Density has units of g/ cm3 or g/ml , or any mass over volume unit. 74 Mercury 13. Volume (cubic meter, liter & milli-liter) metric units conversion worksheet with answers for 6th grade math curriculum is available online for free in printable and downloadable (pdf & image) format. The figure above shows a cube. Volume : the measure of how much space an object occupies. That is: 2300 mm3 ÷ 103 = 2300. You will need a cup, some newspaper and a scale. Volume can be figurestwo different ways. This worksheet features all kinds of problems dealing with volume measurement and conversion. Volume of a Triangular Prism Worksheets. 750 micrograms to g 7. 2 miles/hour 1 foot = 12 inches 1 inch = 2. #N#33 ft = ___ yd ___ ft. Determine the volume of this sample at 760 mm Hg and 37°C. A sample of gas is transferred from a 75 mL vessel to a 500. 3 Maths Worksheet for Class 1. MASS 1 kg = 1,000 g = 1,000,000 mg = 2. 2) 45 grams of glucose, C 6H 12O. Performing a conversion between two different units of volume is very similar. Power 16 P. Determine the metric volume of the cube below. 0 mm Hg = 101. This worksheet will challenge students’ understanding of volume and the equation l x w x h = V. Math worksheet 4th grade pdf 17 best images about 4th grade math also math worksheet 4th grade pdf multiplication math worksheet 4th grade kids activities Math. pdf Metric Conversions Powerpoint. 2 liters at 0. What volume of silver metal will weigh exactly 2500. We can also convert measurements from m into cm. Convert linear measurements within the same system. The Surface Area & Volume Worksheets are randomly created and will never repeat so you have an endless supply of quality Surface Area & Volume Worksheets to use in. Please use all of our printables to make your day easier. If you want to save the entire workbook as a. density relates mass and volume while molar mass relates moles and mass. A cube is a three-dimensional figure with six matching square sides. 0 L of carbon monoxide reacts with oxygen at STP, a. 00 x 10-7 moles B. • Teach students how to convert between pounds, ounces, cups, pints, quarts, and gallons. 2 miles/hour 1 foot = 12 inches 1 inch = 2. 0 moles of O 2 and 3. You can generate the worksheets either in html or PDF format — both are easy to print. How many moles are represented by 11. CONVERTING METRIC UNITS - WEIGHT & VOLUME SHEET 1 ANSWERS 1) 1 kg = 1000 g 2) 2kg = 2000 g 3) 3kg = 3000 g 4) 4kg = 4000 g 5) 5kg = 5000 g 6) 6kg = 6000 g 7) 1 L = = 1000 mL 8) 2 L = 2000 mL 9) 3 L = 3000 mL 10) 4 L = 4000 mL 11) 5 L = 5000 mL 12) 6 L = 6000 mL Which is the most? Circle the largest amount in each box. Exercise 1: The density of aluminum is about 2:7g=cm3. This worksheet provides a lot of practice for converting metric units. Density and Conversion Factors Worksheet Name _____ Answer the following questions and problems. registerednursern. Conversions To convert units of measure (length, weight, volume, time), you must pay attention to the words of the measures. Measuring volume as area times length. docx Day 7--Customary Weight Quiz. Unformatted text preview: Mole Conversions Practice Worksheet Mole-Volume Conversions 1. RaycroftC:\Documents and Settings\stephanie_mckay\Desktop\please convert to pdf thanks\biology 12\2. Number Sequences. 2 lbs 1 lb = 0. Files included (1) 9x3 converting between units of area and volume. Energy, Work, Heat 13 M. ) perform the Average Depth calculation for each. 005 tonnes 5) 0. 9 5 (98) + 32 = 212 F 4. Temperature Conversion Worksheet Answers Fahrenheit Celsius Comments 212º = 100º Water boils 200º = 93º 100º = 38º 80º = 27º 70º = 21º 68º = 20º Typical room temperature 50º = 10º 32º = 0º Water freezes You can convert a temperature from Celsius to Fahrenheit in 3 steps: 1. 1416 and r is the radius. We need your resources! Click here to find out how to contribute! #N#Weights and Weighing (Deb Cadman) PDF. Skip Counting by 10s. 0 mL Pressure V 2 = 50. Here are the statutory requirements: Number and place value. Convert 40 F to Celsius. Speed 11 J. 7th grade/level math worksheets Seventh (7th) grade/level math worksheets to master 7th grade mathematics topics. 4 Plane angle. 2 miles/hour 1 foot = 12 inches 1 inch = 2. 09290304 square meters m 2. Volume III Supplemental Worksheets More worksheets will be added soon Call us if you need one sooner, 888-510-MATH Lesson 1 Supplemental Worksheet: Solve for x. 5 yards to inches. What is the new volume? 2. 9g of HCl? 2. Conversion of mksq Units to Gaussian Units 8 VII. Conversion of Units Capacity Worksheet 3. Math worksheets: Convert volumes including fluid ounces, cups, pints, quarts & gallons. Download Free Density Worksheets With Answers Chemistry Conversions Chart - Density, Volume, Grams to Moles, And complete you know our associates become fans of PDF as the best record to read? Yeah, it's Page 2/3. These are pyramid, truncated pyramid, cylinder, hollow cylinder (pipe), cone, truncated cone, sphere, the segment of a sphere and a barrel. 00133 Aluminum 2. Volume of a rectangular prism. 000 Lead 11. Perimeter, Area, and Volume. 38 mol Ne 2. (example: 8,000 mL or 80 L?) Four critical thinking questions about measurement with liters and. To select non-adjacent sheets, hold down Ctrl while clicking the tabs of each sheet you want to save as PDF. awalker71 2 months ago report. CHALLENGE The Art Club is tiling a wall 8 feet tall by 10 feet long. We need your resources! Click here to find out how to contribute! #N#Weights and Weighing (Deb Cadman) PDF. 25 centimeters into kilometers. Each Fifth grade math worksheet is a PDF printable with an answer key attached. Download Free Density Worksheets With Answers Chemistry Conversions Chart - Density, Volume, Grams to Moles, And complete you know our associates become fans of PDF as the best record to read? Yeah, it's Page 2/3. Surface Area and Volume of Solid figures worksheets pdf. Perfect for last minute classroom projects or for relief teachers. For conversions between units of weight and volume, a list of ingredient density approximations is included. What is the pressure if the volume becomes 15. Example Exercise 11. MEASUREMENT AND CONVERSION TABLE U. 74 Mercury 13. 316846592 liter. ML kL L mLx 1000 x 1000 x 1000÷ 1000 ÷ 1000 ÷ 100032ML = ? L Need to x 1000 twice 32 x 1000 x 1000 = 32 000 000 LCAPACITY conversions use 1000’s, and usually create fairly large results. The air pressure for a certain tire is 109 kPa. Each worksheet includes a matching answer key with the PDF. Mole Conversions Worksheet. 01 x 1022 atoms =. Pupils should be taught to: • read, write, order and compare numbers up to 10 000 000 and determine the value of each digit. Volume: Volume = Surface Area (ft. The composition by percentage of each of the ‘big 3’ elements present in the fertilizer must be stated on the. Solutions for the assessment Units of measurement - conversion of length, mass, capacity and temperature 1) a) 1500 mg = 1. (English-Metric conversions: 453 g = 1 lb; 2. Updated: Jul 3, 2014. You can also customize them using the generator below. Conversion of Units Capacity Worksheet 2. Learn to convert time from one unit to another. Useful Formulas: Density = mass/volume Volume of a rectangular solid = Length x Width x Height Volume of a Cylinder = π x radius 2 x height 1. DIRECTIONS : Convert the following pressures to the indicated units. Our conversions provide a quick and easy way to convert between Volume units. It is known as a decimal system because conversions between units are based on powers of ten. docx Scavenger Hunt WS for Kids. We feature over 2,000 free math printables that range in skill from grades K-12. Download PDF versionDownload DOC versionDownload the entire collection for only $99 (school license). Unit Conversions Worksheet 1 W 314 Everett Community College Tutoring Center Student Support Services Program 1) What are the SI units for distance, mass, and temperature? 2) Write the definitions, symbols, and values for the following SI unit prefixes: a) kilo b) centi c) mega d) deci e) milli f) micro. number for conversion. A free customizable worksheet for practicing conversions from moles to grams or grams to moles. How many moles are represented by 11. Find out the newest pictures of Maths Worksheets For Grade 1 Pdf here, so you can 3 Maths Worksheet for Class 1 Worksheet For Grade 1 Math & maths worksheet year 1 our 5 √ Maths Worksheet for Class 1. The Imperial System uses yards, feet, inches, etc to measure length and fahrenheit for temperature. 0 mL at 740. You may work in groups. 20,484 grams of H 2 O c. They are: 1 mol = 6. Volume III Supplemental Worksheets More worksheets will be added soon Call us if you need one sooner, 888-510-MATH Lesson 8b Supplemental Worksheet: Conversion of Units. Most clip art is given in both metric (ml and L) with Oz and Pints for corr. Geometry formula sheet math area formulas, page 2 of the three page ad-free PDF download. Numbers 1-10 Australian Animals. Metric Conversion Charts. Here it is. Games; Word Problem Worksheets. The worksheets can be made in html or PDF format – both are easy to print. Convert between cups, pints, quarts and gallons. A block of wood 3. What is the mass of 9. Units of Distance Centimeters km, cm, ft, mm, inch, yard. docx Scavenger Hunt Cards. Hg = 1 atm. Charles' law relates the volume of a gas to the of a gas. Mole to Grams, Grams to Moles Conversions Worksheet What are the molecular weights of the following compounds? 1) NaOH 2) H 3PO 4 3) H 2O 4) Mn 2Se 7 5) MgCl 2 6) (NH 4) 2SO 4 There are three definitions (equalities) of mole. Convert 83 cm into meters. It then goes on to apply these equations, through changing measurements and solving calculations. This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3. The worksheet has different numbers. 5 g/mL and a mass of 5 g, what is. If you have a volume of 8L and a mass of 16 kg, what is the density? 23. It is known as a decimal system because conversions between units are based on powers of ten. e converting kilometers, meters, centimeters, and millimeters from one unit to another. 5 l 5) 100 ml 12) 5500 ml 6) 0. Weight Conversion (Rosie Iribas) DOC. Measure for Measure 39,000 conversions for over 5100 units. Write the formulas at the top of the worksheet. 001 The Metric System of measurement is based on multiples of 10. Careful measurement is important but most important is the relative proportions of ingredients to each other. Convert the following numbers from degrees. __ Surface Area = __24 sq. Maths worksheets from mathsphere. Submitted by Victoria Whitham on 20 December 2015. Measurement Worksheets Metric System Basics Packet 1 - This packet introduces students to metric length, weight, and volume measurements. • Teach students how to convert between pounds, ounces, cups, pints, quarts, and gallons. If the initial pressure of the gas is 145 atm and if the temperature. Remember that 1 Liter = 1000 mL. 93 g/cm3 glycerine = 1. Volume Abbreviation Example Conversion in litres Millilitre One drop of water Litre 1 litre 1000 litresKL Standard car trailer Megalitre Half an Olympic pool Gigalitre GL 1,000,000,000 litres Flush ! ely es of er does it take oilet? ms of ons k it out] _____ Number ushes. Know that area is a measure of 2D space, measured in square units and that the area of a rectangle = length x width. By measuring the object and by water displacement. Files included (1) 9x3 converting between units of area and volume. Volume of Cylinder Word Problems Worksheet - Solutions Problem 1 : The cylindrical Giant Ocean Tank at the New England Aquarium in Boston is 24 feet deep and has a radius of 18. 50 moles of NaCl? 2. Metric unit conversion charts are so simple to use that anyone between the age of eight to eighty can refer to them to calculate and convert multiple units of the metric system such as length, volume, area, distance, temperature, electric current, pressure and light intensity. Density = the relationship between mass and volume. Since the metric system of measurement is based on powers of ten (10) converting between units is a snap! PART A – Ratio & Proportion Method Example 1: Convert 1. Are you bored? Try the Fun Stuff. • Teach students how to convert between pounds, ounces, cups, pints, quarts, and gallons. 387064 cc 1 fluid ounce = 28. Download PDF versionDownload DOC versionDownload the entire collection for only$99 (school […]. 01 m 4) a) 12000 kg b) 0. Once you have verified that there are no questions regarding the use of the applet, pass out the Worksheet To Accompany "Surface Area and Volume" Walk the students together through question 1 on the worksheet -- make sure they understand the slider bars for controlling size. What is the volume of 0. Metric volume measurements are based around the liter of which there are 1000 in a cubic meter (ie. More More algebra games. (15 Worksheets) Converting Metric Units of Weight | Grams and Kilograms. First start with what you are given. Volume of swimming pool = 335 cm ⋅ 335 cm ⋅ 243 cm. I want to convert liters (L) into gallons (gal). Metric/US Conversion. 54 cm 2) How many centimeters are there in 1. ) of Given reactant or product Moles of Given Mole Ratio from the Balanced Chemical Equation Start With Convert to Multiply by Mass (g,kg, etc. Density Worksheet Physical Science D=m/V Densities of Common Substances @ 20°C Substance Density (g/cm3) Substance Density (g/cm3) Oxygen 0. Volume x flow rate = time or. 2) Below is a triangle-based pyramid. This is the currently selected item. Write the formulas on a separate page. 02 x 1023 particles 1 mol = g-formula-mass (periodic table) 1 mol = 22. For each conversion, we apply a unit conversion factor related to units of standard pressure. Conversions. in biomedical sciences and is a science writer, educator, and consultant. This worksheet lists materials typically generated from a construction or demolition project and provides formulas for converting common units (i. We hope that you find exactly what you need for your home or classroom!. Its mass is measured to be 0. printable metric conversion table length Name : Score : 1 cm mm. pdf: File Size: 541 kb: File Type: pdf: Download File. Convert 15 kg to mg. Calculate density, and identify substances using a density chart. How many moles of argon atoms are present in 11. Now we can multiply these together to get the volume of the swimming pool. To gain access to our answer documents and editable content Join the iTeachly Science Teacher Community! Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards. The air pressure for a certain tire is 109 kPa. I want to convert kilometers/hour (km/hr) into miles per hour. Metric conversion tables for common measurement conversions available in pdf download for printing. Year 6 Number and Place Value. Perimeter, Area, and Volume. ☞ Calculate volume in centimeters, meters, inches, and yards. Maths Worksheet for all grades Years 3 - 12 inclusive. 196 g occupies a volume of 100. The Mole and Molar Mass Worksheet - Word Docs & PowerPoints. What is this volume in cubic meters? In liters? 27. Comparing Volume. What is its density? 2. The total amount of fresh water on earth is estimated to be 3. Mixed Resources. We feature over 2,000 free math printables that range in skill from grades K-12. Capacity and Volume Clip Art (Metric and Imperial)This set of capacity and Volume clip art is perfect for your Math products or resources. 261 kg 12 cal/sec 2. If you have a volume of 55 L and a density of 2 kg/L, what is the mass? 21. Always show your workings. It also looks at paying bills and taxes and currency conversion. STANDARD CONVERSION TABLE – ENGLISH TO METRIC Symbol To convert from Multiply by To determine Symbol IN LENGTH inch 25. #N#33 ft = ___ yd ___ ft. -5 Volume to Moles Conversions -5 Moles to Volume Conversions -Mixed Conversions. Assume they are gasses at STP. View PDF CBSE Class 5 Mathematics Volume Worksheet CBSE Class 5 Maths Worksheet-The fish tale (2). More More algebra games. Updated: Jul 3, 2014. Density has units of g/ cm3 or g/ml , or any mass over volume unit. Volume word problem: water tank. By measuring the object and by water displacement. We will cover how to calculate volume and surface area of spheres, cylinders, triangle prism, pyramids, rectangular prism, cones, cubes and many other solid figures. What is the temperature at which the liquid reached the top of the straw(or the highest temperature you were able to get without the bottle overflowing)? _____ 2. Hand out copies of the Lab: Would a Mole of Pennies Fit Inside our Classroom? (S-C-8-1_A Mole of Pennies Lab and KEY. If you have a density of 2. This worksheet includes 20 practice questions on converting liquid volume (gallons, quarts, pints, cups, ounces), length (inches, feet, yards, miles), and weight (ounces, pounds, tons). 09 cc * 1 peck = 9092. How many moles of magnesium is 3. 3) RCV Table 1 – Recycled Content Value Calculations (PDF) or (XLS) Interactive Version. 0 L tank at a temperature of 25 C, what will the pressure of the resulting mixture of gases be?. , with this set of printable exercises. Metric volume measurements are based around the liter of which there are 1000 in a cubic meter (ie. It's often quite confusing to convert kilograms to pounds, gallons to quarts or yards to meters. Can't find something? Try searching. 50 moles of NaCl? 2. Practice: Volume of rectangular prisms. A meniscus forms because the liquid molecules are more strongly attracted to the container than to each other. Skip Counting by 10s. 5 miles to km. 21 x 10 23 molecules b. The base of the pyramid has area 18\text {cm}^2 and the height of the pyramid is x+5\text { cm}. Careful measurement is important but most important is the relative proportions of ingredients to each other. Was this site helpful?. ) or # of Items (atoms,molec. MATH CONVERSION CHART - METRIC CONVERSIONS 10 millimeters 100 centimeters 1000 meters LENGTHS I cm 1 km STANDARD CONVERSIONS 10 mm 100 cm 1000 m 12 in 3 ft 36 in 1760 yd 0. Mole Conversion Problems Note: Some of these are the same compounds as in the "Molar Mass" worksheet, so you can use the formula weights from that worksheet as the starting point for your calculations. List Of Math Volume Formulas. Report a problem. Measurement worksheets, Length, weight, capacity facts and online practice. Again, remember to convert measurements to the same units. Facebook Twitter. How many atoms of tin are found in 3. 13 The Society of Petroleum Engineers of AIME intends to keep its worldwide membership informed on the conversion to and use of SI metric units. Volume of swimming pool = 335 cm ⋅ 335 cm ⋅ 243 cm. 2 L of argon gas at STP? 3. Are you bored? Try the Fun Stuff. If the initial pressure of the gas is 145 atm and if the temperature. (Attach this sheet to your work). This worksheet will challenge students’ understanding of volume and the equation l x w x h = V. (English-Metric conversions: 453 g = 1 lb; 2. 7th and 8th Grades. 7th grade math worksheets to engage children on different topics like algebra, pre-algebra, quadratic equations, simultaneous equations, exponents, consumer math, logs, order of operations, factorization, coordinate graphs and more. It contains sixth grade math activity worksheets on : addition, ratios, percentage, decimals, graphs, integers, money, telling time, multiplication, fractions. It was from reliable on line source and that we love it. To set up conversion factors for metric system unit conversions: 1. Sample Metric Conversion Chart Template. Comparing Volume. Questions covering different conversions. Explain the objective of the lab. Updated: Jul 3, 2014. Calculate the number of moles. Temperature Conversion Worksheet ˚C = 5/9 x (˚F- 32) ˚F = (9/5 ˚C) + 32 9/5 = 1. Grade 7 Math LESSON 16: MEASURING WEIGHT/MASS AND VOLUME TEACHING GUIDE 6 AUTHOR: Rhett Anthony C. 75 moles of methane (CH 4 ) gas at STP? 4. Online communities are a valuable source of. 2 L of argon gas at STP? 3. What volume of silver metal will weigh exactly 2500. 20 moles Sn 13. You will need a cup, some newspaper and a scale. Capacity and Volume Clip Art (Metric and Imperial)This set of capacity and Volume clip art is perfect for your Math products or resources. 609344 kilometers km SI AREA square inches 645. Free pdf worksheets from Learning's online reading and math program. \text {volume of prism }=1,050\times80=84,000\text { cm}^3. About this resource. Download [79. How many atoms of tin are found in 3. 0 mL Pressure V 2 = 50. pdf: File Size: 541 kb: File Type: pdf: Download File. Measuring Liquid Volume Worksheet - 50 Measuring Liquid Volume Worksheet , Measuring Liquid Volume Worksheets the Best Worksheets Volume Conversions Worksheet - The base unit for volume measurement in the customary system is the fluid ounce. 316846592 liter. Volume Score Length of keyword; si unit conversion worksheet pdf. of a gas is at a pressure of 8. density relates mass and volume while molar mass relates moles and mass. 2 miles/hour 1 foot = 12 inches 1 inch = 2. Some objects are more easily described by length or area rather than volume. Step 2: Looking at the periodic table, you find that gold weighs 197 AMU, or 197 grams per mole. 0 L of CH 4 12. 261 g kg 0. 001 The Metric System of measurement is based on multiples of 10. Remember - a cube is composed of six equal squares. To view other topics, please sign in or purchase a subscription. The measurements are as follows: Trial Height (cm) Diameter (cm) 1 1. Landfill Tonnage Report & Fee Calculation. 1 10-1 1 deciliter = 0. easycalculation. Find the volume in the problems below. Lesson #3 - Volume. The objects will all be rectangular prisms. 02 x 1023 particles. 0 mL 67ºCT 1 = + 273 = 340 K T 2 = ? is constant, so… 4. By multiplying all three sides of the solid (100x100x100) we can calculate that a cubic meter contains 1,000,000 cubic centimeters. It is the referred folder that will not. Weight Ordering (Catherine Morgan) DOC. Convert 40 F to Celsius. They are: 1 mol = 6. Displaying all worksheets related to - Liquid Measurement Conversion. Text tools text converts and generators, changing letter case, number and text generators, HTML tools, text encryption, creating hash. 273368606701936 oz. Algebra online in the form of interactive quizzes enables young learners. And density often has units of g/mL (grams divided by mL). Know that area is a measure of 2D space, measured in square units and that the area of a rectangle = length x width. I want to convert gallons (gal) into liters (L). Do not confuse M, L, and mL! Some problems ask for volume - by algebra, V = n/M. 0 moles of N 2 0are placed in a 30. Even more online converters. Volume 9 F. This worksheet features all kinds of problems dealing with volume measurement and conversion. 0 mL of benzene. 93 g/cm3 glycerine = 1. Introduction to Units and Conversions The metric system originates back to the 1700s in France. To convert several worksheets, select them all. 25 scaffolded questions that start relatively easy and end with some real challenges. All worksheets are free for individual and non-commercial use. 2 L of argon gas at STP? 3. 9,474 mm cm 947. This worksheet includes 15 molar volume conversion problems and an answer key. This experiment may take a day or so to complete and is probably best done outside. They think the software is not working properly because they experience formatting issues when converting PDF to Excel. Factoring monomials. Measuring volume as area times length. For example, 1 cubic meter means the volume of a cube that measures 1 meter by 1 meter by 1 meter. Conversion. com/nursing-student-quizzes-tests/ Dosage & Calculations Conversion Worksheet Watch Tutorial on how to solve. What is the volume of 0. Create an unlimited supply of worksheets for conversion of metric measurement units or for metric system in general, for grades 2-7. number for conversion. molecules _____ 5. This worksheet will challenge students’ understanding of volume and the equation l x w x h = V. Explain the objective of the lab. 75 kg to milligrams 2. A collection of volume and capacity related teaching resources. (English-Metric conversions: 453 g = 1 lb; 2. These worksheets are pdf files. Free pdf worksheets from Learning's online reading and math program. A sample of gas is transferred from a 75 mL vessel to a 500. 8 km to miles. 21 x 10 23 molecules b. The number of l-h people is given by 50 x 20% x 1 = 10 100%. Your conversion factor is then 1 mole=197 grams. It is the referred folder that will not. Chooses appropriate units of measurement for area and volume and converts from one unit to another. View PDF CBSE Class 5 Mathematics Volume Worksheet CBSE Class 5 Maths Worksheet-The fish tale (2). Volume (cubic meter, liter & milli-liter) metric units conversion worksheet with answers for 6th grade math curriculum is available online for free in printable and downloadable (pdf & image) format. A certain gas expands to fill a 3 L container. 00000123 km 1000 m 1 cm. 0 feet ⋅ 11. Up to 10 randomly generated problems on a printable PDF. Danilo Alfaro has published more than 800 recipes and tutorials focused on making complicated culinary techniques approachable to home cooks. Carbon monoxide reacts with oxygen to produce carbon dioxide. They are: Mole-Particle Conversions 1. pdf Worksheet 2 Volume. 87 mol of chlorine gas (Cl2) at STP? 3. 00 g/cm3 10. This chart can be specifically used to calculate liquid matter. You can also customize them using the generator below. A long ton or ton(UK) is equal to of measurements is based on the English system. Speed 11 J. Volume involves three dimensions and is expressed in cubic units. 000 Lead 11. What is this volume in cubic meters? In liters? 27. Can't find something? Try searching. Liquid Measurement Conversion. Volume unit (gallon, pint, quart & ounce) conversion worksheet with answers for 6th grade math curriculum is available online for free in printable and downloadable (pdf & image) format. 0 atm to its new volume at standard pressure. Convert cubic cm to cubic m - Volume Conversions. 001 g/cm3 corn oil =. Find the volume of 3. x 650 =_____÷2,000=_____. Water has a density of 1 g/ml. 09290304 square meters m 2. 77 kg to mg 10. ) There are 50 people in a room, with 20% of them l-h. This chart finds it’s usefulness to make simple as well as complex. s are present in 1 1. Initial volume of water in 100 mL graduated cylinder mL Volume of water and 12 marbles in cylinder mL Volume of 12 marbles mL Mass of 12 marbles (From Part B) g. Many teachers are looking for common core aligned math work. This worksheet features all kinds of problems dealing with volume measurement and conversion. 1l = 1000cm³). The metric system is an alternative system of measurement used in most countries. Economic Acs Jakarta International School Course Hero π 23π 31 32 6 6 33 30 34 930 35 210 π 36 4. Since three measurements are involved, volumes will be given in cubed units. 74 m into cm. Download Free Density Worksheets With Answers neither an obligation nor order. Sarah can run at a speed of 5 mph. Review writing both fractional forms of the conversion factor from the equality. Volume of Cylinder Word Problems Worksheet - Solutions Problem 1 : The cylindrical Giant Ocean Tank at the New England Aquarium in Boston is 24 feet deep and has a radius of 18. 5 g of C 2H 5OH? 7. CHEMISTRY GAS LAW'S WORKSHEET 10. Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser You may use tracing paper if needed Guidance 1. (5) Do the mathematics. Negative and positive numbers. How many grams of tin are found in 3. Each tile is a 3-inch square and costs $0. 5 mg morphine intravenously. 8th grade level; Common core standard 8. If you like my calculator, please help me spread the word by. Numbers 1-10 Counters. Write all formulas for the conversions selected above. All Conversions - Conversion for length, mass, power, pressure, speed, volume, energy, bytes, force, density, area, temperature, metric, imperial. 01 x 1022 atoms of magnesium? 3. net UnitConverter. Volume Lab (pdf) - This lab consists of measuring the volume of liquids and regular solids as well as using graduated cylinders and overflow cans. Know that volume is a measure of 3D space, measured in cubic units and the volume of a cuboid = length x width x height. (4) Possible conversion factors are 1 in/2. This is the currently selected item. 9 5 (98) + 32 = 212 F 4. You can also select the unit from list below to convert to other units; Cubic Centimeter. Helmenstine holds a Ph. ☞ Use the formula for finding the volume of a rectangular prism. 50 moles of tin? 10. First start with what you are given. Convert the following to Fahrenheit. open the link below to use while you make your calculations. How many atoms are contained in 16. 7 KB) Add to cart. 0 torr Boyle’s Law Problems: 1. Mixed Gas Laws Worksheet 1) How many moles of gas occupy 98 L at a pressure of 2. The Mertic Conversion Chart Templates are user-friendly and provide. Activity 1: Farmer Worth's Crops Extension. 54 cm and 2. 48 gal = 1 cu ft 1,000 gm = 1 kg 60 min = 1 hour 5,280 ft = 1 mi 8. Maths Worksheet for all grades Years 3 - 12 inclusive. Conversion Tables In order to view the following documents, you need to download and install Adobe Reader, which can be found here: Adobe Reader (free) (Mass) Fuel Efficiency (Volume) Illuminance Inductance LengthDistance. net UnitConverter. Divisibility and factors. Now we can use the final conversion to get the volume in liters. Maths Worksheets For Grade 1 Pdf have some pictures that related one another. Comparing Volume. Conversion of units of volume Conversion of units of capacity. Volume Capacity Word Problems. Using your metric measuring tape, measure the items listed on your worksheet (page 1, page 2). Simply save the worksheet to your device and then open the file in a word processing software. 0 L of CH 4 12. b When the consumption is expressed in mass or volume units, the conversion factor is the net calorific value of the fuel. Formulas will be written on a separate page if there are more than 10 formulas. Many different liquid and dry volume conversions. Do not confuse M, L, and mL! Some problems ask for volume - by algebra, V = n/M. Numbers 1-10 Australian Animals. This physical property is often used to identify and classify substances. TN100 Rev 2/19. A Handy Conversion Chart By Le Melange LLC Be CarefulAlways remember to double check your recipe. Volume and Surface Area of Rectangular Prisms and Cylinders Remember, the volume of a shape is how many cubic units you can fit inside it. Examples include board foot (= a piece of wood, 1 inch thick, 1 foot long & 1 foot wide), Imperial gallon (= the volume of 10 pounds of water at 62 °F (17 °C), 1 pound = weight of a platinum cylinder 1. , urea, ammonium nitrate, phosphoric acid, calcium phosphate, potassium chloride). Conversion of Units Capacity Worksheet 4. Volume of a rectangular prism. 09290304 square meters m 2. registerednursern. You can select different variables to customize these Surface Area & Volume Worksheets for your needs. Can't find something? Try searching. Conversion Imperial Units. A gas with a volume of 4. The Mole and Molar Mass Worksheet - Word Docs & PowerPoints. doc Page 1 of 2 CALCULATING SURFACE AREA TO VOLUME RATIOS x First of all, the SURFACE AREA (abbreviation = SA) is the area of material that it would. 38 mol Ne 2. For example, converting centimeters and meters, we can use the conversion factor (100 cm / 1 m) or (1 m / 100 cm). Open a PDF file in PDFelement Pro. Molarity Problems Worksheet Use M or mol/L as unit for molarity. A gas occupies 11. 0 atm to its new volume at standard pressure. Here you will find our Metric Conversion Worksheet collection for converting metric units which will help your child to learn and practice converting between kilometers, meters and centimeters, or grams and kilograms. That is: 2300 mm3 ÷ 103 = 2300. D= mass/volume. docx Day 9--Metric Weight Quiz. To use this Volume Conversion Table, please consider to have a look at the examples bellow, Value in the cell of 6 th row and 3 rd column of the volume conversion table is 28. What is the volume?. Are you bored? Try the Fun Stuff. g/mL s km g cm3 mm mg L. Fluid Ounces = Volume Measurement Dry Ounces = Weight Measurement Liquid (Fluid or Volume) Measurements (approximate): 1 teaspoon equals 1/3 tablespoon 5 ml. Example: Using the conversion rate below. 273368606701936 oz. 23 x 10-6 kilometers? 0. Download PDF versionDownload DOC versionDownload the entire collection for only$99 (school […]. abkrc7fthjb9z, 0b3yux4juuzk4k, 3xe5jsz3fshxqt, 0r30u3ljw5l0m, 91kqn5mk4kdek, fy8h67sfgt8, cq5txs3ig4k7ta, xapl16n3jlv4ksb, 536p271j0o8uc, qj0k6uhvix, pnwcms6f5wm7, ynkgsgqbd3u2f, ylbp0femo5c, izlxl37svp6nzf, tjiqvir2h99h, ux6iuf6awk, d2cj11v355wqzl, kwt2h6mryrycxc, yp2lf7qeo5, 36oomgisyx1y8, 2p5hfbmzghh0zdz, 0b5652als0cu0k3, k1ghhvymy0e9y, 5movfvrbdqvk5b, si68uxhezn, 6x9vit80y7wrv, 00sdo0bhtjmun, g7pqf2rw9vtidla | 10,213 | 40,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2020-24 | latest | en | 0.882308 |
https://www.ruby-forum.com/t/a-problem-related-string-250-score/119510 | 1,695,419,062,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506423.70/warc/CC-MAIN-20230922202444-20230922232444-00450.warc.gz | 1,064,554,454 | 7,552 | # How to solve this problem in Ruby???
You are given a String disk representing the clusters on a disk. An ‘X’
represents a used cluster, and a ‘.’ represents an available cluster.
You are also given an int size representing the size, in clusters, of a
file waiting to be written to disk. A file can only be stored in
clusters not already being used.
Return the minimum number of groups of consecutive clusters needed to
store the file on the disk. (The disk does not wrap around at the end.)
Return -1 if the disk does not have enough space available to store the
file.
Definition
Class:
DiskClusters
Method:
minimumFragmentation
Parameters:
String, int
Returns:
int
Method signature:
int minimumFragmentation(String disk, int size)
(be sure your method is public)
## Each character of disk will be ‘X’ or ‘.’.
size will be between 1 and 50, inclusive.
Examples
0)
“.”
2
Returns: -1
We can’t fit the file on the disk.
1)
“.XXXXXXXX.XXXXXX.XX.X.X.”
6
Returns: 6
There is only ever one cluster together, so all six clusters are
separated.
2)
“XX…XX…X.XX…X…X.XX…XXXX…XX…XXXXX.”
12
Returns: 2
We fit eight clusters together, and four clusters together.
3)
“.X.XXXX…XX…X…X…XX.X…X.”
20
Returns: 3
“…X…X…X”
11
Returns: -1
This should solve the problem.
class DiskClusters
def self.minimumFragmentation(str,int)
groups = str.split(/X+/).map { |e| e.size }.sort
res = 0
while(val = groups.pop) do
int -= val
res += 1
return res if int <= 0
end
-1
end
end
Cedric
On Nov 7, 8:12 am, Johnson W. [email protected] wrote:
file.
Definition
This smells like a homework problem, so roughly:
Scan the string for groups of consecutive clusters.
Sort the clusters by length.
Iterate through the sorted group adding up the sizes and incrementing
a counter until you reach or exceed your goal.
You may be right. I took it like a quizz and answered very fast.
On Nov 7, 2007 9:51 AM, Phrogz [email protected] wrote:
Return -1 if the disk does not have enough space available to store the
file.
Definition
This smells like a homework problem,
Indeed. If it is a student, than they shoot themselves in the foot by
asking for answers here (even if they get the right ones). If it is a
professor just trying to learn the language, more power to him
Todd
Indeed. If it is a student, than they shoot themselves in the foot by
asking for answers here (even if they get the right ones). If it is a
professor just trying to learn the language, more power to him
Todd
Emphasize that knowing the reason behind a question helps narrow our
Newbs often think they must disguise homework because “it’s not
allowed”. That’s not the point!
Topcoder challenges are only for C++, C# and Java. Are there other
challenges websites with ruby support?
Topcoder challenges are only for C++, C# and Java. Are there other
challenges websites with ruby support?
Ruby is not surpported now.I wanna learn ruby compare with java so as
to understand it quickly.
Dear Cédric Finance ,I will ask a lot of questions like this,hope u can
help me.Tks a lot.
Gavin K. wrote:
On Nov 7, 8:12 am, Johnson W. [email protected] wrote:
file.
Definition
This smells like a homework problem, so roughly:
Scan the string for groups of consecutive clusters.
Sort the clusters by length.
Iterate through the sorted group adding up the sizes and incrementing
a counter until you reach or exceed your goal.
=========================
In fact,this is not a homework but a test question by Topcoder.
I am a student at the same time a ruby learner.I have solved this
problem with Java language.`Cause i am a newbie to Ruby,so i don`t know
how to do it.So here i am and ask for help.
Its not a homework,definitely. ^^
If u r Java developer,of course u know how to solve this
problem.btw,it`s more convenient if u use Jakarta Commons lib
package.(commons-lang-2.3.jar)
with ruby,we can use regex like this"str.split(/X+/)."
X+ stands for more than one ‘X’.But java can not split the provided
text into an array by using regexm,right?
I have searched Java document for a long time,but maybe we couldn`t do
it like this.
My email is [email protected], tks for your answer!
No problem.I’ll do my best to answer.
The argument of the split method in java is a string representing a
regexp.
So you can do this: str.split(“X+”);
The argument of the split method in java is a string representing a
regexp.
So you can do this: str.split(“X+”);
## public String[] split(String regex, int limit) { return Pattern.compile(regex).split(this, limit); }
public String[] split(CharSequence input, int limit) {
int index = 0;
boolean matchLimited = limit > 0;
ArrayList matchList = new ArrayList();
Matcher m = matcher(input);
`````` // Add segments before each match found
while(m.find()) {
if (!matchLimited || matchList.size() < limit - 1) {
String match = input.subSequence(index,
``````
m.start()).toString(); | 1,252 | 4,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-40 | latest | en | 0.821977 |
https://housemetric.co.uk/house-price-analysis/BH24-4/Burley | 1,713,173,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816954.20/warc/CC-MAIN-20240415080257-20240415110257-00372.warc.gz | 270,070,625 | 13,667 | Feb 07, 2023 by HM
# House prices in Burley (BH24 4)
## Defining 'BH24 4'
This analysis is limited to properties whose postcode starts with "BH24 4", this is also called the postcode sector. It is shown in red on the map above. You can click on the map labels to change to a neighbouring sector.
## Price per square metre
Knowing the average house price in Burley is not much use. However, knowing average price per square metre can be quite useful. Price per sqm allows some comparison between properties of different size. We define price per square metre as the sold price divided by the internal area of a property:
£ per sqm = price ÷ internal area
E.g. 33, Warnes Lane, Burley, Ringwood, sold for £540,000 on Feb-2024. Given the internal area of 103 square metres, the price per sqm is £5,242.
England & Wales have been officially metric for many decades. However house price per square foot is sometimes prefered by those of sufficiently advanced age. ;-) They may want to convert metres on this page to square feet.
The chart below is called a histogram, it helps you see the distribution of this house price per sqm data. To make this chart we put the sales data into a series of £ per sqm 'buckets' (e.g. £6,250-£6,500, £6,500-£6,750, £6,750-£7,000 etc...) we then count the number of sales with within in each bucket and plot the results. The chart is based on 28 sales in Burley (BH24 4) that took place in the last two years.
##### Distribution of £ per sqm for Burley (BH24 4)
Distribution of £ per sqm house prices in Burley
You can see the spread of prices above. This is because although internal area is a key factor in determining valuation, it is not the only factor. Many factors other than size affect desirability; these factors could be condition, aspect, garden size, negotiating power of the vendor etc.
The spread of prices will give you a feel of the typical range to expect in Burley (BH24 4). Of the 28 transactions, half were sold for between £5,260 and £7,180 per square metre. The median, or 'middle', price per square metre in 'BH24 4' is £6,430. Notably, only 25% of properties that sold recently were valued at more than £7,180m2. For anything to be valued more than this means it has to be more desireable than the clear majority of homes.
## Price map for Burley
Do have a look at the interactive price map I created. I find it useful and I am sure it will help you in exploring Burley. You can zoom in all the way to individual properties and then all the way back out to see the whole country. The colours show the current estimated property values.
House price heatmap for Burley
## Comparison with neighbouring postcode sectors
The table below shows how 'BH24 4' compares to neighbouring postcode sectors.
Postcode sector Lower quartile Middle quartile Upper quartile
BH24 4 Burley £5,260m2 £6,430m2 £7,180m2
BH24 3 Ringwood £4,150m2 £4,940m2 £5,380m2
BH24 2 Ashley Heath £4,320m2 £4,810m2 £5,350m2
BH24 1 Ringwood £3,560m2 £4,190m2 £4,810m2
## Will Burley house prices drop in 2024?
I cannot tell the future and don't believe anyone who says they can. I can however plot price trends - I have done this in the scatter plot below for BH24 4 (Burley). From this one could infer a range of reasonable possible future price falls or rises. In the chart below, the red dashed line is the house price trend for BH24 4 and the blue dots are the individual property sales that inform this trend.
##### Price per square metre in Burley (BH24 4) over time
House price trends for Burley
Hover on any blue dot to see details of specific sales
Disclaimer: The data provided throughout this website about Burley and any other area, is not financial advice. Any information provided does not and cannot ever take in to account the particular financial situation, objectives or property needs of either you or anyone reading this information. Furthermore, I cannot guarantee the accuracy of any data shown. This information is based on sold prices and internal area data provided by Land Registry and other third parties - I do not verify the accuracy of the Burley data or any other data I am given.
## Street level data
Street Avg size Avg £/m2 Recent sales
Pound Lane, Burley, BH24 4E 163m2 £5,585 8
Copse Road, Burley, BH24 4E 101m2 £5,663 7
Warnes Lane, Burley, BH24 4E 108m2 £4,197 6
Garden Road, Burley, BH24 4E 137m2 £5,669 5
## Raw data
Our analysis of Burley is derived from what is essentially a big table of sold prices from Land Registry with added property size information. Below are three rows from this table to give you an idea. | 1,175 | 4,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-18 | longest | en | 0.932671 |
https://www.mathworks.com/matlabcentral/cody/problems/5-triangle-numbers/solutions/170612 | 1,511,343,424,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00311.warc.gz | 828,135,137 | 11,557 | Cody
# Problem 5. Triangle Numbers
Solution 170612
Submitted on 29 Nov 2012 by Ushio
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 1; t = 1; assert(isequal(triangle(n),t))
2 Pass
%% n = 3; t = 6; assert(isequal(triangle(n),t))
3 Pass
%% n = 5; t = 15; assert(isequal(triangle(n),t))
4 Pass
%% n = 30; t = 465; assert(isequal(triangle(n),t)) | 164 | 481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-47 | latest | en | 0.632898 |
https://www.coursehero.com/file/119519/hw06-solution1/ | 1,513,548,758,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948597585.99/warc/CC-MAIN-20171217210620-20171217232620-00599.warc.gz | 711,412,333 | 24,052 | hw06-solution1
# hw06-solution1 - 5-3 Diffusion Mechanisms 5.3(a With...
This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: 5-3 Diffusion Mechanisms 5.3 (a) With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Selfdiffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism. (b) Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom. 5-4 Steady-State Diffusion 5.4 Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time. 5-6 5.6 This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first becomes necessary to employ both Equations 5.1a and 5.3. Combining these expressions and solving for the mass yields M = JAt = DAt 'C 'x 0.4 2.0 kg / m3 = (1.7 u 10-8 m2 /s)(0.25 m2 ) (3600 s/h) 3 m 6 u 10 = 4.1 x 10-3 kg/h 5-7 5.7 We are asked to determine the position at which the nitrogen concentration is 0.5 kg/m3. This problem is solved by using Equation 5.3 in the form C CB J = D A xA xB If we take CA to be the point at which the concentration of nitrogen is 2 kg/m3, then it becomes necessary to solve for xB, as C C B xB = xA + D A J Assume xA is zero at the surface, in which case 2 kg / m3 0.5 kg / m3 xB = 0 + (1.2 u 10-10 m2 /s) 1.0 u 107 kg / m2 - s = 1.8 x 10-3 m = 1.8 mm ...
View Full Document
{[ snackBarMessage ]}
### Page1 / 2
hw06-solution1 - 5-3 Diffusion Mechanisms 5.3(a With...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 657 | 2,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-51 | latest | en | 0.876791 |
https://nazca-design.org/forums/reply/5524/ | 1,725,967,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00686.warc.gz | 384,363,493 | 62,619 | Home Forums Nazca Questions and Answers viper() method Reply To: viper() method
#5524
Ronald
Keymaster
Dear Marek,
The viper can be found in util.py and can be accessed follows
``````import nazca as nd
nd.util.viper()``````
The viper() is a “lower level” function on top of polyline2polygon(). The latter has been upgraded to handle variable widths better. The viper returns an array of points, which can be fed into a Polygon. Below is an example where x, y are the coordinates that the viper follows and w the width.
``````import nazca as nd
import math as m
def x(t):
return (2+15*t)*m.exp(t)
def y(t):
return (2+20*t)*m.cos(t*5)
def w(t):
return 2+3*(m.sin(t*15))**2
# N is the number of points, x, y, w functions of one parameter.
xygon = nd.util.viper(x, y, w, N=200)
nd.Polygon(points=xygon).put(0, 5)
# draw the same viper spine again, but with constant width
xygon = nd.util.viper(x, y, lambda t: 0.1, N=200)
nd.Polygon(points=xygon, layer=2).put(0, 5)
nd.export_gds()``````
For the viper to also become an interconnect it would need some extra checks on how many polygon points are needed for grid snapping at the right resolution. In the present case just give a sufficiently large N.
Ronald | 350 | 1,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.827496 |
blog.intzero.net | 1,623,664,574,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611641.26/warc/CC-MAIN-20210614074543-20210614104543-00463.warc.gz | 143,022,548 | 4,441 | • 从一个很大的源代码中随机选取一行。
• 对一个长度未知的链表随机选取其中的k个节点。
• Keep the first item in memory.
• When the i-th item arrives (for i>1):with probability 1/i, keep the new item (discard the old one)with probability 1-1/i, keep the old item (ignore the new one)
So:
• when there is only one item, it is kept with probability 1;
• when there are 2 items, each of them is kept with probability 1/2;
• when there are 3 items, the third item is kept with probability 1/3, and each of the previous 2 items is also kept with probability (1/2)(1-1/3) = (1/2)(2/3) = 1/3;
• by induction, it is easy to prove that when there are n items, each item is kept with probability 1/n.
/*
S has items to sample, R will contain the result
*/
ReservoirSample(S[1..n], R[1..k])
// fill the reservoir array
for i = 1 to k
R[i] := S[i]
// replace elements with gradually decreasing probability
for i = k+1 to n
j := random(1, i) // important: inclusive range
if j <= k
R[j] := S[i] | 308 | 945 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-25 | latest | en | 0.868775 |
https://www.teachoo.com/2112/1373/Ex-2.2--2---Define-relation-R-on-set-N-of-natural-numbers/category/Finding-Relation---Set-builder-form-given/ | 1,695,805,277,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00566.warc.gz | 1,150,852,619 | 28,109 | Finding Relation - Set-builder form given
Chapter 2 Class 11 Relations and Functions
Concept wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 2.2, 2 Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range. Given y = x + 5 Also, x is a natural number less than 4 Hence, x is 1, 2, and 3. Hence, R = {(1, 6), (2, 7), (3, 8)} Domain of R = Set of all first elements in relation = {1, 2, 3} Range of R = Set of all first elements in relation = {6, 7, 8} | 206 | 652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-40 | latest | en | 0.886903 |
http://www.chegg.com/homework-help/questions-and-answers/wrote-program-beverage-survey--errors-instructions-write-program-performs-survey-tally-bev-q4244868 | 1,398,086,764,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00069-ip-10-147-4-33.ec2.internal.warc.gz | 349,001,038 | 8,116 | # BevTally Program C++
750 pts awarded
I wrote this program for beverage survey . what are my errors? Here are my instructions?
Write a program that performs a survey tally on beverages. The
program should prompt for the next person until a sentinel value of –1 is
entered to terminate the program. Each person participating in the survey
should choose their favorite beverage from the following list:
1. Coffee 2. Tea 3. Coke 4. Orange Juice
Sample Run:
Please input the favorite beverage of person #1: Choose 1, 2, 3, or 4 from the
above menu or -1 to exit the program
4
Please input the favorite beverage of person #2: Choose 1, 2, 3, or 4 from the
above menu or -1 to exit the program
1
Please input the favorite beverage of person #3: Choose 1, 2, 3, or 4 from the
above menu or -1 to exit the program
3
Please input the favorite beverage of person #4: Choose 1, 2, 3, or 4 from the
above menu or -1 to exit the program
1
Please input the favorite beverage of person #5: Choose 1, 2, 3, or 4 from the
above menu or -1 to exit the program
1
Please input the favorite beverage of person #6: Choose 1, 2, 3, or 4 from the
above menu or -1 to exit the program
-1
The total number of people surveyed is 5. The results are as follows:
Beverage Number of Votes
********************************
Coffee 3
Tea 0
Coke 1
Orange Juice
This is what I done , but errors.
#include
using namespace std;
int main()
{
int choice;// choose beverages
int c,co,o,t; // four beverages
int total; // total
int person = 1 ; //
cout << " Enter how many person for survey :" ;
cin >> person;
do
{
// display the menu and get a choice
cout << " Beverages Votes\n\n";
cout << "1. Coffee \n";
cout << "2. Tea \n";
cout << "3. Coke \n";
cout << "4. Orange Juice \n";
cout << "Please input the favorite beverage of person "<< person << ": Choose 1, 2, 3, or 4 from the
above menu or -1 to exit the program :"
cin >> choice;
// respond to the user's menu selection
switch (choice)
{
case 1 :
c++
break;
case 2 :
t++
break;
case 3 :
co++
break;
case 1 :
o++
break;
}
} while (choice != -1);
// display beverages' votes
cout << "The total number of people surveyed is " << total <<. the="" results="" are="" as="" follows:="" endl="" br="" cout="" beverages="" votes="" n="" 1="" coffee="" c="" 2="" tea="" t="" 3="" coke="" co="" 4="" orange="" juice="" o="">
return 0;
} | 690 | 2,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-15 | latest | en | 0.784335 |
http://www.jiskha.com/display.cgi?id=1191348824 | 1,495,839,656,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608686.22/warc/CC-MAIN-20170526222659-20170527002659-00147.warc.gz | 656,924,118 | 4,174 | # math
posted by on .
228 people have registered for a sightseeing at the grand canyon. if a toutr director can take 35 people on each tour how many tour directors will be needed to accomadate everyone?
• math - ,
Rebeca, please scroll down to find the answer to your original question.
We also recommend you book mark each question you ask so that you can easily find it and its answers atain.
• math - ,
would the answer be 6? i divided 228 into 35 i got 6.51 and the choices are 5,6,7,8or,9
• math - ,
I don't think you want six and a half tour directers. What would you do with just half a person?
How can you solve this dilemma?
• math - ,
round it out to seven?
• math - ,
Right, Cristal! :-) You'll need 7 tour guides.
### Answer This Question
First Name: School Subject: Answer:
### Related Questions
More Related Questions
Post a New Question | 218 | 871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-22 | latest | en | 0.931698 |
http://math.stackexchange.com/questions/4582/explanation-of-maclaurin-series-of-x-pi?answertab=oldest | 1,462,398,538,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860124045.24/warc/CC-MAIN-20160428161524-00033-ip-10-239-7-51.ec2.internal.warc.gz | 178,908,590 | 20,173 | # Explanation of Maclaurin Series of $x^\pi$
I am reviewing Calc $2$ material and I came across a problem which asked me to explain why $x^\pi$ does not have a Taylor Series expansion around $x=0$. To me it seems that it would have an expansion but it would just be $0$, so maybe it's not a suitable expansion. It doesn't have any holes and it is infinitely differentiable so I don't know why it couldn't have an expansion.
-
Are you sure it's infinitely differentiable? – Qiaochu Yuan Sep 13 '10 at 22:48
Well I may be missing something but it would go $\pi * x^(\pi-1)$ then $\pi * (\pi -1)x^(\pi-2)$...I don't see where it would have a gap. – Planeman Sep 13 '10 at 22:54
How do you even define $x^\pi$ if $x<0$? The answer to this question should show you why there's a singularity at 0 without having to take derivatives. – Jonas Meyer Sep 13 '10 at 23:05
Consider what happens for higher powered derivatives. $\frac{d^4}{dx^4}x^{\pi}=(\pi)(\pi -1)(\pi -2)(\pi -3)x^{\pi -4}$ You can think of this as $c \frac{1}{x^{k}}$ for some positive real number k which is not defined at 0.
-
Oh, ok. I was forgetting to evaluate the power and therefore I didn't realize that at a certain point it turned negative. Would a function like this have an expansion around something like x=1? – Planeman Sep 13 '10 at 23:04
I think centering it at 1 or any positive number would work fine – WWright Sep 13 '10 at 23:16
Planeman: Sure, $\exp(\pi\ln\;x)$ is differentiable at $x=1$. – J. M. Sep 13 '10 at 23:17
I decided to elaborate on my comment above.
What is $x^\pi$ if $x$ is negative? Trying to approximate with $x^r$ for rational $r$ is problematic, depending on which rational numbers you use to approximate; should the answer be positive, negative, imaginary? Perhaps more reasonable would be to take $x^\pi=e^{\pi \log(x)}$ for some suitably chosen branch of the logarithm on $(-\infty,0)$, the most common choice being $\log x=\ln(-x)+\pi i$. Your function then becomes $$f(x) = \left\{ \begin{array}{lr} e^{\pi \ln x} & : x>0 \\ 0 & : x=0 \\ e^{i\pi^2}e^{\pi \ln(-x)} & :x<0 \end{array} \right.$$
In particular, it is not real-valued. You can still go ahead and try to take its derivatives, but this piecewise representation may make it less surprising that there is going to be a singularity at zero.
-
I'll add that I agree that WWright's answer is a good one: the fourth derivative going to infinity as x decreases to 0 shows that the function can't be 4 times differentiable in a neighborhood of 0, no matter what definition you give it for negative x. In fact, I didn't prove anything. My intent is in part to show that there are problems even in determining what is being asked. – Jonas Meyer Sep 14 '10 at 0:00
I honestly hadn't considered that issue until you brought it up. It's an excellent point! – WWright Sep 14 '10 at 0:17
As a graphical supplement to Jonas's and WWright's answers:
This is a plot of the real and imaginary parts of $(x+iy)^\pi$ in the complex plane. Note the cut running across the negative real axis. This cut is precisely the reason why you cannot have a Maclaurin expansion; polynomials cannot exhibit cuts, and a Maclaurin expansion amounts to approximating your function with a sequence of polynomials.
-
I agree that the existence of a cut is relevant to the singularity at 0, but I'm not sure I follow your last comment. For example, the function $f$ in my answer can be uniformly approximated by polynomials (with complex coefficients) on each interval $[-\epsilon,\epsilon]$, by Weierstrass's approximation theorem. Also, I think the question was just about the existence of the Maclaurin series, not whether it approximates the function; for that the fact that the 4th derivative doesn't exist suffices. – Jonas Meyer Sep 14 '10 at 1:35
(But it's true that the cut implies that the Maclaurin series couldn't both exist and locally approximate the function.) – Jonas Meyer Sep 14 '10 at 1:37
I'm essentially saying that trying to find a Maclaurin expansion is the same as asking "does the function behave like a polynomial in the vicinity of the point of expansion?" The cut makes $z^\pi$ not behave like a polynomial in the vicinity of $z=0$. Alternatively, we can think of $z^\pi$ as $z^3 z^{\pi-3}$, the product of a (known) smooth function and the "unknown" $z^{\pi-3}$. $z^{\pi-3}$ has a vertical tangent at the origin, and polynomials cannot have vertical tangents. – J. M. Sep 14 '10 at 1:52
Perhaps I'm just nitpicking, but I was reacting to what you wrote: unlike on open sets in the plane, on bounded intervals every continuous function, no matter how non-polynomial-like, can be uniformly approximated by polynomials. Approximation by the Maclaurin series is asking for much more than this, namely analyticity, and I believe that this is what you intended. – Jonas Meyer Sep 14 '10 at 2:00
Ah, yes you're right, thanks for the reminder and supplying the additional rigor needed. – J. M. Sep 14 '10 at 2:19 | 1,373 | 4,966 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2016-18 | latest | en | 0.944444 |
https://forum.freecodecamp.org/t/how-to-find-integral-of-polynomail-function-between-two-limits/429740 | 1,660,653,300,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00396.warc.gz | 267,335,201 | 5,675 | # How to find integral of polynomail function between two limits
I need and urgrent response thanks alot dear developers.
Firstly, welcome to the forums.
While we are primarily here to help people with their Free Code Camp progress, we are open to people on other paths, too. Some of what you are asking is pretty trivial in the Free Code Camp context, so you might find that if you’re not getting the instruction and material you need in your current studies, the FCC curriculum will really help you get started. At a modest guess I’d say investing a 4-5 hours working through the curriculum here will really pay off. You can find the curriculum at https://www.freecodecamp.org/learn.
With your current questions, we don’t have enough context to know what you already know or don’t know, so it is impossible to guide you without just telling you the answer (which we won’t do).
It is pretty typical on here for people to share a codepen / repl.it / jsfiddle example of what they have tried so that anyone helping has more of an idea of what help is actually helpful.
Please provide some example of what you’ve tried and I’m sure you’ll get more help.
Happy coding
Hello,
A ratio of polynomials can be reduced to a polynomial plus a sum of fractions. Each of the fractions is a ratio of polynomials where the degree of the numerator is less than the degree of the denominator. If a denominator factorises you can split the fraction into further partial fractions. If you use complex numbers the denominators can all be linear, or you could leave it as a real quadratic factor. Repeated factors cause a bit of a problem, please read your textbook.
A factor of the form Aa+bxAa+bx can be integrated immediately giving Kln(a+bx)Kln(a+bx), differentiate this again to work out what KK should be.
For a factor of the form Bx+Ca+bx+cx2Bx+Ca+bx+cx2 where the denominator does not factorise, complete the square in the denominator. The integral will either be a log or an inverse tangent or a sum of the two. | 452 | 2,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-33 | latest | en | 0.925485 |
http://openstudy.com/updates/510b5b74e4b09cf125bc055c | 1,448,884,556,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398461529.84/warc/CC-MAIN-20151124205421-00175-ip-10-71-132-137.ec2.internal.warc.gz | 175,636,079 | 11,766 | aspenfields 2 years ago im bad at work problems can any one help with this problem ? Four truck drivers who worked for a shipping company reported the amount of fuel used and their individual total mileages for the past month. The first driver made 8trips last month for a total mileage of 7,680miles and used 1,010gallons of fuel; the second driver made 7trips and used 1,330gallons of fuel to drive 9,940miles; the third driver made 12trips and used 1,790gallons of fuel to drive 14,640miles; and the fourth driver drove 13,920miles and used 2,050gallons of fuel during 12trips.Whaic matrix correctly represents this data
1. aspenfields
what*
2. aspenfields
thanks. can u help?
3. aspenfields
aw ok then :(
i can...
5. aspenfields
ok :)
6. aspenfields
how do i figure it out?
do you know the options for the matrix??
8. aspenfields
yea one sec
hey you have to understand the Q first correctly that we have to report something to company.... the things are total mileage and fuel comsumed in the last month... so there will be 3 columns for the matric and rows equal to number of drivers
10. aspenfields
so between option one no. of trips mileage(x1000mi) fuel used(x1000 gal) driver 1 | 8 7.68 1.01 driver 2 | 7 9.94 1.33 driver 3 | 12 1.79 14.33 driver 4 | 12 2.05 13.92 option 2 no. of trips mileage(x1000mi) fuel used(x1000 gal) driver 1 | 8 7.68 1.01 driver 2 | 7 9.94 1.33 driver 3 | 12 14.94 1.79 driver 4 | 12 13.92 2.05
11. aspenfields
those are the only 3 column ones
12. aspenfields
i think its b but im not sure the words mix me up XP
which words??
i think first option is right ..... do you got how to approach to this problem or yet not???
what happen ??
16. aspenfields
im tried reading it and i just skipedover words
hello....let me show you steps : 1. summarize the what is asked and what is given 2.them model the what is given part to get what is being asked....
hello ....got it ??
19. aspenfields
k .. d1 8 trips 7680 miles 1010 gal d2 7 trips 1330 gallons 9940gal d3 12 trips1790 gallons 14640 miles d4 13920 miles 2050 gallons and 12trips
yea ....that great ......you are on the way ....
21. aspenfields
yea | 665 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2015-48 | longest | en | 0.884897 |
http://www.coderanch.com/t/388873/java/java/Loop-Array-Logic | 1,467,210,690,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397748.48/warc/CC-MAIN-20160624154957-00135-ip-10-164-35-72.ec2.internal.warc.gz | 448,440,061 | 21,631 | Win a copy of Mesos in Action this week in the Cloud/Virtualizaton forum!
# Loop/Array Logic
Rick Rodriguez
Ranch Hand
Posts: 44
I am currently having trouble reasoning through some mathematical logic. I have been tasked to write a program that can take 50 coins and note the different combinations that could be used to generate \$1.00.
Only and all 50 coins must be utilized. I have poured over the for/while looping logic, as well as array functionality.
Can anyone give me any "hints" or straight forward logic guidance on this progject? Incidently, since this is a homework project, hinting would be preferred.
Thank you.
[This message has been edited by Rick Rodriguez (edited June 02, 2001).]
Val Dra
Ranch Hand
Posts: 439
Did they tell you what amount of each coins there are ?
[This message has been edited by Val Dra (edited June 02, 2001).]
Junilu Lacar
Bartender
Posts: 7466
50
Originally posted by Rick Rodriguez:
Can anyone give me any "hints" or straight forward logic guidance on this progject? Incidently, since this is a homework project, hinting would be preferred.
It's very commendable of you to say that. I'd give you an A+ for integrity.
I suspect the modulus operator (%) would play a part in the logic and a while loop as well. You have to make some assumputions too, like what the different coin denominations are. Some countries have a 20-cent coin but no 25-cent coin.
Junilu
Rick Rodriguez
Ranch Hand
Posts: 44
The denominations are:
Penny: .01
Nickel: .05
Dime: .10
Quarter: .25
[This message has been edited by Rick Rodriguez (edited June 02, 2001).]
Rick Rodriguez
Ranch Hand
Posts: 44
Thank you for the integrity statement. I know that we can share code, for the most part, but I want to try to understand this logic on my own. Not being able to do this on my own, however, I need to turn to someone for some hints.
I appreciate the feedback. Have you been able to create such a program, with the parameters I have outlined?
Thanks again.
Originally posted by JUNILU LACAR:
It's very commendable of you to say that. I'd give you an A+ for integrity.
I suspect the modulus operator (%) would play a part in the logic and a while loop as well. You have to make some assumputions too, like what the different coin denominations are. Some countries have a 20-cent coin but no 25-cent coin.
Junilu
Bill Norton
Greenhorn
Posts: 27
You only wanted hints so hear goes:
You probably need 4 nested loops, one for each coin. In the innerest(real word?) loop you would need two test conditions:
1. adding up all four loop values to see if they equal 50
2. adding up all four loop values plus mulipling by the value(.1, .5, .10., .25) to see if equals a \$1.00
Note: on the second step you would be much better off using 1,5,10,25 and seeing if it adds up 100
Also Note: If my algebra is correct, there is only one solution, if you have to use all the coins.
Sorry if this was too much or not enough.
But I hope this helps,
Bill
Mike Curwen
Ranch Hand
Posts: 3695
There is a point of confusion for me:
You said "only and all 50 coins must be used." If true, this is very simple. Add all coins up in sequence received, and if they don't make 1.00, then no other combination will do that either. If they do add up, then that must mean that ... it's on the tip of my brain...
i'll make it simpler for myself. Imagine 4 coins passed in a 4-entry array. Imagine further that they are all quarters.
The combinations are:
1,2,3,4
2,3,4,1
3,4,1,2
4,1,2,3
umm...
2,1,4,3
etc....
isn't it 4 to the 4th combinations? But maybe this logic only works because my example has been reduced to a too-simple base case.
Or for the question being asked, is this actually only one combination, and 'order' is not being considered. The problem becomes much easier if that is the case. (ie: once you find a combination of coins, you don't need to list off all possible combinations of THAT combination).
"Using any number of the 50, but only from those 50..." is probably what is meant?
Rick Rodriguez
Ranch Hand
Posts: 44
I appreciate your response Bill. Have you been able to recreate the solution to my project and are now feeding me hints? I just want to verify that it has worked for you, before moving forward.
You mentioned "if your algebra is correct" below. What theorem/formula did you use?
Thanks again Bill and anyone else who may be able to create this piece of code, feeding me hints afterwards.
Thanks again.
Originally posted by Bill Norton:
You only wanted hints so hear goes:
You probably need 4 nested loops, one for each coin. In the innerest(real word?) loop you would need two test conditions:
1. adding up all four loop values to see if they equal 50
2. adding up all four loop values plus mulipling by the value(.1, .5, .10., .25) to see if equals a \$1.00
Note: on the second step you would be much better off using 1,5,10,25 and seeing if it adds up 100
Also Note: If my algebra is correct, there is only one solution, if you have to use all the coins.
Sorry if this was too much or not enough.
But I hope this helps,
Bill
Rick Rodriguez
Ranch Hand
Posts: 44
I appreciate you replying to my posting Mike. To answer your question, I need to utilize all 50 coins. Apparently, there are several combinations, less than 10, of 50 U.S. coins that can add up to \$1.00.
Have you been able to create this code? Thanks again Mike.
Originally posted by Mike Curwen:
There is a point of confusion for me:
You said "only and all 50 coins must be used." If true, this is very simple. Add all coins up in sequence received, and if they don't make 1.00, then no other combination will do that either. If they do add up, then that must mean that ... it's on the tip of my brain...
i'll make it simpler for myself. Imagine 4 coins passed in a 4-entry array. Imagine further that they are all quarters.
The combinations are:
1,2,3,4
2,3,4,1
3,4,1,2
4,1,2,3
umm...
2,1,4,3
etc....
isn't it 4 to the 4th combinations? But maybe this logic only works because my example has been reduced to a too-simple base case.
Or for the question being asked, is this actually only one combination, and 'order' is not being considered. The problem becomes much easier if that is the case. (ie: once you find a combination of coins, you don't need to list off all possible combinations of THAT combination).
"Using any number of the 50, but only from those 50..." is probably what is meant?
Mike Curwen
Ranch Hand
Posts: 3695
When Bill said "if my algebra is correct", he was thinking along the same lines I was.
There is no real 'formula', but just think hard for a second. If you are given a single array of 50 coins, and you must use ALL of them to see if they add up to 1.00, then this is a TRUE/FALSE situation. Do they add up, or don't they? If they don't, then adding them up in a different order will not produce a different result. But what I get from your next comment, is that you want a list of all arrays that successfully add up to \$1.00...
Your last sentence "Apparently, there are several combinations, less than 10, of 50 U.S. coins that can add up to \$1.00" really helped clarify the program requirement.
The brute force method is an easy approach. You have 50 positions to fill (one for each of the 50 coins). There are four possible values in each position (1,5,10,25). That means there are 4 to the 50 combinations you must check. When you add up a certain combination, and it equals \$1.00, then remember that combination. The list of combinations that adds up to \$1.00 will be your answer.
This is different than what I thought you were supposed to do. I thought it was "I've given you 50 random coins. How many ways can you make \$1.00? (you can use as little as 4 coins, and as many as 50)".
I am a geek with nothing better to do today.. sigh...
I think I'll work on this a bit.
Mike Curwen
Ranch Hand
Posts: 3695
I have a program that finds two combinations. Anyone else?
p.s. My earlier musings might lead you to produce a program with 50 nested for loops. Sorry about that. Loops represent coins.. think of it that way. Bill's hints were much better.
Bill Norton
Greenhorn
Posts: 27
Rick you mentioned that there was more than one combination that would work. Do you have to use each type of coin? If you do than I believe there is only one possiblility here is why
p = # of pennies
n = # of nickels
d = # of dimes
q = # of quarters
You have two equations that can now be derived
p+n+d+q = 50
1p + 5n + 10d + 25q = 100 //I multiplied by 100 to make things pretty here
If you solve for p in the second equation you get
p = 100 - 5n - 10d - 25q
Now substituion p into the first equation
100-5n-10d-25q+n+d+q= 50
Simplify
-4n-9d-24q = -50
4n + 9d + 24q = 50
Now since q is the amount of quarters there are only a few possiblities. q= 1,2,3
3 would not work
So for 2-> 4n + 9d + 24(2) = 50
4n + 9d = 2
Can't work!
Therefore the only possible value left is q=1
This leaves us with a new equation
4n + 9d + 24 =50
4n + 9d = 26
Now d would have to equal 2 to make the equation
4n + 9(2) = 26
4n = 8
n=2
Any other values for 9d or 4n will not add up to 26.
Now if you don't have to use all types of coins then you could just have q=0, and get the equation
4n + 9d =50
Ther are some more possibilities there.
Well need to go lawn needs moving and it looks like rain!
Bill
ps I didn't proof read sorry for mistakes
Mike Curwen
Ranch Hand
Posts: 3695
The original question didn't state that you had to use at least one of each type, only that you had to use 50 coins that add up to 100.
q d n p
0 2 8 40 = 100 count = 50
1 2 2 45 = 100 count = 50
Rick, when you are done (or close), post your code and let us see how you did it.
Junilu Lacar
Bartender
Posts: 7466
50
Originally posted by Mike Curwen:
> I have a program that finds two combinations. Anyone else?
I got the same results.
Rick, using brute force, you can find the good combinations using three "for" loops.
Junilu
Bob Graffagnino
Ranch Hand
Posts: 81
I too got the same results using 4 "for" loops. Each "for" loop has the local variables totalCoins and totalCoinAmount. These variables are initalized to the values of the corresponding outer loop variables. For instance:
[This message has been edited by Bob Graffagnino (edited June 04, 2001).]
Junilu Lacar
Bartender
Posts: 7466
50
Originally posted by Bob Graffagnino:
>>I too got the same results using 4 "for" loops
You could probably simplify your solution. 3 nested "for" loops are sufficient really because another loop for pennies would be superfluous. Also, I didn't use any variables other than the three for-loop index variables. I did have a separate 3-line method that determined whether a combination of coins met the criteria.
Junilu
Mike Curwen
Ranch Hand
Posts: 3695
Like most programs I write, I usually start with the longest or most 'brute force' way of doing it.. especially with math or logic problems.
And then come the optimizations.
I've gone through two 'cycles' of this optimization effort, and I admit to thinking there must be a way to kill off the outer loop (the quarters loop). But now thanks to Junilu, I see now why the pennies loop was worse than useless (it's a big timewaster!). So I've got it down to three loops as well.
Hopefully Rick will tell us when the due date is, so I can see if my code is the best it could be.
Roger Graff
Ranch Hand
Posts: 112
Doh! You're right, the pennies loop is not needed. Anyone come up with a recursive solution?
-Sorry moderators. I accidently posted this message as graff48 instead of my corrected user account Bob Graffagnino.
[This message has been edited by graff48 (edited June 05, 2001).]
Mike Curwen
Ranch Hand
Posts: 3695
I try to back away slowly from anything recursive
But I admit to thinking 'there is probably a way to do this with recursion'.
Joel Cochran
Ranch Hand
Posts: 301
In all logic problems some assumptions have to be made. As I understand it, our assumptions here are:
1: We have w pennies, x nickels, y dimes, and z quarters.
2: w + x + y + z MUST equal 50 and there cumulative value MUST equal 100.
This allows us to conclude that only x penny values evenly divisable by 5 could complete our needs, so we can set the penny array like so... { 5, 10, 15, 20, 25, 30, 35, 40, 45 }. 0 and 50 are automatically excluded since they do not meet the minumum requirements.
Now we can write a function to whittle this down a little: we'll use 45 for the 1st example.
1: 45 pennies provides a difference of 5 coins.(50 - 45 = 5).
2: The 5 coins must total 55 (100 - 45 = 55).
*2A: Add a vlidity check here: multiply the # of coins (5) x our denominator (5) = 25. Is this less than the amount needed (55)? Yes, so you can continue. {If NOT then this number of pennies is not valid and you can stop here.}
3: Can 5 coins containing at least one nickel, one dime, and one quarter = 55? To find out we add 5 + 10 + 25 and get 40. (this could actually become a program constant)
4: 55 - 40 = 15.
5: Divide 15 by our denominator (5) = result (3).
6: A Static array of arrays could determine what the corresponding values would be (*hint: a result of 3 would point us to a positional Array[3] which would have subsequent Array[3][0]=5 and Array[3][1]=10. Array[0] and Array[1] would both be null) Deciphering the Array values will tell you which coins you need to add to your assumed coins ( 45 pennies, 1 nickel, 1 dime, 1 quarter ) to achieve a value of 100.
7: So our solution for 45 is "45 pennies, 2 nickels, 2 dimes, and 1 quarter"
Confused? Me too... let's test it with penny array 20...
1: 20 pennies provides a difference of 30 coins.(50 - 20 = 30) .
2: The 30 coins must total 80 (100 - 20 = 80).
*2A: Validity check: multiply the # of coins (30) x our denominator (5) = 150. Is this less than the amount needed (80)?
No, it is not, so this number of pennies cannot represent a valid solution.
Now that you've built the function, loop through the array and call it 9 times, printing out the good results and you should come up with a correct solution. I hope this helps in some way... it may not be complete but should give you plenty to think about!
------------------
I'm a soldier in the NetScape Wars...
Joel
[This message has been edited by Joel Cochran (edited June 05, 2001).]
Joel Cochran
Ranch Hand
Posts: 301
Sorry for the second post, but I wanted to point out that there are a couple of things I intentionally left out of my solution to give Rick a chance to work with some of the ideas since he said this was for a homework assignment.
------------------
I'm a soldier in the NetScape Wars...
Joel
Marcellus Tryk
Ranch Hand
Posts: 64
Bob -
Here's a recursive solution. It prints a solution as a sequence of coins in descending order. If anyone would like to try it out you can compile it and run it at the command line to solve any problem of this form. It could easily be modified to accept coins of arbitrary denomination.
To solve the stated problem: java Coins 100 50
By the way I believe this problem is np-complete. Does anyone have any opinions on this?
Here's the basic idea of this algorithm:
The trivial case is one coin. There's a solution if the amount is equal to a valid denomination and no solution if it isn't.
Otherwise -
Remove a quarter and see if there's a solution for what you have left (amount - 25, coins - 1). Do the same with a dime, nickel, and penny.
-----------------------------------------------------------------
[This message has been edited by Tod Tryk (edited June 07, 2001).]
[This message has been edited by Tod Tryk (edited June 07, 2001).]
Rick Rodriguez
Ranch Hand
Posts: 44
Todd,
I appreciate your help, but we (our class) haven't learned the "Switch" statement yet. We're still on loops.
To let everyone know, tonight our teacher could not understand, after class, why I could not understand this simple concept. When he discovered that the "nested" for loop concept was not explained, either in the book or by himself last week, he reminded me to teach the rest of the class next week.
At any rate, after he explained the logic of how nested for loops work, "Eyes Are Open!"
This project is due next week for me. I will, however post my code within the next two days. Please wait to post your own until such time, as it may be deemed as "cheating" on my part, if it is inferred that I got my code from this message board.
Thank you again everyone, and come back to see my code this Saturday.
Marcellus Tryk
Ranch Hand
Posts: 64
Rick -
I know you didn't want a solution, but I figured since your implementation is supposed to use nested loops you wouldn't mind if I posted the recursive algorithm as a curiosity.
Best of luck on finishing your assignment - I'm looking forward to seeing it!
Tod
Mike Curwen
Ranch Hand
Posts: 3695
Just that I'd raise my hand and say "Holy cow I feel dumb." LOL.
That coding example is why I back away slowly from recursion. And I haven't heard "np-complete" since I dropped out of University.
But thanks for the code, because God help me, I'm going to try and understand it.
Marcellus Tryk
Ranch Hand
Posts: 64
Mike -
I don't even know if it makes sense to ask if this problem is np-complete. I was wondering if there might be a way to discover all the solutions to the coin problem without exhaustive search - and then I thought - naah this problem is probably np-complete. Being very non-expert in this area I was hoping someone could shed some light on it. Anyway I think I'm getting off-topic so let me just say in conclusion - I encourage the use of logic when constructing loops that process arrays.
Rick Rodriguez
Ranch Hand
Posts: 44
Ok Guys! I have finally coded this program. If it were only explained to me, either in my text book or by my teacher, the "logic/flow" of nested 'for' loops, I would have never needed to post questions on this matter in the first place.
I started coding this program at approximately 1:00pm and finished at approximately 2:00pm, ET. Most of that time was spent in my IDE's debug session, looking at how the 'for' loops rotate to find the parameters defined.
At any rate, here is my homework code:
Incidently, the text that my "Intro to Java" course utilizes is 'JAVA: Programming From the Beginnnig', by K.N. King. I think that this is a fantastic book for beginners, and I can only speak from that perspective, since I am a beginner.
Thanks again guys for all of your input, and let me know if I can "fine tune" my code any further.
(edited by Cindy to format code)
[This message has been edited by Cindy Glass (edited June 09, 2001).]
Junilu Lacar
Bartender
Posts: 7466
50
The solution below is pure brute force. I don't know if having "if" statements in each of the for-loop actually makes a whole lot of difference. Finding bottlenecks should really be done with a profiler though. I go by the XP philosophy of going for simple and clear code first, then refactor to optimize for performance using the results of using a profiler.
<pre>
{
public static void main(String[] args)
{
// We don't need a loop for pennies because
// we can derive the penny count from the
// count of other coins we have so far
for (int q = 0; q < 4; q++)
for (int d = 0; d < 10; d++)
for (int n = 0; n < 20; n++)
if ( isGoodCombo(q, d, n) )
System.out.println(
" 25c = " + q +
" 10c = " + d +
" 5c = " + n +
" 1c = " + (50-q-d-n));
}
// chose to move evaluation to a separate method
// to keep the main loop code clear and concise
static boolean isGoodCombo(int q, int d, int n)
{
int p = 50 - q - d - n;
return ((p + q + d + n) == 50) &&
((q*25 + d*10 + n*5 + p) == 100);
}
}
</pre>
[This message has been edited by JUNILU LACAR (edited June 10, 2001).]
Mike Curwen
Ranch Hand
Posts: 3695
Hi Rick,
I'm assuming the empty if() statements between your for loops were used during development for debugging? Otherwise, I can't see what they'd be used for, except perhaps a 'continue' statment? As it is, they are doing nothing...
I would also suggest there is a small (negligable) optimization that can be made to the quarters loop. You can bring the exit condition down to < 3. Also, regard how others have dropped the pennies loop... pennies are worth '1' each, and we can determine how many of them there are by how many of the other coins there are.
The last thing is also a very slight optimization, but if we are talking about efficiency, then lets squeeze everything we can out of math...Your innermost if() statement is coded thus Because Java performs short-circuited ANDs, we can realize (I know, I know) a very small boost, if you place the second test first. That way, it only does 4 additions, instead of 3 multiplications and then 4 additions. Because we need both of them to be true, we should always check the 'cheapest' side first. Once it proves to be false, Java ignores the other side of the AND (short-circuited AND). Congratulations on figuring it out.
fred cook
Greenhorn
Posts: 8
In general, when you are trying to code always look for an algebraic way to make the code work faster. Brute force is a lowsy way to do stuff. Try to stay away from using it. The responses you've been getting could be very useful in the future.
--Freddy
------------------
I have a cute bird | 5,624 | 21,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2016-26 | latest | en | 0.957001 |
http://www.scribd.com/doc/24740311/%D8%A7%D9%84%D8%A7%D8%B1%D8%A8%D8%B9%D9%88%D9%86-%D8%A7%D9%84%D9%86%D9%88%D9%88%D9%8A%D8%A9 | 1,438,189,446,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042986451.45/warc/CC-MAIN-20150728002306-00180-ip-10-236-191-2.ec2.internal.warc.gz | 679,034,875 | 34,335 | Welcome to Scribd. Sign in or start your free trial to enjoy unlimited e-books, audiobooks & documents.Find out more
Standard view
Full view
of .
0 of .
Results for:
P. 1
الاربعون_النووية
# الاربعون_النووية
Ratings: (0)|Views: 11|Likes:
### Availability:
See more
See less
01/03/2010
pdf
text
original
ةــيوـنــلا ــعــبراةــيوـنــلا ــعــبرا
ثيداحل ةمئاق ثيداحل ةمئاق
باطخ ن رمع صفـح أ نـؤم رــأ نعتمس :اق وس :وـي
}
رــك اـمـإ اا اـمـع امـإ ىإ هـرج هـوس ىإ هرج تاـ نم ،وـ ا اكي ـأر ـأ اــي اــ هـرج تـا نـ ،هوــس هإ را ا ىإ هرج
{
. ــر ن ـعامسإ ن م ــع وـأ نم اإ ]:فـج اخ هزدرـ ن رـغم ن 1 نــ وـأ ،:وـاــ رـشـ ـ ناج ن 1907 ض .[ةفم ك أ ام نيذ ام امع
اا امع : ثيرش ةمئاءاإ | 1,888 | 1,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-32 | latest | en | 0.134895 |
https://myessayguru.com/corr-ana-assignment-due-in-24-hours/ | 1,657,135,718,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104676086.90/warc/CC-MAIN-20220706182237-20220706212237-00364.warc.gz | 444,673,821 | 8,450 | CORR ANA ASSIGNMENT DUE IN 24 HOURS
DUE IN 24 HOURS
Correlational Analysis
Before beginning this assignment, please watch the following videos on correlation:Correlation – The Basic Idea Explained (Links to an external site.)Correlation Basics (Links to an external site.)
Get Your Custom Essay Written From Scratch
We have worked on a similar problem. If you need help click order now button and submit your assignment instructions.
Just from \$13/Page
Study Description: A school educator is interested in determining the potential relationship between grade point average (GPA) and IQ scores among ninth graders. The educator takes a random sample of 30 ninth graders aged 14 years old and administers the Wechsler Intelligence Scale for Children-Fourth Edition (WISC-IV). The WISC-IV includes a Full Scale IQ (FSIQ; however, for this assignment we will just call it IQ).
Need assignment help for this question?
Order Now
Output file: See Week 5 SPSS Output.pdf file (Links to an external site.).
Hypotheses – Formulate null and alternative hypotheses. What do you think is the relationship between IQ scores and GPA?
Variables – Describe the scale of measurement (nominal, ordinal, interval, or ratio) for each of the variables.
Correlation – Write an overview of the results of the correlation (at least two paragraphs), including the appropriate and necessary statistical results within sentences and in proper APA formatting. Be sure to provide sufficient explanation for any numbers presented. Consider the following in your overview and conclusions:
Is there a significant correlation between IQ scores and GPA? If so, what does a significant correlation mean?
Using the correlation table and scatterplot, explain whether the relationship is positive, negative, or no correlation.
Describe the strength of the relationship (e.g. very strong, moderate, weak, etc.). | 389 | 1,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-27 | latest | en | 0.867303 |
https://www.bestpfe.com/relaxation-time-scale-of-the-monte-carlo-dynamics/ | 1,656,797,257,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00342.warc.gz | 710,263,034 | 22,413 | # Relaxation time scale of the Monte Carlo dynamics
Get Complete Project Material File(s) Now! »
## Worst-case complexity
Time complexity
From an algorithmic point of view, one would be interested in finding an algo-rithm that solves a CSP (i.e. solves all its possible instances) in a reasonable amount of time. Intuitively, an algorithm will require more time to provide an answer for large size instances. There are several ways to define the size of an instance. A natural definition is to take N the number of variables, but one could also have chosen N + M, or |E| the total number of edges in the graphical representation of a CSP. Most of the time one works with instances for which M = αN, with α a constant independent of N, therefore all these possible def-initions are polynomially related. In computational complexity theory [1] it is therefore equivalent to work with any of them, and we will fix the size of the problem to be N.
One defines the time complexity function T (N) of an algorithm as the largest amount of time needed for this algorithm to solve an instance of given size N. Note that this definition depends on the computer used to implement the algorithm. One will not enter into the details, and just state that one uses a ‘realistic’ computer model. Realistic means that such a model can run at most one (or a finite small number) of operations per unit time, where operations are for instance comparison, sum and multiplication. Some examples of realistic model are the one-tape (or multi-tape) Turing machines, and the random-access machines (RAMs). The algorithms can be split into classes according to the behavior of their time complexity function.
1. A polynomial time algorithm has a time complexity function T (N) = O(p(N)) for some polynomial function p.
2. An exponential time algorithm has a time complexity function that can be lower bounded by 2bn for some constant 0 < b.
This distinction is useful because exponential time algorithm have a time com-plexity that explodes when the size increases, therefore they become inefficient for large size instances. One says that a problem is intractable when it is so hard that no polynomial algorithm can solve it. Note that the time-complexity function is a worst-case measure, in the sense that is corresponds to the time needed for the algorithm to solve the hardest instance of a given size. It could happen that the majority of other instances require much less time, and the algorithm would be ‘efficient’ in a large proportion of the instances it encouters. As an example, the k-SAT problem is solved by an exponential time algorithm called DPLL, that one will describe further. As we shall see in the following, experimental observations and rigorous results show however that for instances such that M = αN , with α small enough, this algorithm finds solutions very quickly for the vast majority of instances, even when N gets large.
Complexity classes
The theory of computational complexity classifies the difficulty of the decision problems according to the existence of an algorithm that solve a problem in polynomial time. This classification uses the notion of polynomial reduction. One says that a problem π1 can be reduced to another problem π2 when there is a function that maps an instance of π1 into an instance of π2. To be more precise, let Dπ be the set of instances of problem π. Let Yπ ⊆ Dπ be the subset of instances for which the answer to the decision problem is ‘yes’. Then a reduction is a function f : Dπ1 → Dπ2 such that for all I1 ∈ Dπ1 , one has I1 ∈ Yπ1 if and only if f (I1) ∈ Yπ2 . A polynomial time reduction is a reduction f that is computable by a polynomial time algorithm in the size of the problem π1. One requires additionally that the size of f(I1) is polynomial in the size of I1. The relation « π1 can be reduced polynomially to π2 » is denoted π1 ∝ π2. This relation is transitive:
Lemma 1 (Transitivity) if π1 ∝ π2 and π2 ∝ π3, then π1 ∝ π3.
One defines the two following classes of problems:
1. The polynomial class P is the set of problems that can be solved by a polynomial time algorithm.
2. The non-deterministic polynomial class NP is the set of problems that have a polynomial time verifiability. This means when one provides a certificate that allows to check a ‘yes’ answer to the problem, the verification itself can be done in polynomial time.
The term non-deterministic comes from an alternative definition of the NP class, as the set of problems that can be solved by a polynomial time non-deterministic algorithm. A non-deterministic algorithm is composed of two stages. In the first stage, a certificate is proposed, and in the second stage, the certificate is verified. A non-deterministic algorithm is said to operate in polynomial time if the second stage is done in polynomial time, disregarding the first stage that might have required more time. Roughly speaking, a non-deterministic polynomial algorithm can pursue alternate paths or branches in parallel, and says ‘yes’ if any of these branches says ‘yes’.
Note that if one has a polynomial reduction from π1 to π2, then π1 cannot be harder than π2. Suppose indeed that one has a polynomial algorithm that solves π2 . One can construct a polynomial algorithm to solve π1 as follows. Given an instance I1 ∈ Dπ , this algorithm just applies the transformation f to get an instance I2 = f(I1)1of π2, and applies the polynomial algorithm to solve the instance I2. Therefore:
Lemma 2 If π1 ∝ π2, then π2 ∈P implies π1 ∈P (and equivalently π1 ∈/P implies π2 ∈/P) Note that a problem π in P is automatically in NP. If there is a polynomial time algorithm that solves π, this algorithm can be converted into a checking algorithm that ignores the certificate that is provided, and simply returns its answer. Therefore P⊆NP.
A natural question is then to ask if P=NP, or if there exist problems in NP that are not in P. This is an open question, but it is widely conjectured that P=6NP. This conjecture is supported by the existence of another class of NP problems gathering the hardest problems in NP. It is called the NP-complete class (NP-c). A problem π is in NP-c if it satisfies the two conditions:
1. π ∈NP
2. Any problem π0 in NP can be polynomially reduced to it: π0 ∝ π
If one problem in NP-c could be solved in polynomial time, then all the problems in NP could be solved in polynomial time thanks to the polynomial reduction, and one would have P=NP. On the other hand, if one problem in NP is in-tractable, then all the problems in NP-c are intractables. Some problems in NP-c, such as the satisfiability problem, have been studied for decades without any proof that they belong to P, that is why one believes that P=6NP. Figure 1.2 illustrates the conjecture on the structure of the sets NP, P and NP-c.
From the definition of the NP-c class, proving that a problem π ∈NP belongs to NP-c requires to prove that any problem π0 ∈NP can be polynomially reduced to π. This seems difficult to achieve, but once one has proved that there is at least one problem in NP-c, the proof can be simplified thanks to this Lemma:
Lemma 3 Let π1 ∈NP-c. If π2 ∈NP, and π1 ∝ π2, then π2 ∈NP-c.
Indeed since π1 ∈NP-c then for any problem π0 ∈NP one has π0 ∝ π1, then by transitivity π0 ∝ π2 . The first problem that has been proved to be NP-complete is the satisfiability problem, by Stephen Cook in 1971 [41]
Theorem 1 (Cook,1971) The satisfiability problem is NP-complete.
We will not give the proof, but solely remark that the SAT problem has a very universal structure, that allows to re-write any decision problem into the SAT problem. Once we know that the SAT problem is in NP-c, one can apply the lemma 3 to prove that another problem π is NP-c. The method should follow these steps:
1. Show that π ∈NP.
2. Select a known problem π0 ∈NP-c
3. Construct a reduction f from π0 to π
4. Prove that f is a polynomial reduction.
Using this method, hundreds of problems have been shown to belong to NP-c.
The table 1.1 gathers the problems that one has introduced above:
### An algorithm for the SAT problem
The Unit Clause Propagation procedure
We have seen that the 2-SAT problem is in P. We present here a polynomial algorithm that solves it. The algorithm is a sequential assignment procedure, which means that at each time step a variable is assigned to a given value. The algorithm ends either when all the variables are assigned, the assignment obtained being SAT, or when it has proven that the formula is UNSAT.
Each time a variable is assigned, one can simplify the initial CNF formula according to the following reduction procedure. Suppose that one has assigned the variable i to xi = 1 (the case xi = 0 is symmetric). Each clause containing the literal xi is satisfied by this assignment, therefore can be removed from the formula. In each clause containing the opposite literal xi, one removes the literal xi since it cannot satisfy the clause. The length of these clauses is then reduced by 1. One denotes F |{xi = 1} the simplified formula obtained with this procedure. This reduction procedure might produce a 0-clause, namely if in the formula there a unit clause (a 1-clause) of the form c = (xi), that is violated by the choice xi = 1. One calls this a contradiction, it means that the simplified formula F |{xi = 1} is UNSAT, therefore to construct a SAT assignment for F , the choice xi = 0 is necessary.
If at some step the formula obtained contains a unit clause, one is forced to satisfy it, and the simplified formula thus obtained might contain new unit clauses that one will have to satisfy in turn. This sequence of forced steps is called Unit Clause Propagation (UCP), and is done with the recursive UCP procedure (see algorithm 1). This procedure takes as input a CNF formula F (possibly containing unit clauses), a partial assignment of variables A, and a set of variables V . In the first call we set F to be the initial CNF formula, and
A = ∅, V = {1,…,N}
Algorithm 1 UCP (F ,A,V )
if there is a 0-clause in F then
return UNSAT
end if
while there exist unit clauses do
Pick a unit clause in F , say c = (li) and satisfy it
Add the assignment of the variable xi in A
if there is a 0-clause in F then
return UNSAT
end if
end while
return A
The UCP procedure (1) returns all the assignments that were forced due to the presence of unit clauses. There is three possible outputs for the UCP procedure:
1. the output is an assignment of all the variables in V , then F is SAT and the assignment produced is a solution to F .
2. the output is a partial assignment. This partial assignment can be ex-tended to a complete SAT assignment if and only if the input formula F is SAT. The simplified formula obtained does not contain unit clauses.
3. the output is UNSAT, therefore the input formula F is UNSAT.
The algorithm to solve 2-SAT uses the UCP procedure. It works as follows: given a 2-SAT formula F over a set of variables V , choose a variable i ∈ V , and fix xi = 1. Then call the UCP procedure UCP(F |{xi = 1},{xi = 1},V \ i). If the output has assigned all the variables, declare F satisfiable, and return the SAT assignment found. If it is only a partial assignment, keep it in memory, and let F 0, V 0 be the simplified formula and the set of not-yet assigned variables. Choose a new i0 variable in V 0, and restart the UCP procedure UCP(F 0|{xi0 = 1},{xi0 = 1},V 0 \ i0).
If the output is UNSAT, it means that the initial choice has to be changed. Thus set xi = 0 and restart the UCP procedure with UCP(F |{xi = 0},{xi = 0},V \ i). If once again the procedure outputs UNSAT, then declare that the initial formula F is UNSAT. If not one can keep the partial assignment found and repeat the procedure. The step in which one chooses arbitrarily the value of a variable in a 2-clause is called a free step. By opposition, the assignment of a variable in a unit clause is called a forced step. The step in which we came back to the arbitrary choice done at the free step xi = 1 to change it to xi = 0 is called a backtracking step.
One can measure the complexity of this algorithm by the number of variable-fixing operations. For a 2-SAT formula, S. Cook showed that this algorithm has a polynomial complexity [41]. Note however that the algorithm presented above does not provide a method for solving the optimization problem MAX-2-SAT. In fact this problem has been shown to be in NP-c by M.R Garey, D.S Johnson and L Stockmeyer in [42] in 1976.
The DPLL algorithm for the SAT problem
The above algorithm can in fact be slightly modified to work on a CNF formula containing clauses of arbitrary length. In 2-SAT formulas, since any free choice produces unit clauses, one only needs to do backtracking steps on the last vari-able assigned on a free choice. All the other variables fixed at previous free steps are fixed, since they belong to a partial assignment compatible with at least one solution (if it exists). When there are clauses of length greater than 2 instead, one might have to do several free steps before having the possibility to apply the UCP procedure. If these choices lead to a contradiction, the backtracking then have to explore all the possible choices for these free variables before concluding that the formula is UNSAT. The number of backtracking steps might explode, leading to an exponential time complexity. This algorithm is called the Davis Putnam Logemann Loveland (DPLL) algorithm, from the authors names, and has been introduced in 1962 [43]. It is described by the algorithm 2 (see [3] p.726).
Algorithm 2 DPLL (INPUT: a SAT formula F , OUTPUT: is F satisfiable ?)
If F is empty then return SAT
If F contains and empty clause then return UNSAT
Select an unset variable xi
If F |{xi = 1} = 1 then return SAT
If F |{xi = 0} = 1 then return SAT
Return UNSAT
The iterations of DPLL can be represented by a decision tree, as shown in figure 1.3 (the example is inspired from [44]).
It starts at the root of the tree, with the initial formula F containing all the variables to be assigned. A first variable is chosen and assigned to a value (x1 = 1 in the example). The left-node of the first generation represent the simplified formula obtained with this choice. Since no unit clauses are produced, one needs to make another arbitrary choice: x2 = 1 in the example. If there is a unit clause, the algorithm fixes the value of one variable present in a unit clause. When the algorithm finds a contradiction, it backtracks to the last variable assigned during a free step, and restarts the search from this node.
From this representation, one can measure the running time of DPLL on the formula F , by counting the number of nodes of the tree. It is intuitively clear why DPLL have an exponential complexity on SAT formulas with clauses of length greater than 2. If most of the variables have been fixed during a free step, and all the free choices need to be backtracked, the number of steps is roughly the number of nodes of a binary tree of depth N , which is O(2N ). On a 2-SAT formula instead, each free choice is followed by a cascade of forced steps (possibly O(N ) of them). Even if all the variables have been fixed during a free step, and all the free steps have been backtracked, the complexity is still polynomially bounded by N2.
In the algorithm 2, one has not specified how to choose the variable during a free step. One can use heuristics that aim at increasing the performances of DPLL. For instance, one can decide to pick a variable that will generate many unit clauses, or a variable that will satisfy the largest number of clauses. Since there is no good criterion to decide which heuristic to use, one would have to experiment them on the formula one has to solve.
READ Laying the backing cloth: a postcolonial and diasporic fabric
#### Performances of the algorithms and random CSP ensembles
Complete and incomplete algorithms
An algorithm that solves a CSP, i.e. that provides an answer to all the possible instances of the CSP, is called a complete algorithm. DPLL is an example of complete algorithm for the SAT problem. By opposition, an incomplete algorithm is not guaranteed to provide an answer to any possible instances. In a decision problem, proving that a instance has a solution is often easier than proving that there is no solution. In the SAT problem for instance, a certificate for a ‘yes’ answer (to the question ’Is there a solution ?’) is provided by a solution to the problem, that is an assignment of the N variable, therefore having size N. A certificate for the answer ‘no’ instead might have an exponential length, if for instance one tries to exhibit all the possible assignments, showing that they all are UNSAT. The display of this certificate by itself have an exponential time complexity. An incomplete algorithm only tries to find a solution, and answers ’I don’t know’ whenever it is unable to provide one, or to prove that the problem has no solution.
We will present in the following chapters several incomplete algorithms for CSPs. Although they do not solve all the possible instances of the problem, it is observed experimentally that they are more efficient than the complete algorithms, on typical instances. To characterize their performances, one needs to precise what we mean by typical instances. This can be done by introducing random ensembles of instances. One then study the performances of algorithms on instances drawn from this ensemble.
Random CSP ensembles
We have seen that an instance of CSP can be represented with a hypergraph. Therefore one can use random graph ensembles to define the random CSP en-sembles. The Erdős Rényi (ER) ensemble GN (k, M) is an example of random hypergraph ensemble. An instance of GN (k, M) is drawn by choosing indepen-dently for each clause a k-tuple of variables uniformly at random among the Nk possible ones. Note that while the hyperedges a ∈ C have a fixed degree k, the degree of the vertices i ∈ V is not fixed.
We will also use another random hypergraph ensemble, called the k-uniform (l + 1)-regular random hypergraph ensemble (the choice (l + 1) is made here for convenience when using the cavity method, see chapter 7). In this ensemble both the hyperedges and the vertices have a fixed degree k and l + 1. All the hypergraphs with this property are equiprobable in this ensemble.
To define a random ensemble for the SAT problem, we need to specify how to draw the signs Jia. The random k-SAT ensemble is one of the most studied random ensembles for the k-SAT problem. An instance is obtained by drawing a hypergraph from GN (k, M), then independently for each variable in each k-tuple, one chooses a sign Jia with probability 1/2.
It is useful to introduce the density of constraints α = M/N, and to work with the ensemble GN (k, αN). For the k-uniform (l + 1)-regular random hy-pergraph ensemble, since N and M must satisfy the relation N(l + 1) = M k, the density of constraints is related to the degrees as l + 1 = αk. Intuitively, in-creasing the density of constraints, the number of solutions should shrink, since it is harder to satisfy an overconstrained instance. At high density, we will see that most of the instances are in fact UNSAT.
We define the thermodynamic (large size) limit as N, M → ∞ with a fixed ratio α = M/N. In this limit, several properties of the random k-SAT en-semble concentrate around their mean value. This is called the self-averaging phenomenon. In particular, in this limit many random CSPs (including the ran-dom k-SAT ensemble) exhibit threshold phenomena. The probability of some property jumps abruptly from 1 to 0 when varying the control parameter α. We say that a property of the instances is typical when its probability goes to 1 in the thermodynamic limit.
The most prominent threshold is the satisfiability threshold αsat(k). Be-low αsat(k), typical instances are SAT, while above αsat(k), typical instances are UNSAT. In the satisfiability phase α < αsat(k), many other phase tran-sitions concerning the properties of the set of solutions of a random instance are predicted by the so-called cavity method, that we shall describe in the next chapters.
Among them, the clustering transition describes a drastic change in the structure of the set of solutions in the space of configuration. Below the cluster-ing threshold, the set of solutions is rather well-connected, any solution can be reached from any other one by nearby intermediate solutions, while above the clustering threshold the solution set splits in a large number of distinct groups of solutions, called clusters, which are internally well-connected but well separated one from the other.
Performances of DPLL on the random k-SAT en-semble
Our aim is to study and compare the performances of algorithms on a random ensemble of instances. We start by giving the experimental study of the DPLL algorithm on the random k-SAT ensemble.
For a fixed value k, one can plot the fraction of UNSAT instances found by DPLL as a function of α for several sizes N. The result for k = 3 is shown in figure 1.4 taken from the study of S. Kirkpatrick an B. Selman in 1994 [45] One can see that it is a increasing function of α, that goes from 0 to 1. As N increases, the drop becomes sharper around a threshold value. This is a numerical evidence of the satisfiability threshold.
One is interested in the running time of DPLL on random instances of k-SAT. On figure 1.5 taken from the study of D. Mitchell, B. Selman, H. Levesque in 1992 [46], the median running time is plotted against α. One can see that the running time has a peak around the satisfiability threshold. The height of this peak increases rapidly with N . At smaller and larger values of α instead, the running time is much smaller, and grows with N at a smaller rate. These observations indicate that the instances drawn from the random k-SAT ensemble are harder when α is close to αsat(k ). In the region of smaller α, the instances are underconstrained and therefore easy to solve. In the region of larger α, the instances are so overconstrained that DPLL finds quickly a contradiction showing that the instance is UNSAT.
In [47], P. Beame, R.Karp, T. Pitassi, and M. Saks show that the size of a certificate for an unsatisfiable instance of random 3-SAT is with high probability (w.h.p.) bounded from above by 2cN/α, with c some contant, where « with high probability » means with probability going to one in the large size limit N, M → ∞ at a fixed ratio α = M/N. This results confirms the decreasing complexity observed in the unsatisfiable phase when α increases above the satisfiability threshold αsat. In [48], A. Frieze and S. Suen show that in the satisfiable phase a modified version of DPLL without backtracking (with the unit clause branching rule) can find solutions efficiently for small enough densities: up to α ≤ 3.003 for k = 3, which is strictly smaller than the satisfiability threshold prediction αsat(k = 3) = 4.267 ([7]). This confirms the experimental observation that DPLL works in polynomial time for small enough densities. In [49], S. Cocco and R. Monasson provide a theoretical estimation of the algorithmic threshold for DPLL on random 3-SAT, introducing the random (2 + p)-SAT to study the evolution of the formula under the sequential assignments, with p the fraction of 3-clauses in the formula. They predict that above the density α ‘ 3.003, the running time of DPLL with the unit clause branching rule becomes typically exponential.
The algorithmic barrier
The numerical study of the performances of DPLL on the random k-SAT en-semble indicates that in the region of α close to the satisfiability threshold the typical instances are hard to solve. In the next chapter we will introduce sev-eral incomplete algorithm that are designed to search for solutions, and we will compare their performances on the random ensembles. In particular, one is in-terested in determining the interval of α where it is possible to find an algorithm that is successful on typical instances. As we have seen this interval has to be in the satisfiable region α < αsat(k), but one could ask whether there exists a range in which typical instances have solutions, but no algorithm is able to find one. More precisely, one is interested in the putative algorithmic barrier αalg(k), above which no algorithm is able to find a solution in polynomial time, assuming P=6NP, and for typical instances. Is it possible to show that αalg(k) coincides with αsat(k), by exhibiting an algorithm efficient in the entire satisfiable region? Or are there limitations that force the strict inequality αalg(k) < αsat(k)? The structure of the set of solutions of typical instances undergoes a series of phase transitions in the satisfiable phase, that are predicted by the cavity method. It is therefore interesting to know if some of these phase transitions affect the per-formances of algorithms. For instance, algorithms such as Monte Carlo Markov Chains for the resolution of CSPs are affected by the clustering transition.
Introduction
1 Definitions
1.1 Constraint Satisfaction Problems
1.1.1 Definitions
1.1.2 The satisfiability problem
1.1.3 A graphical representation
1.1.4 More examples of CSP
1.1.5 Other combinatorial optimization problems
1.2 Worst-case complexity
1.2.1 Time complexity
1.2.2 Complexity classes
1.3 An algorithm for the SAT problem
1.3.1 The Unit Clause Propagation procedure
1.3.2 The DPLL algorithm for the SAT problem
1.4 Performances of the algorithms and random CSP ensembles
1.4.1 Complete and incomplete algorithms
1.4.2 Random CSP ensembles
1.4.3 Performances of DPLL on the random k-SAT ensemble
1.4.4 The algorithmic barrier
1.4.5 Some properties of the random hypergraph ensembles
1.5 Statistical physics and constraint satisfaction problems
1.5.1 Boltzmann probability measure
1.5.2 Statistical physics models
1.5.3 Graphical models
2 Phase transitions in random CSPs
2.1 The satisfiability transition
2.1.1 Upper bounds
2.1.2 Lower bounds
2.2 Quenched and annealed averages
2.3 Overview of the phase transitions in random CSPs
2.3.1 The clustering transition
2.3.2 Cluster decomposition
2.3.3 Complexity and the condensation transition
2.3.4 Computing the complexity function
2.3.5 Computing c and sat from the complexity
2.3.6 Rigidity and freezing transitions
2.3.7 Some values of the thresholds
3 Local Search Algorithms
3.1 Simulated Annealing
3.1.1 Monte Carlo method
3.1.2 Metropolis algorithm
3.1.3 Heat bath algorithm
3.1.4 Relaxation time scale of the Monte Carlo dynamics
3.1.5 Cooling scheme
3.2 Focused algorithms
3.2.1 RandomWalkSAT algorithm
3.2.2 WalkSAT algorithm
3.2.3 Focused Metropolis algorithm
4 Message Passing Algorithms
4.1 Belief propagation
4.1.1 BP messages
4.1.2 BP iterations
4.1.3 Marginals
4.1.4 Bethe free entropy and Entropy
4.1.5 Hard-fields
4.1.6 Warning Propagation
4.1.7 Survey Propagation
4.2 Algorithms
4.2.1 Sampling procedure
4.2.2 BP-guided decimation
4.2.3 SP-guided decimation
5 Performance of the algorithms on random Constraint Satisfaction Problems
5.1 Small connectivity k
5.1.1 Two numerical experiments on local search algorithms at small k
5.1.2 Overview of the algorithmic performances for small values of k
5.1.3 Frozen variables and whitening dynamics
5.1.4 Analytical study of the BP-guided decimation
5.2 Large connectivity k
5.2.1 Asymptotic expansion of the thresholds
5.2.2 Algorithmic performances in the large k limit
6 Biased Measures for random Constraint Satisfaction Problems
6.1 Definitions
6.1.1 Biased measure over the set of solutions
6.1.2 Intra-clause bias
6.1.3 Bias with interactions at distance 1
6.1.4 Bias with interactions at larger distance
6.2 Biased measures studied in the literature
6.2.1 Bias counting the frozen variables
6.2.2 Local entropy
6.2.3 Biased interactions in the hard sphere packing problem
7 Cavity Method
7.1 Definition of the model
7.1.1 Graphical model
7.1.2 BP equations and Bethe free-energy
7.2 Replica symmetric cavity method
7.2.1 RS cavity equations
7.2.2 Intra-clause bias
7.2.3 Bias with interaction at distance 1
7.3 The dynamic transition
7.3.2 The reconstruction problem and its recursive distributional equations
7.4 Hard fields and the naive reconstruction
7.4.1 Intra-clause bias
7.4.2 Bias with interactions at distance 1
7.5 The distribution of soft-fields
7.5.1 Uniform measure
7.5.2 Bias with interaction at distance 1
7.6 1RSB formalism
7.6.1 1RSB cavity equations
7.6.2 Simplifications for X = 1
7.6.3 Complexity (X = 1) and condensation threshold
7.7 Kesten-Stigum bound
7.7.1 Intra-clause bias
7.7.2 Bias with interactions at distance 1
8 Finite k results
8.1 Numerical resolution for finite k
8.2 On the numerical determination of the dynamic transition
8.2.1 Scalar bifurcations
8.2.2 Discontinuous functional bifurcations
8.2.3 The stability parameter in the functional case
8.3 Results of the cavity method
8.3.1 The existence of a RS phase for > d,u
8.3.2 More detailed zero temperature phase diagrams
8.4 Results of Simulated Annealing
8.4.1 Estimating the algorithmic threshold for Simulated Annealing
8.4.2 Performances of Simulated Annealing with optimal RS parameters
8.4.3 Performances of Simulated Annealing with the biased measure
8.5 Comparison with the biased measure with interactions at distance
9 The asymptotics of the clustering transition for the uniform measure
9.1 The large k limit for a finite distance n
9.1.1 Evolution of the hard fields
9.1.2 The reduced order parameter
9.1.3 Evolution of the soft fields distribution
9.2 The limit of large distance n
9.2.1 The regime
9.2.2 The difficulties for < 1
9.2.3 Reweighted probability distributions
9.2.4 Numerical resolution
9.2.5 The fixed point equation and the determination of d
9.2.6 An analytic lower bound on d
10 The asymptotics of the clustering transition for biased measures
10.1 A first upper-bound for the intra-clause bias
10.2 The asymptotics of the clustering threshold
10.2.1 Setting
10.2.2 A specialization of some formulas
10.3 The large k limit for a finite distance n
10.3.1 Evolution of the hard fields
10.3.2 Evolution of the soft fields distribution
10.4 The limit of large distance n
10.4.2 A reweighting scheme
10.4.3 A Gaussian approximation for the quasi-hard fields
10.4.4 Algorithmic implementation
10.5 Results
Conclusion
Bibliography
GET THE COMPLETE PROJECT | 7,241 | 30,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-27 | latest | en | 0.919555 |
https://kerodon.net/tag/045S | 1,722,752,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00703.warc.gz | 267,701,430 | 5,753 | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Theorem 8.2.5.1. Let $\lambda : \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}^{\operatorname{op}}_{-} \times \operatorname{\mathcal{C}}_{+}$ be a coupling of $\infty$-categories which is representable by a functor $G: \operatorname{\mathcal{C}}_{+} \rightarrow \operatorname{\mathcal{C}}_{-}$. Then a functor $F: \operatorname{\mathcal{C}}_{-} \rightarrow \operatorname{\mathcal{C}}_{+}$ corepresents the coupling $\lambda$ if and only if it is left adjoint to $G$.
Proof of Theorem 8.2.5.1. Let $\lambda : \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}^{\operatorname{op}}_{-} \times \operatorname{\mathcal{C}}_{+}$ be a coupling of $\infty$-categories which is representable by a functor $G: \operatorname{\mathcal{C}}^{\operatorname{op}}_{-} \rightarrow \operatorname{\mathcal{C}}_{+}$. By virtue of Proposition 8.2.5.2, the functor $G$ admits a left adjoint if and only if the coupling $\lambda$ is corepresentable. If this condition is satisfied, then there exists a functor $F: \operatorname{\mathcal{C}}_{-} \rightarrow \operatorname{\mathcal{C}}_{+}$ which corepresents the coupling $\lambda$; moreover, $F$ is uniquely determined up to isomorphism (Theorem 8.2.0.4). We will complete the proof by showing that $F$ is a left adjoint of $G$.
Choose a diagram
8.50
$$\begin{gathered}\label{equation:adjunction-and-duality0} \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{ \widetilde{G} } \ar [d]^{\lambda } & \operatorname{Tw}( \operatorname{\mathcal{C}}_{-} ) \ar [d] \\ \operatorname{\mathcal{C}}^{\operatorname{op}}_{-} \times \operatorname{\mathcal{C}}_{+} \ar [r]^-{ \operatorname{id}\times G } & \operatorname{\mathcal{C}}_{-}^{\operatorname{op}} \times \operatorname{\mathcal{C}}_{-}, } \end{gathered}$$
which exhibits $\lambda$ as represented by $G$ (see Definition 8.2.3.5). Then, for each universal object $C \in \operatorname{\mathcal{C}}$, the image $\widetilde{G}(C) \in \operatorname{Tw}( \operatorname{\mathcal{C}}_{-} )$ corresponds to an isomorphism in the $\infty$-category $\operatorname{\mathcal{C}}_{-}$ (Proposition 8.2.3.7). Using Proposition 8.2.4.3, we can choose a commutative diagram
8.51
$$\begin{gathered}\label{equation:adjunction-and-duality} \xymatrix@R =50pt@C=50pt{ \operatorname{Tw}( \operatorname{\mathcal{C}}_{-} ) \ar [r]^-{ \widetilde{F} } \ar [d] & \operatorname{\mathcal{C}}\ar [d]^{ \lambda } \\ \operatorname{\mathcal{C}}_{-}^{\operatorname{op}} \times \operatorname{\mathcal{C}}_{-} \ar [r]^-{ \operatorname{id}\times F} & \operatorname{\mathcal{C}}^{\operatorname{op}}_{-} \times \operatorname{\mathcal{C}}_{+} } \end{gathered}$$
which exhibits $\lambda$ as corepresented by $F$ (see Variant 8.2.4.5). It follows that the composite functor $\widetilde{F} \circ \widetilde{G}: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}$ carries universal objects of $\operatorname{\mathcal{C}}$ to couniversal objects of $\operatorname{\mathcal{C}}$. Applying Theorem 8.2.4.6, we deduce that the diagram
$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r]^-{ \widetilde{F} \circ \widetilde{G} } \ar [r] \ar [d]^{\lambda } & \operatorname{\mathcal{C}}\ar [d]^{\lambda } \\ \operatorname{\mathcal{C}}^{\operatorname{op}}_{-} \times \operatorname{\mathcal{C}}_{+} \ar [r]^-{ \operatorname{id}\times (F \circ G)} & \operatorname{\mathcal{C}}^{\operatorname{op}}_{-} \times \operatorname{\mathcal{C}}_{+} }$
is couniversal when viewed as an object of the $\infty$-category $\operatorname{Fun}_{\pm }(\operatorname{\mathcal{C}}, \operatorname{\mathcal{C}})$.
In particular, there exists an (essentially unique) morphism
$\widetilde{\epsilon }: (\operatorname{id}_{\operatorname{\mathcal{C}}_{-}}, \widetilde{F} \circ \widetilde{G}, F \circ G) \rightarrow ( \operatorname{id}_{\operatorname{\mathcal{C}}_{-}}, \operatorname{id}_{ \operatorname{\mathcal{C}}}, \operatorname{id}_{ \operatorname{\mathcal{C}}_{+} } )$
in the $\infty$-category $\{ \operatorname{id}_{ \operatorname{\mathcal{C}}_{-}} \} \times _{ \operatorname{Fun}( \operatorname{\mathcal{C}}_{-}, \operatorname{\mathcal{C}}_{-} )^{\operatorname{op}} } \operatorname{Fun}_{\pm }(\operatorname{\mathcal{C}}, \operatorname{\mathcal{C}})$. Let $\epsilon : F \circ G \rightarrow \operatorname{id}_{ \operatorname{\mathcal{C}}_{+} }$ denote the image of $\widetilde{\epsilon }$ under the forgetful functor $\operatorname{Fun}_{\pm }( \operatorname{\mathcal{C}}, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( \operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{C}}_{+} )$. We will show that $\epsilon$ is the counit of an adjunction between $F$ and $G$.
Using Example 8.2.2.12, we see that the diagram
$\xymatrix@R =50pt@C=50pt{ \operatorname{Tw}( \operatorname{\mathcal{C}}_{-} ) \ar [r]^-{ \operatorname{id}} \ar [d] & \operatorname{Tw}( \operatorname{\mathcal{C}}_{-} ) \ar [d] \\ \operatorname{\mathcal{C}}_{-}^{\operatorname{op}} \times \operatorname{\mathcal{C}}_{-} \ar [r]^-{ \operatorname{id}\times \operatorname{id}} & \operatorname{\mathcal{C}}_{-}^{\operatorname{op}} \times \operatorname{\mathcal{C}}_{-} }$
is couniversal when viewed as an object of the $\infty$-category $\operatorname{Fun}_{\pm }( \operatorname{Tw}(\operatorname{\mathcal{C}}_{-}), \operatorname{Tw}(\operatorname{\mathcal{C}}_{-} ) )$. In particular, there exists an (essentially unique) morphism
$\widetilde{\eta }: ( \operatorname{id}_{ \operatorname{\mathcal{C}}_{-}^{\operatorname{op}} }, \operatorname{id}_{ \operatorname{Tw}(\operatorname{\mathcal{C}}_{-})}, \operatorname{id}_{ \operatorname{\mathcal{C}}_{-} } ) \rightarrow ( \operatorname{id}_{ \operatorname{\mathcal{C}}_{-}^{\operatorname{op}} }, \widetilde{G} \circ \widetilde{F}, G \circ F )$
in the $\infty$-category $\{ \operatorname{id}_{ \operatorname{\mathcal{C}}_{-} } \} \times _{ \operatorname{Fun}( \operatorname{\mathcal{C}}_{-}, \operatorname{\mathcal{C}}_{-} )^{\operatorname{op}} } \operatorname{Fun}_{\pm }(\operatorname{Tw}(\operatorname{\mathcal{C}}_{-}), \operatorname{Tw}(\operatorname{\mathcal{C}}_{-}))$. Let $\eta : \operatorname{id}_{ \operatorname{\mathcal{C}}_{+} } \rightarrow G \circ F$ denote the image of $\widetilde{\eta }$ under the forgetful functor $\operatorname{Fun}_{\pm }( \operatorname{Tw}(\operatorname{\mathcal{C}}_{-}), \operatorname{Tw}(\operatorname{\mathcal{C}}_{-}) ) \rightarrow \operatorname{Fun}( \operatorname{\mathcal{C}}_{-}, \operatorname{\mathcal{C}}_{-} )$. We will complete the proof by showing that $\eta$ is compatible with $\epsilon$ up to homotopy, in the sense of Definition 6.2.1.1. For this, we must verify the following:
$(Z1)$
The identity isomorphism $\operatorname{id}_{ F}$ is a composition of the natural transformations
$F = F \circ \operatorname{id}_{ \operatorname{\mathcal{C}}_{-} } \xrightarrow {\operatorname{id}\circ \eta } F \circ G \circ F \xrightarrow { \epsilon \circ \operatorname{id}} \operatorname{id}_{ \operatorname{\mathcal{C}}_{+} } \times F = F$
in the $\infty$-category $\operatorname{Fun}( \operatorname{\mathcal{C}}_{-}, \operatorname{\mathcal{C}}_{+} )$.
$(Z2)$
The identity isomorphism $\operatorname{id}_{ G }$ is a composition of the natural transformations
$G = \operatorname{id}_{ \operatorname{\mathcal{C}}_{-}} \circ G \xrightarrow {\eta \circ \operatorname{id}} G \circ F \circ G \xrightarrow { \operatorname{id}\circ \epsilon } G \times \operatorname{id}_{\operatorname{\mathcal{C}}_{+}} = G$
in the $\infty$-category $\operatorname{Fun}( \operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{C}}_{-} )$.
Using Theorem 8.2.4.6, we deduce that the diagram (8.50) is a couniversal object of $\operatorname{Fun}_{\pm }( \operatorname{\mathcal{C}}, \operatorname{Tw}(\operatorname{\mathcal{C}}_{-} )$: that is, it is initial when viewed as an object of the $\infty$-category $\operatorname{\mathcal{E}}= \{ \operatorname{id}_{ \operatorname{\mathcal{C}}_{-} } \} \times _{ \operatorname{Fun}( \operatorname{\mathcal{C}}_{-}, \operatorname{\mathcal{C}}_{-} )^{\operatorname{op}}} \operatorname{Fun}_{\pm }( \operatorname{\mathcal{C}}, \operatorname{Tw}(\operatorname{\mathcal{C}}_{-}) )$. It follows that the diagram
$\xymatrix@R =50pt@C=50pt{ & ( \operatorname{id}_{\operatorname{\mathcal{C}}_{-}}, \widetilde{G} \circ \widetilde{F} \circ \widetilde{G}, G \circ F \circ G) \ar [dr]^-{ \operatorname{id}\circ \widetilde{\epsilon }} & \\ ( \operatorname{id}_{\operatorname{\mathcal{C}}_{-}}, \widetilde{G}, G) \ar [ur]^{\widetilde{\eta } \circ \operatorname{id}} \ar [rr]^{ \operatorname{id}} & & ( \operatorname{id}_{\operatorname{\mathcal{C}}_{-}}, \widetilde{G}, G) }$
commutes up to homotopy in $\operatorname{\mathcal{E}}$. Assertion $(Z2)$ follows by applying the forgetful functor $\operatorname{Fun}_{\pm }( \operatorname{\mathcal{C}}, \operatorname{Tw}(\operatorname{\mathcal{C}}_{-})) \rightarrow \operatorname{Fun}( \operatorname{\mathcal{C}}_{+}, \operatorname{\mathcal{C}}_{-} )$. Assertion $(Z1)$ follows by a similar argument, using the observation that the diagram (8.51) is a couniversal object of $\operatorname{Fun}_{\pm }( \operatorname{Tw}(\operatorname{\mathcal{C}}_{-}), \operatorname{\mathcal{C}})$ (Theorem 8.2.2.11). $\square$ | 3,151 | 9,280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.431538 |
https://www.cristos-vournas.com/448083807 | 1,686,125,114,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00197.warc.gz | 768,806,057 | 12,765 | # The Planet Surface Rotational Warming Phenomenon
### The Planet Surface SPECIFIC HEAT "Cp" Warming PHENOMENON
I’ll try here in few simple sentences explain the very essence of how the planet surface SPECIFIC HEAT "cp"warming Phenomenon occurs.
Lets consider two identical planets "H" and "L" at the same distance from the sun. Let’s assume the planet "H" has a Higher average surface specific heat, and the planet "L" has a Lower average surface specific heat.
Both planets "H" and "L" get the same intensity solar flux on their sunlit hemispheres. Consequently both planets receive the same exactly amount of SOLAR RADIATIVE ENERGY.
Depending on the specific heat capacity of bodies containing the heat energy, the temperatures can be quite different.
For Lower average surface specific heat planet "L" the sunlit hemisphere surface gets warmed at higher temperatures than for Higher average surface specific heat planet "H" the sunlit hemisphere.
The surfaces emit at σT⁴ intensity – it is the Stefan-Boltzmann emission law.
Thus the planet "L" emits more intensively from the sunlit hemisphere than the planet "H".
So there is MORE ENERGY LEFT for the planet "H" to accumulate then. That is what makes for Higher surface specific heat planet "H" to be a WARMER PLANET.
That is how the Planet Surface SPECIFIC HEAT "cp" Warming PHENOMENON occurs.
### .
04.03.2021 16:07
Yes, and thank you. So once L cools to planet H temperature, it continues to emit more radiation, even when it is cooler.
04.03.2021 18:12
IR emission = σTΛ4.
Planet LTday =200K, LTnight =100K
To balance the same energy in = out
Planet HTnight grows higher (100+7,5) K, than HTday goes down (200-1) K.
So the HTmean > LTmean
04.03.2021 17:55
Important - Rough example:
LTday>HTday
LTday - HTday = 20 oC
But
HTnight - LTnight = 50 oC
So HTmean>LTmean
Thanks
04.03.2021 17:43
Important topic H on average is always warmer than L because L lit surface emits more intense24/7
For L Tmean=(Tnight+Tday)/2
For H the Tnight↑↑→T↑mean ←T↓day
(Tnight↑↑+T↓day)/2>(Tnight+Tday)/2 Thanks
03.03.2021 13:22
Once the warmer planet L emitted adequet radiation to reach the same T as planet H, would not any further emissions then be equal?
04.03.2021 16:04
Yes, and thank you. Just trying to understand. I should have specified planet L as sunlight side being warmer. Can you be more specific as to your answer. Once planet L cools to planet H temperatur
03.03.2021 15:27
Thank you for asking. The planet L is not the warmer planet on average. The planet L is warmer on the sunlit side. The L is on average colder, because during the day it has already emitted more. | 711 | 2,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-23 | latest | en | 0.804984 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.