text stringlengths 1 1.11k | source dict |
|---|---|
cr.crypto-security, one-way-function, key-exchange
If yes, how is that proven? Your question suggests the following tentative reduction to obtain a OWF from a secure KE: Given an input, interpret it as the private random coins of two simulated parties to the KE protocol. Based on their private randomness, there will be some sequence of public messages between them, and at some point they will stop and agree on a private shared key (with some high probability). Then the public transcript of their communications is the output of the OWF. | {
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electric-circuits, electrical-resistance
I observe this every time I do an experiment on parallel resistors or solve a parallel combination problem.
How can we prove $R_{ev}<R_{1},R_{2},R_{3},...R_{n}$ or that $R_{ev}$ is less than the Resistor $R_{min}$, which has the least resistance of all the individual resistors? Think about current flow.
If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$
Dividing everything by the voltage give us:
$$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$
Which is the same as:
$$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$
Since there is more current flowing in all the resistors than through just one resistor, then the equivalent resistance must be less than the individual resistors. | {
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Part 1 of 3. and inside x2 +y2 = 4. Single Integral - the domain is the integral I (a line). set up the triple integral in terms of single integrals, but do not evaluate it). Z b a Z g 2(x) g 1(x) Z u 2(x;y) u 1(x;y) F(x;y;z)dzdydx : Now evaluate that iterated integral by go-ing from the inside to the outside. Example Use cylindrical coordinates to find the volume of a curved wedge cut out from a cylinder (x − 2)2 + y2 = 4 by the planes z = 0 and. Usually, one direction is considered to be positive, the other negative. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each \(\Delta x\times\Delta y\times\Delta z. Problems: 5,7,9,13,17,33 3. Step 1: Draw a picture of E and project E onto a coordinate plane. Triple integral is an integral that only integrals a function which is bounded by 3D region with respect to infinitesimal volume. Triple Integrals in Cylindrical Coordinates A point in space can be located by using | {
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"url": "http://cghb.bollola.it/triple-integral-pdf.html"
} |
photography, gaia, instruments, image-processing, charge-coupled-devices
Title: What was the first use of time-delay integration in Astronomy? Are there instances before GAIA? In the context of GAIA, time-delay integration is the clocking of a CCD's charge shift register at a frequency such that it matches the linear speed across the focal plane of the actual image. So even though the image and data are moving continuously, their rates are locked together and the data coming out the end of the row is un-smeared.
As background only, the first time I'd seen this speed-matching technique on a 2D imaging sensor used was in semiconductor wafer and mask inspection tools produced by KLA. Here is a patent filed in 1988 US4877326A | {
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demodulation
$$Smoothed(k) = \frac{1}{2-2\cos(\omega)}( Sig(k) - 2\cos(\omega) Sig(k-1) + Sig(k-2) )$$
where $\omega = 2\pi\frac{F_i}{F_s}$ (note that, if $F_i=16000$ and $F_s=48000$, we get $\omega=\frac{2\pi}{3}$ and hence $\cos(\omega)=-\frac{1}{2}$. If you substitute this into the second equation, you will simply get the first equation again.
This method is simply creating a $2^{nd}$ order FIR filter with zeros at $\pm F_i$, and with the gain at DC normalised to unity. | {
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ros, roslaunch, turtlebot, turtlebot-bringup, icreate
[ 1090.506012] init: turtlebot main process ended, respawning
[ 1090.654396] init: turtlebot main process (10367) terminated with status 1
[ 1090.654522] init: turtlebot main process ended, respawning
[ 1090.806787] init: turtlebot main process (10386) terminated with status 1
[ 1090.806871] init: turtlebot main process ended, respawning
[ 1090.950699] init: turtlebot main process (10404) terminated with status 1
[ 1090.950777] init: turtlebot main process ended, respawning
[ 1091.095898] init: turtlebot main process (10422) terminated with status 1
[ 1091.096100] init: turtlebot main process ended, respawning
[ 1091.250507] init: turtlebot main process (10440) terminated with status 1
[ 1091.250590] init: turtlebot respawning too fast, stopped
[ 1093.583821] init: turtlebot main process (10477) terminated with status 1
[ 1093.583907] init: turtlebot main process ended, respawning
[ 1093.710527] init: turtlebot main process (10499) terminated with status 1 | {
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dirac-equation, dirac-matrices, clifford-algebra
Thoughts?
Cheers The Euler-Lagrange-Equation is given by:
$$\frac{\partial\mathcal{L}}{\partial {\psi}} - \partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi)} = 0$$
Let us take both derivatives separately. We treat $\psi$ and $\bar{\psi}$ as independent fields. This gives
$$\frac{\partial}{\partial\psi}\bar{\psi}(i\gamma^\mu\partial_{\mu}-m)\psi = -m \bar{\psi}$$
and
$$\partial_{\mu}\frac{\partial}{\partial(\partial_{\mu}\psi)}\bar{\psi}(i\gamma^\mu\partial_{\mu}-m)\psi = i\partial_{\mu}\bar{\psi}\gamma^\mu$$
So by plugging this into the first equation we get:
$$-m\bar{\psi} - i\partial_{\mu}\bar{\psi}\gamma^\mu = 0 $$
Now, to make this look nicer (and bring this in the usual form) we say, that the differential operator operates to the left:
$$\bar{\psi}(i\gamma^{\mu} \overleftarrow{\partial}_{\mu}+m) = 0$$ | {
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kinematics
Title: Elementary question about velocity When we talk about the nonvanishing velocity of an object, are we assuming we are static relative to the object or not? Velocity is always relative; there is no such thing as an absolute velocity. If you measure an object to have a non-zero velocity that automatically means it is moving with respect to you. Conversely, if you are static relative to the object that automatically means its velocity is zero relative to you. | {
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python, object-oriented, game, rock-paper-scissors
Title: Rock-Paper-Scissors in procedural and object-oriented I am comfortable with the procedural style, and learning the object oriented, so I have done a small Rock Paper Scissors game in both styles.
This script is in procedural style and this one is in object oriented style. Did I implement this little game correctly object oriented wise? And if no, where did I mess things up? And how can I improve them?
This is the code in procedural style:
import random, sys
def computer_choice():
computer_choice = random.choice("rps")
return computer_choice
def user_choice():
user_ch = raw_input(" Your Choice ? ")
if user_ch != "r" and user_ch != "p" and user_ch != "s":
print " Wrong input, Please try again.\n"
user_ch = user_choice()
return user_ch | {
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java
both if and else block is setting a boolean. But I don't find any use of it in your code. boolean prodPresent = false
How do you get access to items ? is it instance variable ?
The if and else block has lot of common code which can be written outside of the condition.
Here I am not concerned with your logic but tried to organize the second version of the code which you posted. My intention is to avoid repetitive code in both the if-else block
public void setOrderedProduct(String prodName, int eoq) {
PurchaseOrderDTO itemData = new PurchaseOrderDTO();
String productCd = getProdCd(prodName);
itemData.setProdname(prodName);
itemData.setQty(eoq);
itemData.setProdcd(productCd);
double costPrice = ProductView.getProductCost(vendorno, productCd);
itemData.setPrice(costPrice);
double extPrice = ProductView.getProductMSRP(vendorno, productCd);
itemData.setExt(extPrice); | {
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Similar to the example above, the returned value for results is the same:
> $\mathrm{results}\left[1\right]$
${\mathrm{DataSeries}}{}\left(\left[\begin{array}{ccccccc}3.2& 0.8366600265340756& -0.30734449954313& 1.4775510204081639& 2.0& 4.0& 5.0\end{array}\right]{,}{\mathrm{labels}}{=}\left[{\mathrm{mean}}{,}{\mathrm{standarddeviation}}{,}{\mathrm{skewness}}{,}{\mathrm{kurtosis}}{,}{\mathrm{minimum}}{,}{\mathrm{maximum}}{,}{\mathrm{cumulativeweight}}\right]{,}{\mathrm{datatype}}{=}{\mathrm{anything}}\right)$ (10)
The tableweights option controls the width of columns in an embedded table.
> $\mathrm{interface}\left(\mathrm{displayprecision}=4\right):$
> $\mathrm{DataSummary}\left(\mathrm{df},\mathrm{summarize}=\mathrm{embed},\mathrm{tableweights}=\left[4,2,2,2\right]\right):$ | {
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rosmake, ogre
cmake_minimum_required(VERSION 2.4.6)
include($ENV{ROS_ROOT}/core/rosbuild/rosbuild.cmake)
rosbuild_init()
set(EXECUTABLE_OUTPUT_PATH ${PROJECT_SOURCE_DIR}/bin)
set(LIBRARY_OUTPUT_PATH ${PROJECT_SOURCE_DIR}/lib)
rosbuild_genmsg()
rosbuild_gensrv()
set(CMAKE_MODULE_PATH "/usr/local/lib/OGRE/cmake/")
find_package(OIS REQUIRED)
find_package(OGRE REQUIRED)
include_directories(${OIS_INCLUDE_DIRS})
include_directories(${OGRE_INCLUDE_DIRS})
set(OGRE_LIBRARIES ${OGRE_LIBRARIES} ${OGRE_Terrain_LIBRARIES})
set(LIBS ${LIBS} ${OIS_LIBRARIES} ${OGRE_LIBRARIES} ${OGRE_TERRAIN_LIBRARIES})
set(Sim1Sources ./src/sim1.cpp ./src/BaseApplication.cpp)
set(Sim1Headers ./src/sim1.h ./src/BaseApplication.h)
rosbuild_add_executable(talker src/talker.cpp)
rosbuild_add_executable(listener src/listener.cpp)
rosbuild_add_executable(Sim1 ${Sim1Sources} ${Sim1Headers})
target_link_libraries(Sim1 ${LIBS})
Originally posted by eqzx on ROS Answers with karma: 13 on 2011-06-22
Post score: 1 | {
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fluid-mechanics, homework
What is the density and the speed of the mixture coming out of the outlet?
The answer in the back of the textbook says the correct answer is 2.6 m/s for the velocity and 1461 kg/m3 for the density. Does anyone know why my numbers are slightly off, is it possibly due to the fact that fluid B enters normal to the surface. I have attached the question below.
Thanks in advance for any help The answer rounded the exit velocity to 2.6 instead of 2.67, then, again, dropped the fraction (0.54) after the decimal point:
$\gamma_c = \dfrac{(1200*0.1*2)+(1500*0.2*3)}{(0.3*2.6)} = 1461(.54)$ | {
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all you have is a fair coin. coin toss probability calculator,monte carlo coin toss trials. An unfair coin having probability of showing Head p is flipped 6 times. It can be calculated by dividing the number of possible occurrence by the total number of options. In general, the probability vanishes, pn(M) = 0, for M < n since it's impossible to have n consecutive heads with fewer than n total flips. The coin is tossed four times. Flipping and Spinning Coins --- The Simplest Probability Models. 7: THEORETICAL PROBABILITY 41 A suburban high school has a population of 1376 students. "Likewise, if Mr. Then, the probability of getting Heads on any. This feature is not available right now. This number is always between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. What is the probability of selecting a girl. In our sampling distribution we defined 100 values from 0 to 1 for our parameter p. All of the experiments above involved independent events with a small population | {
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c#
return result;
}
/// <summary>
/// Implicit conversion operator from bool to Pin.
/// </summary>
/// <param name="val">
/// The bool to convert to Pin.
/// </param>
/// <returns>
/// The corresponding Pin.
/// </returns>
/// <remarks>
/// TRUE corresponds to High.
/// FALSE corresponds to Low and Undefined.
/// </remarks>
public static implicit operator Pin(bool val)
{
return new Pin(val ? PinValue.High : PinValue.Low);
} | {
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c#, wpf
if (first10.Text != empty & score10.Text != empty & last10.Text != empty)
{
student10 = new Student(first10.Text, last10.Text, Convert.ToInt32(score10.Text));
if (first11.Text != empty & score11.Text != empty & last11.Text != empty)
{
student11 = new Student(first11.Text, last11.Text, Convert.ToInt32(score11.Text));
if (first12.Text != empty & score12.Text != empty & last12.Text != empty)
{
student12 = new Student(first12.Text, last12.Text, Convert.ToInt32(score12.Text)); | {
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c++, c, makefile
make[2]: *** [report/test] Error 2
/home/travis/build/Loki-Astari/ThorsSerializer/build/tools/Project.Makefile:49: recipe for target 'ThorsStorage.dir' failed
make[1]: *** [ThorsStorage.dir] Error 2
/home/travis/build/Loki-Astari/ThorsSerializer/build/tools/Project.Makefile:49: recipe for target 'src.dir' failed
make: *** [src.dir] Error 2
The command "export PATH=${PATH}:$(pwd)/build/bin;make" exited with 2. | {
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quantum-mechanics, quantum-information
Title: Maximum number of bits in one single qubit? How many bits can a qubit hold?
I know there are probabilities involved in a qubit, but there must be some limit to how many bits a qubit can take. If different qubits can have different amounts of information then we can think of these as different sizes of processor-registers: $8$-bit, $16$-bit, $32$-bit or $64$-bit, can't we?
But my question is really, is there a limit to how much information content (discreete-wise) a qubit can store? The answer to your question is "1, but with some qualifications". I start by saying "1" because there is a bound called Holevo bound which says the following.
If you have $N$ classical bits of information, and wish to send them using qubits from $A$ to $B$, in such a way that $B$ can confidently reconstruct what the classical information was, say by writing it down in a book, then you will need $N$ qubits. | {
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ros, c++
Originally posted by lfr on ROS Answers with karma: 201 on 2016-05-30
Post score: 0
Original comments
Comment by BennyRe on 2016-05-30:
Everything looks good to me. Can you please post your complete CMakeLists.txt?
Comment by lfr on 2016-05-31:
I updated the question (with the new CMakeLists.txt)
Comment by BennyRe on 2016-06-06:
In your question you write that you do add_executable(main_test src/main_test.cpp src/functions_test.cpp) which is correct. In your CMakeLists.txt you don't do this. You write that you updated your file, was it this what you updated?
Comment by lfr on 2016-06-06:
Yes, it was what I updated because it didn't works and I wanted to try what spmaniato advises in his answer.
Hello !
I found my mistake. I forgot that templates functions must be defined inside the header file. It works properly now.
Thank you to all of you for helping me.
lfr.
Originally posted by lfr with karma: 201 on 2016-06-14
This answer was ACCEPTED on the original site
Post score: 0 | {
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java, beginner, game, javafx
winNote.setText("O Wins!");
}
// ENDS THE GAME IF SOMEONE WINS
if (winNote.getText().equals("X Wins!") || winNote.getText().equals("O Wins!"))
gameCont = false;
// BELOW IS THE DRAW CHECK, ENDS THE GAME IF PASSED
else if (!(t11.text.getText().equals("")) && !(t12.text.getText().equals("")) && !(t13.text.getText().equals("")) && !(t21.text.getText().equals("")) && !(t22.text.getText().equals("")) && !(t23.text.getText().equals("")) && !(t31.text.getText().equals("")) && !(t32.text.getText().equals("")) && !(t33.text.getText().equals(""))) {
winNote.setText("Draw!");
gameCont = false;
}
}
}
}
});
}
} | {
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points P 1, P 2, and P 3 is given by = | → − → | , where θ is the angle ∠P 1 P 2 P 3.This formula uses the law of sines.If the three points are given by their coordinates (x 1,y 1), (x 2,y 2), and (x 3,y 3), the radius can be expressed as The area of the clock is 50 in2. Drag either orange dot at the ends of the diameter line. lessons in math, English, science, history, and more. {{courseNav.course.topics.length}} chapters | Also find the Area of a circle. That is, the exercise will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember (and apply) the Formula. What is the standard form equaton of a circle? Circles. If the area of a circular orange slice is 12 in2 , what is its radius? A circle can have many radii (the plural form of radius) and they measure the same. Radius is 1 2 the diameter. This website uses cookies to ensure you get the best experience. Example: Find the diameter of the | {
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• It is an iteration of a holomorphic function near an attracting fixed point, and the asymptotics should be determinable from the leading term or two of the power series near the fixed point. I would not expect that there is anything special about iterating radicals other than the power series behavior near $x$. Thus, it is very believable that there is a formula as you wrote, including for non-integer $k$. – zyx Nov 24 '13 at 20:12
• From the definition of $C_k$, as $k\to\infty$, is it expected that $C_k$ is finitely bounded (instead of getting smaller and smaller)? Furthermore, why is the bound apparently $\ln2$? – Tito Piezas III Nov 24 '13 at 20:24
• I suppose it might be worth knowing that $$x = 1 + \frac{\log 2}{k} + \frac{\log 2 + (\log 2)^2}{2k^2} + O\left(\frac{1}{k^3}\right).$$ – Antonio Vargas Nov 24 '13 at 20:36
• Hm. Interesting that the appropriate root of $x^k=x+1$ can be approximated by that formula... – Tito Piezas III Nov 24 '13 at 20:43 | {
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"lm_q2_score": 0.84997116805678,
"openwebmath_perplexity": 425.64481926582175,
"openwebmath_score": 0.9718883037567139,
"tags": null,
"url": "https://math.stackexchange.com/questions/579541/on-the-paris-constant-and-sqrtk1-sqrtk1-sqrtk1-sqrtk1-dots"
} |
organic-chemistry, reaction-mechanism, carbonyl-compounds, carbocation, rearrangements
As shown by the products, all of these reactions have undergone a rearrangement to give perticular product except for the nitrile formation from the aldehyde.
The mechanism of the Schmidt reaction is well understood (Ref.2). In the case of a carboxylic acid reaction with hydrazoic acid, the initial formation of an acylium ion is followed by the addition of $\ce{HN3}$ to form an acyl azide intermediate. The possible migration of alkyl or acyl group produces the corresponding isocyanate, which undergoes hydrolysis to yield an amine (not an amide) and carbon dioxide:
References:
Karl Friedrich Schmidt, “Process of making derivatives of hypothetical imines including amines and their substitution products,” U.S. Patent 1,564,631, 1925.
Peter A. S. Smith, “The Schmidt Reaction: Experimental Conditions and Mechanism,” J. Am. Chem. Soc. 1948, 70(1), 320–323 (DOI: https://doi.org/10.1021/ja01181a098). | {
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geography, geomorphology, satellites, satellite-oddities
The U.S. science team is located at NASA's Jet Propulsion Laboratory, Pasadena, Calif. The Terra mission is part of NASA's Science Mission Directorate.
Image Credit: NASA/GSFC/METI/ERSDAC/JAROS, and U.S./Japan ASTER Science Team
Image Addition Date: 2008-08-08 The pattern shown in this image has an amazing amount of approximate repetition of shapes; long straight segments with a periodicity of about 2 kilometers, "stair steps" northward and sharp, pointy extensions southward. Is there any understanding how such a distinct pattern could be formed? | {
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"tags": "geography, geomorphology, satellites, satellite-oddities",
"url": null
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electromagnetism, electromagnetic-radiation, photons, polarization, antennas
Title: Exotic types of polarization I am aware that in the classical approximation of electromagnetic waves, waves can be linearly polarized (so that the B-field oscillates in one dimension as $B=B_0\cos(\omega t)$), circularly polarized (so that the field is $B=B_0(\cos(\omega t)\hat i+\sin(\omega t)\hat j)$, or elliptically polarized. My question is: why are these the only options? For instance, I know that circularly/elliptically polarized light can be generated by putting two antenna perpendicular to each other - couldn't putting three antenna perpendicular to each other generate a more complex pattern? I'm especially interested in patterns that involve the magnetic field vector moving in three dimensions, rather than just one or two. | {
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gravity, celestial-mechanics
$$\begin{aligned}
1 + q &\approx 1 \\
\left(1 - z\right)^{-2} &\approx 1 + 2z
\end{aligned}$$
Where the second line is the binomial approximation. This gives:
$$1 - z \approx 1 + 2z - qz^{-2}$$
Rearrange to solve for $z$:
$$z^3 \approx \frac{q}{3}$$
And then using the definitions of $z$ and $q$ this becomes
$$h \approx r \left(\frac{m_2}{3 m_1}\right)^{1/3}$$
Which is the usual formula for the size of the Hill sphere. | {
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statics
Title: Can you assume that the forces in the vertical direction are evenly distributed for an off level rigging problem? This is an off-level rigging problem. When solving for the tension forces in the red slings (lifting from point A), can the assumption be made that the y force vector at points B and C are each half of the weight of the object? Or must one calculate the force vectors at point A? In which case the y vectors are not symmetrical. No.
To convince yourself, why an equal distribution of the vertical component is wrong, try looking at the corresponding horizontal components (Kamran has calculated these values in his answer): The horizontal component at C is twice as large as the horizontal component at B, which results in a net horizontal load on the object, giving it a constant, horizontal acceleration off to the side. (And there is no moment equilibrium, so it would also rotate.) | {
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javascript, performance
oLAZB4o9hqgqopATY3x0mlhPO/lNtj/s19G3TDfwN+scHilCYWMsm3hOc60WyBDIarCYSMcR3i2fVzA5jYYfKCnWbH9HX4t4c8S1sh4XOZzOWTxWeafO7RtE4pE8NJpreGYhm1/gG8f9HEqgQAQBsICwsaPwdBr8NsGWBaFBgmRwn7bGbhyBb7dC72AI0Ab/0KdKFysxYR43x4QhcCvKpAJFggFC+PFuJSFeKMs6dnF2qOsHNCTgfwfl5gsFvMgXIgAAAAASUVORK5CYII=", | {
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"tags": "javascript, performance",
"url": null
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thermodynamics, energy, pressure, temperature
Title: Can someone explain (visually) how a throttling device works? I know that the throttling device reduces pressure of the gas but that's it. I can't somehow imagine how the change in cross-sectional area can affect how the particles of gas to lose energy and see a decrease in pressure?! Can someone perhaps visually, using a diagram, even if it's just doodles, explain how the individual particles are affected? I searched all over and can't get a satisfying explanation. A throttling device is basically a flow restrictor which, when flow is asserted through it, exhibits a pressure drop the same way pushing current through an electrical resistor develops a voltage drop across it.
Take the example of a throttling valve on a compressed air tank that is exhausted to ambient pressure. | {
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hydrogen-bond, fats
Or upon reflection maybe my question is, why are trans fats considered so much worse than other unsaturated fats? If all hydrogenation does is break up double bonds and insert hydrogen, straightening out any cis-kinks in the chain, doesn't this just generate the same acid with fewer double bonds and straighter chains? How is this any different than going out in nature and finding the same unsaturated fat, rather than going to the trouble of converting?
What's causing the difference, here? Where is my understanding breaking down? Someone may very well give a more thorough answer but from what I understand, most naturally occurring unsaturated fats have their double bonds in the cis conformation which is generally higher in energy than the trans conformation. | {
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beam, plastic
Assume elastic behavior, $\sigma_b = \dfrac{6M}{bh^2} \leq \dfrac{F_y}{n}$, or
assume plastic behavior, $\sigma_b = \dfrac{4nM}{bh^2} \leq F_y$
The last step is to determine $P$:
Since $M = P*a$, plug $M$ into the two equations above, you will get the $P$ that satisfies the limit of the bending stress with the desired safety factor. However, you also need to back-check/compare the $P$ derived from steps two (deflection) and three (shear stress) before making the conclusion.
If you still couldn't get satisfactory force with the desired safety factor, you will have to increase the depth/thickness of the member or adjust the deflection limit.
Hope this helps. | {
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thermodynamics, energy, statistical-mechanics, entropy, reversibility
Title: $dU=dQ$ and $dU=TdS$, but $dQ$ not always equal to $TdS$? Why? $$ dU = dQ+dW $$ $$ dU=TdS-pdV $$ The equations above are always true for a thermodynamic state of a certain system. Now let's say that we have a situation where $dW=0$, this tells us that $$ dU=dQ $$ $$ dU=TdS $$But still I can't write $ dQ=TdS $, since this only works for a reversible change of my system. So if I don't have a reversible system I can work with $ dU=dQ $ and $ dU=TdS $, but I can't work with $ dQ=TdS $.
I get this, but I have been trying to figure out why this is, and I just can't seem to get it. In $dS = \frac{dQ}{T}$, the $dQ$ is the heat exchange on a reversible path from the initial state to the final state, irrespective of how the process is actually carried out. | {
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"tags": "thermodynamics, energy, statistical-mechanics, entropy, reversibility",
"url": null
} |
quantum-mechanics, newtonian-mechanics, momentum, heisenberg-uncertainty-principle, stability
If the center of mass is off by $\Delta x$, then the angle is $\Delta \theta = \frac{2\Delta x}{\ell}$.
Using the same equations as before, we find $C_1 + C_2 = \Delta \theta$. Let's assume the initial velocity is zero - because "on average" it will be, given that the direction of initial velocity is equally likely to point back towards equilibrium and away from it - so we get $C_1 = C_2$, and the solution is a $\cosh$ function:
$$\theta = 2 C_1 \cosh(\alpha t)$$
Where $C_1 = \frac{\Delta\theta}{2} = \frac{\Delta x}{\ell}$.
We now have the time to fall (time to reach a certain $\theta$) as
$$t = \frac{1}{\alpha}\cosh^{-1}\left(\frac{\theta\ell}{2\Delta x}\right)$$
The equation we derived earlier for the time taken with a given initial velocity can be rewritten as
$$t = \frac{1}{\alpha}\sinh^{-1}\left(\frac{\theta\ell\alpha}{\Delta v}\right)$$
and we know that
$$\Delta x \Delta p = \hbar$$ | {
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voice, reconstruction
However in practice the attenuation at most frequencies is quite severe. Unless the recording room is extremely quiet and the recording extremely low noise, the remaining speech signal will be below the noise floor and hence impossible to recover. | {
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be any one of its angles... Two-Dimensional shape with at least one set of parallel sides which is also a!... Quadrilateral before we look at the hierarchy congruent sides our maps knows the value of sources... Will never intersect Pythagorean theorem the C^2 is always an example of a parallelogram with a right angle can!, i.e., 90 degrees Perhacs, see, area in terms of Cartesian coordinates of vertices, arising. Parallelogram with all its interior angles are right angles are rectangles, and some are not '' ) the! Geography and finally learn what countries a parallelogram is always a rectangle if in the same area or false: in any triangle less... Of educational and reference sites for parents, teachers and students the same area look better have right angles and. Parallelogram has two pairs of parallel sides is independent of the angles constituting the figure Perhacs, see, in. Faq ; Contact the diagonal of a parallelogram with congruent diagonals ¯AC and.... Can also be a right angle | {
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"tags": null,
"url": "http://safetruck.by/pxi0yj76/a-parallelogram-is-always-a-rectangle-if-162f45"
} |
electrostatics, electric-fields, conductors
Title: Induced charge in a conductor My question here is based on the following question in Physics for Scientists and Engineers (3rd edition) by Randall Knight:
An electroscope is positively charged by touching it with a positive glass rod. The electroscope leaves spread apart and the glass rod is removed. Then a negatively charged plastic rod is brought close to the top of the electroscope, but it doesn’t touch. What happens to the leaves?
a. The leaves get closer together.
b. The leaves spread farther apart.
c. One leaf moves higher, the other lower.
d. The leaves don’t move.
The answer given is a. Let's say that the total charge on the electroscope is +Q. I'm guessing that as you bring the negatively charged rod near to the electroscope, the positive charge on the electroscope "shifts" towards the negative rod, as follows:
But it seems to me that there could (in theory?) be other possible configurations of the total charge, such as this: | {
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classification, keras, tensorflow, convolutional-neural-network, mnist
# MNIST dataset contains black and white images, so we threshold the images
localised_grid_gray = cv.cvtColor(localised_grid, cv.COLOR_BGR2GRAY)
adaptive_thresh = cv.adaptiveThreshold(localised_grid_gray, 255, cv.ADAPTIVE_THRESH_GAUSSIAN_C, cv.THRESH_BINARY_INV, 11, 2)
for i in range (0,2):
adaptive_thresh = cv.medianBlur(adaptive_thresh, 3)
cv.imshow("Thresholded", adaptive_thresh)
sudoku_grid_stored = np.empty((9,9,46,46)).astype("uint8")
for y in range(0,9):
for x in range(0,9):
cropped_digit = adaptive_thresh[50*y+2:(50*y)+48,50*x+2:(50*x)+48].astype("uint8") # Crop slightly to remove any sudoku grid outlines that may exist
sudoku_grid_stored[y][x] = cropped_digit
# cv.imshow("7", sudoku_grid_stored[2][0]) # 7 lies in 3rd row, 1st column (zero indexed)
# Inputs to CNN require must have three colour channels
sudoku_grid_stored = np.repeat(sudoku_grid_stored[..., np.newaxis], 3, -1) | {
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space amount of variation in a dataset we! First of the units of observations are devoid of a good measure dispersion! The term statistic first came into use as late as 1817 there is a site that makes learning Statistics by! How spread out ” the elements of a good job of describing this dataset of math. Unit as the mean from their average are called the dispersion data i.e of! New planets measures of dispersion examples values ; in terms of the units of observations from their average are called.! These are the variance is a relative measure average of deviations from the highest and the standard deviation is =! This lesson, you will read about the scatter of the data statistical dispersion are expressed in the.. Interpret it the distribution of the data from one another and gives a clear idea about the scatter of average... We analyze a dataset into four equal parts rangeis the difference between largest! You subtract the lowest scores in a test devoid of a Variable such as height | {
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Thus we get, $\underbrace{r\times r \times ... \times r}_{\text{n times}}=r^n$
Then I thought of ways in which n distinct things can be distributed to r different persons so that every person gets at least one should be $r^n-({}^rP_1+{}^rP_2+...+{}^rP_{r-1})$, where ${}^rP_x$ is the number of ways $x$ persons does not get any item. However, later I felt that I am not correct with "${}^rP_x$ is the number of ways $x$ persons does not get any item". It should be ${}^rP_x\times (r-x)^n$ as there are ${}^rP_x$ ways to choose persons who don't get any item and we can distribute the $n$ items to the remaining persons in $(r-x)^n$ ways. So the final solution can be:
$r^n-({}^nP_1 \times (r-1)^n +{}^nP_2 \times (r-2)^n+...+{}^nP_{n-1}\times (r-(r-1))^1)$
This looks very bad to me. Am I correct with this? Is there any better solution? | {
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ros, ros2, ardent, rosparam, parameter
Originally posted by William with karma: 17335 on 2018-06-27
This answer was ACCEPTED on the original site
Post score: 1
Original comments
Comment by Hakaishin on 2020-12-02:
Is there a way to achieve this by just using this->declare_parameter<float>("foo", 15.0); this->get_parameter("foo", foo); syntax? Or in this example how come that the parameter doesn't need to be declared? From reading a long issue involving some 2 booleans with undeclared_auto_param_something_something and another boolean I just got that you always should declare your params, but this example doesn't do that? And is the documentation 2years later still missing? | {
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biochemistry, radicals, organic-oxidation
Title: How does the human body deal with free radicals it creates itself? I know that free radicals are created all over your body all the time as a byproduct of its chemical processes, (or maybe I do not and that is false). However I am confused about the distinction between free radicals formed from UV photons that will affect DNA base pairings, and the ones that the human body creates on a consistent basis.
How does the body deal with the free radicals it created by itself?
And how are those free radicals different from the ones that cause oxidative damage (e.g., from UV rays)? Reactive oxygen species, primarily superoxide, may result from one-electron reduction of molecular oxygen. Given that aerobic metabolism involves loooooong chain of electron transfers, it isn't all that rare.
Human body uses specialized enzymes to defang reactive oxygen species. The most well known are peroxidase and superoxide dismutase (see wiki for details). | {
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quantum-mechanics, special-relativity, vectors, mathematics, linear-algebra
So, that is what is actually going on in the second equality of the special relativity formula
\begin{align}
g(A,B)=g_{\mu\nu}A^{\mu}B^{\nu}=A_{\mu}B^{\mu}.
\end{align}
You’re first taking the vector $A$, and using the metric $g$ to convert it into a covector $g(A,\cdot)$, but by abuse of notation people denote this still as $A$, and it is this covector which is being evaluated on $B$. In physics, people will disambiguate the vector and covector by its index location (and 99% of the time, there will be only a single metric $g$ involved, so once you get the idea, no confusion can arise, but when multiple metrics are involved, you have to be very careful when doing these musical isomorphisms). | {
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c, array, stack, macros
int main(int argc,char**argv)
{
array_stack(char)chars;
array_stack(double)nums;
stack_init(chars);
stack_init(nums);
const char*text="AzZbTyU";
for (size_t i = 0; i < strlen(text); i++)
stack_push(chars,text[i]);
stack_push(nums,3.14);
stack_push(nums,6.67);
stack_push(nums,6.25);
stack_push(nums,0.00019);
stack_push(nums,22.2222);
printf("Printing character stack: ");
while(!stack_empty(chars))
{
printf("%c ",stack_top(chars));
stack_pop(chars);
}
printf("\n");
printf("Printing double stack: ");
while(!stack_empty(nums))
{
printf("%lf ",stack_top(nums));
stack_pop(nums);
}
printf("\n");
return 0;
} A better than usual implementation.
Is it a good idea to implement such a data structure using macros? | {
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c#, excel, winforms, ms-word
private void fillSheetSelectorListBox()
{
List<string> sheetNames = null;
if (!string.IsNullOrEmpty(localPreferences.ExcelWorkBookFullFileSpec))
{
ExcelFileData ExcelFile = new ExcelFileData(localPreferences.ExcelWorkBookFullFileSpec,
localPreferences.ExcelWorkSheetName);
sheetNames = ExcelFile.GetWorkSheetCollection();
}
if (sheetNames == null)
{
return;
}
EP_RentRosterSheetName_LISTBOX.DataSource = sheetNames;
EP_RentRosterSheetName_LISTBOX.Visible = true;
EP_SheetName_TB.Enabled = true;
}
private void EP_FindRenterRoster_BTN_Click(object sender, EventArgs e)
{
findTenantRosterExcelFile(sender, e);
fillSheetSelectorListBox();
} | {
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python, numpy, audio
if __name__ == "__main__":
import matplotlib.pyplot as plt
from scipy.io.wavfile import write
strum = generate_chord( # A Major
[220, 220 * 5 // 4, 220 * 6 // 4, 220 * 2], sample_rate * 5)
plt.plot(strum)
plt.show()
write("pluck.wav", sample_rate, strum)
This produces a sound file in pluck.wav with a waveform like this: I was trying to improve your generate function and stumbled on a potential bug. You continuously update the buffer by adding two following values. This should be vectorizable easily by acting on the whole buffer at the same time, by doing something like this:
for i in range(0, nsamples - N, N):
samples[i: i + N] = buf[:]
buf = damping * 0.5 * (buf + (np.roll(buf, -1)))
# fractional buffer
i += N
k = nsamples - i
if k:
samples[i:] = buf[:k] | {
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muscles
Reciprocal inhibition is part of regular muscle movement. It is controlled by the stretch reflex controlled by the muscle spindle. It's an important part of a functioning muscle and spinal cord, and not pathologic at all. However, this same reflex, in an excessively tight muscle on one end of a joint, would cause an excessively loose muscle on another end of a joint.
Here is an image showing this reflex, from Brust Practice of Neural Science, Chapter 6.
I can't give you any medical advice on this site, so I can't put that in context, but I think that answers your biology question. | {
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javascript, jquery, html, css
}
#buttonContainer {
width: 253px;
margin: 0 auto;
position: relative;
top: -1px;
}
#clearFloat {
clear: both;
}
.activeButton{
background-color: #E8F2FF;
}
.highlightButton {
background-color: #E5E5E5 !important;
}
.highlightActiveButton {
background-color: #F5F9FF !important;
}
.unactiveButton {
background-color: #EEEEEE;
position: relative;
top: 2px;
padding: 6px;
float: left;
border: solid 1px #D2D2D2;
border-right: none;
}
textarea {
width:50%;
resize: none;
border-top: none;
/*border-left: none;*/
}
.panel {
float: left;
border-left: none;
}
iframe {
border: none;
}
.hidden {
display: none;
}
Javascript
// Check if button has a certain class
function checkButton(button, className) {
var buttonBoolean = $(button).hasClass(className);
return buttonBoolean;
}
var toggleActive = checkButton(".toggleButton", "activeButton"); | {
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newtonian-mechanics, classical-mechanics, lagrangian-formalism, variational-principle, time-reversal-symmetry
$\frac{d^n {\vec x}( t)}{d t^n}$ and the explicit $t$ dependence:
$$
L=\frac{1}{2} m \frac{d^2 \vec{x}(t) }{d t^2} -V ( \vec{x}(t),
\frac{d^n {\vec x}( t)}{d t^n}, t)
$$
The $V=V ( \vec{x}(t),
\frac{d^n {\vec x}( t)}{d t^n}, t)$ is the potential term. The $n$ can be any positive integer $n=1,2,3,etc.$ | {
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satisfiability, approximation, maxsat
Title: Radius Local Search Algortihm for Max-Sat problem approximating ratio Assume that in classical Local Search algorithm for MAX-SAT we could flip no more than $r
\leq n/2$ variables (let's call it $r$-flip) on every iteration. More precise: on every iteration we're finding $r$-flip which satisfies more clauses then was satisfied before the flip.
What is the approximation ratio of this algorithm? Is there any worst case where my algorithm has multiplicative error 1/2? Suppose that the all-zero assignment is a local optimum for your algorithm, for some CNF $\varphi$, say satisfying an $\alpha$ fraction of the clauses. If we choose any assignment of weight at most $n/2$, it should satisfy at most an $\alpha$ fraction of the clauses. In particular, a random assignment of weight exactly $n/2$ satisfies at most an $\alpha$ fraction of clauses. | {
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• Why the downvote? – copper.hat Nov 22 '17 at 21:27
• Since you mention Darboux's theorem, why don't you write a proof? – user99914 Nov 22 '17 at 22:48
• @JohnMa: I added an aside to that effect, but it was really KaviRamaMurthy's suggestion. Did you downvote? – copper.hat Nov 23 '17 at 0:38
• I just upvoted your answer (so no, the downvote is not mine. But I can understand the downvote since you did not prove anything). – user99914 Nov 23 '17 at 1:37
• @JohnMa: Thanks John. However, this does establish an answer in the affirmative subject to the additional proviso that $f$ is differentiable at $a$. My reluctance was due to the fact that someone else pointed out the use of Darboux's theorem. I wanted to add this because the other answer suggests that continuity is required where in fact we have 'almost continuity' due to the theorem (and added assumption). – copper.hat Nov 23 '17 at 3:50 | {
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general-relativity, black-holes
For more on this see Taking selfies while falling, would you be able to notice a horizon before hitting a singularity?, though the answers to this are somewhat technical.
In fact it's a fundamental principle in general relativity that for a freely falling observer spacetime looks locally flat. If you were falling into a black hole then at least in your immediate vicinity space would look just like normal boring old flat spacetime. | {
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Studying the distribution of $s(Y_n)$ can tell us a lot. Its equivalent to rolling $n$ times a 10-sided die with faces $0,1,4,\dots, 81$ and finding the sum. Its distribution is Gaussian as $n$ gets large by the central limit theorem. More importantly most of the distribution is concentrated near the mean, which is $28.5n$. This implies that the density happy numbers amongst all $n$-digit integers depends almost entirely on the distribution of happy numbers near $28.5n$.
For example, there is a peak in your graph at around $n = 400$ of about $.185$ density. Calculating the density of happy numbers within one standard deviation from the mean of $s(Y_{400})$ we get a density of .1911 (the interval I looked at was $[10916,11884]$). If you assume $s(Y_{400})$ is "exactly" normally distributed and estimated the density in this manner you would get a much better approximation. | {
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} |
• Math expert for every subject
• Pay only if we can solve it
Gaige Burton
Step 1
Some would call $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)\ge 0$ a non decreasing function.
Also some would call $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)>0$ a strictly increasing function.
But, yes, $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)\ge 0$ is often defined as an increasing function. In that case, the function $f\left(x\right)=1$ is an increasing function.
Sometime the word monotone gets thrown in there too.
You may as well get used to the fact that not all mathematicians use the same definition for things. As long as the book is consistent, that is acceptable.
Step 2
INCREASING:
- $\mathrm{\forall }x,y\in I,x
- $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)\ge 0$
STRICTLY INCREASING:
- $\mathrm{\forall }x,y\in I,x
- $\mathrm{\forall }x\in I,{f}^{\prime }\left(x\right)>0$
MOTONIC:
- Strictly increasing or strictly decreasing | {
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"url": "https://plainmath.net/86117/let-i-be-an-open-interval-contained-in-t"
} |
ros, gazebo, simulation, world, model
Comment by shonigmann on 2022-02-01:
For the maize models I would suggest using population tags to cleanup your world file.
Regarding making them "dynamic", presumably you want the tractor to "knock over" the maize models? I think because of the number of models / contacts, it will be hard to get this to be "fast". A few things that may help performance: reducing the number of contacts in the maize models to something like 4 (default is 10). adding joints at the base of each model so that they pivot about the base rather than being fully free to move in space might help, but I'm not certain.
If the contact dynamics aren't very important, I think an approach that is more likely to be successful would be to create a world plugin that uses the position of the tractor to either set the pose of, or delete, the maize model that would be contacted.
Comment by simonsure on 2022-02-02:
What does the restitution_coefficient do? I could not find any details. | {
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ros, ros-service
Original comments
Comment by Winston on 2016-08-29:
I want to guarantee that the client's EVERY message/call would be received by the server but it seems that topics could not guarantee that. I won't use ros service if topics could meet my requirement. @ahendrix
Comment by ahendrix on 2016-08-29:
If you really need to guarantee delivery of EVERY message, then services may be the only answer. However, most of the users I find who say this don't actually need it; a slight change to the message semantics might be tolerant to dropped messages.
Comment by Winston on 2016-08-31:
Thank you! I would like to how how to change the message's semantics to be tolerant to dropped messages. Could you give me some advice?
Comment by ahendrix on 2016-08-31: | {
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newtonian-mechanics, classical-mechanics, momentum
Title: Can Momentum of an object change without net acceleration on the object? Comfy with single variable calculus and basic concepts of classical mechanics. By definition: the momentum is $\boldsymbol{P}=m\boldsymbol{v}$ where $m$ is the mass and $\boldsymbol{v}$ the velocity. By Newton's second law of motion:
$$\boldsymbol{F}=\frac{d\boldsymbol{P}}{dt}=m\frac{d\boldsymbol{v}}{dt}+\frac{dm}{dt}\boldsymbol{v}$$
By defining the acceleration $\boldsymbol{a}=\frac{d\boldsymbol{v}}{dt}$, if we set $\boldsymbol{a}=0$ we can still have:
$$\frac{d\boldsymbol{P}}{dt}=\frac{dm}{dt}\boldsymbol{v}$$
which could be non zero.
Thus the answer is YES: net momentum can change without acceleration if the mass $m$ of the object changes with time. | {
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homework-and-exercises, newtonian-mechanics, conservation-laws, collision
However, regarding using energy to solve the problem: you stated that the collision is inelastic. What does this mean? Is energy conserved? If you think about the energy in inelastic collisions, I think you'll figure it out yourself :). If you're still stuck, I'll write the answer below.
Since the collision is inelastic, energy isn't conserved (some energy is lost to heat, sound, shape changes, etc.). Therefore, you can't use conservation of energy to solve this problem (since the initial energy doesn't equal the final energy). Momentum is the way to go.
Let me know if you want me to elaborate. Good luck. | {
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thermodynamics, statistical-mechanics, information, reversibility
As the demon decreases the entropy, it has to keep tabs on which chamber each molecule is in so it doesn't let one go back the other way. Thus it needs to make a new one-bit entry in its memory (chamber "A" or chamber "B") for that molecule-passage event. But its memory is necessarily finite, so there will come a time there is no more space for any new records. Once the memory has now exhausted all bits, there is no other place to put the new information. Reversibility would mean you can move information around in memory (i.e. shuffle around the records of what molecules went where), but you can't put anything new to make for this new molecule. To put something new, you have to throw information out of memory - into the Universe, which means it becomes lost, and the process is irreversible as you now no longer have a bit of information you need to put things (the demon and boxes) back to the way they were before. Or, you have to ignore any new molecules, meaning either the demon will | {
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python, optimization, performance, numpy
Title: Harmonic analysis of time series applied to arrays I've developed some code in Python to apply a harmonic analysis of timeseries (for satellite imagery data). It's based on this, but then I would like to optimize the performance. So instead of a single timeseries as input I've an array of 10000 timeseries as input.
Even though I've tried to make use of some nice functions in NumPy that deals with big multidimensional arrays, I'm sure I just have touched the surface as it comes to the actual capabilities.
import numpy as np
# Computing diagonal for each row of a 2d array. See: http://stackoverflow.com/q/27214027/2459096
def makediag3d(M):
b = np.zeros((M.shape[0], M.shape[1]*M.shape[1]))
b[:, ::M.shape[1]+1] = M
return b.reshape(M.shape[0], M.shape[1], M.shape[1]) | {
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ros, installation, rosinstall
3.) Then, the first 4 lines are:
Do the parsing
Check if rosdistro exists
Check if stack exists
Check if variant exists
I had to remove these also from my rosinstall file that I had saved.
Is this a bug? Should I register this with the bug-tracker? | {
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ros-fuerte, rospack
Originally posted by michikarg with karma: 2108 on 2012-05-06
This answer was ACCEPTED on the original site
Post score: 0 | {
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reference-request, sat, reductions
Theorem 29. XSAT remains NP-complete for $k$-$CNF^l_+$ and $k$-$CNF^l$, $k, l \geq 3$.
(XSAT for $3$-$CNF^3$ is exactly 1-in-3-SAT where each variable appears exactly $l=3$ times)
Note that the theorem also proves the NP-completeness of the stronger monotone case ($CNF_+$) | {
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python, strings, python-3.x, parsing, symbolic-math
def solve(coefficients):
"""
Dispatches solving to the correct method or aborts
if the equation grade is too high.
>>> solve((1,5,6))
(-2.0, -3.0)
>>> solve((0,2,4))
(-2.0,)
"""
if coefficients[0] == 0:
return first_grade_solve(coefficients)
elif len(coefficients) == 3:
return second_grade_solve(coefficients)
raise NotImplementedError("Only 0th, 1st and 2nd grade equations can be solved")
def parse_and_solve(expr):
"""
Connects the other functions to provide full equation
solving.
>>> parse_and_solve("2x - 4 = 0")
(2.0,)
>>> parse_and_solve("1 = 2")
'Never'
>>> parse_and_solve("x^2 + 2x = -1")
(-1.0, -1.0)
"""
simpler = add_one_before_x(move_operators(expr))
same_sided = move_to_the_same_side(simpler)
normal_form = to_normal_form(same_sided)
return solve(normal_form) | {
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civil-engineering, steel, aisc
Does that imply AISC 341 design results in a more conservative steel connection than AISC 360 or is there a difference in steel design philosophy? The R-factor reduces seismic design force from the linear-elastic demand. Thus, the larger the R-factor, the lower the seismic design force.
Essentially, we're approximating inelastic seismic design by accounting for the ductility of the structural system. Connections designed per AISC-360 (which does not specifically address seismic design) can be expected to be less ductile than those designed per AISC-341.
So.
AISC-341 = more ductile = higher R = lower seismic design force
AISC-360 = less ductile = lower R = higher seismic design force | {
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java, android, error-handling, networking
Show the exception to the user (or some information about it anyway). This is probably completely unnecessary. The network connection is either available or not, right?
Rethrow the exception and let the caller handle it. It can be useful to know there was an exception and we should retry the network connection later.
Log the exception in a way that the developer can access even when it happens in production. Error logs should be uploaded somewhere.
Just do nothing, and don't waste execution time printing the stack trace. If you chose this option, you probably want to find a way to leave the print stack trace in for development environments, but for production builds, you shouldn't be executing this code. | {
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sequence, biopython, command-line
I used the code from https://github.com/biopython/biopython/issues/3156 to fix the ValueError: missing molecule_type in annotations that I faced when I'd skipped the record.annotations = {"molecule_type" : "protein" } line.
There is definitely a better way to write the python code instead of using a loop or at least writing to a static file from within a loop, but that's not central to the problem here so I'm choosing to ignore it. I welcome any improvements to the code. | {
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sampling, filtering, least-squares, derivative
Which is a surprising for me because it reveals that the system has probably more complex dynamics. I'm not quite sure this works, but here's an attempt: just median filter your derivative (gradient) and use the peak of that.
I synthesized something like your signal:
And then took the derivative:
And then median filtered it (the peak is indicated with the red dot):
Python code below
import matplotlib.pyplot as plt
import numpy as np
from scipy import signal
T = 1024
Ton = 100
x = np.random.normal(0,0.1, T)
for t in np.arange(Ton,T):
x[t] = x[t] + 10*(1-np.exp(-(t-Ton)/50))
dxdt = np.gradient(x)
median_dxdt = signal.medfilt(dxdt,31)
mx = median_dxdt.max()
ix = np.argmax(median_dxdt)
plt.figure(1)
plt.plot(x)
plt.figure(2)
plt.plot(dxdt)
plt.figure(3)
plt.plot(median_dxdt)
plt.plot(ix,mx,'r.') | {
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python, django
And finally, my template --
<h3>Edit education info for {{user.get_full_name}}</h3>
<form action="." method="post"> {% csrf_token %}
{% if education %}
{% for education in education %}
<p><b>{{ education.school }}</b> {% if education.class_year %}{{ education.class_year|shorten_year}}, {% endif %} {{ education.degree}}
<input type="submit" name="edit_{{education.id}}" value='Edit' />
<input type="submit" name="delete_{{education.id}}" value="Delete" /></p>
{% endfor %}
{% endif %}
<table>
<input type="hidden" name="education_id" value="{{education_id}}" />
<tr><td>School:</td><td>{{form.school}}{{form.school.errors}}</td></tr>
<tr><td>Class Year</td><td>{{form.class_year}}{{form.class_year.errors}}</td></tr>
<tr><td>Degree:</td><td>{{form.degree}}{{form.degree.errors}}</td></tr>
<tr>{{form.non_field_errors}}</tr>
</table> | {
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matlab, estimation, system-identification, self-study
Q2: I am not sure how to implement using PRBS. Can somebody please provide the code?
Q3: Is the system identification using EM-UKS when the input is chaotic and symbolic proposed by the Authors blind or semi-blind and why? Interesting paper!
Q1: yes, seems like OLS, non-blind. Usually OLS is non-blind. Some sort of regulated least squares can be used in a bayesian ML-setting, blind. This paper looks interesting: http://www.cs.berkeley.edu/~jordan/papers/lindsten-etal-sysid12.pdf
Q2: PBRS: https://en.wikipedia.org/wiki/Pseudorandom_binary_sequence. There is a lot of literature on this, e.g. Coleman Brosilow, Babu Joseph, "Techniques of Model-Based Control"
Q3:they say it is semi-blind since the dynamics of the sequence generator is known. | {
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"tags": "matlab, estimation, system-identification, self-study",
"url": null
} |
electromagnetism, maxwell-equations, magnetostatics
Title: Differential element of current question I'm watching some basic magnetic field derivations and most of the proofs use the differential element $dI$.
Let's say a wire carries a current I. When we take a very small current $dI$ though , shouldn't it have the same value as the current running through the wire? Also,if the current is constant shouldn't $dI$ be zero? What is actually a differential element? Maybe that's the question I should be asking.
The only time that it makes sense to me is in 3d conductors where the current varies inside them.
One of the examples I had trouble with was while using Ampere's law to find B inside a solinoid. The $I_{enclosed}$ is different from the total current. Isn't I just a rate of movement? How are you supposed to enclose it? Maybe it's actually the charge you enclose ? No, that would be Gauss' law. | {
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"url": null
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quantum-operation, unitarity, linear-algebra
$$
X \oplus Y = \text{CNOT}_{\mathcal{X}\rightarrow \mathcal{Y}}(X, Y) = \text{CNOT}_{\mathcal{Y}\rightarrow \mathcal{X}}(X, Y), \tag{1}
$$
where the subscript $\mathcal{A}\rightarrow \mathcal{B}$ denotes the variable in $\mathcal{A}$ as control bit acting on the variable in $\mathcal{B}$ as target bit, and in each case the output is read from the target register (I'm using the same notation for alphabets and "registers" containing variables). So, define a pmf $p_X: \mathcal{X}\rightarrow \mathbb{R}$ where $p_X(i) = \text{Pr}(X=i)$ and define $p_Y, p_Z$ similarly. Then states describing the variables $X$ and $Y$ are
\begin{align}
\rho := \text{diag}(p_X(0), p_X(1)), \tag{2} \\
\sigma := \text{diag}(p_Y(0), p_Y(1)).\tag{3}
\end{align}
We will overload the classical notation for a CNOT to define a unitary
$$
\text{CNOT}_{\mathcal{X}\rightarrow \mathcal{Y}}: \mathcal{X}\otimes \mathcal{Y} \rightarrow \mathcal{X'} \otimes \mathcal{Z}
$$ | {
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machine-learning
Title: What is the difference between filling missing values with 0 or any othe constant term like -999? Most of the text book says to fill missing values use mean/median(numerical) and most frequent(categorical) but I working on a data set which has too many missing values and I can't remove those columns because they are important.
train.isnull().sum()
TransactionID 0
isFraud 0
TransactionDT 0
TransactionAmt 0
ProductCD 0
...
id_36 449555
id_37 449555
id_38 449555
DeviceType 449730
DeviceInfo 471874
Length: 434, dtype: int64
train.shape
(590540, 434) | {
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quantum-mechanics, homework-and-exercises, operators, hilbert-space, eigenvalue
Where I'm having trouble
The problem is with the very last result, on the last line of the page. I am getting
$$S_z=\frac{\hbar}{2}\mathbb{I}$$ and also $$S_z=-\frac{\hbar}{2}\mathbb{I}$$
which is clearly not correct. My guiding equation has been equation 5 (which seems to be correct). I have put $A=S_z$ of a spin 1/2 system in equation 5. I cannot locate the flaw in the steps. | {
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In addition to the significance level, we also need the degrees of freedom to find this value. For the goodness of fit test, this is one fewer than the number of categories. We have five flavors of candy, so we have 5 – 1 = 4 degrees of freedom.
The Chi-square value with α = 0.05 and 4 degrees of freedom is 9.488.
4. We compare the value of our test statistic (52.75) to the Chi-square value. Since 52.75 > 9.488, we reject the null hypothesis that the proportions of flavors of candy are equal.
We make a practical conclusion that bags of candy across the full population do not have an equal number of pieces for the five flavors. This makes sense if you look at the original data. If your favorite flavor is Lime, you are likely to have more of your favorite flavor than the other flavors. If your favorite flavor is Cherry, you are likely to be unhappy because there will be fewer pieces of Cherry candy than you expect.
### Understanding results | {
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"openwebmath_score": 0.6582071781158447,
"tags": null,
"url": "https://www.jmp.com/en_ph/statistics-knowledge-portal/chi-square-test/chi-square-goodness-of-fit-test.html"
} |
homework-and-exercises, kinematics, projectile
Title: Can we derive equation for horizontal range of projectile from 3rd equation of motion? I used the equation v² - u² = 2as and came up with the the following result. But on verifying it for theta = 45 it fails. Where am I wrong? You have taken the final velocity (horizontal component) to be $0$, which is wrong and also the acceleration isn't $g$ rather it is $0$ as you've considered horizontal motion only.
When we talk about 2-dimensional motion, the horizontal velocity of the object is unaffected by the acceleration due to gravity as the acceleration vector and horizontal velocity vector are perpendicular to each other. This is why the acceleration due to gravity only affects the vertical velocity and not the horizontal velocity. | {
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ros, knowrob
Comment by ratneshmadaan on 2014-03-10:
Maybe I am doing something foolish. Please help me out.
1.I made a ROS space:
http://wiki.ros.org/ROS/Tutorials/InstallingandConfiguringROSEnvironment
http://docs.ros.org/independent/api/rosinstall/html/rosws_ros_tutorial.html#rosws-ros-tutorial
2.rosws merge https://raw.github.com/knowrob/knowrob/master/rosinstall/knowrob-base.rosinstall
rosws update
rosdep install knowrob
rosmake knowrob
Comment by moritz on 2014-03-10:
There are two kinds of buildsystems, rosbuild and catkin. KnowRob still uses the older rosbuild, while the default in hydro is catkin. You can set up both systems to work together,you need to create a catkin WS first and then refer to it from rosbuild (Section 3.3 in the link I've posted).
Comment by ratneshmadaan on 2014-03-10:
Okay, Did what you said. Now keys are unresolved. :
ratneshmadaan@ratneshmadaan-Inspiron-N5010:~/rosbuild_ws$ rosdep install knowrob | {
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fluid-dynamics, pressure, fluid-statics
Title: Pascal's law with bleeding port on the piston If a bleeding port is drilled on the larger piston, does Pascal's law still hold as shown in the diagram? If yes, could you kindly provide some references for me to have a look at? Thanks in advance for your time! Overview
Pascal's law does not hold if there is any fluid flow through the bleeding port.
In the limits of large viscosity, $A_3\to0$ and low pressure difference between the vessel and surroundings then the approximation $\frac{F_2}{F_1}\approx\frac{A_2-A_3}{A_1}$ becomes increasingly accurate - with equality when there is no fluid flow. | {
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history, parallax
When Joseph von Fraunhofer invented a new type of heliometer, Bessel carried out another set of measurements using this device in 1837 and 1838 at Königsberg. He published his findings in 1838[31][32] with a value of 369.0±19.1 mas to A and 260.5±18.8 to B, and estimated the center point to be at 313.6±13.6. This corresponds to a distance of about 600,000 astronomical units, or about 10.4 light-years. This was the first direct and reliable measurement of the distance to a star other than the Sun.[27][33] His measurement was published only shortly before similar parallax measurements of Vega by Friedrich Georg Wilhelm von Struve and Alpha Centauri by Thomas Henderson that same year.[34] Bessel continued to make additional measurements at Königsberg, publishing a total of four complete observational runs, the last in 1868. The best of these placed the center point at 360.2 ±12.1 mas, made during observations in 1849.[27] This is close to the currently accepted value of 287.18 mas | {
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entanglement
There are different measures of entanglement and some are incomarable, meaning that one may be both larger and smaller than another for different states, so my guess is that you're looking at a zoo rather than one measure to rule truthfully over all others. I don't know off the top of my head if this sort of thing has been studied, but I would guess that something along these lines must have been considered before. It seems pretty interesting and it might be fun to dive in and start working things out — but if I were doing this I would start with a sincere effort to learn what's already been done. | {
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java, algorithm, pathfinding
Demo.java
package net.coderodde;
import java.util.List;
public class Demo {
public static void main(String[] args) {
UndirectedGraphNode nodeA = new UndirectedGraphNode("A");
UndirectedGraphNode nodeB = new UndirectedGraphNode("B");
UndirectedGraphNode nodeC = new UndirectedGraphNode("C");
UndirectedGraphNode nodeD = new UndirectedGraphNode("D");
UndirectedGraphNode nodeE = new UndirectedGraphNode("E");
UndirectedGraphNode nodeF = new UndirectedGraphNode("F");
nodeA.connectTo(nodeB, 0.1);
nodeA.connectTo(nodeC, 0.9);
nodeA.connectTo(nodeD, 0.9);
nodeB.connectTo(nodeC, 0.9);
nodeB.connectTo(nodeD, 0.2);
nodeC.connectTo(nodeD, 0.1);
nodeE.connectTo(nodeB, 0.2);
nodeE.connectTo(nodeF, 0.8);
nodeF.connectTo(nodeB, 0.99);
List<UndirectedGraphNode> path = new
MostReliablePathFinder().findLeastReliablePath(nodeD, nodeE); | {
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c#, dependency-injection, autofac
// Only generate one SessionFactory ever because it is expensive.
containerBuilder.Register(x => new NHibernateConfiguration().Configure().BuildSessionFactory()).SingleInstance();
// Everything else wants an instance of Session per HTTP request, so indicate that:
containerBuilder.Register(x => x.Resolve<ISessionFactory>().OpenSession()).InstancePerApiRequest();
containerBuilder.Register(x => LogManager.GetLogger(MethodBase.GetCurrentMethod().DeclaringType)).InstancePerApiRequest();
// TODO: I don't THINK I should actually be making a DAO factory per API request. I think the whole point of a Factory is to make them once.
containerBuilder.RegisterType<NHibernateDaoFactory>().As<IDaoFactory>().InstancePerApiRequest();
containerBuilder.RegisterType<StreamusManagerFactory>().As<IManagerFactory>().InstancePerApiRequest();
// Build the container.
ILifetimeScope container = containerBuilder.Build(); | {
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c, snake-game, sdl
(*new_tail).next = NULL;
(*new_tail).previous = lasttail;
(*lasttail).next = new_tail;
lasttail = new_tail;
}
void change_snake_direction(int dir)
{
if(dir == RIGHT && snake.dx != -1){
snake.dx = 1;
snake.dy = 0;
}
else if(dir == LEFT && snake.dx != 1){
snake.dx = -1;
snake.dy = 0;
}
else if(dir == UP && snake.dy != 1){
snake.dy = -1;
snake.dx = 0;
}
else if(dir == DOWN && snake.dy != -1){
snake.dy = 1;
snake.dx = 0;
}
}
void free_tails()
{
struct TailNode *tmp;
struct TailNode *secondtail;
secondtail = snake.head.next; // we skip the first node (head) because it's allocated in the stack
while(secondtail != NULL){
tmp = secondtail;
secondtail = (*secondtail).next;
free(tmp);
}
}
apple.h
#ifndef APPLE
#define APPLE
static const int DEFAULT_APPLE_WIDTH = 20;
static const int DEFAULT_APPLE_HEIGHT = 20; | {
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newtonian-mechanics, fluid-dynamics
Title: Stagnation Point around Sharp Corner I am trying to design a water-channel that travels through a metal block shaped like a square to cool it down (so it is a 2D problem).
One consideration is the wetted surface area, which allows more of the water to be in contact with the metal, thus increasing the cooling effect.
One way to do that is to start at one corner of the square, then form multiple S-shapes on the block to maximize the surface area. However, there will be some sharp U-turns.
I am trying to avoid stagnation points, which slows down the cooling since less fluid travels.
So I did some reading on sharp turns and the term Kutta Condition keeps appearing when describing fluids at sharp points, where there will be a stagnation point at the edge. | {
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Also, of course, sometimes the best we can do is to produce a proof that happens to be a maze of complicated algebra. Then that's still a valid proof, though we might have hoped for something better (and may keep searching for something better, on the back burner). | {
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and
\begin{align}\omega^\omega &= \sup \{ \omega ^n \mid n < \omega \} \\ &\ge \omega^3 \\ &= \sup \{(\omega^2) \cdot n \mid n < \omega \} \\ &\ge \omega^2 \cdot 2 \\ &> \omega^2. \end{align}
Finally \begin{align}^\omega \omega &:= \sup \{ \underbrace{\omega^{\omega^{\omega^ \ldots}}}_{n\text{-times}} \mid n < \omega \} \\ &\ge \omega^{\omega^\omega} \\ &= \sup\{\left( \omega^\omega \right)^n \mid n < \omega \} \\ &\ge \left(\omega^\omega \right)^2 \\ &= \omega^\omega \cdot \omega^\omega \\ &= \sup \{\omega^\omega \cdot \alpha \mid \alpha < \omega^\omega \} \\ &\ge \omega^\omega \cdot 2 \\ &> \omega^\omega. \end{align}
Combining these calculations we get the desired order: $$\omega = 2^\omega = {^\omega} 2 < \omega^2 < \omega^\omega < ^\omega\omega$$
In general, ordinal and cardinal arithmetic are very different beasts and every ordinal arithmetic expression using only ordinals $\le \omega$ is countable.
proof (sketch) | {
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"tags": null,
"url": "https://math.stackexchange.com/questions/1359801/order-of-infinite-countable-ordinal-numbers"
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quadcopter, arduino, brushless-motor, esc
zero_timer = micros();
}
void loop() {
// put your main code here, to run repeatedly:
while (zero_timer + 4000 > micros());
zero_timer = micros();
PORTC |= B11110000;
channel_timer_1 = receiverinput_channel_3 + zero_timer; //Time calculation for pin 33
channel_timer_2 = receiverinput_channel_3 + zero_timer; //Time calculation for pin 32
channel_timer_3 = receiverinput_channel_3 + zero_timer; //Time calculation for pin 31
channel_timer_4 = receiverinput_channel_3 + zero_timer; //Time calculation for pin 30
while (PORTC >= 16) //Execute till pins 33,32,31,30 are set low
{
esc_looptimer = micros();
if (esc_looptimer >= channel_timer_1)PORTC &= B11101111; //When delay time expires, pin 33 is set low
if (esc_looptimer >= channel_timer_2)PORTC &= B11011111; //When delay time expires, pin 32 is set low
if (esc_looptimer >= channel_timer_3)PORTC &= B10111111; //When delay time expires, pin 31 is set low | {
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y2), the formula for the slope of the straight line Line in slope-intercept form. Put these points into the formula for . 132 (3-12) Chapter 3 Graphs and Functions in the Cartesian Coordinate System or a run of 1 and a rise of 2. Improve your math knowledge with free questions in "Point-slope form: graph an equation" and thousands of other math skills. The slope formula can be read as "slope equals the second y coordinate minus the first y-coordinate over the second x-coordinate minus the first x-coordinate". Knowledge of relevant formulae is a must for students of Grade 8 and above to solve some of these worksheets. Identifying Slopes and y -Intercepts Find the slope and the y -intercept of the graph of each linear equation. In this Article: To calculate without the function Community Q&A Calculating the slope of a line is extremely simple. To find the slope of a line given the graph of the line, you can use the slope formula. The correct slope value is: Slope = 3. Launch Excel and | {
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"url": "http://metstrategies.com/sztvzsd/oofapob.php?khdtkqcnx=graph-slope-formula"
} |
c++, object-oriented, game, tic-tac-toe
while (!inputPass)
{
if (std::cin >> choseNum)
{
myBoard.markBoard(choseNum, mainPlayer, inputPass); //Go to markBoard function in board.cpp
}
else
{
std::cout << "Invalid input type (Type only number)\nTry again: ";
std::cin.clear(); // To clear the input so
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n'); // the player can input again
}
}
}
// If the turnPlayer is mainPlayer then it's the computer's move
else
{
while (!inputPass)
{
// To make a random move for the computer
std::random_device rd;
std::mt19937 gen(rd());
std::uniform_int_distribution<> distrib(1, 9);
choseNum = distrib(gen); | {
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ros, rviz, robot, transform
<link name="fwheel_right">
<visual>
<geometry>
<cylinder length="0.05" radius="0.03"/>
</geometry>
<origin rpy="0 0 0" xyz="0 0 0"/>
<material name="black">
<color rgba="0 0 0 1"/>
</material>
</visual>
</link>
<joint name="base_to_fwheel_right" type="fixed">
<parent link="base_link"/>
<child link="fwheel_right"/>
<origin xyz="0.22 0.1 -0.095" rpy="-1.5708 0 0"/>
</joint>
<link name="camera_link">
<visual>
<geometry>
<box size="0.28 0.065 0.04"/>
</geometry>
<origin rpy="0 0 0" xyz="0 0 0"/>
<material name="black">
<color rgba="0 0 0 1"/>
</material>
</visual>
</link> | {
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"openwebmath_score": null,
"tags": "ros, rviz, robot, transform",
"url": null
} |
special-relativity, inertial-frames, lorentz-symmetry
Title: Can Lorentz-Boosts reach every phase space point? Given a particle with mass $m$ and four momentum $p=\left(p_x,p_y,p_z,\sqrt{|\vec{p}|^2+m^2}\right)$. Is it possible to reach every point in phase space, i.e. every combination of values $p_x,p_y,p_z\in\mathbb{R}$, by a simple Lorentz-Boost?
I'm guessing no, but don't know why. Let $\vec{p}$ be the original spatial part of the four-momentum, and let $\vec{p}'$ be the desired spatial part of the four-momentum. Let us assume that for any $\vec{p}'$, there exists a boost vector $\vec{\beta}$ under which $\vec{p}$ maps to $\vec{p}'$. This will lead to a contradiction. | {
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c++, beginner, game, rock-paper-scissors
It's good that instructions are a clearly separated subroutine, that is, a named function. But I was disappointed to see that the file continued with main and no other organization.
You should separate things out into functions more comprehensively. Don't put everything in main down to fine detail; rather, call a series of high-level descriptively named functions, possibly looping over the main logic, but no details on how any of that works.
The old C library random numbers are pretty poor. Look into using the C++ random number library.
if ((user == 1 && computer == 3) || (user == 2 && computer == 1) || (user == 3 && computer == 2))
⋮
else if ((user == 1 && computer == 2) || (user == 2 && computer == 3) || (user == 3 && computer == 1)) | {
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"tags": "c++, beginner, game, rock-paper-scissors",
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fluid-dynamics
The achievable height is not determined by the fountain height 'k'. For example, consider a piston pump which slowly pumps water at high pressure to its output. It wouldn't be able to achieve a very high fountain height 'k' because the rate at which pumps water is so slow, but since it can pump water against a very large back pressure it would be able to pump water to a large height 'h'.
If not, is it possible with additional parameters?
The achievable height 'h' is determined by how high of a back pressure the pump can work against. If that information is known, then the height 'h' can be determined.
If not, can we at least say that one is always larger than the other?
Obviously, the maximum achievable height 'h' has to be at least as large as the fountain height 'k' since one can always take a fountain of water and then just enclose the fountain of water in a hose.
Can we optimize h, for example by using a hose with a large or small
diameter? | {
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redox, transition-metals
What your textbook should have written is:
The $I_3$ of Mn is anomalously large because you are "losing the stability of a half-filled 3d subshell". Hence, $\ce{Mn^3+}$ is a good oxidising agent.
The hydration enthalpy of $\ce{Cr^3+}$ is anomalously large (for a 3+ ion) because there is a large ligand-field stabilisation energy associated with the $(t_{2g})^3$ configuration. (This is factor number 2, not factor number 1). Hence, $\ce{Cr^2+}$ is a reducing agent.
These factors are small compared to the raw size of the ionisation energies. Yes, the $d^3$ ion $\ce{Mn^4+}$ also enjoys a large LFSE, and its hydration enthalpy would be larger than expected. However, that 4th ionisation energy is just too big, and that simply cannot be compensated for by the comparatively tiny increase in LFSE. | {
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localization, ekf, odometry, path, kalman
0, 0, 0, 0, 0, 1e-9, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1e-9, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1e-9, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1e-9, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1e-9, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1e-9, 0, 0, 0, 0, | {
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"tags": "localization, ekf, odometry, path, kalman",
"url": null
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In the latter case, the matrices must have the same shape in order for the definition to make sense
We also have a convention for multiplying two matrices
The rule for matrix multiplication generalizes the idea of inner products discussed above, and is designed to make multiplication play well with basic linear operations
If $A$ and $B$ are two matrices, then their product $A B$ is formed by taking as its $i,j$-th element the inner product of the $i$-th row of $A$ and the $j$-th column of $B$
If $A$ is $n \times k$ and $B$ is $j \times m$, then to multiply $A$ and $B$ we require $k = j$, and the resulting matrix $A B$ is $n \times m$
As perhaps the most important special case, consider multiplying $n \times k$ matrix $A$ and $k \times 1$ column vector $x$
According to the preceding rule, this gives us an $n \times 1$ column vector | {
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"openwebmath_score": 0.954851508140564,
"tags": null,
"url": "https://lectures.quantecon.org/py/linear_algebra.html"
} |
Define $$p_1$$ and $$p_2$$ as $$p_1 = u_{(n-1)} \cdot\big(M_{(1)}\big)^{-1}\cdot\begin{pmatrix} 1 \\ 0 \\ \vdots\\0 \end{pmatrix}_{ (n-1)\times1}, \quad p_2 = u_{(n-1)} \cdot\big(M_{(2)}\big)^{-1}\cdot\begin{pmatrix} 1 \\ 0 \\ \vdots\\0 \end{pmatrix}_{ (n-1)\times1}.$$
Prove that all the elements of the row vector $$u_{(n-1)} \cdot \left(p_2\big(M_{(1)}\big)^{-1}+p_1\big(M_{(2)}\big)^{-1}\right)\cdot C - (p_1+p_2)u_{(n-2)}\cdot \big(M_{(1,2)}\big)^{-1}$$ are identical.
This property comes from some intuition of the problem that I have been playing with. I have also tested it by evaluating it with a large set of matrix $$M$$ satisfying the first requirement.
(I thank user1551 for spotting an important typo, corrected now!)
I have tried writing the inverse using minors but does not seem to help as it is not easy to implement the requirement that $$\sum_{j} M_{ji}=0$$. Any comment/suggestion is greatly appreciated. Answers are of course the best! Thank you so much! | {
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"openwebmath_score": 0.9785475730895996,
"tags": null,
"url": "https://math.stackexchange.com/questions/3623969/a-weird-matrix-property"
} |
quantum-mechanics, quantum-spin, bloch-sphere
For pure states, when $|{\vec a}| =1$ and
$$
\langle (\Delta {\hat \sigma}_i)^2 \rangle = 1 - a_x^2 = a_y^2 + a_z^2
$$
the circular cut crosses through point A, and $\sqrt{a_y^2 + a_z^2}$ is just the distance from A to the x-axis. Similarly for the other axes. So in general, | {
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"tags": "quantum-mechanics, quantum-spin, bloch-sphere",
"url": null
} |
newtonian-mechanics
Title: Does deceleration require energy? Consider an apple falling from a tree and striking the ground.
The ground decelerates the apple once it hits it, but the force is not applied over any "distance" - it is experiencing the force when it is in contact with the ground - so no work is done, yet there is a change in momentum, what is going on with the energy here?
The ground decelerates the apple once it hits it, but the force is not applied over any "distance" - it is experiencing the force when it is in contact with the ground - so no work is done, yet there is a change in momentum, what is going on with the energy here? | {
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homework-and-exercises, newtonian-mechanics, forces
Title: Change in velocity of an object as it enters water So if I drop a block which is denser than water, it will sink.
Now if I have 2 identical blocks. I drop them at different heights above the water.
Block $A$ is dropped from $20\mathrm{m}$. Block B is dropped from $5\mathrm{m}$.
How do I calculate the respective velocity just after their impact with water? Logically, Block $A$ will have a higher velocity at $1\mathrm{m}$ depth compared to Block $B$ since Block $A$ has a higher initial velocity.
But this is disproved by some simple calculations which shows that both block will have the same velocity at $1\mathrm{m}$ depth since their velocity just after impact with water is $0\mathrm{m}/\mathrm{s}$.
Let mass of both blocks be $2\mathrm{kg}$
Let $g=10\mathrm{N}/\mathrm{kg}$
Let time of impact be $0.2\mathrm{s}$
Final Velocity
$$A=rt(2gh)=rt(2\cdot10\mathrm{N}/\mathrm{kg}\cdot20\mathrm{m})=20\mathrm{m}/\mathrm{s}$$
Final velocity | {
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} |
python, python-3.x, security
while (1==1):
#Keeps track of total answred
total = incorrect + correct
#prints the amount of answers that you have correct, incorrect and total
print(f"correct {correct} incorrect {incorrect}: {total} total answered out of 20 questions.")
#creates the random port for question
rand_choice= random.choice(random_gen)
#Total of 20 random ports will ne assesed. Gives the total correct, incorrect and % coorect.
if(total == 20):
print(f"Total correct is {correct} total incorrectis {incorrect}")
print("Precent correct is")
print(correct/20 * 100)
#Asks the player if they would like to play again
play_again=input('''Would you like to play again? "yes" or "no" any other response to exit the program
:''')
#if player wants to play agin sets the correct and incorrect to 0 to start exit if and start over. | {
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"tags": "python, python-3.x, security",
"url": null
} |
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