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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
You are not logged in.
## #1 2013-07-05 06:07:32
mukesh
Member
Registered: 2010-07-18
Posts: 31
### solution of triangle
if the median 'AD' of a triangle 'ABC' is perpendicular to side AB then prove that 'tanA+2tanB=0'
Offline
## #2 2013-07-06 01:25:05
bob bundy
Registered: 2010-06-20
Posts: 8,085
### Re: solution of triangle
hi mukesh,
Here's an outline of a way to prove this. see diagram below.
There's no right angle to get tanA easily so I used the sine and cosine rules:
and
Put these together to get tanA and simplify.
work on this expression for tanA, making use of the following:
After much simplification you can get this equal to -2tanB, from which the required result follows.
It's a tough one so expect it to take 2 or 3 pages. If you get stuck post back where you've got to, and I'll compare your answer with mine.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
## #3 2013-07-06 01:33:08
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: solution of triangle
Well, I have a solution that takes 2 or 3 lines to write up, but requires quite a bit of inspection.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
## #4 2013-07-06 03:05:22
bob bundy
Registered: 2010-06-20
Posts: 8,085
### Re: solution of triangle
hi Stefy,
What does 'quite a bit of inspection' mean exactly.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
## #5 2013-07-06 03:44:24
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: solution of triangle
I do not have an exact definition. Roughly, it means that some times you will think of it, sometimes you won't, and it mostly depends on luck, not unlike many other geometry problems.
My solution extends the line AB and names E the foot of the perpendicular from C to that line. Then I used The basic trig equations to get the result.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
## #6 2013-07-06 04:45:27
bob bundy
Registered: 2010-06-20
Posts: 8,085
### Re: solution of triangle
hi Stefy,
That is a brilliant way to do it. Short and no complicated trig stuff. I am in awe. ( no dazzled-smiley-face available)
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Offline
## #7 2013-07-06 04:52:18
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037
### Re: solution of triangle
The thing I dislike about those kinds of ideas are that they can be tough to get. It does come down to luck and experience a lot.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline
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# Please How I can get the figure like in the Picture below
1 view (last 30 days)
Abdelkader Hd on 12 Aug 2022
Commented: Rik on 12 Aug 2022
hello community I look to get the figure like in the picture below using contourf and Thank you
clc
clear all
x=[]; y=[]; z=[];
for n=1:1001
x1=0.01*(n-1);
x2=0.01*(n-4);
B=[0,x1,4,x2;x1,3,x2,x2;x1,0,5,x2;0,5,x2,x1];
Pd=eig(B);
if max(real(Pd))<0
disp('fail');
disp(n);
end
B1=[B(1,1),B(1,2);B(2,1),B(2,2)];
B2=[B(3,3),B(3,4);B(4,3),B(4,4)];
B3=[B(1,3),B(1,4);B(2,3),B(2,4)];
Sum=det(B1)+det(B2)+2.*det(B3);
Et=sqrt(Sum-sqrt(Sum.^2-4.*det(B)))./sqrt(2);
E=2*max(0,real(2*Et));
x(n)=x1; y(n)=x2; z(n)=E;
n=n+1;
end
[X,Y] = meshgrid(x,y);
contourf(X,Y,Z,100, 'edgecolor','none');
plot(x,y)
##### 2 CommentsShowHide 1 older comment
Abdelkader Hd on 12 Aug 2022
@Rik thank you for your response, Please if you can write how I can do this, because I beginner use of matlab
Rik on 12 Aug 2022
You first need to define your variables:
n=(1:1001);
x=0.01*(n-1);
y=0.01*(n-4);
Now we have vectors, but you want the 2D grid they define:
[X,Y] = meshgrid(x,y);
Now we can create a Z array of the correct size to hold the output and loop through all elements of these arrays by using linear indexing.
Z=zeros(size(X));
for n=1:numel(X)
Z(n)=YourCode(X(n),Y(n));
end
contourf(X,Y,Z,100, 'edgecolor','none');
function E=YourCode(x1,x2)
% Don't forget to write comments to explain what this code does. You will
% have forgotten in 6 months, making it impossible to find any bugs.
B=[0,x1,4,x2;x1,3,x2,x2;x1,0,5,x2;0,5,x2,x1];
Pd=eig(B);
if max(real(Pd))<0
disp('fail');
disp(n);
end
B1=[B(1,1),B(1,2);B(2,1),B(2,2)];
B2=[B(3,3),B(3,4);B(4,3),B(4,4)];
B3=[B(1,3),B(1,4);B(2,3),B(2,4)];
Sum=det(B1)+det(B2)+2.*det(B3);
Et=sqrt(Sum-sqrt(Sum.^2-4.*det(B)))./sqrt(2);
E=2*max(0,real(2*Et));
end
##### 2 CommentsShowHide 1 older comment
Rik on 12 Aug 2022
You're welcome
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## 03453660 3 years ago cm, mm, inches, yards all are base units or derived???? please explain
1. akash123
base units...
2. imron07
i dont think there's base and derived unit
3. UnkleRhaukus
$0.9144[\text{m}]\equiv91.44[\text{cm}]\equiv914.4[\text{mm}]\equiv36[\text{in}]\equiv3[\text{ft}]\equiv1[\text{yd}]$
4. 03453660
@UnkleRhaukus i agree with you but mm,cm yards ,inches all were declared as base units by proffesor walter lewin from MIT.
5. UnkleRhaukus
what does base units mean ?
6. 03453660
base units are the units of base quantities eg lenght, time, mass etc
7. UnkleRhaukus
the SI base unit of length is the meter
8. 03453660
in SI can i consider millimeter(mm) as base unit???
9. UnkleRhaukus
nope $1[\text{mm}]=10^{-3}[\text m]$
10. 03453660
ok what about kilogram(kg) in SI system $1[kg]=10^3[g]$
11. 03453660
but still kg is considered as base unit for mass
12. UnkleRhaukus
yep, there are only seven Si base units
13. Vincent-Lyon.Fr
There are seven SI base units. http://www.bipm.org/en/si/base_units/ In case of doubts, always refer to the full SI official brochure: http://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf
14. UnkleRhaukus
and i do agree that the base unit for mass is a confusing choice
15. Vincent-Lyon.Fr
cm and mm are decimal submultiples of the base unit "metre". As for the kilogram, the brochure states: "As an exception, the name of the kilogram, which is the base unit of mass, includes the prefix kilo, for historical reasons. It is nonetheless taken to be a base unit of the SI. The multiples and submultiples of the kilogram are formed by attaching prefix names to the unit name “gram”, and prefix symbols to the unit symbol “g” (see 3.2, p. 122). Thus 10−6 kg is written as a milligram, mg, not as a microkilogram, μkg."
16. 03453660
i asked from my physics professor that in SI system can i consider mm,cm,inches , as base units for length in SI system. he told me no they are derived units because these units are derived from meter. now i m confused that kilogram is a base unit in SI system so its also derived from gram then why not the gram the base unit for length
17. Gowthaman
I think.... The weight of 1 lit volume of water at zero degree Celsius is taken as reference for fixing base unit of weight at initial time, which is equivalent to 1 kg present SI unit. and 1 kg is easily feel-able weight comparing to 1 gm which is very less weight to feel...
18. 03453660
@Vincent-Lyon.Fr thank you so much sir.
19. Vincent-Lyon.Fr
@03453660 Derived units are defined as products of powers of the base units. E.g. m/s or kg/m³ So cm and mm are not derived units. Some of the derived units received special names: kg.m/s² is the coherent derived unit for force, and received the name newton, symbol N.
20. 03453660
so far i was concern about this issue was because i was in doubt because someone told me its base unit and others told me no its derived unit, but i was in strong belief that once a quantity is considered as base or fundamental quantity then all of its units are base units
21. Vincent-Lyon.Fr
This is because there was a confusion between "base quantities" and "base units". Length is a "base quantity", but among all units of length, only the metre is its "base unit".
22. 03453660
@Vincent-Lyon.Fr thus base units depends upon the system i use eg in SI system the base unit for mass is kilogram and in CGS system the base unit for mass is gram
23. Vincent-Lyon.Fr
Yep, I should have written: Length is a "base quantity", but among all units of length, only the metre is its SI-"base unit".
Find more explanations on OpenStudy
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# 10/22/2016 math | Mathematics homework help
Assignment 1: Discussion—The Traveling Salesman Problem
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Some problems in mathematics can be stated very simply but may involve complex solutions. One of the most famous of these is the Traveling Salesman Problem or, as it is known to mathematicians, the TSP.
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Respond to at least two posts contributed by your peers and comment on the problem they demonstrated and the steps they employed to reach a solution. What would you have done the same or different? Do you agree with the solution? Can you suggest a different approach to solving the same problem?
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Vous êtes sur la page 1sur 19
# EXERCISES 3.6 AND 3.
13
CLAIRE S. COLINA B. University of Cartagena Chemical Engineer Program Process Control March 2014
EXERCISE 3.6
Process wastewater (density = 1000 flows at 500 000 kg/h into holding pond with a volume of 5000m3 and then flows from the pound to a river. Initially, the pond is at steady state with a negligible concentration of the pollutants [x(0)=0]. Because of a malfunction in the wastewater treating water treating process, the concentration of pollutants in the inlet stream suddenly increases to 500 mass ppm (kg pollutant per million kg of water) and stays constant at the value (step change)
EXERCISE 3.6
Assuming a perfectly mixed pound, obtain the transfer function of the pollutant concentration of the inlet stream, and determinate for how long can the process malfunction go undetected before the outlet concentration of pollutants exceeds the regulated maximum value of 350 ppm.
Repeat parts (a) assuming that the water flows in plug flow (without mixing) through the
pond. Notice that this means the pond behaves as a pipe and the response of the
concentration is a pure transportation lag.
In both parts (a) and (b), it is assumed that the entire volume of the pond is active. How would your answers be affected if portions of the pond were stagnant and not affected by the flow of water in and out.
EXERCISE 3.6
Obtain the transfer function of the pollutant concentration of the inlet stream. For how long the process malfunction can go undetected (350 ppm)
Assuming: perfect mixing, constant flow, volume and density.
1 1 2 2 =
## 2 1 1 2 2 = Steady State (Eq. 1 on S.S.) 1 1, 2 2, = 2,
(. 1)
Linear Eq.
(. 2)
EXERCISE 3.6
Obtain the transfer function of the pollutant concentration of the inlet stream. For how long the process malfunction can go undetected (350 ppm)
It was assumed constant flow, 1 = 2 = 2 1 2 = (. 3)
## Subtracting Eq.3 Eq.2 1 1, 2 2 = 2 2, (. 4)
Deviation variable 1 2 = 2 (. 5) 1 2 = 2 (. 6)
EXERCISE 3.6
Obtain the transfer function of the pollutant concentration of the inlet stream. For how long the process malfunction can go undetected (350 ppm) Applying Laplace Transform
1 2 = 2
Mathematical Model 1 2 = + 1 . 8
(. 7)
Transfer Function 2 1 = 1 + 1 . 9
It is applied Inverse Laplace Transform to know how long can the process malfunction go undetected (scale function) 1 = 500 1 = 500 Applying Laplace 500 1 = 500 1 + 1
2 =
. 10
EXERCISE 3.6
Obtain the transfer function of the pollutant concentration of the inlet stream. For how long the process malfunction can go undetected (350 ppm) 500 1 2 = ( + ) . 11 = 1
It is applied Inverse Laplace Transform to know how long can the process malfunction go undetected (scale function) 500 1 2 = 1
350 1 = 500 10
2 = 350 = 1
(. 12)
= =
1000
## 50003 3 = 10 500 000
= 0.3
Natural Logarithm
= 10 ln 0.3 = .
EXERCISE 3.6
Repeat (a) assuming water flows in plug flow. The pond behaves as a pipe and the response of the concentration is a pure transportation lag. Because of the Real Translation Theorem, where 0 2 = 0 1 . 13 = 0 1 2 = 500 0 . 14
2 = 350
0 = 0 = = 1000 50003 3 = 10 500 000
300 = 500 0
0.7 + 10 = = .
EXERCISE 3.6
In both parts (a) and (b), it is assumed that the entire volume of the pond is active. How would your answers be affected if portions of the pond were stagnant and not affected by the flow of water in and out. The active volume is represented by the following equation = This reduces the time it takes for the outlet concentration to reach the limit. 2
1 2 =
(. 6)
EXERCISE 3.13
3 3
,
, , 3
EXERCISE 3.13
## The kinetics of the reaction is expressed by the following zeroth-order
expression = 0 Where
0 , 3
()
, ,
EXERCISE 3.13
Determinate the transfer function for the reactor. Express the time constant and gain in terms of the physical parameters.
Under what conditions can the time constant be negative? What would be the consequences of a negative time constant?
EXERCISE 3.13
Determinate the transfer function for the reactor
Rate of moles of components into control volume
## Rate of change of moles components acumulated in control volume
Assume: adiabatic (no heat losses), perfect mixing, constant flow, volume, specific heats and density,
Tref = 0. = 0
= + + ;
= 2
= ( )
1 ,1 1 2 ,2 2 =
1 ,1 1 2 ,2 2 0
2 =
(. 1)
## Non Linear Eq.
EXERCISE 3.13
Determinate the transfer function for the reactor
Applying Taylor Serie to nonlinear terms. 0
0 + 2 2, 2 2,
(. 2)
1 ,1 1 2 ,2 2 0
0 2 2,
2 2, =
(. 3)
## Steady State for Non Linear Eq. (Eq. 1) 1 ,1 1,
2, 2 ,2 2, 0
2, =0
(. 4)
EXERCISE 3.13
Determinate the transfer function for the reactor
Subtracting (Eq. 3 - Eq. 4) and applying deviation variable 1 ,1 1 1 ,1 1, 2 ,2 2 + 2 ,2 2, 0
0 2 2,
2 2,
2, 0
2 2,
2
0 1 ,1 1 2 ,2 + 2 2, 1 ,1 0 2 ,2 + 2 2,
2 =
(. 5)
1 2 =
0 2 ,2 + 2 2,
EXERCISE 3.13
Determinate the transfer function for the reactor
2 1 2 = Applying Laplace Transform 1 2 = 2 The Transfer Function for the reactor is (. 7)
Mathematical Model
(. 6)
2 1 = + 1 1
. 8
EXERCISE 3.13
Express the time constant and gain in terms of the physical parameters.
1 ,1 2 ,2 + 0 2 2,
Gain
(. 9)
Time constant
0 2 ,2 + 2 2,
(. 10)
EXERCISE 3.13
Under what conditions can the time constant be negative? What would be the consequences of a negative time constant?
It is known that the chemical reactor is adiabatic and exothermic. For exothermic reactions < 0 = 0 2 ,2 + 2 2,
0
(. 10)
If 2 ,2 +
THANKS A LOT!
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A060288 Distinct (non-overlapping) twin Harshad numbers whose sum is prime. 3
3, 7, 11, 19, 41, 401, 419, 449, 881, 1021, 1259, 1289, 1471, 1601, 1607, 1871, 1999, 2029, 2281, 2549, 2609, 2833, 3041, 3359, 3457, 4001, 4049, 4481, 4801, 5641, 6329, 7499, 7561, 8081, 8849, 8929, 9613, 9619, 10321, 11131, 12401, 12799, 13033 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Suggested by Puzzle 129, The Prime Puzzles and Problems Connection LINKS EXAMPLE a(3)=19, a prime, because the first Harshad number is 9 and the second is 10 and 9+10=19. To find the Harshad numbers take H1=(p-1)/2 as the first Harshad and then the second Harshad, H2=H1+1. Harshad numbers are those which have integral quotients after division by the sum of their digits. Note that 2+3=5 is not included because 1+2=3 are the first twins whose sum is prime and the next twins, 3+4=7, must not overlap the preceding pair. PROG (UBASIC) 20 A=0; 30 inc A; 40 if Ct=2 then Z=(A-1)+(A-2): if Z=prmdiv(Z) then print A-2; "+"; A-1; "="; Z; "/"; :inc Pt; 50 if Ct=2 then Ct=1:A=A-1; 60 X=1; 70 B=str(A); 80 L=len(B); 90 inc X; 100 S=mid(B, X, 1); 110 V=val(S):W=W+V; 120 if XDt+1 then Ct=0:Dt=0; 150 Dt=Ct:W=0; 160 if A<10000001 then 30; 170 print Pt; CROSSREFS A005349, A060159, A060289 etc. Sequence in context: A132449 A132453 A033871 * A191245 A282914 A284027 Adjacent sequences: A060285 A060286 A060287 * A060289 A060290 A060291 KEYWORD easy,nonn,base AUTHOR Enoch Haga, Mar 23 2001 STATUS approved
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Last modified June 15 20:50 EDT 2019. Contains 324145 sequences. (Running on oeis4.)
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https://nerdutella.com/q29542-Ruby-program-to-print-multiplication-table-of-a-number
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Q:
# Ruby program to print multiplication table of a number
This requires a very simple logic where we only have to multiply the number with digits from 1 to 10. This can be implemented by putting the multiplication statement inside a loop. We have mentioned two ways: one is by using while loop and the second one is by making use of for loop. When you are using while loop, first you will have to initialize i with 1 and increment it by 1 inside the loop. for loop, the method is simpler as it only requires the specification of for keyword along with the range on which the loop is going to work.
Methods used:
• puts: This is a predefined method which is used to put the string on the console.
• gets: This is also a predefined method in Ruby library which is used to take input from the user through the console in the form of string.
• *: This is an arithmetic operator commonly known as multiplication operator which takes two arguments and process them by giving out their product as a result.
Variables used:
• num: This variable is used to store the integer provided by the user.
• mult: This is storing the result for i*num.
• i: This is a loop variable which is initialized by 1.
### Ruby code to print multiplication table of a number
``````=begin
Ruby program to print multiplication table of
a number(by using for loop)
=end
puts "Enter the number:"
num=gets.chomp.to_i
for i in 1..10
mult=num*i
puts "#{num} * #{i} = #{mult}"
end
``````
Output
```Enter the number:
13
13 * 1 = 13
13 * 2 = 26
13 * 3 = 39
13 * 4 = 52
13 * 5 = 65
13 * 6 = 78
13 * 7 = 91
13 * 8 = 104
13 * 9 = 117
13 * 10 = 130
```
Method 2:
``````=begin
Ruby program to print multiplication table of a
number(by using while loop)
=end
puts "Enter the number:"
num=gets.chomp.to_i
i=1
while (i<=10)
mult=num*i
puts "#{num} * #{i} = #{mult}"
i+=1
end
``````
Output
```Enter the number:
16
16 * 1 = 16
16 * 2 = 32
16 * 3 = 48
16 * 4 = 64
16 * 5 = 80
16 * 6 = 96
16 * 7 = 112
16 * 8 = 128
16 * 9 = 144
16 * 10 = 160
```
Code explanation:
The logic of code is pretty simple. In the first method, we are using while loop for the process and in the second one, we are using for loop. We have a variable mult in which we are multiplying the number with the i. The loop will terminate when i becomes equal to 10.
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# Indefinite Integrals
• Aug 4th 2010, 08:07 PM
john-1
Indefinite Integrals
Can someone show me the steps to finding the integral of (x+1)/(x^2+2x) [ln(x^2+2x)] dx
I'm not sure where to begin because I don't have any integration rules with the lnx
• Aug 4th 2010, 08:12 PM
pickslides
Quote:
Originally Posted by john-1
Can someone show me the steps to finding the integral of (x+1)/(x^2+2x) [ln(x^2+2x)] dx
Do you mean 'show that' $\displaystyle \int \frac{x+1}{x^2+2x}~dx = \ln(x^2+2x)$ or 'find' $\displaystyle \int \frac{x+1}{x^2+2x} \ln(x^2+2x)~dx$
??
• Aug 4th 2010, 08:14 PM
john-1
Quote:
Originally Posted by pickslides
Do you mean 'show that' $\displaystyle \int \frac{x+1}{x^2+2x}~dx = \ln(x^2+2x)$ or 'find' $\displaystyle \int \frac{x+1}{x^2+2x} \ln(x^2+2x)~dx$
??
the second one please
• Aug 4th 2010, 08:17 PM
Also sprach Zarathustra
The second one, the first on is not so true...
in second integral use integration by parts...
• Aug 4th 2010, 08:18 PM
john-1
could you show me the steps please? I only have the answer with me in my text but no steps
• Aug 4th 2010, 08:54 PM
TheCoffeeMachine
Using the substitution $\displaystyle u = \ln(x^2+2x)$, we have:
$\displaystyle \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{2}\int\left(\frac{x+1}{x^2+2x}\right)\lef t(\frac{x^2+2x}{x+1}\right)\cdot u \;{du} = \frac{1}{2}\int{u}\;{du} = \frac{1}{4}u^2+k.$
Therefore $\displaystyle \displaystyle \int\frac{(x+1)\ln(x^2+2x)}{x^2+2x}\;{dx} = \frac{1}{4}\ln^2(x^2+2x)+k$.
• Aug 5th 2010, 11:31 AM
john-1
TheCoffeeMachine, in your second step, why did you put (x^2 + 2x) / (x+1)? Where did that come from? mostly confused about 2nd step
• Aug 5th 2010, 11:58 AM
AllanCuz
If $\displaystyle u = ln(x^2 + 2x)$ then $\displaystyle \frac{du}{dx} = \frac{2x +2}{x^2+2x} \to dx = (\frac{1}{2}) \frac{x^2 + 2x}{x+1}du$
He then subbed this into the integral for dx!
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CYCLIDE
From the Greek Kuklos: circle, wheel and eidos: appearance.
2 equal a_i : dupin berger 20.7.3
The cyclides are the envelopes of spheres (C) the centers of which describe a curve or a surface (G0) (the deferent) and such that a fixed point O has a constant power p with respect to these spheres (this notion is therefore analogous to that of cyclic curve in the plane).
They are therefore circled surfaces.
The cyclides with a parabola or a paraboloid as a deferent are the spherical cubic surfaces and the cyclides with a conic or a quadric of another kind are the bispherical quartic surfaces, also called "Darboux cyclides".
General equation: . When the deferent is a conic, the cyclide is called "Dupin cyclide".
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https://planetpailly.com/2020/06/05/sciency-words-closed-timelike-curves/
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# Sciency Words: Closed Timelike Curves
Hello, friends! Welcome to Sciency Words, a special series here on Planet Pailly where we talk about those weird words scientists like to use. Today on Sciency Words, we’re talking about:
## CLOSED TIMELIKE CURVES
Austrian-born logician and mathematician Kurt Gödel was one of Albert Einstein’s closest friends. At Princeton’s Institute for Advanced Study, the two were known to take long walks together, discussing all sorts of strange and wonderful things, no doubt.
As science historian James Gleick tells the story in his book Time Travel: A History, Gödel presented Einstein with a very special gift for Einstein’s 70th birthday. It was the kind of gift only a person like Einstein would appreciate: a series of mathematical calculations. Specifically, these were calculations based on Einstein’s own theory of general relativity which showed that yes, time travel is possible.
Gödel’s calculations were officially published in this 1949 paper. Now I won’t try to explain Gödel’s math because a) I don’t really understand it and b) it’s not really important for the purposes of a Sciency Words post. What is important for our purposes is that Gödel’s 1949 paper introduced a new concept called “closed timelike curves.”
Well, technically speaking, Gödel used the term “closed time-like lines,” not “closed timelike curves.” But as Google ngrams shows us, the hyphen quickly dropped out of “time-like,” and by the 1990’s, “curves” beat out “lines.” So today, closed timelike curves is the most broadly accepted way to say what Gödel was trying to say. The term is also commonly abbreviated at C.T.C.
In short, a closed timelike curve is a path through space and time that circles back to its own beginning. As I understand it, it would take a stupendous amount of force to twist space-time around itself in this way. You’d need the extreme gravitational force of a black hole—or perhaps something even more extreme than that—in order to make a closed timelike curve happen.
But it could happen. As Gödel demonstrated in 1949, general relativity would allow a closed timelike curve to exist, or at least relativity does not forbid such things from existing.
So time travel is possible. It may not be anywhere near practical, but it is at least possible.
Speaking of time travel, are you a fan of time travel adventure stories? The kinds of stories you might see on Doctor Who or The Twilight Zone? Then please check out my new book, The Medusa Effect: A Tomorrow News Network Novella, featuring time traveling news reporter Talie Tappler and her cyborg cameraman, Mr. Cognis.
## 3 thoughts on “Sciency Words: Closed Timelike Curves”
1. It’s a good one! It’s one of those things you’d think a science fiction writer dreamed up, but no, this comes from the work of an actual scientist.
Liked by 1 person
1. A good hook in a story-line will always grab a readers attention.
Liked by 1 person
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http://www.madsci.org/posts/archives/nov2001/1006378023.Ph.r.html
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MadSci Network: Physics Query:
### Re: At what Temperature does water freeze on a road?
Date: Tue Nov 13 11:05:18 2001
Posted By: John Link, Physics
Area of science: Physics
ID: 1005627154.Ph
Message:
I assume you want to know if radiation, evaporation, and wind, etc., can contribute to the cooling of the water layer and/or the road surface, and that we are assuming there are no chemicals (such as salt) which could depress the freezing point of the water. Given those assumptions, there are some ways that ice can form on a road surface even if the measured air temperature is above 32 degrees F.
You can find some previous answers in our archives (use our search engine) which discuss frost, and that frost can form when the measured are temperature is substantially higher than 32 degrees F (as high as 38 or 39 degrees F) due to the boundary layer's cooling due to the (mostly infrared) radiation of the road surface toward the clear sky above. (The boundary layer is a very thin insulating layer of air just above the surface. If the air is very still [very little wind] the boundary layer can insulate the surface enough so that it can cool below the air temperature above the boundary layer. When there is wind, however, there is not much of a boundary layer.)
Similarly, if there is almost no wind and the sky is very clear, the road surface can cool below 32 degrees F even if the measured air temperature (above the road) is as warm as 38 or 39 degrees F. However, the ground below the road will usually be warmer, and so the final answer depends on lots of factors. A bridge (or overpass) whose roadbed is exposed to the air will usually be closer to the temperature of the air, and so the radiative cooling will be less important.
Also, evaporation can cool the water layer, but usually when the road surface is wet the relative humidity is very high and so evaporation is not very substantial.
The answer, as you can tell, depends on lots of factors.
John Link, MadSci Physicist
Current Queue | Current Queue for Physics | Physics archives
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© 1995-2001. All rights reserved.
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# A Bit Puzzle
I love a good bit-level puzzle. Today’s is one I learned about from a comment in a Google+ discussion:
Given two words a and b, return the high bits that they have in common, with the highest bit where they differ set, and all remaining bits clear. So 10101 and 10011 yields 10100. Then do the same thing, only flipped, so we keep low-order identical bits and mask out the high ones.
The problem comes from trie implementation. Here’s code that does it the slow way:
```uint64_t high_common_bits_64_slow (uint64_t a, uint64_t b)
{
uint64_t output = 0;
int i;
for (i=63; i>=0; i--) {
} else {
goto out;
}
}
out:
return output;
}
uint64_t low_common_bits_64_slow (uint64_t a, uint64_t b)
{
uint64_t output = 0;
int i;
for (i=0; i<64; i++) {
} else {
goto out;
}
}
out:
return output;
}
```
The problem is to do much better than these. To keep things simple, let’s say we’re interested primarily in speed on newish x86-64 implementations. Choose any programming language, or just write assembly. In a few days I’ll post my solutions in addition to any good ones that show up here before then. Beware WordPress’s mangling of source code in the comments — feel free to just mail me.
## 34 Replies to “A Bit Puzzle”
1. I’ll just add that I was bummed not to be able to create branch-free x64 code for these. I’m sure it’s possible.
2. You would love the first lab assignment from the _Computer Systems: A Programmer’s Perspective_ book. Having TA’d the course several years now, the bit-twiddling assignments almost always result in some very interesting solutions from the students:
http://csapp.cs.cmu.edu/public/labs.html
3. Andrew Hunter says:
We’ll see how WordPress takes this, but here’s a solution (I think, I haven’t checked all corner cases.);
uint64_t high_common(uint64_t a, uint64_t b) {
uint64_t x = (a ^ b);
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
x |= x >> 32;
return (a & ~x) | ((x >> 1) + 1);
}
Fairly long dependency chain, but all simple ops and entirely branch free. I think there might be a faster way to implement the middle section (“set all bits under any currently set bits….”) but I’d have to check the assembly spec.
Also, is this useful for something? Seems like a fairly odd spec to just cut from whole cloth.
4. Hi Lars, I was an instructor for that course 3 or 4 times and always loved that lab. But too many of the problems have solutions that are easily available on the Internet.
5. Is there a way to link to a G+ comment? I don’t see a way to do that. But anyway, it was posted by Jan-Willem Maessen.
6. Andrew Hunter says:
Lars, yeah, I remember liking that lab in undergrad (HMC).
7. Hi Andrew, your version fails on my tester 🙂
8. Mike G. says:
Offhand, what I’m thinking of is decomposing this into known bit-twiddling problems:
unsigned diffIndex=findHighestBitSet(a^b);
// needed to mask off the top bits. There’s probably a
// more efficient way to do it – too lazy to dig up ]
// “Hacker’s Delight”. The ‘1<<diffIndex'
// shift could also be replaced with a table lookup if
// more efficient.
That's without any testing, but I think that would work, and it should certainly be faster than the naive version.
9. Pascal Cuoq says:
@Andrew
> is this useful for something?
It looks like exactly what a particular big-endian (resp. little-endian) implementation of Patricia trees might need.
One day, I will have the time to measure which is faster between the two sequences below.
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
or
y = x >> 1;
z = x >> 2;
x |= y;
x |= z;
y = x >> 3;
z = x >> 6;
x |= y;
x |= z;
y = x >> 9;
z = x >> 18;
x |= y;
x |= z;
The latter is longer but has shorter dependency chains.
10. Hi John,
If you replace Andrew’s last line with “return (a & ~x) | (x & ~(x >> 1));”, does it pass?
11. Wow, that’s good serendipity; I’ve been thinking of this exact same operation for the last week or so. I use this exact operation in a spatial data structure built around z-order space-filling curves.
(To answer Andrew’s question): If you think of a binary number as a sequence of binary tree decisions on the full range of possible numbers, this bit operation gives you the most specific common ancestor on this tree.
With z-order curves you get a cheap way to compute fairly tight bounding boxes between points in n dimensions.
12. uint64_t high(uint64_t a, uint64_t b) {
uint64_t db = a ^ b, hdb;
db |= (db >> 1);
db |= (db >> 2);
db |= (db >> 4);
db |= (db >> 8);
db |= (db >> 16);
db |= (db >> 32);
return ((a & b) | (db >> 1)) + 1;
}
uint64_t low(uint64_t a, uint64_t b) {
uint64_t x = (a ^ b) & -(a ^ b);
return (a & (x – 1)) | x;
}
13. Eddie, your low code looks good but the high one fails in (I think) the same way as Andrew’s.
14. Tony– it works!
high_common(uint64_t a, uint64_t b)
{
uint64_t x = a ^ b;
int n = 63 – clz(x);
return (a >> n | 1) << n;
}
Fixing this for a == b is left as an exercise.
16. Benjamin Kramer says:
A version with gcc builtins.
uint64_t high(uint64_t a, uint64_t b) {
uint64_t bit = 1ULL << 63-__builtin_clzll(a^b);
return a == b ? a : (a & ~(bit-1)) | bit;
}
uint64_t low(uint64_t a, uint64_t b) {
uint64_t bit = 1ULL << __builtin_ctzll(a^b);
return a == b ? a : (a & bit-1) | bit;
}
Sadly this is not branchless on x86 because bsf/bsr are undefined for 0. The upcoming BMI instructions lzcnt and tzcnt solve this ussue
17. BTW, with GCC __builtin_clz() provides the count-leading-zeros instruction if available.
18. Ryan Fox says:
Darn, I thought I was going to be the only one clever enough to remember about clz.
Anyway, here’s my solution. I think I’m depending on undefined behaviour with shifting UINT32_MAX, so that probably counts as cheating. 😛
uint32_t puzzle(uint32_t a, uint32_t b)
{
uint32_t and = a&b;
uint32_t xor = a^b;
uint32_t result = and & and_mask | 1 << (8*sizeof(result) – leading_zeros – 1);
return result;
}
19. Is the spec defined when a == b? I.e. high(a,b) == low(a,b) == a == b?
20. Ryan, any unsigned value x may be safely shifted by n in either direction provided 0 <= n < CHAR_BIT * sizeof(x).
21. Another variant for high_common() would be Eddie’s low() combined with a bit-reversal instruction if the CPU has one. ARM does, don’t know about x86. I don’t know of any CPU that has bit-reverse but not CLZ though, so this is probably not very useful.
22. Eddie, that’s what I was assuming.
Carlos, is it correct to just return the argument when a==b? It probably doesn’t matter in practice since you’re not going to call the function unless you have two different tree nodes.
23. Andrew Hunter says:
Regehr,
Yeah, I noticed the all-same behavior while walking to work. Serves me right for not doing better testing. (How are you testing these, by the way? Hand-chosen interesting test cases, or one of the tools you’ve mentioned for generating failures? I ran mine through a fuzzer, but that’s not likely to generate any number of “interesting” cases.)
24. Andrew, my tests are made by:
– pick a random 64-bit int and call it a
– initialize b to a, then flip 0..63 random bits
I’m not sure if this is good, but clearly just making a and b separately random is wrong.
25. Benjamin, these are basically my solutions. Your branch is the same one I failed to eliminate.
26. Carlos, hope we can help make your code faster :).
27. Andrew Hunter says:
Regehr, yeah, that looks better than my method (pick a,b iid uniform.) I’d be curious what would happen if you threw a SAT solver at these functions (searching for a setting of bits where they differ). I’ve heard of good results from (similar) techniques from, for one, the code sketching people at MIT, but I don’t have their infrastructure handy, so…
28. Andrew, bit puzzles like this pretty much scream for being attacked by SAT or SMT.
About a dozen times I’ve started on this kind of project before realizing (1) it’s really hard to do a good job, and maybe impossible if there are loops and (2) lots of smart people already work on this kind of thing.
29. MSN says:
For the lower bit function, you don’t need to bother with the tree:
uint64_t d= a ^ b;
uint64_t low_d= (d & -d);
30. An (untestet) asm idea based on the bsf instruction without branches/loops:
Assume a in eax and b in ebx.
andl %eax,%ebx # put a&b in eax
bsfl %ecx,%eax # index of least significant bit in ecx
shr %eax, %cl # shift a&b right by index (cl == lower 8bit of ecx)
shl %eax, %cl # shift left again, so lower bits now cleared
Result in eax now
I’m not sure, if the bsf instruction is more efficient than a table-based findHighestBitSet, though.
31. Correction, because i misunderstood the puzzle:
Assume a in eax and b in ebx.
xorl %eax,%ebx # put a^b in eax
bsfl %ecx,%eax # index of least significant bit in ecx
dec %ecx # index–
shr %eax, %cl # shift a^b right by index (cl == lower 8bit of ecx)
and %eax, 1 # set differing bit
shl %eax, %cl # shift left again, so lower bits now cleared
Result in eax now
32. Here are two branchless versions that I think work even for a==b. high2() assumes that __builtin_clzl(0) returns some arbitrary integer value (but does not trap or otherwise explode).
uint64_t high(uint64_t a, uint64_t b) {
uint64_t db = a ^ b;
db |= (db >> 1);
db |= (db >> 2);
db |= (db >> 4);
db |= (db >> 8);
db |= (db >> 16);
db |= (db >> 32);
return ((a & b) | (db >> 1)) + (db > 0);
}
uint64_t high2(uint64_t a, uint64_t b) {
uint64_t x = a ^ b, nz = x > 0;
uint64_t db = (uint64_t) nz << (63 – (__builtin_clzl(x) & 63));
return (a & ~(db – nz)) | db;
}
33. qznc, bsf is fast on Intel Pentium-2 and later (except Atom), and AMD K10 and later. On Intel Pentium, Intel Atom, and AMD <= K8 it is very slow, perhaps microcoded.
34. A bit off-topic, but has anyone found a way to post code (fixed-width font) on Google+?
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# FET and BJT Difference
## FET and BJT Difference:
FET and BJT Difference (CS, CD, and CG Circuit Comparison) – Table 11-1 compares Zi, Zo and Av for CS, CD, and CG circuits. As already discussed, the CS circuit has voltage gain, high input impedance, high output impedance, and a 180° phase shift from input to output. The CD circuit has high input impedance, low output impedance, a voltage gain of approximately 1, and no phase shift. The CG circuit offers low input impedance, high output impedance, voltage gain, and no phase shift.
### Impedance at the FET Gate:
Consideration of each type of circuit shows that the input or output impedance depends upon which device terminal is involved. In both the CS and CD circuits, the input signal is applied to the FET gate terminal, so Zi is the impedance looking into the gate. Figure 11-25 shows that, for CS and CD circuits the gate input impedance is,
and the circuit input impedance is,
where RG is the equivalent resistance of the bias resistors. Because the reverse-biased gate-source resistance (Rgs) is so large, an unbypassed source resistor has no significant effect on Zi at the gate. The circuit input impedance is largely determined by RG in both the CS and CD circuits.
### Impedance at the FET Source:
The FET source is the output terminal for a CD circuit and the input terminal for a CG circuit. The device impedance in each case is the impedance looking into the source, (see Fig. 11-26).
The circuit impedance at the source terminal must include the source resistor.
### Impedance at the FET Drain:
The output for CS and CG circuits is produced at the FET drain terminal. So, the impedance looking into the drain is the device output impedance (rd). As illustrated in Fig. 11-27, the circuit output impedance for any circuit with the output taken from the FET drain terminal is,
### Voltage gain:
In a circuit with an unbypassed source resistor, the ac voltage at the source terminal follows the ac input at the gate, (see Fig. 11-28). A CD circuit (a source follower) has a voltage gain of approximately 1. A CS circuit with an unbypassed source resistor has vi developed across RS, and so,
In a CG circuit, and in a CS circuit with RS bypassed, the ac input (vi) is developed across the gate-source terminals, and the ac output (vo) is produced at the drain terminal. Thus, the voltage gain is the same for CS and CG circuits with similar component values and FET parameters. The CS voltage gain, equation can be used for the CG circuit, with the omission of the minus sign that indicates CS phase inversion.
Although the voltages gains are equal for similar CS and CG circuits, the low input impedance of the CG circuit can substantially attenuate the signal voltage, and result in a low amplitude output.
### FET-BJT circuit Comparison:
Table 11-2 compares Zi, Zo and Av for the basic FET and BJT circuits. The BJT CE and CB circuits have much higher voltage gains than the corresponding FET CS and CG circuits, while the FET CS and CD circuits have much higher input impedances than the BJT CE and CC circuits. Apart from these FET and BJT Difference, BJTs and FETs can generally perform similar functions. High-frequency, fast-switching, and high-power devices of both types are available. In some switching applications, FETs have the advantage of a smaller voltage drop (VDS(on)) than that across BJTs (VCE(sat)).
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# 5 Model comparisons and testing for lack of fit
## 5.1 F-tests for comparing two models
### 5.1.1 Example:
Model A: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1$$
Model B: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3$$
i.e. in Model A, $$\beta_2=\beta_3=0$$.
Model A is the reduced or simpler model and model B is the full model.
The $$\mbox{SSE}$$ for Model B will be smaller than the $$\mbox{SSE}$$ for Model A but is the reduction enough to justify the two extra parameters?
We have:
Model A:
$\mbox{SST} = \mbox{SSR}(A) + \mbox{SSE}(A)$
Model B:
$\mbox{SST} = \mbox{SSR}(B) + \mbox{SSE}(B)$
Note:
$\mbox{SSE}(A)-\mbox{SSE}(B)=\mbox{SSR}(B)-\mbox{SSR}(A)$
### 5.1.2 F-test to compare models:
Model A: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + ... + \beta_q x_q$$
Model B: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + ... + \beta_k x_k$$
where $$q<k$$ and Model A is nested within Model B.
$$H_0$$: $$\beta_{q+1} = \beta_{q+2} = ... = \beta_k = 0$$
$$H_A$$: At least one $$\beta_{q+1}, ... , \beta_k \neq 0.$$
$F =\frac{(\mbox{SSE}(A)-\mbox{SSE}(B))/(k-q)}{\mbox{SSE}(B)/(n-p)}.$
Under $$H_0$$, $F \sim F_{(k-q),(n-p)},$ where $$p = (k+1).$$
Note: Equivalently, the F-test can be written as:
$F =\frac{(\mbox{SSR}(B)-\mbox{SSR}(A))/(k-q)}{\mbox{SSE}(B)/(n-p)}.$
Note: Models A and B must be hierarchical for the F-test to be valid.
### 5.1.3 Example: Steam data
This data is from a study undertaken to understand the factors that caused energy consumption in detergent manufacturing over a 25 month period. Example from Draper and Smith (1966).
The data variables are:
y = STEAM Pounds of steam used monthly.
x1 = TEMP Average atmospheric temperature ($$^o$$F).
x2 = INV Inventory: pounds of real fatty acid in storage per month.
x3 = PROD Pounds of crude glycerin made.
x4 = WIND Average wind velocity (in mph).
x5 = CDAY Calendar days per month.
x6 = OPDAY Operating days per month.
x7 = FDAY Days below $$32^o$$F.
x8 = WIND2 Average wind velocity squared.
x9 = STARTS Number of production start-ups during the month.
## The following objects are masked from steamdata (pos = 9):
##
## CDAY, FDAY, INV, OPDAY, PROD, STARTS, STEAM, TEMP, WIND, WIND2
## The following objects are masked from steamdata (pos = 16):
##
## CDAY, FDAY, INV, OPDAY, PROD, STARTS, STEAM, TEMP, WIND, WIND2
## The following objects are masked from steamdata (pos = 23):
##
## CDAY, FDAY, INV, OPDAY, PROD, STARTS, STEAM, TEMP, WIND, WIND2
## The following objects are masked from steamdata (pos = 30):
##
## CDAY, FDAY, INV, OPDAY, PROD, STARTS, STEAM, TEMP, WIND, WIND2
## The following objects are masked from steamdata (pos = 37):
##
## CDAY, FDAY, INV, OPDAY, PROD, STARTS, STEAM, TEMP, WIND, WIND2
## The following objects are masked from steamdata (pos = 44):
##
## CDAY, FDAY, INV, OPDAY, PROD, STARTS, STEAM, TEMP, WIND, WIND2
## The following objects are masked from steamdata (pos = 51):
##
## CDAY, FDAY, INV, OPDAY, PROD, STARTS, STEAM, TEMP, WIND, WIND2
## The following objects are masked from steamdata (pos = 59):
##
## CDAY, FDAY, INV, OPDAY, PROD, STARTS, STEAM, TEMP, WIND, WIND2
Model A: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1$$
Model B: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3$$
where $$x_1$$ = TEMP, $$x_2$$ = INV, $$x_3$$ = PROD.
modelA <- lm(STEAM ~ TEMP)
modelB <- lm(STEAM ~ TEMP + INV + PROD)
summary(modelA)
##
## Call:
## lm(formula = STEAM ~ TEMP)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.6789 -0.5291 -0.1221 0.7988 1.3457
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 13.62299 0.58146 23.429 < 2e-16 ***
## TEMP -0.07983 0.01052 -7.586 1.05e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.8901 on 23 degrees of freedom
## Multiple R-squared: 0.7144, Adjusted R-squared: 0.702
## F-statistic: 57.54 on 1 and 23 DF, p-value: 1.055e-07
summary(modelB)
##
## Call:
## lm(formula = STEAM ~ TEMP + INV + PROD)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.2348 -0.4116 0.1240 0.3744 1.2979
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 9.514814 1.062969 8.951 1.30e-08 ***
## TEMP -0.079928 0.007884 -10.138 1.52e-09 ***
## INV 0.713592 0.502297 1.421 0.17
## PROD 0.330497 3.267694 0.101 0.92
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.652 on 21 degrees of freedom
## Multiple R-squared: 0.8601, Adjusted R-squared: 0.8401
## F-statistic: 43.04 on 3 and 21 DF, p-value: 3.794e-09
anova(modelA)
## Analysis of Variance Table
##
## Response: STEAM
## Df Sum Sq Mean Sq F value Pr(>F)
## TEMP 1 45.592 45.592 57.543 1.055e-07 ***
## Residuals 23 18.223 0.792
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(modelB)
## Analysis of Variance Table
##
## Response: STEAM
## Df Sum Sq Mean Sq F value Pr(>F)
## TEMP 1 45.592 45.592 107.2523 1.046e-09 ***
## INV 1 9.292 9.292 21.8588 0.0001294 ***
## PROD 1 0.004 0.004 0.0102 0.9203982
## Residuals 21 8.927 0.425
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
$$H_0$$: $$\beta_2 = \beta_3 = 0$$
$$H_A$$: At least one $$\beta_2, \beta_3 \neq 0$$
\begin{align*} F_{obs} & = \frac{(\mbox{SSE}(A)-\mbox{SSE}(B))/(k-q)}{\mbox{SSE}(B)/(n-p)}\\ & = \frac{(18.223-8.927)/(3-1)}{8.927/(25-4)}=10.93.\\ \end{align*}
$$F_{(0.05,2,21)} = 3.467$$, $$F_{(0.01,2,21)} = 5.780$$
P-value $$<0.01$$, we reject $$H_0$$ and conclude that at least one of $$\beta_2$$, $$\beta_3$$ differ from 0.
anova(modelA, modelB)
## Analysis of Variance Table
##
## Model 1: STEAM ~ TEMP
## Model 2: STEAM ~ TEMP + INV + PROD
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 23 18.223
## 2 21 8.927 2 9.2964 10.934 0.0005569 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 5.2 Sequential sums of squares
### 5.2.1 Example:
Model A: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1$$
Model B: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + \beta_2 x_2+ \beta_3 x_3$$
As noted earlier, the reduction in $$\mbox{SSE}$$ going from Model A to B, is equivalent to the increase in $$\mbox{SSR}$$, i.e. $\mbox{SSE}(A)-\mbox{SSE}(B)=\mbox{SSR}(B)-\mbox{SSR}(A).$
We can denote: $\mbox{SSR}(B|A)=\mbox{SSR}(B)-\mbox{SSR}(A).$
These are the sequential sums of squares. We can write: \begin{align*} \mbox{SST} & = \mbox{SSR}(B) + \mbox{SSE}(B)\\ & = \mbox{SSR}(A) +\mbox{SSR}(B) - \mbox{SSR}(A) + \mbox{SSE}(B)\\ & = \mbox{SSR}(A) + \mbox{SSR}(B|A) + \mbox{SSE}(B).\\ \mbox{SST} - \mbox{SSE}(B) &= \mbox{SSR}(A) + \mbox{SSR}(B|A)\\ \mbox{SSR}(B) &= \mbox{SSR}(A) + \mbox{SSR}(B|A).\\ \end{align*}
If model A is appropriate, $$\mbox{SSR}(B|A)$$ should be small.
### 5.2.2 Example: Steam data
Model A: $$\mathbb{E}[y] = \beta_0$$
Model B: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1$$
Model C: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + \beta_2 x_2$$
Model D: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3$$
SOURCE df seqSS Notation
TEMP 1 45.592 SSR(B|A)
INV 1 9.292 SSR(C|B)
PROD 1 0.004 SSR(D|C)
From the ANOVA table,
\begin{align*} \mbox{SSR}(D)& =54.889\\ & = \mbox{SSR}(B|A) + \mbox{SSR}(C|B) + \mbox{SSR}(D|C)\\ \end{align*}
We can use the F-test for comparing two models to test Seq SS.
1):
Model A: $$\mathbb{E}[y] = \beta_0$$
Model D: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3$$
$$H_0$$: $$\beta_1 = \beta_2 = \beta_3 = 0$$
$$H_a$$: Not all $$\beta_i$$ are 0
$$\mbox{SSR}(A) = 0$$
$$\mbox{SSR}(D|A) = \mbox{SSR}(D) = 54.889.$$
$F_{obs} = \frac{\mbox{SSR}(D|A)/(k-q)}{\mbox{SSE}(D)/(n-p)} = \frac{54.889/(3-0)}{8.927/(25-4)}=43.04$
P-value $$< 0.001$$, we reject $$H_0$$ and conclude that not all $$\beta_i$$ are 0.
2):
Model C: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + \beta_2 x_2$$
Model D: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3$$
$$H_0$$: $$\beta_3 = 0$$
$$H_a$$: $$\beta_3 \neq 0$$
$$\mbox{SSR}(D|C) = 0.004$$
$F_{obs} = \frac{\mbox{SSR}(D|C)/(k-q)}{\mbox{SSE}(D)/(n-p)} = \frac{0.004/1}{8.927/21} = 0.01$
$$F_{(0.1,1,21)} = 2.96096$$, so P-value $$>0.05$$.
We fail to reject $$H_0$$ and conclude there is no evidence $$\beta_3 \neq 0$$, i.e. $$x_3$$ is not needed in the model.
This F-test is equivalent to a t-test for $$\beta_3$$: $T = 0.1$ $F = (0.1)(0.1) = 0.01$
The p-value for both tests $$= 0.92$$.
Note: The Seq SS values depend on the order in which the variables are added to the model (unless the variables are uncorrelated).
### 5.2.3 Example: Steam cont’d in MTB
Regression Analysis: STEAM versus TEMP
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 1 45.5924 45.5924 57.54 0.000
TEMP 1 45.5924 45.5924 57.54 0.000
Error 23 18.2234 0.7923
Lack-of-Fit 22 17.4042 0.7911 0.97 0.680
Pure Error 1 0.8192 0.8192
Total 24 63.8158
Model Summary
S R-sq R-sq(adj) R-sq(pred)
0.890125 71.44% 70.20% 66.32%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 13.623 0.581 23.43 0.000
TEMP -0.0798 0.0105 -7.59 0.000 1.00
Regression Equation
STEAM = 13.623 -0.0798TEMP
Regression Analysis: STEAM versus TEMP, INV, PROD
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 3 54.8888 18.2963 43.04 0.000
TEMP 1 43.6895 43.6895 102.78 0.000
INV 1 0.8580 0.8580 2.02 0.170
PROD 1 0.0043 0.0043 0.01 0.920
Error 21 8.9270 0.4251
Total 24 63.8158
Model Summary
S R-sq R-sq(adj) R-sq(pred)
0.651993 86.01% 84.01% 79.77%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 9.51 1.06 8.95 0.000
TEMP -0.07993 0.00788 -10.14 0.000 1.05
INV 0.714 0.502 1.42 0.170 9.51
PROD 0.33 3.27 0.10 0.920 9.55
Regression Equation
STEAM = 9.51 -0.07993TEMP +0.714INV +0.33PROD
Regression $$>$$ Options $$>$$ sum of squares tests $$>$$ sequential
Regression Analysis: STEAM versus TEMP, INV, PROD
Analysis of Variance
Source DF Seq SS Seq MS F-Value P-Value
Regression 3 54.8888 18.2963 43.04 0.000
TEMP 1 45.5924 45.5924 107.25 0.000
INV 1 9.2921 9.2921 21.86 0.000
PROD 1 0.0043 0.0043 0.01 0.920
Error 21 8.9270 0.4251
Total 24 63.8158
Model Summary
S R-sq R-sq(adj) R-sq(pred)
0.651993 86.01% 84.01% 79.77%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 9.51 1.06 8.95 0.000
TEMP -0.07993 0.00788 -10.14 0.000 1.05
INV 0.714 0.502 1.42 0.170 9.51
PROD 0.33 3.27 0.10 0.920 9.55
Regression Equation
STEAM = 9.51 - 0.07993TEMP + 0.714INV + 0.33PROD
Regression $$>$$ Options $$>$$ sum of squares tests $$>$$ sequential, but change the order in which the predictors are input in MTB. Output below is MTB17, MTB18 rearranges them in the order TEMP INV PROD.
Regression Analysis: STEAM versus PROD, INV, TEMP
Analysis of Variance
Source DF Seq SS Seq MS F-Value P-Value
Regression 3 54.889 18.2963 43.04 0.000
PROD 1 5.958 5.9577 14.02 0.001
INV 1 5.242 5.2415 12.33 0.002
TEMP 1 43.690 43.6895 102.78 0.000
Error 21 8.927 0.4251
Total 24 63.816
Model Summary
S R-sq R-sq(adj) R-sq(pred)
0.651993 86.01% 84.01% 79.77%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 9.51 1.06 8.95 0.000
PROD 0.33 3.27 0.10 0.920 9.55
INV 0.714 0.502 1.42 0.170 9.51
TEMP -0.07993 0.00788 -10.14 0.000 1.05
Regression Equation
STEAM = 9.51 + 0.33PROD + 0.714INV - 0.07993TEMP
The anova and aov functions in R implement a sequential sum of squares (type I). Function Anova(, type= 2) in library(car) gives the adjusted SS (type II)
modelB <- lm(STEAM ~ TEMP + INV + PROD)
anova(modelB)
## Analysis of Variance Table
##
## Response: STEAM
## Df Sum Sq Mean Sq F value Pr(>F)
## TEMP 1 45.592 45.592 107.2523 1.046e-09 ***
## INV 1 9.292 9.292 21.8588 0.0001294 ***
## PROD 1 0.004 0.004 0.0102 0.9203982
## Residuals 21 8.927 0.425
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(lm(STEAM ~ PROD + INV + TEMP))
## Analysis of Variance Table
##
## Response: STEAM
## Df Sum Sq Mean Sq F value Pr(>F)
## PROD 1 5.958 5.958 14.015 0.001197 **
## INV 1 5.242 5.242 12.330 0.002076 **
## TEMP 1 43.690 43.690 102.776 1.524e-09 ***
## Residuals 21 8.927 0.425
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#library(car)
Anova(modelB, type= 2)
## Anova Table (Type II tests)
##
## Response: STEAM
## Sum Sq Df F value Pr(>F)
## TEMP 43.690 1 102.7760 1.524e-09 ***
## INV 0.858 1 2.0183 0.1701
## PROD 0.004 1 0.0102 0.9204
## Residuals 8.927 21
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 5.3 Testing for lack of fit
When replicate values of response are available at some or all of the $$X$$ values, a formal test of model adequacy is available.
The test is based on comparing the fitted value to the average response for that level of $$X$$.
NOTATION: Suppose there are $$g$$ different values of $$X$$ and at the $$i^{th}$$ of these, there are $$n_i$$ observations of $$Y$$.
Let $$\bar{y}_{i.}=\frac{1}{n_i}\sum_{j=1}^{n_i} y_{ij}$$, $$\quad i=1, ..., g.$$
Note: this is the estimate of the group means in the 1-way ANOVA model (means model): $$y_{ij} = \mu_{i} + \epsilon_{ij}$$, where $$\epsilon_{ij}$$ iid $$N(0, \sigma^2)$$.
Then the pure error sums of squares,
\begin{align*} \mbox{SS}_{\mbox{PE}}& =\sum_{i=1}^g \sum_{j=1}^{n_i} (y_{ij}- \bar{y}_{i.})^2\\ df_{PE} & = \sum_{i=1}^g (n_i-1)=n-g, \hspace{1cm} \mbox{where } n=n_1+...+n_g.\\ \end{align*}
Therefore
$\frac{\sum_{i=1}^g \sum_{j=1}^{n_i} (y_{ij}- \bar{y}_{i.})^2}{n-g}$
is an estimator of $$\sigma^2$$.
NOTE:
• Here we use the replicates to obtain an estimate of $$\sigma^2$$ which is independent of the fitted model (SLR).
*This estimator of $$\sigma^2$$ corresponds to the $$\mbox{MSE}$$ in the ANOVA table for the 1-way ANOVA model.
• The 1-way ANOVA model has $$g$$ parameters. The SLR model has $$2$$ parameters. The latter is more restrictive as it requires linearity.
• $$df_{PE} = n-g$$,
• $$df_{SLR} = n-2$$.
The SLR model has a residual SS which is $$\geq$$ residual SS from the means model, i.e. $$\mbox{SSE} \geq \mbox{SS}_{\mbox{PE}}$$.
A large difference $$\mbox{SSE} - \mbox{SS}_{\mbox{PE}}$$ indicates lack of fit of the regression line.
$$\mbox{SS}(\mbox{lack of fit})= \mbox{SSE} - \mbox{SS}_{\mbox{PE}} = \sum_{i,j} (\hat{y}_{i,j} - \bar{y}_i)^2$$, the sum of squared distances of between the SLR estimate and the means model estimate of $$\mathbb{E}(Y_{i,j})$$.
Lack of fit is tested by the statistic:
$F_{obs}=\frac{\left ( \mbox{SSE}-\mbox{SS}_{\mbox{PE}} \right )/(g-2)}{\mbox{SS}_{\mbox{PE}}/(n-g)}.$
$$H_0$$: Regression model fits well
$$H_A$$: Regression model displays lack of fit
Under $$H_0$$, $$F_{obs} \sim F_{g-2,n-g}$$.
Note: This generalises to multiple predictors - the pure error estimate of $$\sigma^2$$ is based on SS between $$y_i$$ for cases with the same values on all predictors. $$df_{SLR} = p$$ instead of 2.
Reject for large values of $$F_{obs}$$.
### 5.3.1 Example: Voltage
Example from Ramsey and Schafer (2002) (case0802 in library(Sleuth3)).
Batches of electrical fluids were subjected to constant voltages until the insulating properties of the fluid broke down.
$$Y$$: time to breakdown
$$X$$: Voltage
The scatterplot of $$Y$$ vs. $$X$$ shows evidence of non-linearity and non-constant variance. The response was log transformed to resolve this.
$$H_0: \beta_1=0$$
$$H_A: \beta_1 \neq 0$$
$$F = 78.4$$, $$p<0.001$$. We reject $$H_0$$ and conclude that $$\beta_1 \neq 0$$.
$$H_0:$$ S.L.R model is appropriate/correct model
$$H_A:$$ S.L.R model has lack of fit.
$F=\frac{(180.07-173.75)/(7-2)}{173.75/(76-7)}=0.5$
$$F=0.50, p=0.773$$. We conclude that there is no evidence of lack of fit.
One-way ANOVA: LOG_TIME versus CODE
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
CODE 6 196.5 32.746 13.00 0.000
Error 69 173.7 2.518
Total 75 370.2
Model Summary
S R-sq R-sq(adj) R-sq(pred)
1.58685 53.07% 48.99% 38.72%
Regression Analysis: LOG_TIME versus VOLTAGE
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 1 190.151 190.151 78.14 0.000
VOLTAGE 1 190.151 190.151 78.14 0.000
Error 74 180.075 2.433
Lack-of-Fit 5 6.326 1.265 0.50 0.773
Pure Error 69 173.749 2.518
Total 75 370.226
Model Summary
S R-sq R-sq(adj) R-sq(pred)
1.55995 51.36% 50.70% 48.50%
Regression Equation
LOG_TIME = 18.96 - 0.5074 VOLTAGE
R code
anova(lm(log(TIME)~VOLTAGE))
## Analysis of Variance Table
##
## Response: log(TIME)
## Df Sum Sq Mean Sq F value Pr(>F)
## VOLTAGE 1 190.15 190.151 78.141 3.34e-13 ***
## Residuals 74 180.07 2.433
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(lm(log(TIME)~as.factor(VOLTAGE)))
## Analysis of Variance Table
##
## Response: log(TIME)
## Df Sum Sq Mean Sq F value Pr(>F)
## as.factor(VOLTAGE) 6 196.48 32.746 13.004 8.871e-10 ***
## Residuals 69 173.75 2.518
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 5.4 Added variable plots
In simple linear regression we can assess the importance of a predictor by:
• t-statistic
• $$\mbox{SSR}$$
• $$R^2$$
• $$Y$$-$$X$$ plot.
The analogues in multiple regression for assessing the importance of a predictor in the presence of other predictors are:
• t-statistic
• Seq/Extra SS
• partial $$R^2$$
• added variable plot.
### 5.4.1 Example: STEAM vs. TEMP, INV, PROD
Model A: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1+ \beta_2 x_2$$
Model B: $$\mathbb{E}[y] = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3$$
where $$x_1$$ = TEMP, $$x_2$$ = INV, $$x_3$$ = PROD.
• The t-statistic for PROD is small: $$T=0.10, p=0.920$$
• $$\mbox{SSR}(B|A) = 0.004$$ is also small.
• The partial $$R^2$$ for PROD is the proportion of variability in the response unexplained by TEMP and INV that is explained by PROD
\begin{align*} R^2(\mbox{PROD|TEMP, INV})& =\frac{\mbox{SSR}(B|A)}{\mbox{SSE}(A)} & = \frac{0.004}{8.931} = 0.00045=0.045\%\\ \end{align*}
• The added variable plot shows the relationship between a response and a predictor, adjusting for other predictors in the model.
‘Adjusting’ $$Y$$ for predictors $$X_1,...,X_k$$ is achieved by computing the residuals from the regression of $$Y$$ on $$X_1,...,X_k$$. The resulting residuals can be thought of as $$Y$$ with the effect of $$X_1,...,X_k$$ removed.
### 5.4.2 Example: Added variable plot for PROD.
i.e. should we add PROD to the model containing the predictors TEMP and INV? (Response is STEAM).
• Compute $$e$$(STEAM$$|$$ TEMP, INV), i.e. the residuals from regression of STEAM on TEMP and INV.
• Compute $$e$$(PROD$$|$$ TEMP, INV), i.e. the residuals from regression of PROD on TEMP and INV.
• AVP for PROD: Plot $$e$$(STEAM$$|$$ TEMP, INV) vs. $$e$$(PROD$$|$$ TEMP, INV).
We can also do:
AVP INV: Plot $$e$$(STEAM$$|$$ TEMP, PROD) vs. $$e$$(INV$$|$$ TEMP, PROD) AVP TEMP: Plot $$e$$(STEAM$$|$$ INV, PROD) vs. $$e$$(TEMP$$|$$ INV, PROD)
### 5.4.3 Example: Steam data cont’d
fit1 <- lm(STEAM ~ TEMP + INV)
fit2 <- lm(PROD ~ TEMP + INV)
summary(lm(resid(fit1)~ resid(fit2)))
##
## Call:
## lm(formula = resid(fit1) ~ resid(fit2))
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.2348 -0.4116 0.1240 0.3744 1.2979
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.487e-17 1.246e-01 0.000 1.000
## resid(fit2) 3.305e-01 3.122e+00 0.106 0.917
##
## Residual standard error: 0.623 on 23 degrees of freedom
## Multiple R-squared: 0.0004869, Adjusted R-squared: -0.04297
## F-statistic: 0.0112 on 1 and 23 DF, p-value: 0.9166
Alternatively you can use the avPlots function in the library(car). In minitab use STORAGE option to save the residuals of both models and make a scatterplot.
### 5.4.4 Properties of AVPs:
• Estimated intercept is 0.
• Slope of the line in AVP for PROD equals $$\hat{\beta}$$ (the coefficient of PROD in the model with TEMP, INV and PROD as predictors.
• Residuals in AVP equal residuals from regression of STEAM on TEMP, INV and PROD.
• $$R^2$$ in AVP for PROD is the partial $$R^2$$ for PROD, i.e. $$R^2$$(PROD$$|$$TEMP,INV).
• $$\hat{\sigma}^2$$ from AVP for PROD $$\approx \hat{\sigma}^2$$ from full model.
$\hat{\sigma}^2_{AVP}(n-2) = \hat{\sigma}^2_{full}(n-p)$
The points in an AVP are clustered tightly around a line if and only if the variable is important.
AV plots may also show outliers, or if the apparent adjusted association between $$Y$$ and $$X_j$$ is due to an influence point.
## 5.5 Visualising Models in Hdim: added variable plots for the bodyfat data.
Bodyfat data from assignment 3: http://rpubs.com/kdomijan/431176
### References
Draper, Norman Richard, and Harry Smith. 1966. Applied Regression Analysis. Wiley Series in Probability and Mathematical Statistics. Wiley.
Ramsey, Fred, and Daniel Schafer. 2002. The Statistical Sleuth: A Course in Methods of Data Analysis. 2nd ed. Duxbury Press.
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# Search by Topic
#### Resources tagged with Rectangles similar to Expanding Pattern:
Filter by: Content type:
Stage:
Challenge level:
### There are 16 results
Broad Topics > 2D Geometry, Shape and Space > Rectangles
### Pebbles
##### Stage: 2 and 3 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Lying and Cheating
##### Stage: 3 Challenge Level:
Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!
### Hallway Borders
##### Stage: 3 Challenge Level:
A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.
### Sorting Logic Blocks
##### Stage: 1 and 2 Challenge Level:
This interactivity allows you to sort logic blocks by dragging their images.
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Rati-o
##### Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
### Transformations on a Pegboard
##### Stage: 2 Challenge Level:
How would you move the bands on the pegboard to alter these shapes?
### Can They Be Equal?
##### Stage: 3 Challenge Level:
Can you find rectangles where the value of the area is the same as the value of the perimeter?
### Torn Shapes
##### Stage: 2 Challenge Level:
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
### Fitted
##### Stage: 2 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
### Hidden Rectangles
##### Stage: 3 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
### 2001 Spatial Oddity
##### Stage: 3 Challenge Level:
With one cut a piece of card 16 cm by 9 cm can be made into two pieces which can be rearranged to form a square 12 cm by 12 cm. Explain how this can be done.
### AP Rectangles
##### Stage: 3 Challenge Level:
An AP rectangle is one whose area is numerically equal to its perimeter. If you are given the length of a side can you always find an AP rectangle with one side the given length?
### Fencing
##### Stage: 2 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
### Fault-free Rectangles
##### Stage: 2 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Two by One
##### Stage: 2 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
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# physics
A rocket burns 5kg per ejecting it as a gas with a velocity of 1600m/s relative to the rocket. How much force is exerted on the rocket. Also find the velocity exerted when its mass reduces to 1/200 of its initial mass.
1. 👍
2. 👎
3. 👁
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# Gmat Problem Solving Practice
## Ace My Homework Closed
What should be a value between 1 and 9 (perhaps 1.15, of course)? To solve this question, we will first consider the five cases that occurred in this exam. _Case 1. If the numbers 5 and 10 are in \$GF(50)\$, and therefore 4 is a square or a matrix variable. If the numbers 3 and 5 are in \$GF(30)\$, and therefore 4 is not square or matrix variable. If 31 is a square or a matrix variable. If the numbers other than 3 are in \$GF(27)\$ and therefore 7 is square or a matrix variable. If they are all in \$GF(10)\$ or \$GF(21)\$ they should be written as 5 = 5 = 2 and so on…__ When you see these five possibilities, just do not ask the Mathematicians if they have a answer to the question! Maybe, you’ll come up with some more interesting results. The Mathematicians will think of any number and class the Möbius functions that are defined by the Möbius function definition from Möbius to linear functions and matrices. The Möbius functions define matrices, while the linear functions that are defined by the linear function definition correspond to number functions. This is equivalent to saying that the number of the leftmost column of the following Möbius function is 1. The square of its first column can also be represented as the Möbius function. Our Möbius functions do not contain any constant. So, we will try to represent these numbers as matrices rather than numbers. So we will use number functions and linear functions that are defined by the linear function definition instead of that by the Möbius functions. By using the Mathematicians to represent these and various other numbers, we will see how to define these other functions in the following problem: Let’s assume that we have answered this questions for some times: “Why does the numbers appearing in this problem be in \$GF(10)\$ or in \$GF(21)\$?” Let’s take a look at what the numbers appearing at the first block of the puzzle: [ ] + [ ], which actually gives us a list of quadratic and real numbers, as shown in Figure 22-4. The numbers 12 – 8, which represents the eight quadratic expressions at a total square we have just given, give the following list.
## Pay Someone To Take My Ged Test
Notice that these numbers arise to describe the least integers that we can say for a given number. This “list” works well for the number 14 but comes into question when we present more numbers! Notice, of course, that the reason why we did not use the Möbius functions is that we only know integers. _Case 2. If we assign 24 to some values which do not contain any numbers. Why is this case equal to 15? If 24 is a square, or a matrix variable, then clearly the only possible case. For example: if 24 is a square, then we need just to change the value to 8 8 13 2 3 6 5 2 5 2 4 20 2 1 2 7 4 19 4 3 11 10 16 4 5 6 6 6 8 6 9 8 9 8 8 9
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# Why do we use the log in gradient-based reinforcement algorithms?
I've been reading some papers on reinforcement learning.
$$\Delta w=\frac{\partial ln\ p_w}{\partial w}r$$
I often see expressions, similar to the above one, where the weights (denoted by $$w$$) are updated following the partial derivative of the policy function (denoted by $$p_w$$) with respect to its weights.
But why do we take the $$\log$$? What is its purpose?
## 1 Answer
We often take the logarithm because:
1. Maximizing $\log \Phi(x)$ is equivalent to maximizing $\Phi(x)$, so in maximum-likelihood problems, we can maximize the log of the likelihood instead of maximizing the likelihood directly and the result will be equivalent.
2. The logarithm converts multiplication to addition, and the derivative of a sum is "nicer" than the derivative of a product. Products often arise when maximizing likelihoods, because the probability of something can often be written as the product of other probabilities (e.g., when we are calculating the probability of that multiple independent events all happen). The derivative of the product is ugly, but the derivative of the log of the product is nice and simple.
3. Gradient descent often involves maximizing a likelihood, so we'll need to take derivatives. The expressions get simpler/cleaner if we can take the derivative of the log likelihood instead.
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# geometry
Daniel sees a lighthouse in the harbor. He estimates the angle of elevation is 60°. If the lighthouse is 120 feet tall, what is the approximate distance between Daniel and the top of the lighthouse? (Assume the lighthouse meets the ground at a right angle
1. 👍
2. 👎
3. 👁
1. tan 60 = 120/x;
x = 120/tan60
1. 👍
2. 👎
2. dude
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2. 👎
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• # Background knowledge
### 1.Basic concepts
Motor is an energy conversion device that converts electric energy into mechanical energy or mechanical energy into electric energy, using magnetic field as media.
PMDC motor is an energy conversion device that converts electric energy into mechanical energy, usingpermanent magnetic field as media provided by permenant magnets like ferrite magnets and neodymium magnets.
Every motor needs two basic conditions to function: magnetic field and current.
### 2.Classification of motor
There are many ways to classify the motors. Traditional classification is as follows.
The motors Kinmore makes belong to brush type strontium ferrite permanent magnet DC motor.
### 3.Basic theories
Research to the motors is based on the following five scientific laws. In order to have a preliminary acquaintance to motor principles, we need to known these laws first.
#### (1) Law of electromagnetic induction (Faraday 1831)
Conductors (of finite dimensions) moving through a uniform magnetic field will have currents induced within them.
The direction of the current is judged by right hand rule and follows the equation:
E=B*L*V
E: Electromotive force (Unit: V)
B: Magnetic flux density of magnetic field (1 Tesla=104 Gauss)
L: Effective length of conductor (Unit: m)
V: Velocity of the conductor (Unit: m/s)
See figure 1 to the right, if we connect a lead wire to the conductor,induced current will be generated.
#### (2) Biot-Savart Law
Conductors with current within them will generate electromagnetic force in a magnetic field. The direction is judged by left hand rule, (see figure 2) and follows the equation:
F=B*I*L
F: Electromagnetic force (Unit: N)
I: Current in the inductor (Unit: A)
B: Magnetic flux density of the magnetic field (Unit: Tesla)
L: Effective length of the conductor (Unit: m)
Left hand rule is also called as motor rule.
Right hand rule isalso called as generator rule.
#### (3) Kirchhoff's circuit laws (See figure 3)
KCL ΣI=0: At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node
KVL ΣU=0: The directed sum of the electrical potential differences (voltage) around any closed network is zero.
#### (4) Law of conservation of energy
The total amount of energy in an isolated system remains constant over time.
#### (5) Ampère's circuital law
In short, conductors with current within them generate magnetic field around them. The direction of the magnetic filed is judged by right hand thrumb rule and follows the equation. (See figure 4)
∮H×dL=∑I=IA+IB+IC+…
H: magnetic field intensity (Unit: A/M)
L: Length of conductor (Unit: M)
I: Current (Unit: A)
### 4.Basic principles
2-pole PMDC motor
2-bar commutator
2-conductors (1-loop coil) simple armature.
According to Biot-Savart Law and left-hand rule,armature runs in CCW direction.
Disadvantage:Dead points exist.
It is a simple but unpractical motor.(Figure 5)
### 5.Electric potential, torque and energy equation
#### (1) Electric potential (Figure 6)
From V=E+2△U+I*r we get E=V-2△U-I*r
Meanwhile E=KE*Φ*n(armature back EMF)
V: power supply voltage (Unit: V)
2△U: brush voltage drop (Unit: V)
I: armature current (Unit: A)
R: rotor resistance (Unit: Ω)
KE: EMF constant = Z/60 (for a 2-pole motor.
Z: number of conductors)
Φ: magnetic flux (Unit: Weber) = average magnetic flux density B * width of magnetic pole *effective length of rotor
N: speed (Unit: rpm)
#### (2) Torque
TE=KTΦ*I（electromagnetic torque: N.M）
KT: torque constant = Z/2π
Φ: magnetic flux (unit: Weber)
I: armature current (unit: A)
#### (3) Relationship between power and torque：
P=T*n/97500 P: power（unit: W）
T: torque (unit: g.cm)
n: speed (unit: rpm)
When the unit of T is “N?m”, P=T*n/9.55（unit: W）
#### (4) Energy equation(Figure 7)
P1=2△U*I+I2r+PE
PE=P2+PFe+Pmec
PE: electromagnetic power P2: output power
Pmec: mechanical loss PFe: iron loss
P2=P1-2△U*I-I2r-PFe-Pmec (unit: W)
Efficiency: η=P2/P1*100%
PFe+Pmec is also called no load power
P0=PFe+Pmec
PE=P2+P0 and TE=T2+T0
### 6.Performance characteristic (Figure 9)
n=f(T2) relationship between speed & torque.
I=f(T2) relationship between current & output power
η=f(T2) relationship between efficiency & torque
P2=f(T2) relationship between output power & torque
#### (1) I=f(T2)
I=TE/KT*Φ=(T0+T2)/KT*Φ=T0/KT*Φ+T2/KT*Φ=I0+[1/KT*Φ]*T2 (liner equation)
I0: no load current Φ: constant
At stall, n=0, E=0, according to Figure 6, current Ist=(U-2△U)/r
#### (2) n=f(T2)
E=V-2△U-I*r=KEΦ*n
n=(V-2△U-I*r)/KE*Φ={U-2△U-[(I0+T2)/KT*Φ]*r}/KE*Φ
=(U-2△U-I0*r)/KE*Φ-r/KE*KT*Φ2*T2
= n0-[r/KE*KT*Φ2]*T2（equation of lines）
#### (3) P2=f(T2)
P2=T2*n/9.55=[n0-(V/KE*KT*Φ2)*T2]/9.55=[n0*T2-(r/KE*KT*Φ2)*(T2)2]/9.55
P2 is a second-degree parabola (Figure 10)
#### (5) Energy transmission graph: (Figure 8)
(Equation iscomplicated thus is omitted here.)
### 7.Analysis of major parameters
#### (1) Turns of coil and magnet wire diameter (other parameters remain unchanged)
We know from 5.1 that the potential constant KE increases when the turns of coil increase. Motor speed n is therefore lowered. On the contrary, when the turns of coil decrease, the motor speed increases.
When the diameter of the magnet wire increases, the rotor resistance r reduces. Back EMF of the rotor increases (E=V-2△U-I*r). The motor speed n therefore increases. On the contrary, when the diameter of the magnet wire decreases, the motor speed n decreases.
The current at stall is in inverse proportion to the resistance r.Turns of the coil and diameter of the magnet wire restrict each other under the space limit of the lamination slot. We should clearly understand such relationship when we try to adjust the motor parameters.
#### (2) Magnetic flux (other parameters remain unchanged)
Magnets with higher magnetic flux density and longer lamination sheets will both increase the magnetic flux Φ.
From 5.1 and 6.2 we know that speed n decreases. At the same time, load (T2) has less influence over speed n.
The characteristic of the motor is thus called hard. On the contrary, if we use magnets with lower magnetic flux density and shorter lamination sheets, the characteristic of the motor is called soft.
#### (3) Air gap
See figure 12, the magnetization curve of the air gap
Φδ=-μ0*(Sδ/δ)*Fδ
Φδ: Air gap flux
Sδ: Air gap area
Δ: Air gap length
Fδ: Air gap magnetomotive force(magnetic EMF)
Permeance angle: α=tg-1[μ0*(Sδ/δ)].
We can see that when δ is longer, α is smaller, air gap flux Φδ is smaller. Motor speed will increase if the other parameters remain unchanged. On the contrary, when δ is shorter, α is larger, air gap flux Φδ is larger. Motor speed will decrease. We will see the same result as we see in 7.2. We usually pursue the maximum possible value of (Φδ*Fδ) in motor design.
#### (4) Effective volume D2*L
Motor torque is proportional to D2*L.
[D: diameter of the rotor L: length of the rotor]
Motor power is proportional to D2*L *n.
• # Evaluation of motor
How to evaluate a motor? Typically industrial products can be evaluated with the following aspects. The most important characteristic of industrial products is low deviation.
#### (1) Full dimensions
The fundamental characteristics that customers require are assembling dimensions and outline dimensions.
The dimensional deviation of a good product should meet product standard requirement. (GB standard, industrial standard or enterprise standard)
#### (2) Basic performance
a. Rated voltage: known parameter (unit: V)
b. No load current: I0 (unit: A)
c. No load speed: n0 (unit: rpm)
d. Rated current: IL (unit: A)
e. Rated torque: TL (Unit: g.cm)
f. Rated speed: NL (Unit: rpm)
g. Current in stall: Ist (unit: A)
h. Torque in stall: Tst (unit: g.cm)
i. Other parameters such as efficiency, power, electric potential constant, torque constant etc. can be calculated from the above data.
#### (3) Special characteristics
a. Vibration: amplitude (unit: mm), vibration velocity (unit: mm/s), vibration acceleration (unit: mm/s2)
b. Noise: sound pressure LP (unit: dB(A) and acoustical power LW (unit: dB(A). They are both relative values.
c. EMC: This index is to evaluate the ability of the motor resisting the radio interference or the radio interference level that the motor generates.
d. Environment test: This is to judge the load capability of the motor under high and low temperature. Alternating temperature test is the common test. Alternating temperature and humidity test is more severe test. Magnetic field of ferrite magnet decreases by 5-7% under -80 ℃. The motor electric performance is therefore deviated.Mechanical shock, external alternating magnetic field, aging (long time) storage will also weaken the magnetic field.
e. Others: such as safety clearance, safety creepage distance, protection class, type of cooling etc.
### 9.Winding type and carbon brush placement principle
3-pole rotor/commutator winding graph (Figure 13)
a.Angel between slots of rotor and slots of commutator is 60°
b.Coil C is in the process of commutating. It is being shortcut by brush at negative terminal.
c.Brush locates at the centre line of the magnetic poles.
5-pole rotor/commutator winding graph(Figure 14)
a.Angle between slots of rotor and slots of the commutator is 0°
b.Coil B is in the process of commutating. It is being shortcut by brush at positive terminal
c.Brush locates at the centre line of the magnetic poles.
#### ● Principle of carbon brush placement
a.Try to get maximum effective conductors. In other words, make the direction of current the same in as many conductors under the same pole as possible. In some cases (such as 12-pole rotor), we will sacrifice the number of effective conductors to improve commutation. Such cases will not be discussed here.
b.Minimize the electric potential of the commutating coil (the one that is shortcut). Typically the sides of that coil are placed at the edges of the magnetic poles or between the magnetic poles. So the carbon brushes are usually placed in the middle of the commutator bars that are connected to the coil.
c. Electric angle between positive and negative brushes is 180°.Conclusion: There is not a sole way to connect coil and commutator segments.
### 10.Typical application
According to the characteristics of our motor models, we hereby describe in details by means of types of power supply and motor load.
#### (1)Classified by input power supply
a.Dry battery, rechargeable battery, small capacity AC/DC adaptor.
The characteristic of such power supply is that they have large internalresistance, There is large voltage drop when the motor is applied with load.The output power of the motor is limited by the capacity of the power supply.When designing such motor, motor efficiency is not the only point to be considered.How to get the largest output power from the power supply is the most important. That is to say, try to get maximum value of P1=V*I.Practically, it is hard to achieve this target considering only the motor parameters. How to properly evaluate such motors is also a subject to us.
b.Input power supply from voltage divider as resistor or resisor/capacitor.
In such circuit, when the current changes, the output voltage changes. In actual application, AC input voltage is usually 120V-240V. Out of various reasons when the current increases, the resistor R1 or capacitor C will take more voltage drop.The output voltage is less. Motor speed is therefore lowered. Its operating point and characteristics will all deviate. The result will be different according to different application of the motors.Take the well-known hair dryer for example, if the above change happens, air flow decreases. The temperature of the resistor R1 rises. The resistor gets more voltage drop. The output voltage is lower, making the situation a vicious circle. The motor will lose its function quickly.
c. Regulated power supply
This is the ideal power supply. The input voltage doesn’t change with the environment or motor load. The motor’s characteristic is decided by the motor parameter itself. The motor performance data we provide to our customers are tested under such power supply. In practical application, high capacity accumulator battery and AC/DC adaptor (variation of V less than 5%) are deemed as regulated power supply.
#### (2)Classified by motor load
a.Fan load
Startup of the motor with fan load is similar to the startup of the motor with no load. So there is no requirement to motor’s startup or stall torque. Sometimes we even need to restrict its torque from being too large. The most important feature of the motor with fan load is the stability and discreteness of its speed in mass production. The output power is in proportional with the motor speed. If the motor speed deviates a lot, the motor working characteristics will also deviate a lot. So the characteristic of the load at working point is the major point we look at.
b.Winching load
Examples are cable retrieving devices for vacuum cleaners and tube retrieving fixtures for Irrigation machines.Similar to winching devices, the motor starts to work at its full load. The most important characteristic of such motor is its stall torque. The consistency of the stall torque is the key point during motor design and fabrication.Central door lock actuator also belongs to such winching load. Motors with such load usually work at short time working cycle.
c.Linear load
Torque of such load is stable during work. The motor power increases linearly with the motor speed. It may reach its full load at startup. But in most cases it starts up with partial load. Usually it works under rated load for a very long time. We should consider various aspects including temperature rise in motor design. Reciprocating pump is the typical linear load.
d.Other load
There are still other loads like eccentric wheel, gearboxes that we are not going to discuss in this article.
• # Instruction for use
### Overload or Stall Condition
The motor temperature would rise gradually due to the internal energy conversion between the windings and iron core during running. The windings will not be burnt under rated load because of the balance between the produced and vented heat. But if it’s overloaded or stalled for a long time, the insulation film of copper wires might be dissolved due to high temperature. This will short-circuit the winding which causes high current even damages the motor and driving board. Besides, under overloaded condition, the strength of the gear or other parts attached on the shaft will be affected (tooth broken or wore out). So please make sure that motors are operated under rated working conditions.
### Motors Working at Lower Speed
For most of the dc motors, we use carbon brushes. When a motor runs, spark occurs in the contact area because of the friction between the brushes and commutator at the timing of the commutation. Carbon dust will accumulate in the commutator slots which might cause short circuit, burn the motor or the driving board if the motor runs at lower speed and the dust couldn’t be burned in time. Please kindly pay attention to this condition.
### Motors Working at Lower Speed
For most of the dc motors, we use carbon brushes. When a motor runs, spark occurs in the contact area because of the friction between the brushes and commutator at the timing of the commutation. Carbon dust will accumulate in the commutator slots which might cause short circuit, burn the motor or the driving board if the motor runs at lower speed and the dust couldn’t be burned in time. Please kindly pay attention to this condition.
### Remarks about PWM Controller
The lifetime of the brushes is shorter when the motor is powered with PWM controller not by rated voltage or constant voltage. And the carbon brushes might wear out easily under certain frequency of the PWM controller. Normally the frequency used for dc motors is 10～20KHZ. Heating might also occur because of sympathetic vibration if the frequency of the PWM switch is close to the motor components’. Besides, please be noted that the motor might not run if with integrated electrolytic capacitor under certain frequency. So we suggest motors with varistor inside if the motor is powered by PWM controller.
### About Inertia and Brake
It’s very common that after power off, the motor shaft will still rotate for a while because of the inertia. If instant brake needed, you can short-circuit the positive and negative poles then the power generated by the motor (reverse current) can stop it quickly. But this might increase the motor current and even shorten the lifetime.
### Lifetime
A motor’s lifetime is related to the operating conditions such as the power supplier, duty cycle, and load conditions etc. The lifetime data on our spec is based on the rated testing conditions and motor running in one direction without any stop. It’s just for reference only. For actual products, please make full testing to ensure the lifetime is long enough.
### Assembly
There are screw holes designed for motor assembly. Please kindly refer to the outline and make sure the screw length is in the recommended range. As for the allowable torque, please kindly refer to the related technical standards. Over that range, it might slip the screw.
### Motor terminals
The motor terminal structures and inner parts might be broken when the soldering temperature is too high. The recommended operating way is using soldering iron 40W, 380℃, and less than 3 seconds. Besides, force on the terminals will also break the terminal structure.
### Axial Force
When you press gear or other parts on the output shaft, support for the other side shaft end will be needed. If it’s not possible to apply the support, the press force should be no more than the max allowable force.
### Shock and Drop
When you press gear or other parts on the output shaft, support for the other side shaft end will be needed. If it’s not possible to apply the support, the press force should be no more than the max allowable force.
### The Use of Binding Material
If binding material like glue is used during the assembly, please make sure it will not be added to the output shaft bearing. For some volatile glue, it might also stain the commuatator which affect the motor performance.
Please pay fully attention to the items mentioned above. If other issues during application, please kindly contact Kinmore for further information.
©2009-2018 Shenzhen Kinmore Motor Co., Ltd. All Rights Reserved. ICP:17128224
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## ABSTRACT
If an element of area DS is located horizontally at a depth h below the surface of a liquid, the force acting downward on it equals the weight of
the column of liquid above DS. Thus if the density of the liquid is r, this weight is F/rhDS. Pascal’s law asserts that this force is independent of the orientation of the element of area DS so it may be horizontal, vertical or at an arbitrary inclination, and the fluid force will be the same. Forces due to
liquid pressure are called hydrostatic forces, and the force per unit area F /DS is called the hydrostatic pressure in the liquid at the depth h . This means that if the element of area is completely immersed in liquid, the forces on either
side of it will be equal and opposite. If, however, DS only has liquid on one side, the hydrostatic force will not be balanced by an equal and opposite one
on the other side.
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# 3 machines have a productivity ratio of 1 to 2 to 5. All 3
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3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven, and begins working again. The job is done in nine hours. What was the ratio of the work performed by the fastest machine as compared to the slowest?
A. 5
B 7
C 15/2
D 17/2
E 12
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3 machines have a productivity ratio of 1:2:5. all 3 machines are working on a job for 3 hours. at the beginning of the 4th hour, the slowest machine breaks. it is fixed at the beginning at hour 7, and begins working again. the job is done in 9 hours. what was the ratio of the work performed by the fastest machine to the work performed by the slowest?
"3 machines have a productivity ratio of 1:2:5" means that in a time interval (1 hour, 1 year, ...) the first machine can produce 1 part (so it's the slowest one), the second can produce 2 parts and the third and fastest one can produce 5 parts.
Now, the fastest machine worked for whole 9 hours thus produced 9*5=45 parts but the slowest machine worked only for 6 hours thus produced 6*1=6 parts, so the ratio of the work performed by the fastest machine to the work performed by the slowest is 45/6=15/2.
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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16 Feb 2011, 04:49
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1:2:5
Let's consider 1 work is done by the slowest machine in 1 hour.
in first 3 hours; Work performed by 3 machines
1*3:2*3:5*3
From beginning of 4th hour until start of 7th hour, 3 hours elapsed. Work performed by 3 machines
0:2*3:5*3. First one is 0 work as it is broken.
Job is done in nine hours; from start of 7th hour until end of 9th hour, 3 hours elapsed. Work performed by 3 machines
1*3:2*3:5*3
Total work performed by slowest machine: 3+3=6
Total work performed by fastest machine: 15+15+15=45
Ratio = 45/6 = 15/2
Ans: 15/2
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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11 Aug 2011, 06:02
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ruturaj wrote:
please explain how slowest machine works for 6 hours
slowest machine worked
1st hour
2nd hour
3rd hour
Got broken
got fixed and returned to work at the beginning of the 7th hour so:
7th hour
8th hour
9th hour
Total of 6 hours.
So the slowest machine made 6 pieces, while the fastest machine made 45
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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17 Feb 2011, 12:45
thanks for posting +1
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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11 Aug 2011, 05:27
please explain how slowest machine works for 6 hours
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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11 Aug 2011, 08:02
ruturaj wrote:
please explain how slowest machine works for 6 hours
3 machines have a productivity ratio of 1:2:5. all 3 machines are working on a job for 3 hours. at the beginning of the 4th hour, the slowest machine breaks. it is fixed at the beginning at hour 7, and begins working again. the job is done in 9 hours. what was the ratio of the work performed by the fastest machine to the work performed by the slowest?
Productivity ratio of 1:2:5 means;
In 1 hour:
Slowest machine just finishes 1 work
Medium fast machine finishes 2 work
Fastest machine finishes 5 work
3 hours work:
Slowest: 3*1=3 work
Medium: 3*2=6 work
Fastest: 3*5=15 work
THEN SLOWEST BREAKS.
It misses(3 hours):
4th hour
5th hour
6th hour
During which:
Medium fast completes: 3*2=6 work
Fastest completes: 3*5=15 work
Then
7th, 8th, 9th hour(3 hours)
Slowest: 3*1=3 work
Medium: 3*2=6 work
Fastest: 3*5=15 work
Work by fastest:Work by slowest
15+15+15:3+3
45:6
15:2
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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12 Aug 2011, 00:22
0-1 fastest = 5 slowest = 1
1-2----5----------1 ----
2-3 fastest = 5 ----1---
6-7----5--------1--
7-8----5---------1--
8-9----5---------1--
hence total = 45 and 6
thus 15/2
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12 Aug 2011, 02:03
+1 for explanation..
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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21 Aug 2011, 10:37
let's suppose that in 1 hour each machine produces 1+2+5 = 8 units
in 3 hrs = 24 units
next 3 hrs = 21 units (slowest isnt working)
further 3 hrs = 24 untis
fastest worked for 9 hours, while the slowest worked for 6 hrs.
units produced by fastest / units produced by slowest = 9*5/6*1 = 15/2
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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21 Aug 2011, 14:29
Let M1 be the slowest and M3 be the fastest machine.
<---first 3 hours--><----3 hours----><---3 hours---->
M1 M1
M2 M2 M2
M3 M3 M3
productivity ratio = 1:2:5 (which is per hour)
x: 2x: 5x
work done by M3/ work done by M1 = (9*5x) / (6*x) = 45/6 = 15/2
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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02 Dec 2012, 22:49
The productivity ratio of the fastest machine to the slowest machine is 5:1.
The fastest machine worked for 9 straight hours. Thus, W = 5 x 9 = 45
The slowest machine brokedown in the 4th hour and was revived on the 7th hour. Thus, W = 1 x 6 = 6
The Output Ratio of fastest to slowest is 45 is to 6 ==> 15 : 2
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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05 Jan 2017, 15:19
Let's say the productivity ratio refers to how many units each produces in 1 hour. The fastest (most productive) machine works for 9 hours and makes 5 per hour, totally 45 units. The slowest (least productive) machine works for 6 hours and produces 1 per hour for a total of 6. Thus, the ratio of units produced fast/slow is 45/6 or 15/2. No more math required
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05 Jan 2017, 17:19
GMATD11 wrote:
3 machines have a productivity ratio of 1 to 2 to 5. All 3 machines are working on a job for 3 hours. At the beginning of the 4th hour, the slowest machine breaks. It is fixed at the beginning of hour seven, and begins working again. The job is done in nine hours. What was the ratio of the work performed by the fastest machine as compared to the slowest?
A. 5
B 7
C 15/2
D 17/2
E 12
fastest machine's work=9hrs*5 units per hr=45 total units
slowest machine's work=6 hrs*1 unit per hr=6 total units
45/6=15/2
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3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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09 Apr 2017, 14:25
Common sense solution:
Ratio of work is 1:2:5
Meaning: As Bunuel defined "in a time interval (1 hour, 1 year, ...) the first machine can produce 1 part (so it's the slowest one), the second can produce 2 parts and the third and fastest one can produce 5 parts."
Time progression:
Hour 1: 1:2:5 parts produced
Hour 2: 2:4:10 parts produced
Hour 3: 3:6:15 parts produced
Hour 4: 3:8:20 parts produced //Machine 1 breaks, produces no more parts
Hour 5: 3:10:25 parts produced
Hour 6: 3:12:30 parts produced
Hour 7: 4:14:35 parts produced //Machine 1 is fixed, continues to produce parts
Hour 8: 5:16:40 parts produced
Hour 9: 6:18:45 parts produced //Job done
So when looking at the final tally of parts produced in 9 hours, the fastest machine produced 45 parts and the slowest produced only 6. Reducing this ratio we get 15/2.
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Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink]
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12 Apr 2017, 09:55
The productivity ratios are 1:2:5. Let's assume that the basic rate is x. So rates are 1x:2x:5x.
The fastes machine worked 9 hours, therefore it did 9/1x work. The slowest machine didn't work from 4th to 6th hours so were broken for 3 hours. Therefore the work done by this machine is 6/5x.
Let's compare work done by both machines: (9/1x)/(6/5x)=(9/x)*(5x/6)=15/2
Re: 3 machines have a productivity ratio of 1 to 2 to 5. All 3 [#permalink] 12 Apr 2017, 09:55
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# Zeerust Is Binomial An Example Of Continous Distribution
## Binomial distribution Define Binomial distribution at
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Example of a Binomial distribution — astroML 0.2 documentation. I have a data set which is a set of continuous distances from some origin. I originally modeled this as a negative binomial distribution by rounding the data and, You might recall that the binomial distribution describes the behavior of a discrete random variable X, Example. Suppose n = 20 Continuous Distributions;.
Discrete distributions empirical Bernoulli binomial. See how to use the normal approximation to a binomial distribution and is continuous whereas the binomial distribution is above example,, This makes Figure 1 an example of a binomial distribution. The calculation of cumulative binomial probabilities can be quite tedious..
### Can a negative binomial distribution be used to model a
CONTINUOUS COUNTERPARTS OF POISSON AND BINOMIAL. Common examples of discrete probability distributions are binomial distribution, distribution. As seen from the example, distribution and a continuous https://en.wikipedia.org/wiki/Multinomial_distribution The Bernoulli and Binomial Distributions . continuous distributions as, The Bernoulli Distribution is an example of a discrete probability distribution..
This makes Figure 1 an example of a binomial distribution. The calculation of cumulative binomial probabilities can be quite tedious. Binomial Distribution die is one example of a discrete uniform distribution; for finding the probability for a continuous uniform distribution: P(X)
Even though the normal is a continuous distribution. For example consider a game of dice In a binomial distribution the random variable can take continuous, including the binomial, normal, Poisson, geometric, We can also п¬Ѓnd the quantiles of a binomial distribution. For example, here is the 90th
continuous, including the binomial, normal, Poisson, geometric, We can also п¬Ѓnd the quantiles of a binomial distribution. For example, here is the 90th One key reason for this is that the Normal is a continuous distribution In principle such problems can be reduced to a Binomial by grouping, for example
26/10/2013В В· An introduction to the binomial distribution. I discuss the conditions required for a random variable to have a binomial distribution, discuss the binomial Probability Distributions: Discrete vs. Continuous The weight of a fire fighter would be an example of a continuous variable; Negative binomial distribution
This shows an example of a binomial distribution with various parameters. computes the Probability Density Function at values x in the case of continuous Binomial distribution definition, a distribution giving the probability of obtaining a specified number of successes in a finite set of independent trials in which
Probability Distributions: Discrete vs. Continuous The weight of a fire fighter would be an example of a continuous variable; Negative binomial distribution 26/10/2013В В· An introduction to the binomial distribution. I discuss the conditions required for a random variable to have a binomial distribution, discuss the binomial
• For Binomial Distribution with large n, (for Continuous) and Mass • So for our example with 11 students: On Continuous Versions of Poisson and Binomial Distributions By continuous Poisson distribution with continuous Poisson and binomial distributions still
There are many differences between binomial and poisson distribution, and F-distribution are the types of continuous difference between Binomial and Poisson You might recall that the binomial distribution describes the behavior of a discrete random variable X, Example. Suppose n = 20 Continuous Distributions;
You might recall that the binomial distribution describes the behavior of a discrete random variable X, Example. Suppose n = 20 Continuous Distributions; The binomial distribution model is an important probability model that is used when there are two possible outcomes Examples of Use of the Binomial Model 1.
In order for a continuous distribution discrete one (like the binomial), a continuity correction should be used. Here is an example of an Np chart. In order for a continuous distribution discrete one (like the binomial), a continuity correction should be used. Here is an example of an Np chart.
Stats True/False study guide by The number of defective pencils in a lot of 1000 is an example of a continuous The mean of the binomial distribution is See how to use the normal approximation to a binomial distribution and is continuous whereas the binomial distribution is above example,
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Search a number
43781 is a prime number
BaseRepresentation
bin1010101100000101
32020001112
422230011
52400111
6534405
7241433
oct125405
966045
1043781
112a991
1221405
1316c0a
1411d53
15ce8b
hexab05
43781 has 2 divisors, whose sum is σ = 43782. Its totient is φ = 43780.
The previous prime is 43777. The next prime is 43783. The reversal of 43781 is 18734.
It is a happy number.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 43681 + 100 = 209^2 + 10^2 .
It is a cyclic number.
It is not a de Polignac number, because 43781 - 22 = 43777 is a prime.
It is a super-2 number, since 2×437812 = 3833551922, which contains 22 as substring.
Together with 43783, it forms a pair of twin primes.
It is a Chen prime.
It is equal to p4559 and since 43781 and 4559 have the same sum of digits, it is a Honaker prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (43783) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21890 + 21891.
It is an arithmetic number, because the mean of its divisors is an integer number (21891).
243781 is an apocalyptic number.
It is an amenable number.
43781 is a deficient number, since it is larger than the sum of its proper divisors (1).
43781 is an equidigital number, since it uses as much as digits as its factorization.
43781 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 672, while the sum is 23.
The square root of 43781 is about 209.2390976849. The cubic root of 43781 is about 35.2448142196.
The spelling of 43781 in words is "forty-three thousand, seven hundred eighty-one".
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# How to write the union of sets
This is just a question about notation(and I can not write it pretty well in Latex either).
Is $X=(0,+\infty)\subset\Bbb{R}$ and $Y=\Bbb{R}$.
Then $X\times Y= (0,+\infty)\times \Bbb{R} =$ ?
I've tried with:
1. $\bigcup_{x>0,x\in\Bbb{R},y\in\Bbb{R}}[x,y]$ (Sorry about the subindex)
2. $\{ \cup[x,y] : x,y\in\Bbb{R}, x>0 \}$ (I don't think it's correct)
3. $A =\{ [a,b] : a,b\in\Bbb{R}, a>0\}$ then $X\times Y=\cup_{[x,y]\in A}[x,y]$
4. Just $(0,+\infty)\times \Bbb{R}$ (but I need one more explicit)
How would you write it? Thanks!
• Option 4 is correct, and most succinct. Or use the definition of $\times$ to write it out explicitly, $\{ (x,y) \colon x,y \in \mathbb R, x>0\}$. Commented Aug 24, 2014 at 3:15
• When intervals are involved, I prefer to use $\langle x,y\rangle$ for ordered pairs, and not $(x, y)$. Commented Aug 24, 2014 at 3:29
• @AsafKaragila, do you mean $\{\langle x,y \rangle : x,y\in\Bbb{R}, x>0\}$? Commented Aug 24, 2014 at 3:36
• Yes. That is what I mean. Because writing $(x, y)$ can be confused be the interval. Is $(1,3)$ an ordered pair or an open interval? Commented Aug 24, 2014 at 3:38
• It is an ordered pair. I was thinking in intervals, that's why I did that mess. I like your way, I think I didn't see it before in these cases. Commented Aug 24, 2014 at 3:44
Just writing $(0,+\infty)\times \Bbb{R}$ or, using the definition of Cartesian Product, $\{ (x,y) : x,y\in\Bbb{R}, x>0 \}$ is correct.
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## 79.10 Final bootstrap
The following result goes quite a bit beyond the earlier results.
Theorem 79.10.1. Let $S$ be a scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Any one of the following conditions implies that $F$ is an algebraic space:
1. $F = U/R$ where $(U, R, s, t, c)$ is a groupoid in algebraic spaces over $S$ such that $s, t$ are flat and locally of finite presentation, and $j = (t, s) : R \to U \times _ S U$ is an equivalence relation,
2. $F = U/R$ where $(U, R, s, t, c)$ is a groupoid scheme over $S$ such that $s, t$ are flat and locally of finite presentation, and $j = (t, s) : R \to U \times _ S U$ is an equivalence relation,
3. $F$ is a sheaf and there exists an algebraic space $U$ and a morphism $U \to F$ which is representable by algebraic spaces, surjective, flat and locally of finite presentation,
4. $F$ is a sheaf and there exists a scheme $U$ and a morphism $U \to F$ which is representable by algebraic spaces or schemes, surjective, flat and locally of finite presentation,
5. $F$ is a sheaf, $\Delta _ F$ is representable by algebraic spaces, and there exists an algebraic space $U$ and a morphism $U \to F$ which is surjective, flat, and locally of finite presentation, or
6. $F$ is a sheaf, $\Delta _ F$ is representable, and there exists a scheme $U$ and a morphism $U \to F$ which is surjective, flat, and locally of finite presentation.
Proof. Trivial observations: (6) is a special case of (5) and (4) is a special case of (3). We first prove that cases (5) and (3) reduce to case (1). Namely, by bootstrapping the diagonal Lemma 79.5.3 we see that (3) implies (5). In case (5) we set $R = U \times _ F U$ which is an algebraic space by assumption. Moreover, by assumption both projections $s, t : R \to U$ are surjective, flat and locally of finite presentation. The map $j : R \to U \times _ S U$ is clearly an equivalence relation. By Lemma 79.4.6 the map $U \to F$ is a surjection of sheaves. Thus $F = U/R$ which reduces us to case (1).
Next, we show that (1) reduces to (2). Namely, let $(U, R, s, t, c)$ be a groupoid in algebraic spaces over $S$ such that $s, t$ are flat and locally of finite presentation, and $j = (t, s) : R \to U \times _ S U$ is an equivalence relation. Choose a scheme $U'$ and a surjective étale morphism $U' \to U$. Let $R' = R|_{U'}$ be the restriction of $R$ to $U'$. By Groupoids in Spaces, Lemma 77.19.6 we see that $U/R = U'/R'$. Since $s', t' : R' \to U'$ are also flat and locally of finite presentation (see More on Groupoids in Spaces, Lemma 78.8.1) this reduces us to the case where $U$ is a scheme. As $j$ is an equivalence relation we see that $j$ is a monomorphism. As $s : R \to U$ is locally of finite presentation we see that $j : R \to U \times _ S U$ is locally of finite type, see Morphisms of Spaces, Lemma 66.23.6. By Morphisms of Spaces, Lemma 66.27.10 we see that $j$ is locally quasi-finite and separated. Hence if $U$ is a scheme, then $R$ is a scheme by Morphisms of Spaces, Proposition 66.50.2. Thus we reduce to proving the theorem in case (2).
Assume $F = U/R$ where $(U, R, s, t, c)$ is a groupoid scheme over $S$ such that $s, t$ are flat and locally of finite presentation, and $j = (t, s) : R \to U \times _ S U$ is an equivalence relation. By Lemma 79.8.1 we reduce to that case where $s, t$ are flat, locally of finite presentation, and locally quasi-finite. Let $U = \bigcup _{i \in I} U_ i$ be an affine open covering (with index set $I$ of cardinality $\leq$ than the size of $U$ to avoid set theoretic problems later – most readers can safely ignore this remark). Let $(U_ i, R_ i, s_ i, t_ i, c_ i)$ be the restriction of $R$ to $U_ i$. It is clear that $s_ i, t_ i$ are still flat, locally of finite presentation, and locally quasi-finite as $R_ i$ is the open subscheme $s^{-1}(U_ i) \cap t^{-1}(U_ i)$ of $R$ and $s_ i, t_ i$ are the restrictions of $s, t$ to this open. By Lemma 79.7.1 (or the simpler Spaces, Lemma 64.10.2) the map $U_ i/R_ i \to U/R$ is representable by open immersions. Hence if we can show that $F_ i = U_ i/R_ i$ is an algebraic space, then $\coprod _{i \in I} F_ i$ is an algebraic space by Spaces, Lemma 64.8.4. As $U = \bigcup U_ i$ is an open covering it is clear that $\coprod F_ i \to F$ is surjective. Thus it follows that $U/R$ is an algebraic space, by Spaces, Lemma 64.8.5. In this way we reduce to the case where $U$ is affine and $s, t$ are flat, locally of finite presentation, and locally quasi-finite and $j$ is an equivalence.
Assume $(U, R, s, t, c)$ is a groupoid scheme over $S$, with $U$ affine, such that $s, t$ are flat, locally of finite presentation, and locally quasi-finite, and $j$ is an equivalence relation. Choose $u \in U$. We apply More on Groupoids in Spaces, Lemma 78.15.13 to $u \in U, R, s, t, c$. We obtain an affine scheme $U'$, an étale morphism $g : U' \to U$, a point $u' \in U'$ with $\kappa (u) = \kappa (u')$ such that the restriction $R' = R|_{U'}$ is quasi-split over $u'$. Note that the image $g(U')$ is open as $g$ is étale and contains $u$. Hence, repeatedly applying the lemma, we can find finitely many points $u_ i \in U$, $i = 1, \ldots , n$, affine schemes $U'_ i$, étale morphisms $g_ i : U_ i' \to U$, points $u'_ i \in U'_ i$ with $g(u'_ i) = u_ i$ such that (a) each restriction $R'_ i$ is quasi-split over some point in $U'_ i$ and (b) $U = \bigcup _{i = 1, \ldots , n} g_ i(U'_ i)$. Now we rerun the last part of the argument in the preceding paragraph: Using Lemma 79.7.1 (or the simpler Spaces, Lemma 64.10.2) the map $U'_ i/R'_ i \to U/R$ is representable by open immersions. If we can show that $F_ i = U'_ i/R'_ i$ is an algebraic space, then $\coprod _{i \in I} F_ i$ is an algebraic space by Spaces, Lemma 64.8.4. As $\{ g_ i : U'_ i \to U\}$ is an étale covering it is clear that $\coprod F_ i \to F$ is surjective. Thus it follows that $U/R$ is an algebraic space, by Spaces, Lemma 64.8.5. In this way we reduce to the case where $U$ is affine and $s, t$ are flat, locally of finite presentation, and locally quasi-finite, $j$ is an equivalence, and $R$ is quasi-split over $u$ for some $u \in U$.
Assume $(U, R, s, t, c)$ is a groupoid scheme over $S$, with $U$ affine, $u \in U$ such that $s, t$ are flat, locally of finite presentation, and locally quasi-finite and $j = (t, s) : R \to U \times _ S U$ is an equivalence relation and $R$ is quasi-split over $u$. Let $P \subset R$ be a quasi-splitting of $R$ over $u$. By Lemma 79.9.1 we see that $(U, R, s, t, c)$ is the restriction of a groupoid $(\overline{U}, \overline{R}, \overline{s}, \overline{t}, \overline{c})$ by a surjective finite locally free morphism $U \to \overline{U}$ such that $P = U \times _{\overline{U}} U$. Note that $s$ admits a factorization
$R = U \times _{\overline{U}, \overline{t}} \overline{R} \times _{\overline{s}, \overline{U}} U \xrightarrow {\text{pr}_{23}} \overline{R} \times _{\overline{s}, \overline{U}} U \xrightarrow {\text{pr}_2} U$
The map $\text{pr}_2$ is the base change of $\overline{s}$, and the map $\text{pr}_{23}$ is a base change of the surjective finite locally free map $U \to \overline{U}$. Since $s$ is flat, locally of finite presentation, and locally quasi-finite and since $\text{pr}_{23}$ is surjective finite locally free (as a base change of such), we conclude that $\text{pr}_2$ is flat, locally of finite presentation, and locally quasi-finite by Descent, Lemmas 35.27.1 and 35.28.1 and Morphisms, Lemma 29.20.18. Since $\text{pr}_2$ is the base change of the morphism $\overline{s}$ by $U \to \overline{U}$ and $\{ U \to \overline{U}\}$ is an fppf covering we conclude $\overline{s}$ is flat, locally of finite presentation, and locally quasi-finite, see Descent, Lemmas 35.23.15, 35.23.11, and 35.23.24. The same goes for $\overline{t}$. Consider the commutative diagram
$\xymatrix{ U \times _{\overline{U}} U \ar@{=}[r] \ar[rd] & P \ar[r] \ar[d] & R \ar[d] \\ & \overline{U} \ar[r]^{\overline{e}} & \overline{R} }$
It is a general fact about restrictions that the outer four corners form a cartesian diagram. By the equality we see the inner square is cartesian. Since $P$ is open in $R$ (by definition of a quasi-splitting) we conclude that $\overline{e}$ is an open immersion by Descent, Lemma 35.23.16. An application of Groupoids, Lemma 39.20.5 shows that $U/R = \overline{U}/\overline{R}$. Hence we have reduced to the case where $(U, R, s, t, c)$ is a groupoid scheme over $S$, with $U$ affine, $u \in U$ such that $s, t$ are flat, locally of finite presentation, and locally quasi-finite and $j = (t, s) : R \to U \times _ S U$ is an equivalence relation and $e : U \to R$ is an open immersion!
But of course, if $e$ is an open immersion and $s, t$ are flat and locally of finite presentation then the morphisms $t, s$ are étale. For example you can see this by applying More on Groupoids, Lemma 40.4.1 which shows that $\Omega _{R/U} = 0$ which in turn implies that $s, t : R \to U$ is G-unramified (see Morphisms, Lemma 29.35.2), which in turn implies that $s, t$ are étale (see Morphisms, Lemma 29.36.16). And if $s, t$ are étale then finally $U/R$ is an algebraic space by Spaces, Theorem 64.10.5. $\square$
Comment #7111 by F. Liu on
1) A typos in the 4th paragraph of the proof: in the statement "... the image $g(U′)$ is open as $g$ is étale and contains $u′$. ", $u'$ should be $u$. 2) In the 5th paragraph it was claimed that $s,t$ are the base changes of the morphisms $\bar{s}, \bar{t}$ by $U\to \overline{U}$. This seems incorrect. Anyway, the desired property of $\bar{s},\bar{t}$ can be deduced as follows. Note that $s$ admits a factorization $R=U\times_{\overline{U},\bar{t}}\bar{R}\times_{\bar{s},\overline{U}}U\xrightarrow{\text{pr}_{23}} \bar{R}\times_{\bar{s},\overline{U}}U \xrightarrow{\text{pr}_2} U$. The second map is the mentioned base change of $\bar{s}$, and the first map is a base change of the surjective finite locally free map $U\to \overline{U}$. By 'descent' we get the desired property of $\bar{s}$ from the same property of $s$.
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## How much interest will she have made after 4 years, Mathematics
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Celine deposited \$505 into her savings account. If the interest rate of the account is 5% per year, how much interest will she have made after 4 years?
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > asclghm Structured version Unicode version
Theorem asclghm 16397
Description: The algebra scalars function is a group homomorphism. (Contributed by Mario Carneiro, 4-Jul-2015.)
Hypotheses
Ref Expression
asclf.a algSc
asclf.f Scalar
asclf.r
asclf.l
Assertion
Ref Expression
asclghm
Proof of Theorem asclghm
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eqid 2436 . 2
2 eqid 2436 . 2
3 eqid 2436 . 2
4 eqid 2436 . 2
5 asclf.l . . . 4
6 asclf.f . . . . 5 Scalar
76lmodrng 15958 . . . 4
85, 7syl 16 . . 3
9 rnggrp 15669 . . 3
108, 9syl 16 . 2
11 asclf.r . . 3
12 rnggrp 15669 . . 3
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CAT > Practice Questions Level 2: Clocks & Calendars - 2
# Practice Questions Level 2: Clocks & Calendars - 2 - Notes | Study Level-wise Practice Questions for CAT Preparation - CAT
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This EduRev document offers 20 Multiple Choice Questions (MCQs) from the topic Clocks & Calendars (Level - 2). These questions are of Level - 2 difficulty and will assist you in the preparation of CAT & other MBA exams. You can practice/attempt these CAT Multiple Choice Questions (MCQs) and check the explanations for a better understanding of the topic.
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:At what time between 5 and 6 o'clock will the hands of a clock be 3 minutes apart?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:In the year 1648, if February had 5 Sundays, then what was the day on February 13, 1750?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:At what time between 4 o'clock and 5 o'clock will the hands of a clock be at a right angle?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:At what time between 4:15 a.m. and 5:05 a.m. will the angle between the hour hand and the minute hand of a clock be the same as the angle between the hands at 8:45 p.m.?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:Alex turned a clock on at 3:00 pm. But the clock is defective, due to which it lags behind by 9 minutes after each day (24 hours). What will be the real time when the clock indicates 6:00 am on the 4th day of it's successive working?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:If 15th March, 2013 was a Friday, then 10th July, 2013 will be a
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:At what time between 7 o'clock and 8 o'clock will the hands of a clock be in a straight line but not together?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:How many days will there be from 23rd January, 2011 to 31st July, 2013 (both days included)?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:Two clocks are set correctly at 10 a.m. on Friday. The first clock gains 2 minutes per hour, which is twice as much as gained by the second clock. What time will the second clock register when the correct time is 2 p.m. on the following Monday?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:A man went outside between 7 o'clock and 9 o'clock at such a time that the minute hand and the hour hand were found to be coinciding before 8 o'clock; and when he returned, again he found both the hands to be coinciding, but after 8 o'clock. What was the time when he returned to the house?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:A clock gains 2 minutes in an hour and an other clock loses 4 minutes in an hour. If both these clocks were set at 8 a.m, what will be the time in the first clock, if the second clock shows 10 p.m?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:At what time between 9 and 10 will the hands of a clock be together?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:What day of the week was it on 15th August, 1987?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:How many times during a day will the hour hand and the minute hand of a clock be six minutes apart?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:In a college, a 24-hour watch loses 5 minutes in 3 hours. If it is set correct on Tuesday midnight, then when will the watch show the correct time next?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:A clock shows the time as 6 am. If the minute hand gains 2 minutes every hour, then how many minutes will the clock gain by 9 pm?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:A tutor teaches 8 days consecutively and then takes off on the ninth day. If he starts teaching on Monday, then on what day of the week will he get his 12th off day?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:I purchased some clocks from a second hand goods store. There were some problems in them. When the actual time passed 1 hour, the wall clock was 10 minutes behind it. When 1 hour was shown by the wall clock, the table clock showed 10 minutes ahead of it. When table clock showed 1 hour, the alarm clock was 5 minutes behind it. When alarm clock showed 1 hour, the wrist watch was 5 minutes ahead of it. Assuming that all clocks are correct with actual time at 12 noon, what will be shown by the wrist watch after 6 hours?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:Today is Friday. A person wants to take a doctor's appointment. As the doctor is busy, he asks him to come three days after the day before the day after tomorrow. On which of the following days does the doctor ask the person to come?
Question for Practice Questions Level 2: Clocks & Calendars - 2
Try yourself:If the current time is 6:20, then after how much time will the hands of the clock make the same angle as they do at present?
The document Practice Questions Level 2: Clocks & Calendars - 2 - Notes | Study Level-wise Practice Questions for CAT Preparation - CAT is a part of the CAT Course Level-wise Practice Questions for CAT Preparation.
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# What's the benefit of 24V over 12V in solar and charging?
What are all the benefits of 24V solar systems over 12V solar systems in regards to solar panels and battery charging?
All I've been able to find is that you can use smaller gauge wire with 24V, there's got to be more benefits than that.
Does 24V charge the batteries faster, etc?
• Don't underestimate how useful the smaller gauge wire is. Commented Jun 17, 2019 at 4:53
• It is only really relevant to the transmission of solar power. If you're making something with an integrated PV panel, then it's of less interest. Commented Jun 17, 2019 at 8:57
The benefit of a higher voltage is relative. It depends on distance, and power level.
If your panels are right next to your tent, the batteries inside, and you're just running a few LEDs, then 12v is just fine.
If you want to run several hundred watts, and you have distances of 10m, then the difference is very significant. Not only smaller cables, but smaller connectors, and often smaller and cheaper equipment. Why cheaper equipment? It has cables, fuses, PCB traces, connectors inside, and active devices at 100v are not 2x as expensive as 50v devices.
To put some numbers on that. Let's say you want to run 120 watts over 10m, this is 10A at 12v. Let's say you're using 1mm$$\^2\$$ cable. The resistance of 20m (out and back) is 0.34 ohms, giving you a 3.4v drop at the end, or 28% of your power. Running 5A at 24v, you'd lose 1.7v, which is 7% of your power.
Other things being equal, you can run 4x the distance, or use 4x the power, or have much more scope for future expansion, by running the higher voltage.
48v is a common industrial voltage, used for building-sized DC power. It's the highest voltage you can go to and stay 'touch safe' 'low voltage'.
Of course when loads get to kW and distances to 100m to 1km, then 240v gets useful, and at MW and 10s of km then 100kV + is the norm.
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# Algebra - Polynimial
### vocabulary
Monomial A real number, variable, or a product of a real number and one or more variables. ( NO + or - signs) 18, K, 2.5xy
Binomial A polynomial of TWO terms. Examples: 4c - 2, 8k + 25
Polynomial A monomial OR the sum or difference of two or more monomials. Examples: 2x OR 3x - 7
Trinomial A polynomial of THREE terms Examples: 2x² - 4x + 5 3b³ + 7b² - b
degree of a monomial Sum of the exponents of the variables. −4x³ y² has a degree of 5
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# Logic Design - Chapter 9: Registers
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- SHIFT REGISTER BASICS.
- SERIAL IN/SERIAL OUT SHIFT REGISTERS.
- Operation Of Shift Registers.
- PARALLEL IN/SERIAL OUT SHIFT REGISTERS.
- 4-bit parallel in/serial out shift register.
- BIDIRECTIONAL SHIFT REGISTER.
- The function table of the register.
- Ring Shift Counter.
- Ring Shift Counter Operation.
- Johnson Shift Counter.
- Johnson Shift Counter Operation.
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### Logic Design - Chapter 9: Registers
1. 1. CHAPTER 9 REGISTERS 1
2. 2. REGISTER WITH PARALLEL LOAD Load Clock Clear Function 0 x 1 No change 1 ↓ 1 Load x x 0 clear 2
3. 3. SHIFT REGISTER BASICS 3
4. 4. SERIAL IN/SERIAL OUT SHIFT REGISTERS CP 4
5. 5. Operation Of Shift Registers Clock Serial I/P State of register Serial pulse Bit ( parallel outputs ) 0/P bit Initial 1 0 1 1 1 1 1 1 1 0 1 1 1 2 0 1 1 0 1 1 3 1 0 1 1 0 0 4 x 1 0 1 1 1 5
6. 6. PARALLEL IN/SERIAL OUT SHIFT REGISTERS 6
7. 7. 4-bit parallel in/serial out shift register 7
8. 8. BIDIRECTIONAL SHIFT REGISTER No change Shift right Shift left Parallel load 8
9. 9. The function table of the register S1 S0 Operation 0 0 No change. 0 1 Shift right. 1 0 Shift left. 1 1 Parallel load. 9
10. 10. Ring Shift Counter 10
11. 11. Ring Shift Counter Operation Q3 Q2 Q1 Q0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 11
12. 12. Johnson Shift Counter 12
13. 13. Johnson Shift Counter Operation 13
14. 14. T0 T1 T2 T3 T4 T5 T6 T7 T0 Q3 0 0 0 0 1 1 1 1 0 Q2 0 0 0 1 1 1 1 0 0 Q1 0 0 1 1 1 1 0 0 0 Q0 0 1 1 1 1 0 0 0 0 14
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# LSAT Logic Games : Solving two-variable logic games
## Example Questions
### Example Question #71 : Solving Two Variable Logic Games
Eight students – Alice, Ben, Carl, Daisy, Earl, Frank, Gretchen, and Hailey – are sitting at the front of a school bus in four seats, with one student sitting on the right side and one student on the left side of each seat. The seats are numbered sequentially 1 through 4, with Seat 1 at the front of the bus and Seats 2, 3, and 4 immediately behind, in that order. The following conditions apply to the seating arrangement:
Ben must sit on the right side of whatever seat he sits in.
Frank and Gretchen sit closer to the front of the bus than Daisy.
Carl and Daisy do not sit in the same seat.
Earl and Hailey sit exactly one seat apart from each other, and on the same side of the seat.
If Carl and Hailey sit in the same seat, Frank and Ben also sit in the same seat.
If Hailey sits in Seat 1 on the right side of the seat, which one of the following could be true?
Carl and Hailey sit in the same seat.
Alice and Ben sit in the same seat.
Daisy and Earl sit in the same seat.
Daisy and Hailey sit in the same seat.
Ben and Earl sit in the same seat.
Carl and Hailey sit in the same seat.
Explanation:
Carl and Hailey can sit in the same seat in this scenario. With Hailey sitting on the right side of Seat 1, Earl must sit in seat 2 (on the right side). Further, since Carl and Hailey are sitting together, Ben and Frank must sit together, and it must be in Seat 3. This means Daisy must sit in Seat 4 with Alice, and Gretchen can take the remaining spot in Seat 2. The remaining answer choices violate one or more conditions under these circumstances.
### Example Question #71 : Solving Two Variable Logic Games
A consultant has agreed to see each of his nine clients-- L, M, N, O, P, Q, R, S, T-- once in the next six days, from Monday through Saturday. He arranges his schedule so that he can see at least one of his clients each day, while maintaining the following conditions:
O is always scheduled on a day before R and M.
P is not scheduled for Saturday.
If T is scheduled on a day after O, then S is scheduled on a day after N.
If T is scheduled on a day before O, then R is scheduled on a day before L.
The consultant always sees fewer clients on Friday and Saturday combined than he sees on any other two days of the week combined.
If the consultant sees O on Wednesday, what is the maximum number of clients that he can see after Wednesday?
5
2
3
4
1
4
Explanation:
The correct answer is 4 clients. Placing O on Wednesday does not lead to any particularly unique diagrams. This game is fairly loose-- there are a lot of possible diagrams. The key insight here is the fact that Friday and Saturday can accommodate, at most, two clients total and any other day accommodates at most two.
Here is a possible diagram that maximizes clients after Wednesday:
Mon: T
Tues: Q, P
Wed: S, O
Thurs: R, N
Fri: M
Sat: L
### Example Question #73 : Solving Two Variable Logic Games
A gym teacher wants his 6 students to line up in height order, from shortest to tallest. Corrin and Theresa are girls. Ben, Jonathan, Will, and Dan are the boys.
- Will is neither the tallest nor the shortest
- No girl is taller than Jonathan
- Dan is shorter than Corrin, but taller than Theresa
- Ben is the tallest
If Jonathan is the 2nd tallest, what must be true?
Dan is the 2nd tallest.
Will is 4th tallest.
Will is 2nd shortest.
Will is 5th tallest.
Corrin is not the shortest or second shortest.
Corrin is not the shortest or second shortest.
Explanation:
If Jonathan is 2nd tallest, that means Will can go anywhere but 6th, 2nd or 1st tallest positions. Since Corrin is taller than Dan, and Theresa is shorter than Dan, each of their positions can be altered by Will. Thus, the only certainty, is that Corrin cannot be the first or second in line, because Theresa and Dan must precede her.
### Example Question #74 : Solving Two Variable Logic Games
Daniel, Fiona, Ivette, Kevin, Marge and Polly are in line to get their cars washed, and all of their cars are washed only once. The order in which their cars recieve a wash is restricted by the following stipulations:
-Ivette's wash is separated from Daniel's wash by 3 other washes
-Fiona's wash is separated from Kevin's wash by one other wash
-Polly's car is washed before Daniel's
How many different spots in the line can Fiona occupy?
6
4
2
3
5
6
Explanation:
This is a time-consuming problem in that you must test to see if F can be in any position. There are two things to consider as far as time restraints: first, use the test thus far to your advantage. look at all your old sequences and see where F has occupied successfully. Second, every spot K occupies can be occupied by F.
See if F can be first (and consequently third, the spot K must occupy)
Place F first and follow though all the restrictions
(F, _, _, _, _, _) --> (F, _, K, _, _, _) --> (F, I, K, _, _, D) I choose to put D last to leave more flexibility with the placement of P. See rule 3. --> (F, I, K, (M/P),(M/P) , D). F and K are replaceable so F can occupy spot 1 and 3.
Place F second and work it through
(_, F, _, _, _, _) --> (_, F, _, K, _, _) --> (I, F, _, K, D, _) --> (I, F, P, K, D, _) --> (I, F, P, K, D, M). F and K are interchangeable so F works in both 2 and 4.
Next test F in 5.
(_, _, _, _, F, _) --> (_, _, K, _, F, _) --> (_, (I/D), K, _, F, (I/D)) --> (P, (I/D), K, _, F, (I/D)) --> (P, (I/D), K, M, F, (I/D)). F can work in 5.
Next test F in 6.
(_, _, _, _, _, F) --> (_, _, _, K, _, F) --> (_, (I/D), _, K, (I/D), F) --> (P, (I/D), _, K, (I/D), F) --> (P, (I/D), M, K, (I/D), F). F can work in 6.
F can work in any spot, so since there are 6 spots, the answer is 6.
### Example Question #341 : Lsat Logic Games
A chef is arranging spices on a shelf. Four of the spices are in large jars: garlic, oregano, pepper, and salt. Three of the spices are in small jars: basil, cumin, and mint. The following conditions apply:
A large jar must be first or fourth
Pepper must come after cumin but before basil
The jar of salt must be the first large jar in the line
Basil cannot be immediately before or after garlic
If the rules are changed so that basil must come directly after garlic, and if all other rules remain the same, which of the following cannot be true?
pepper is fifth
garlic is fourth
mint is first
cumin is fourth
mint is second
garlic is fourth
Explanation:
If garlic and basil must be next to each other, then there are now five spices that must come after salt. Only four spices can come after salt when salt is third, so salt cannot be third.
### Example Question #71 : Solving Two Variable Logic Games
Three circus performers—Flo, Genie, and Kyle—will perform the following six circus acts: acrobatics; ball juggling; clowning; dunking; elephant demonstrations; fire twirling. Each act is performed only once in any given show, and only one act is performed at a time. The following conditions dictate the sequencing of these acts:
• Flo performs exactly one act before Genie performs any of her acts
• Flo does not perform first and he does not perform last
• Genie does not do acrobatics and ball juggling
• Kyle does not do acrobatics and elephant demonstrations
• Ball juggling is performed immediately after the elephant demonstrations
Which one of the following is an acceptable list of the performers and their circus acts, in order from first to last in the show?
Kyle: fire twirling; Flo: elephant demonstrations; Genie: dunking; Kyle: ball juggling; Flo: acrobatics; Genie: clowning
Kyle: clowning; Flo: acrobatics; Flo: elephant demonstrations; Kyle: ball juggling; Genie: dunking; Genie: fire twirling
Kyle: clowning; Flo: dunking; Genie: elephant demonstrations; Kyle: ball juggling; Flo: acrobatics; Genie: fire twirling
Kyle: dunking; Kyle: fire twirling; Flo: clowning; Genie: acrobatics; Flo: elephant demonstrations; Genie: ball juggling
Flo: fire twirling; Genie: elephant demonstrations; Kyle: ball juggling; Kyle: dunking; Flo: acrobatics; Genie: clowning
Kyle: clowning; Flo: dunking; Genie: elephant demonstrations; Kyle: ball juggling; Flo: acrobatics; Genie: fire twirling
Explanation:
This is a sequence and matching game. The following are deductions we can make from the list of conditions (and thus are added to those conditions to guide our handling of the questions):
Kyle must perform first because Flo is eliminated from the first slot by virtue of an explicit rule and Flo must go before Genie by virtue of an explicit rule, (which means Genie can't go first, as well).
Flo must do acrobatics because neither Kyle nor Genie can do it.
If we go rule by rule, including these new rules we have deduced, and apply each of them one by one to the answer choices, we can eliminate all but the credited response.
### Example Question #77 : Solving Two Variable Logic Games
Three circus performers—Flo, Genie, and Kyle—will perform the following six circus acts: acrobatics; ball juggling; clowning; dunking; elephant demonstrations; fire twirling. Each act is performed only once in any given show, and only one act is performed at a time. The following conditions dictate the sequencing of these acts:
• Flo performs exactly one act before Genie performs any of her acts
• Flo does not perform first and he does not perform last
• Genie does not do acrobatics and ball juggling
• Kyle does not do acrobatics and elephant demonstrations
• Ball juggling is performed immediately after the elephant demonstrations
Which of the following must be true?
Flo does ball juggling.
Genie does fire twirling.
Flo does elephant demonstrations.
Genie does clowning.
Flo does acrobatics.
Flo does acrobatics.
Explanation:
Because neither Genie nor Kyle can do acrobatics, by virtue of explicit rules, the only performer left to do acrobatics is Flo.
### Example Question #78 : Solving Two Variable Logic Games
Three circus performers—Flo, Genie, and Kyle—will perform the following six circus acts: acrobatics; ball juggling; clowning; dunking; elephant demonstrations; fire twirling. Each act is performed only once in any given show, and only one act is performed at a time. The following conditions dictate the sequencing of these acts:
• Flo performs exactly one act before Genie performs any of her acts
• Flo does not perform first and he does not perform last
• Genie does not do acrobatics and ball juggling
• Kyle does not do acrobatics and elephant demonstrations
• Ball juggling is performed immediately after the elephant demonstrations
If Kyle performs fourth in the circus show, then when could acrobatics be performed?
Second
First
Fourth
Third
Sixth
Second
Explanation:
Since Flo is the only one who does acrobatics, and Flo cannot go first, we can immediately eliminate "first" as an answer choice. Likewise, we can eliminate "sixth," since Flo can't go last. "Fourth" is also out since Kyle doesn't do acrobatics. "Second" is the only viable option because Flo must come before Genie, which means acrobatics is second.
### Example Question #71 : Solving Two Variable Logic Games
Three circus performers—Flo, Genie, and Kyle—will perform the following six circus acts: acrobatics; ball juggling; clowning; dunking; elephant demonstrations; fire twirling. Each act is performed only once in any given show, and only one act is performed at a time. The following conditions dictate the sequencing of these acts:
• Flo performs exactly one act before Genie performs any of her acts
• Flo does not perform first and he does not perform last
• Genie does not do acrobatics and ball juggling
• Kyle does not do acrobatics and elephant demonstrations
• Ball juggling is performed immediately after the elephant demonstrations
If Genie does clowning immediately before Flo does elephant demonstrations, which one of the following must be true?
Kyle does fire twirling as first act.
Genie performs fourth.
Genie does dunking as the fifth act.
Flo does acrobatics as the second act.
Kyle performs sixth.
Flo does acrobatics as the second act.
Explanation:
We can deduce the following sequence: Kyle . . . Flo (acrobatics) . . . then :Genie (clowning); Flo (elephant demonstrations); Kyle (ball juggling)—the latter three forming a block.
Because Flo (acrobatics) must come before any act by Genie, it must come early in the sequence (earlier than the block of three—GFK). If Flo (acrobatics) does not occupy slot two, then the sequencing necessarily breaks down. That means Flo doing acrobatics must be the second act performed.
### Example Question #80 : Solving Two Variable Logic Games
Three circus performers—Flo, Genie, and Kyle—will perform the following six circus acts: acrobatics; ball juggling; clowning; dunking; elephant demonstrations; fire twirling. Each act is performed only once in any given show, and only one act is performed at a time. The following conditions dictate the sequencing of these acts:
• Flo performs exactly one act before Genie performs any of her acts
• Flo does not perform first and he does not perform last
• Genie does not do acrobatics and ball juggling
• Kyle does not do acrobatics and elephant demonstrations
• Ball juggling is performed immediately after the elephant demonstrations
Which one of the following could be true?
Ball juggling is the second act performed.
Fire twirling is the first act performed.
Acrobatics is the first act performed.
Elephant demonstrations is the last act performed.
Elephant demonstrations is the first act performed.
Fire twirling is the first act performed.
Explanation:
Flo does acrobatics, so it cannot come first.
For ball juggling to go second, elephant demonstrations would have to first. But elephant demonstrations can't go first because Kyle must go first and he doesn't do elephant demonstrations.
Finally, elephant demonstrations can't go last because ball juggling must follow.
Consequently, the only option left is fire twirling going first---and that is the credited response.
Tired of practice problems?
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# Choropleth mapping
Choropleth Mapping uses a thematic map displaying a quantitative attribute using ordinal classes which are assigned a symbolize feature and is applied over an area.They are based on the same data and spatial units but can represent different patterns due to the choice of class interval. They usually present population data at a large or small scale for example sex ratio in a province would be classified as small while population of school children in census tracts would be large.Choropleth mapping can be abused, by using different techniques to create a map and can be easily misrepresented. User must be aware and observe critically before concluding to a decision. There are unfortunately three common problems with choropleth mapping:
Choice of shading pattern, Choice of classification system, and Choice of spatial unit.
Use the universal code of 'darker is greater' when specifying shading patterns, and with colors bright colors represent higher values.
Classification System:
Must specify number of classes and class intervals in the map.
-The greater the number of classes used, makes the map more confusing to intrepid.
-Max of 5 classes is sufficient to display data.
-There are four common ways to specify class intervals: equal interval, percentile, and natural breaks.
-Equal Interval: Splits data into user-specified number of classes of equal width.
-Percentile: Data divided so that equal number of observations fall into each class.
-Natural Breaks: Splits data into classes based on natural breaks using a commonsense method when interpreting data.
-Manual: Gives the user freedom to set classes and number of breaks within the data.
Choice of Spatial Unit:
Usually out of users control. Extremely hard to go beyond minimum resolution or translate data in other non matching units.
Choice does exist with change of data into larger spatial units. census tracts --> census districts.
Amalgamation of data can lose information bu masking internal pattern and variation that could be potentially important to the user.
Part of the Spatial Unit is MAUP (Modifiable areal unit problem) where boundaries of spatial units act to hide underlying patterns in data.
Ecological fallacy problem associates with MAUP. And occurs when it is inferred that a data for study area applies to indv. within an area.
For a Choropleth map a user needs a legend to make sure correct interpretation is seen. It tells the user what shading patterns, colors, lines, and point symbols mean. As well it gives information about the class interval system used. Also annotation should be used to bring the map to life by giving information by textual or graphical techniques that labels the map and features within it. Annotation consists of a map title, legend, North arrow, and other specific labels such as names to points on a map. Choropleth mapping is representing science as art. Using technical decisions about scale, generalization, and reference systems by representing them by Artistic decisions to make the map appear appeasing. Using this user friendly method helps communicating spatial information to others such as for government policy making, or business management.
In relation to the case study you could use it for showing population density's of grizzly bears kills in different regions of BC by representing them in color. Using Natural breaks to represent the population in intervals.
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2.99
# Question: Next week, a student worker will be
Next week, a student worker will be assigned one morning of kitchen duty. The sample space has seven elementary outcomes e1 , e2, .. , e7 where e1 represents Sunday, e2 Monday, and so on. Two events are given as A = {e4 , e5, e6 , e7) and B = {e1, e6 , e7).
(a) Draw a Venn diagram and show the events A and B.
(b) Determine the composition of the following events:
(i) AB (ii) B¯ (iii) AB¯ (iv) A U B.
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# A block of mass 4kg placed on a horizontal surface is connected to another mass of 3 kg by a massless inextensible string going over a Pulley as shown. A force of F = 50 newton is applied horizontally on A as shown. At the instant, theta is 60° and the blocks are at momentary rest the tension in the string is(take g=10 m/s^2)
Arun
25757 Points
3 years ago
Dear student
T = m2 *g = 3 *10 = 30 Newton.
Hope it helps.
Thanks and regards
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https://www.physicsforums.com/threads/sliding-on-the-surface-of-a-rough-sphere-ploting-theta-in-excel.293789/
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# Sliding on the surface of a rough sphere (ploting theta in excel)
#### chart2006
1. The problem statement, all variables and given/known data
I have determined the solution to my problem however now the problem asks me to plot theta as the block flies off the sphere when Mu k ranges from 0.0 to 0.5 in increments of 0.05 I don't know a great deal about working with Microsoft Excel. Well I do just not enough to know how to plot theta from this equation. The information is as follows.
r = 1.5m
$$V_0 = 3m/s$$
$$\mu$$ = 0.0 thru 0.5 in increments of 0.05
$$\theta = ?$$
2. Relevant equations
$$V^2 = \frac{(2 - 4 \mu^2) (e^{2\mu\theta} - cos(\theta)) - 6 \mu sin(\theta)} {(1 + 4\mu^2)}+ V_0^2 e^{2\mu\theta}$$
$$V = rgcos(\theta)$$
3. The attempt at a solution
As for the attempt at the solution as I said I've already established the solution now I only need to plug into excel in order to solve for theta. Any help would be appreciated.
Last edited:
Related Advanced Physics Homework News on Phys.org
#### Brian_C
You need to state the entire problem. I don't think anyone understands what you're trying to do.
"Sliding on the surface of a rough sphere (ploting theta in excel)"
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# Acceleration direction on the MPU6050 sensor
Hello, I have a question about acceleration around the axes. Let's take the Y axis as a basis to clear my doubts. In the development of my project I am using the MPU6050 sensor direction exactly as represented in image A, in this case (image A) when I read the acceleration values and do the conversions for m/s² it is around 9.8m/s² ( gravitational acceleration), but when inverted as in image B the value is negative -9.8m/s² (gravitational acceleration). My question arises on the following line, am I using the sensor position correctly? And if so, why does the equipment (image A) have a representation of the direction of the Y axis (taken as a basis for clearing doubts) pointed upwards by means of an arrow, that is, against gravity and the value is positive?
(A)
(B)
Thanks for listening!
Hugs!
The response of an accelerometer to gravity can be confusing, because when held still, the accelerometer is actually measuring the upward acceleration, produced by the force exerted by the surface that is preventing a fall.
This is consistent with the direction of acceleration reported in the horizontal direction, i.e. due to the motion of your hand, etc.
How confused! Thank you!
Can I use the equipment (MPU6050 sensor) as in image A?
The accelerometer can be used in any orientation.
The difference between acceleration due to gravity and acceleration due to other applied forces is that gravity attracts all components of the accelerometer chip equally, whereas other external forces are applied only to the sensor framework.
Ok! Thank you!
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# algebra 2
Sonia is selling her house through a real estate agent whose commission is 6% of the selling price. What should be the selling price so Sonia can get \$84,600?
1. 👍 0
2. 👎 0
3. 👁 178
1. 1.06 * 84,600 = ?
1. 👍 0
2. 👎 0
2. 89,676
but I thought I had to set up problem as an algebraic?
1. 👍 0
2. 👎 0
posted by Kat
3. 0.96 x = 84,600
x = 84,600/0.96
x = \$ 88,125 selling price
1. 👍 0
2. 👎 0
posted by Kat
4. It should be 0.94x
1. 👍 0
2. 👎 0
5. wait I think it would be 0.94 not 0.96 . the answer would be 90,000
1. 👍 0
2. 👎 0
posted by Kat
Take 0.06 of both numbers and see which one equals 84,600
1. 👍 0
2. 👎 0
## Similar Questions
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Distance between Kōchi (KCZ) and Shonai (SYO)
Flight distance from Kōchi to Shonai (Kōchi Airport – Shonai Airport) is 499 miles / 802 kilometers / 433 nautical miles. Estimated flight time is 1 hour 26 minutes.
Driving distance from Kōchi (KCZ) to Shonai (SYO) is 618 miles / 994 kilometers and travel time by car is about 11 hours 43 minutes.
Map of flight path and driving directions from Kōchi to Shonai.
Shortest flight path between Kōchi Airport (KCZ) and Shonai Airport (SYO).
How far is Shonai from Kōchi?
There are several ways to calculate distances between Kōchi and Shonai. Here are two common methods:
Vincenty's formula (applied above)
• 498.533 miles
• 802.312 kilometers
• 433.214 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 498.555 miles
• 802.346 kilometers
• 433.232 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
Airport information
A Kōchi Airport
City: Kōchi
Country: Japan
IATA Code: KCZ
ICAO Code: RJOK
Coordinates: 33°32′45″N, 133°40′8″E
B Shonai Airport
City: Shonai
Country: Japan
IATA Code: SYO
ICAO Code: RJSY
Coordinates: 38°48′43″N, 139°47′13″E
Time difference and current local times
There is no time difference between Kōchi and Shonai.
JST
JST
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 98 kg (217 pounds).
Frequent Flyer Miles Calculator
Kōchi (KCZ) → Shonai (SYO).
Distance:
499
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
499
Round trip?
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# The work-energy theorem states that the work done on an object is equal to a change in which quantity? kinetic energy displacement potential energy mass
###### Question:
The work-energy theorem states that the work done on an object is equal to a change in which quantity?
kinetic energy
displacement
potential energy
mass
### You and your friend were taking turns rolling a die when you thought of a game: If a one is rolled by either player, you pay your friend 50 cents, but when a different number is rolled, your friend pays you 10 cents. After 30 rolls, it turned out that you were even, and neither of you won anything. How many times was a one rolled? (PLEASE NO LINKS I REALLY NEED HELP)
You and your friend were taking turns rolling a die when you thought of a game: If a one is rolled by either player, you pay your friend 50 cents, but when a different number is rolled, your friend pays you 10 cents. After 30 rolls, it turned out that you were even, and neither of you won anything. ...
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### What was calvins idea of the elect and their place in society
What was calvins idea of the elect and their place in society...
### Here this is the image
Here this is the image...
### The difference between 2 numbers is 8. The sum of the two numbers is 20. Find the number.
The difference between 2 numbers is 8. The sum of the two numbers is 20. Find the number. ...
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### Which claim below best describes the effects of the invention of the telephone? (1) Organizing business deals became more cumbersome in the 1880s. (2) Its facilitation of communication contributed to business growth (3) Its facilitation of communication led to political controversy (4) It led to the passage of reforms to the patent system
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The Super Discount store (open 24 hours a day, every day) sells 8-packs of paper towels, at the rate of approximately 420 packs per week. Because the towels are so bulky, the annual cost to carry them in inventory is estimated at $.50. The cost to place an order for more is$20 and it takes four day...
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## A community for students. Sign up today
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## qwertyfreak 3 years ago Add and simplify if possible. x-7 x+5 _____ + _____ 2 2 a.) 2x+12 _________ 2 b.) x-1 c.) 2x-2 ________ 2
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1. annas
b is the answer
2. annas
x+x-7+5/2 = 2x-2/2 = 2(x-1)/2= x-1
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Page 1 of 1
### [HELP] How to make a portal with parallax
Posted: 06/27/2015, 10:42 pm
Hi everyone, I want to ask how to make a portal with parallax like this:
https://youtu.be/4HdnfPHCIGQ?t=3m13s
from 3:13 to 3:40
Does he mentioned something about the track matte?
Thank you so much for helping!
(Does the effect similar to the 3D camera projection tutorial ?)
### Re: [HELP] How to make a portal with parallax
Posted: 06/28/2015, 12:46 am
The key point there is to use moving camera and 3D camera tracking and set the other scene that you see through the portal at a distance behind the portal so that you get parallax. Yes, a track matte is used to mask out the scene seen through the portal. The track matte would simply be a black and white version of the hole in the wall.
### Re: [HELP] How to make a portal with parallax
Posted: 06/28/2015, 4:12 am
star+circle wrote:The key point there is to use moving camera and 3D camera tracking and set the other scene that you see through the portal at a distance behind the portal so that you get parallax. Yes, a track matte is used to mask out the scene seen through the portal. The track matte would simply be a black and white version of the hole in the wall.
Thank you again star+circle for helping!!
But could you explain more specifically about the track matte issue ? I`ve got no idea how to use that in the effect, luma matte or alpha matte?
### Re: [HELP] How to make a portal with parallax
Posted: 07/11/2015, 4:43 am
Let's see if I can explain it better with this example picture.
You need two tracked planes:
1) one for the surface where the portal is, in this case the column. Create Null 1 on this surface
2) second at a distance where the "scene" seen through the portal is. Create Null 2 there.
For the matte, you can create a solid with a mask (the green ellipse), make it a 3D plane and parent it to Null 1 (Alt-drag pick whip Null 1 to move plane to null's position)
Parent the "scene" plate (3D layer) to Null 2 the same way.
Place "scene" layer directly below matte layer and select Alpha Matte from the TrkMat options of the "scene" layer.
Result: The "scene" will be seen through the portal matte, but it will appear to be at a distance behind the column, not right on the column. This gives the shot the parallax it needs to be realistic.
(Obviously the "scene" will need to be appropriately scaled to the distance it is from the portal, so needs to be shot from such a distance. I used a still image from the video to show the principle, the "scene" would have to be wide enough to fill the the portal hole from any angle seen in a given shot.)
Same Null 1 can then be used to place your portal frame on the surface.
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0
# How to calculate total bases?
Updated: 12/19/2022
Wiki User
15y ago
Total bases is the sum of the player's number of doubles times 2, plus the player's number of triples times 3, plus the player's number of home runs time 4, plus the player's number of singles. Example: A player, for a season, has 40 singles, 20 doubles, 3 triples, and 11 home runs. You would calculate the number of total bases as 40 + (20 * 2) + (3 * 3) + (11 * 4) which works out to 40 + 40 + 9 + 44 or 133 total bases.
Wiki User
15y ago
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https://photon.libretexts.org/The_Science_of_Solar/Solar_Basics/A._Introductory_Physics_for_Solar_Application/II._Electricity/4._Circuits_(Resistors_and_Capacitors)
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# 4. Circuits (Resistors and Capacitors)
• Page ID
324
• Now that electric fields, voltage, and current have been explained, we can introduce circuits. Circuits are networks that connect various electrical elements such as voltage sources (i.e. batteries), resistors, and capacitors. Below are listed the various parts of a circuit which may be crucial for understanding solar technology.
• A voltage source, or source of emf (electromotive force),is some device that creates a potential difference between two points, thus generating a current through a circuit. A common example is the battery, which converts chemical energy (involved in chemical reactions within the battery) into electrical energy.
• Capacitors are places in the circuit where at least two conductor surfaces are separated by some insulator. If a voltage source is present in the circuit, a capacitor will store charge, gradually charge up, and eventually prevent current flow upon full charge. A common example is two parallel metal plates, where opposite charges accumulate on each plate.
• Resistors are circuit elements which can be added to limit current flow, following the equation
where V is voltage, I is current (measured in Amps, A) and R is resistance (measured in Ohms, Ω). In other words, the lower the resistance and current, the lower the voltage in the circuit. Resistance in a circuit can also represent the natural resistance of the wires or other components themselves.
Above: A circuit diagram, including a battery (emf), capacitor (C), a resistor (R), and a switch that redirects current and therefore switches off the source of voltage.
Source: onlinephys.com. 4 March 2012.
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• 3. NATURA RAZELOR X Razele X sunt radiaţii de natură electromagnetică (ca şi razele luminoase, ultraviolete, razele gamma) a căror energie este definită prin relaţia: E = h υ  unde: E = energia cuantei fotonului h = constanta lui Planck υ= frecvenţa undei electromagnetice Tipurile de radiaţii se deosebesc între ele prin frecvenţă, lungime de undă şi perioadă Lungimea de undă ( λ ) este distanţa minimă între două puncte consecutive situate pe direcţia de propagare a undei  Razele X au λ = 8 - 0,06 A  Razele ultraviolete au λ = 3900 – 136 A  Razele gamma au λ = 6,06 – 0,001 A ( 1 A = 1mm/10 ) Deosebirea între diferitele tipuri de radiaţii constă în locul de origine a fiecărui tip (radiaţiile X sunt emise la nivelul învelişului electronic; radiaţiile gamma sunt emise la nivelul nucleului)
• 4. TUBUL RÖNTGEN - PRODUCEREA RADIAŢIILOR X În drumul său un electron incident acţionând asupra unui alt electron, îl pune în mişcare, transferându-i o anumită cantitate de energie; în urma interacţiunii dinre cei doi electroni se produc radiaţiile X, ca urmare a transferului de energie În röntgendiagnostic, radiaţiile X iau naştere în urma interacţiunii dintre electronii plecaţi de la nivelul catodului, care au viteze mari şi cei de la nivelul anodului
• 5. Componentele unui aparat de radiodiagnostic convenţional  Tubul Röntgen  Generatorul de tensiune  Dispozitive de comandă şi control  Accesorii
• 6. Tubul Röntgen Constituie partea principală a unei instalaţii de radiodiagnostic, unde energia electrică de mare tensiune se transformă în radiaţii X Componentele tubului Röntgen  Tubul de sticlă  Catodul  Anodul  Sistemul de răcire a tubului  Învelişul tubului (cupola)
• 7. PROPRIETĂŢILE RADIAŢIILOR X Divergenţa – are implicaţii în alegerea tehnicilor de examinare, în protecţia faţă de radiaţii precum şi în înţelegerea formării imaginii radiologice Penetrabilitatea - este direct proporţională cu Kv fasciculului radiant; pentru a modifica penetrabilitatea în scopul propus pentru o examinare, fasciculul de radiaţii trebuie filtrat, având ca scop îndepărtarea din fascicul a fotonilor cu energie joasă Atenuarea – este principalul fenomen fizic prin care materia diminuă sau atenuează intensitatea unei radiaţii, absorbţia razelor X în ţesuturile examinate este determinată de o serie de factori care ţin de regiunea examinată.  Numărul atomic al structurilor exmainate  Densitatea structurilor examinate  Graosimea structurilor anatomice examinate  Calitatea fasciculului de raze X Luminiscenţa - fluorescenţa, fosforescenţa Efecte chimice – impresionează emulsia fotografică a filmelor radiologice Efecte biologice
• 8. PARTICULARITĂŢILE IMAGINII RADIOLOGICEEfectul de penumbră  cu cât distanţa Ob - ecran este mai mare cu atât imaginea este de dimensiuni mai mari şi contur mai şters (efect de penumbră)  cu cât distanţa film – Ob e mai mare cu atât dimensiunile imaginii sunt mai mici
• 9. • cu cât focarul este mai mic cu atât imaginea este mai clară, conturul este mai net iar penumbra dispare
• 10. paralaxa – fenomenul prin care se pot disocia două formaţiuni care se suprapun stabilindu-se distanţa la care sunt situate faţă de film pentru că obiectele situate aproape de film se deplasează mai puţin iar cele la distanţă mare se deplasează mai mult
• 11.  Legea incidenţelor tangenţiale – când raza este tangentă la suprafaţa unui corp opac conturul rezultat este net Fenomenul de sumaţie – pe imaginea radiologică se sumează formaţiunile traversate rezultând o imagine unică. Când peste o opacitate se sumează o transparenţă scade intensitatea opacităţii prin substracţie motiv pentru care se folosesc radiografii în incidenţe perpendiculare
• 12. CONDIŢII PENTRU O IMAGINE RADIOLOGICĂ DE BUNĂ CALITATE Razele X să fie produse de un focar cît mai mic Distanţa tub- obiect să fie cât mai mare Raza centrală să fie perpendiculară pe film şi să treacă prin mijlocul regiunii explorate Planul obiectului să fie paralel cu filmul Eliminarea radiaţiilor secundare
• 13. IMAGINEA RADIOLOGICĂ este reprezentarea bidimensională a unui obiect tridimensional fiind un complex de opacităţi şi transparenţe care tind să redea situaţia, forma, dimensiunile, structura şi uneori funcţiile componentelor anatomice. OPACITATEA este rezultatul trecerii razelor X printr-un corp absorbant (cu număr atomic mare- osul) TRANSPARENŢA este rezultatul trecerii razelor X printr-un mediu neabsorbant , aerul.
• 14. ASPECTE TEHNICO- RADIOLOGICE ALEINVESTIGAŢIEI DENTO-ALVEOLARE
• 15.  Executarea corectă a unei radiografii dentare sau de maxilar constituie condiţia esenţială pentru punerea unui diagnostic corect si pentru aplicare aunui tratament adecvat
• 16. METODELE PRINCIPALE DE RADIOGRAFIERE DENTO ALVEOLARĂ Metode de radiografiere intraorală Radiografii periapicale (dentoalveolare glogale, totale) cu film retroalveolar (Dieck) cu film ocluzal (Belot, Simpson) Radiografii cu film „bite-wing” Metode de radiografiere extraorală
• 17. INCIDENŢA ENDOBUCALĂ, RETROALVEOLARĂ, IZOMETRICĂ ŞI ORTORADIALĂ – DIECK 1907- Cieszyinschi – Dieck este considerată ca fiind incidenţa capabilă să furnizeze cele mai numeroase şi mai complete date despre dinţi, crestele alveolare şi formaţiunile anatomice vecine
• 18. RETROALVEOLARĂ aşezarea filmului se face endobucal în spatele alveolelor unui grup de 2-3 dinţi vecini, marginea filmului trebuie să depăşească cu 2 mm. planul cuspidian – examinarea dinţilor este completă;se utilizează filme de ¾ cm. o Back Cu un strat de gelatinobromură de Ag. mai dens U filmele au o „faţă” si un „dos” – filmul se aşează totdeauna cu „faţa” spre fasciculul de radiaţii X pe dosul filmelor există un semn,punct sau incizură unde se practică perforaţia şi care trebuie identificată pentru prinderea filmelor, poziţionarea corectă şi recunoaşterea dinţilor
• 19. examinarea completă a întregii dentiţii necesită folosirea a : 11-14 filme la adult 6 filme la copil
• 20. IZOMETRIA - totalitatea manevrelor indicate pentru a obţine pe film oimagine de aceeaşi dimensiune cu cea reală a dintelui. Pentru a obţine o imagineizometrică fasciculul de radiaţii trebuie să fie perpendicular pe bisectoareaunghiului dat de axa film-dinte
• 21.  deoarece în practică este destul de greu de realizat principiul izometriei introdus de Cieszyinschi, Dieck a introdus un sistem standard de realizare a acestei incidenţe: planul ocluzal al maxilarului respectiv mandibulei trebuie să fie orizontale (pentru arcada superioară capul va fi poziţionat în uşoară flexie, iar pentru arcada inferioară în uşoară extensie pe tegument vârful conului localizator va fi aplicat în dreptul apexului – linia de proiecţie a apexurilor  pe maxilar aripa nasului – tragus  pe mandibulă – de la menton la lobul urechii cu 1 cm. deasupra liniei bazilare
• 22. ORTORADIAL reprezintă localizarea în spaţiu a centrării în plan orizontal în secţiune maxilarele au formă asemănătoare cu o potcoavă; în interiorul acesteia se înscrie un cerc imaginar principiul ortoradialităţii cere ca fasciculul de radiaţii să fie orientat în plan orizontal astfel încât raza centrală să prelungească razele cercului imaginar
• 23. Avantajele radiografiei periapicale Dieck  rapidă  nu necesită manevre laborioase  centrarea nu crează probleme deosebiteDezavantajele radiografiei periapicale Dieck  abordarea oblică a filmului şi structurilor alveolare  abordarea excentrică  lipsa de paralelism dinte-film
• 24. Tipuri particulare de radiografi periapicale tehnica bisectoarei cu susţinere manuală a filmului de către pacient tehnica modernizată a bisectoarei cu susţinere retroalveolară printr-un suport
• 25. TEHNICA PARALELISMULUI – caracteristici  paralelismul dintre dinte şi film  centrarea pe mijlocul dintelui şi filmului  centrarea perpendiculară pe axul longitudinal al dintelui şi filmului
• 26. Avantajele tehnicii paralelismului • fidelitatea imaginii • limitarea erorilor de centrare • este o metodă curentă în tratatele moderne de radiologie stomatologicăDezavantajele tehnicii paralelismului • dotarea tehnică • introducerea filmului
• 27. INCIDENŢA INTERPROXIMALĂ CU FILME CU ARIPIOARE- „BITE-WING” ( TEHNICA RAPER)
• 28.  prin această tehnică se pun în evidenţă dinţii aparţinând ambelor arcade (dinţii antagonişti în ocluzie)  coroanele în întregime  coletul  rebordul alveolar  partea ocluzală a septului interdentar  jumătatea coronară a rădăcinii  parodonţiul segmentului dentoalveolar
• 29. Indicaţii decelarea precoce a cariilor coronare, de colet, localizate subgingival, sub coroanele de înveliş modificări de contur ale camerei pulpare procese de calcificare ale pulpei decelarea leziunilor osoase în parodontopatiile marginale fisurile smalţului fracturi coronoradiculare
• 30. RADIOGRAFIA PANORAMICĂ A ÎNTREGII DENTIŢII (ORTOPANTOMOGRAFIA) reprezintă obţinerea pe un film a imaginii desfăşurate a întregii dentiţii • tipuri de aparate aparate la care sursa de radiaţii este introdusă în gura pacientului în centrul cercului imaginar înscris în potcoava maxilarului; din acest punct se emit radiaţiile X de-a lungul razelor cercului venind dinăuntru în afară şi impresionează un film extraoral aplicat pe tegumentul labiojugal d-a lungul arcadei de radiografiat; pe film sunt proiectate elementele anatomice întâlnite de radiaţii, mărite de volum şi uşor deformate.
• 31.  Ortopantomograful – are la bază deplasarea simultană şi antagonică a tubului de radiaţii situat în spatele capului pacientului şi a unui film de 15/30 într-o casetă semicirculară situat în faţa pacientuluimişcarea fiind astfel reglată încât să se realizeze o măturare a întregii dentiţii – expunerea durează în medie 15 secunde iar pe film se proiecteazăarticulaţiile temporomandibulare, întreaga dentiţie, sinusurile maxilare, fosele nazale şi mandibula în întregime – flow-ul de mişcare scade netitatea elementelor de structură osoasă
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https://caveofpython.com/machine-learning/r-squared-what-is-it/
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# Cave of Python
»»» R Squared: What Is It?
## R Squared: What Is It?
If we fit a line to some data, one way to measure the “goodness of fit” is to use a measure known as R squared. However, this isn’t the full story, so it’s important to use other techniques as well.
For example, if your model diverges from the data at one end, and that’s the bit you intend to use for predictions, R squared won’t alert you to that.
• R squared typically varies from 0 to 1, where 0 indicates a very poor fit with the data, while 1 is a perfect fit.
• R squared is also known as the coefficient of determination.
• It is possible for R squared to be negative. This indicates that your model predictions are worse than if you had just predicted your values to always have their average value.
Here’s an example that uses the cooling.csv data that we saw last time. This data forms almost a straight line if we take the inverse square of temperature. Then we use a scikit-learn linear regression model to model (approximate) the data.
Finally we calculate an R squared score.
``````import pandas as pd
from sklearn.metrics import r2_score
from sklearn.linear_model import LinearRegression
X = df['minute'].values.reshape(-1, 1)
y = 1.0/df['temperature']**2
model = LinearRegression()
model.fit(X, y)
y_predicted = model.predict(X)
score = r2_score(y_predicted, y)
print("R2 score:", score)``````
``R2 score: 0.9995632665411859``
R squared here is almost 1, indicating a very good fit with the data.
#### Calculating R Squared
How is R squared actually calculated? Here we add on code that calculates it “from scratch”.
``````import pandas as pd
from sklearn.metrics import r2_score
from sklearn.linear_model import LinearRegression
import numpy as np
X = df['minute'].values.reshape(-1, 1)
y = 1.0/df['temperature']**2
model = LinearRegression()
model.fit(X, y)
y_predicted = model.predict(X)
score = r2_score(y_predicted, y)
print("R2 score:", score)
variance_of_residuals = np.var(y_predicted - y)
total_variance = np.var(y)
r2 = 1 - (variance_of_residuals/total_variance)
print("Calculated R2:", r2)``````
``````R2 score: 0.9995632665411859
Calculated R2: 0.9995634571940355``````
#### Interpretation
R squared can be thought of as telling us how much of the variance in the data can be explained by the model.
First we calculate the variance of the data. This is simply a measure of how far apart, or how widely scattered, the y-values are. It measures how far the values are from the average value.
Now we calculate the residuals. These are all the distances between the actual y-values and the y-values predicted by the model.
We then find the variance of the residuals. We expect this to be smaller than the total variance; we expect the y-values to be on average much closer to the values predicted by the model than they are to the average y-value.
Dividing the variance of the residuals by the total variance gives us the fraction of the total variance that the model does not predict. We then subtract from 1 to get the fraction of the total variance that the model does predict.
The only way R squared can be negative is if the predictions made by the model are actually worse than if we simply estimated the values using their average. This would obviously indicate a ‘goodness of fit’ that’s worse than useless.
Blog at WordPress.com.
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http://onlinepsa.in/index.php/2015/07/25/psa_practice_qnr10/
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# Quantitative Reasoning – Practice – Take a break!
PSA_PRACTICE_QNR10.
Ans: The entire test can be thought to be divided into 4 parts each divided by a break of 12-minutes. The below figure should give the idea:
Total duration= 3 hrs= 180 minutes.
Break time = 12*3= 36 minutes.
Remaining time= 180 -36= 144 minutes.
Required part for each part of the test = 144/4 = 38 minutes. Option C.
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https://answerriddle.com/answer-the-age-of-a-person-in-1998-became-equal-to-the-sum-of-the-digits-of-the-year-of-their-birth-how-old-is-this-person/
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# Answer: The age of a person in 1998 became equal to the sum of the digits of the year of their birth. How old is this person?
Riddle: The age of a person in 1998 became equal to the sum of the digits of the year of their birth. How old is this person?
This riddle has been around for some time on internet platforms such as Instagram, Facebook, and WhatsApp, and many people still argue over what the answer might be, as there are many possibilities.
The answer to The age of a person in 1998 riddle is below.
Suppose you are born in 19xy then
1998-1900-10x-y=1+ 9+x+y
Left is the 1998 minus the year of birth
Right expression is the sum of the digits of year of birth
Simplify to get
88=11x+2y
The only digits that satisfy this are
X=8
Y=0
So birth year is 1980
Current age is 41
or
I think most people completely misunderstood the question lmfao.
You can’t do the math if you can’t even get the question right.
The year is 1998 (NOT THE BIRTH YEAR). The info that are not given is the person’s age in 1998, and their birth year.
The birth year is 19XY (where X and Y can take values of 0-9, capped at year 1998)
Possible sum is > 10 and < 28 as it cant be 10 or else it’ll be 1900 and the person would be 98 years.
1998 – 28 = 1971
1998 – 10 = 1988
So Possible birth year is > 1971 and < 1988
Which means X is either 7 or 8.
If X is 7 and Y is odd, the sum would be even but age would be odd. ❌
If X is 7 and Y is even, the sum would be odd but age would be even.❌
Thus, X cant be 7.
If X is 8, and Y is even, sum would be even, and age would also be even. ✅
If X is 8, and Y is odd, sum would be odd, but age would be even. ❌
So possible years would be
1980, 1982, 1986, 1988.
Thus, the birth year is 1980 (1+9+8+0 = 18)
1998 – 1980 = 18
1. Noli says:
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https://ohbug.com/uva/1338/
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Crossing Prisms
Prof. Bocchan is a mathematician and a sculptor. He likes to create sculptures with mathematics. His style to make sculptures is very unique. He uses two identical prisms. Crossing them at right angles, he makes a polyhedron that is their intersection as a new work. Since he finishes it up with painting, he needs to know the surface area of the polyhedron for estimating the amount of pigment needed. For example, let us consider the two identical prisms in Figure 14. The definition of their cross section is given in Figure 15. The prisms are put at right angles with each other and their intersection is the polyhedron depicted in Figure 16. An approximate value of its surface area is 194.8255. Figure 14: Two identical prisms at right angles Given the shape of the cross section of the two identical prisms, your job is to calculate the surface area of his sculpture. Input The input consists of multiple datasets, followed by a single line containing only a zero. The first line of each dataset contains an integer n indicating the number of the following lines, each of which contains two integers ai and bi (i = 1, . . . , n). A closed path formed by the given points (a1, b1), (a2, b2), . . . , (an, bn), (an+1, bn+1) (= (a1, b1)) indicates the outline of the cross section of the prisms. The closed path is simple, that is, it does not cross nor touch itself. The right-hand side of the line segment from (ai, bi) to (ai+1, bi+1) is the inside of the section. Youmayassumethat3≤n≤4,0≤ai ≤10and0≤bi≤10 Figure15: Outlineofthecross (i = 1,...,n). One of the prisms is put along the x-axis so that the outline section of its cross section at x = ξ is indicated by points (xi,yi,zi) = (ξ,ai,bi) (0 ≤ ξ ≤ 10, i = 1,...,n). The other prism is put along the y-axis so that its cross section at y = η is indicated by points (xi,yi,zi) = (ai,η,bi) (0 ≤ η ≤ 10, i = 1,...,n).
Output The output should consist of a series of lines each containing a single decimal fraction. Each number should indicate an approx- imate value of the surface area of the polyhedron defined by the corresponding dataset. The value may contain an error less than or equal to 0.0001. You may print any number of digits below the decimal point. Sample Input 4 50 0 10 75 10 5 4 75 10 5 50 0 10 4 0 10 10 10 10 0 00 3 00 0 10 10 0 4 0 10 10 5 00 95 4 50 0 10 55 10 10 4 05 5 10 10 5 50 4 71 41 01 95 0 2/3 Figure 16: The intersection
3/3 Sample Output 194.8255 194.8255 600.0000 341.4214 42.9519 182.5141 282.8427 149.2470
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https://onlineroulette.org.uk/systems/columns-to-dozens-switcher/
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# Columns to Dozens Switcher
Here's a little secret for you- this is one of our favourite roulette systems. We normally play it with a Triple Martingale progression so be warned, it's pretty aggressive. So you are going to need to start off with small bets, or you'll need deep pockets and a table with a decent sized betting limit on the outside bets
Play the Columns to Dozens Switcher at Betfair (Exclusive £500 Bonus)
We sometimes even play this with the Even Money Switcher system, switching between the two systems when we feel the need.
Let's get one thing straight before we start. This roulette strategy will not improve your odds at the roulette table. No system will do that. We play it because it gives us a good system to pick our next bets- it take the emotion out of the game and it has worked for us in some sessions. You cannot get rid of the house edge. The answer to the question: How to Win at Roulette? is the following: choose your table wisely (don´t play on a wheel with 2 zeros, always play European Roulette), extend your winning streaks and cut short your losses.
OK, with that out of the way, let´s look at the Columns to Dozens Switcher.
How to Play the Columns to Dozens Switcher.
This is a hedging strategy that aims to cover a decent amount of the table. The flip side of that is that a losing bet is expensive. You pays your money and you takes your chance in other words (to coin a phrase).
Start off by setting a profit target for your session (the amount of money that you want to win, and the point at which you will leave the table). You also need to set a stop loss- that level at which, again, you walk away. This is a useful exercise for any system by the way.
You are going to be covering just under 2 thirds of the table on the Outside Bets, specifically the Dozens Bet and the Columns Bet. Start off how you like and cover 2 of the 3 available, so 2 dozens or 2 columns. Another variant which is similar to this is the Six Line Quattro system.
If you Win the Bet
On your next bet, you are going to flip your bet (from columns to dozens or from dozens to columns). All you are doing here is mixing your play up a bit. Theoretically it makes no difference as each spin is a mutually exclusive event, but we prefer to try and randomise our play, at least a little bit. Call us superstitious! Follow a flat betting profile, continue to bet the same amount as your initial bet.
Bet on the 2 columns or dozens that didn´t come in, and continue like that, switching between the columns and dozens each time. You can also bet on the dozen or column that came in, plus one other- it´s up to you, just stick to some simple rules so that you do not have to make any decisions on what to bet during the session- you just follow your system- all the decisions are made before the first spin of the wheel.
If You Lose the Bet
You will lose a bet at some point. It´s extremely unlikely that you don´t unless you play a very short session. When you lose a bet, this is where the aggressive part of the system comes into play, the Triple Martingale.
After a loss, repeat your bet on the previous spin (so stick with the same column or dozen), but triple the bet on each outside bet. So, if you were betting £10 en each column (£20 in total), you are now betting £30 on each column or dozen, so a total of £60. We told you losing was expensive! What you are trying to do here, is to claw back your loss from the previous spin. But because you are paying to cover just under two thirds of the table, you need to triple your bet rather than double your bet as you would do playing the standard Martingale on say the even money bets.
Careful!
If you are tripling your bet after a loss on 2 bets, you can imagine that your bets ramp up very quickly! We don´t recommend that you progress more than 3 steps along this Triple Martingale, or at least keep your initial bets low. This is why it is so important to set a stop loss on this system- don´t just blindly triple your bets until you hit the table limits unless this was your plan all along and you are happy to risk this amount.
You are hoping that you win the very first bet after a loss, or at least the one after that. Once you have won, reduce your bet back down to your starting bet level.
Examples
Let´s have a look at some examples.
1. Roger starts betting with £10 units.
He places £10 on the first and third columns. He wins.
Amount bet £20.
The number 13 comes in (lucky for some!) in the first column.
Payout £20, plus he gets one £10 bet back = £30.
Profit: £10.
Roger switches to £10 on the first and last dozens, as the middle dozens hit last.
2. Anna is betting £10 on the 2nd and last dozens.
Red 5 comes in and she loses the first bet, so she triples her bets on the same dozens (£30 on each).
32 hits- a win!
So here´s her cash flow:
1st bet: £10 + £10 = £20. She´s Twenty quid down.
2nd bet: £30 + £30 = £60. She´s £80 pounds down.
One bet wins. The payout is £60 plus she gets the winning bet back of £30, giving her £90.
Her overall profit is £10.
Now she reverts back to her original bet of £10, and switches to the 1st and third columns which didn´t hit last time.
3. Juan loses 2 bets and wins the third one.
Initial bet £20.
2nd bet £60
3rd bet: £180. Total bet is now £260
Juan wins this one, so makes £180 plus he gets one £90 bet back = £270
He is now £10 up.
You can see, just like the Martingale, that if you lose twice, you will end up betting £260 to win £10. Your payout at this point is only 3.8%. The longer you progress up the Triple Martingale, the worse this ratio becomes. You do not want to be spending much time on this escalator.
Having said that, you are covering just under 2/3 of the table each time. That´s the trade off. The important thing is to understand how this roulette strategy works and you can then make an informed decision based on your budget and appetite for risk. Always play within your means.
You can also "Randomise"
Normally we would switch between columns and dozens after every bet. You can also throw in some additional "randomisation" and choose whichever 2:1 outside bet takes your fancy.
Columns to Dozens versus Even Money Switcher
I guess you could say that this system is a variant of the Even Money Switcher system that we have also shared in this strategy section. So how do they compare?
Well, the main difference is the table coverage and the betting profile after a loss. In this system, you are covering a third more numbers, but it is more expensive trying to claw back your profits after a loss.
Also, in the Even Money Switcher, if you can find a French Roulette table that plays La Partage, you can get the house edge down to 1.3% because the casino will give you half your money back if the ball drops into zero. That rule doesn´t apply to the columns or dozens bet. Sign up at UK Casino Club and play the French roulette with their 700 bonus and you will benefit from La Partage.
Which one do we prefer? Guess what? We switch between both of them ;)
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http://www.gurufocus.com/term/Book%20Value%20Per%20Share/UPL/Book%2BValue%2Bper%2BShare/Ultra%2BPetroleum%2BCorporation
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Switch to:
Ultra Petroleum Corp (NYSE:UPL)
Book Value Per Share
\$-19.52 (As of Dec. 2015)
Ultra Petroleum Corp's book value per share for the quarter that ended in Dec. 2015 was \$-19.52.
During the past 12 months, Ultra Petroleum Corp's average Book Value Per Share Growth Rate was -1510.50% per year. During the past 3 years, the average Book Value Per Share Growth Rate was 72.90% per year. Please click Growth Rate Calculation Example (GuruFocus) to see how GuruFocus calculates Wal-Mart Stores Inc (WMT)'s revenue growth rate. You can apply the same method to get the book value growth rate using book value per share data.
During the past 13 years, the highest 3-Year average Book Value Per Share Growth Rate of Ultra Petroleum Corp was 73.70% per year. The lowest was -49.00% per year. And the median was 42.75% per year.
Ultra Petroleum Corp's current price is \$0.31. Its book value per share for the quarter that ended in Dec. 2015 was \$-19.52. Hence, today's P/B Ratio of Ultra Petroleum Corp is .
During the past 13 years, the highest P/B Ratio of Ultra Petroleum Corp was 759.71. The lowest was 0.00. And the median was 10.85.
Definition
Ultra Petroleum Corp's Book Value Per Share for the fiscal year that ended in Dec. 2015 is calculated as:
Book Value Per Share = (Total Equity - Preferred Stock) / Shares Outstanding (EOP) = (-2,991.9 - 0.0) / 153.3 = -19.52
Ultra Petroleum Corp's Book Value Per Share for the quarter that ended in Dec. 2015 is calculated as:
Book Value Per Share = (Total Equity - Preferred Stock) / Shares Outstanding (EOP) = (-2,991.9 - 0.0) / 153.3 = -19.52
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Theoretically it is what the shareholders will receive if the company is liquidated. Total equity is a balance sheet item and equal to total assets less total liabilities of the company.
Book value may include intangible items which may come from the companys past acquisitions. Book value less intangibles is called Tangible Book.
Explanation
Usually a companys book value and Tangible Book Value per Share may not reflect its true value. The assets may be carried on the balance sheets at the original cost minus depreciation. This may underestimate the true economic values of the assets. It also may over-estimate their true economic value because the assets can become obsolete.
For financial companies such as banks and insurance companies, their assets may be reported in current market value of the assets owned. Book values of financial companies are more accurate indicator of the economic value of the company.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Ultra Petroleum Corp Annual Data
Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Book Value Per Share 4.14 5.62 7.25 4.27 7.47 10.45 -3.78 -2.17 1.38 -19.52
Ultra Petroleum Corp Quarterly Data
Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Book Value Per Share -2.46 -2.17 -1.49 -0.81 0.03 1.38 1.55 1.40 1.40 -19.52
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# Monthly Archives: July 2017
## On Curriculum: What Doesn’t Work
For some reason I feel compelled to write about teaching even though I just quit teaching. Quit? Yes, I resigned from my job this year and am taking a break from teaching high school math. Why did I do it? Honestly, I don’t think I can adequately articulate it, and I don’t owe anyone an explanation, but quite simply, I needed a break. I have a sneaking suspicion that I shall return to teaching again some day (probably sooner than I realize), but in the meantime, I’ve been pursuing some of my other passions and working on acquiring some new skills.
I’m still very interested in being part of the conversation on teaching high school math, and I still use Twitter every day to keep up with it. In fact, I have so much to say on the topic that I figured I might as well blog about it. I hope that blogging will be cathartic for me, helpful to other teachers out there, and helpful for me if/when I return to teaching.
So that was quite an introduction to a post in which I wanted to talk about curriculum.
I’m inspired to write about curriculum because the school where I taught had such a horrible, out-dated curriculum, and it was a huge burden for me. I’m pretty sure the curriculum pre-dated my own high school years, so I was shocked that I was required to teach it to my students. It was the most rote, procedural, and repetitive mathematics that I have ever come across. It made me think of the Cold War era, which I actually don’t really know anything about as I was born after that time, but if I could imagine it, I imagine different countries putting their young people in little school factories to see who could solve equations by hand the fastest.
Such was the imagery in my head because the entire curriculum at my school was built around solving equations algebraically. Here is the procedure for solving quadratic equations. Here is the procedure for solving exponential equations. Here is the procedure for solving trigonometric equations. And so on. Naturally, this led to an incredibly teacher-centered classroom. For each lesson, there were pages of notes that the teacher talked about. Then the teacher did some examples. Then the students were supposed to mimic the teacher exactly on a worksheet of 25 identical problems. It was brutal. I felt so sorry for… everyone involved.
Now, I don’t mean to say that we shouldn’t teach solving equations. The concept of what it means to solve an equation is a fundamental part of mathematics. During my first year of teaching I quickly realized the lack of conceptual understanding my students had as a result of our pathetic curriculum. Our assessments would be filled with equations to solve, but not a single student could answer the questions: What does it mean to solve an equation? What does it mean if a number is a solution to an equation?
When I discovered this discrepancy, I just felt terrible. Why were we making students do something that they didn’t understand? Hey kids, memorize exactly what the teacher did, regurgitate it on an exam, and then do it again. There’s no need to understand it. Heck, you can get an A+ grade without actually understanding anything. I quickly realized that no genuine learning was happening. It was sad.
My last two years of teaching I incorporated the two italicized questions from above into the first non-review unit almost every day. (Yikes, don’t get me started on how our curriculum wasted the first unit of every year on “review”.) Last year, I finally had more students answer the first one with something along the lines of “find the values that make the equation true” than students who said “IDK” or “get the answer”.
Besides a lack of understanding, our curriculum lacked efficiency and modern technology. Before becoming a teacher, I was first and foremost a mathematician, and I assure you that no mathematician was solving by hand some of the equations we made our students solve by hand. Mathematicians use technology. If I were to come across an equation that I knew I could solve by hand, but that would take me more than 60 seconds, I would turn to my computer or pick up my Iphone and use Wolfram or Desmos to find the solutions and then carry on from there. I don’t waste my precious time doing a rote procedure when a computer can do it so much faster. I spend my time on bigger and better, more important and more relevant mathematical ideas. Our students should be doing likewise.
Again, I’m not saying that students shouldn’t know how to solve equations or that mathematics is purely conceptual. However, telling students to memorize a specific procedure isn’t that important or that useful. Rather, let students explore equations, find methods that work for them, and develop fluency. As they do this, they will acquire and practice important mathematical skills such as: manipulate equations, model with equations, create different representations of equations, and solve equations efficiently.
Finally, this type of curriculum is also very boring. Be silent. Watch teacher. Work in isolation. No creativity. No thinking. No discussion. There is a complete lack of genuine student engagement. Those poor kids. It’s no surprise that so many dislike math/school. And poor teachers! Our curricula can set us up for failure or, at least, prevent us from seeing the successes our classrooms are capable of.
Wow, am I still complaining about my old curriculum? It appears so. I guess I had to get something off my chest. I should probably stop complaining and maybe write a post called On Curriculum, Part Two: Making it Better. Although I could probably write a whole book on that topic. Now there’s an idea…
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https://www.numbersaplenty.com/3176
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Search a number
3176 = 23397
BaseRepresentation
bin110001101000
311100122
4301220
5100201
622412
712155
oct6150
94318
103176
112428
121a08
1315a4
14122c
15e1b
hexc68
3176 has 8 divisors (see below), whose sum is σ = 5970. Its totient is φ = 1584.
The previous prime is 3169. The next prime is 3181. The reversal of 3176 is 6713.
It can be written as a sum of positive squares in only one way, i.e., 2500 + 676 = 50^2 + 26^2 .
It is a tau number, because it is divible by the number of its divisors (8).
It is a plaindrome in base 14.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 191 + ... + 206.
It is an amenable number.
3176 is a deficient number, since it is larger than the sum of its proper divisors (2794).
3176 is a wasteful number, since it uses less digits than its factorization.
3176 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 403 (or 399 counting only the distinct ones).
The product of its digits is 126, while the sum is 17.
The square root of 3176 is about 56.3560112144. The cubic root of 3176 is about 14.6991931931.
Adding to 3176 its reverse (6713), we get a palindrome (9889).
It can be divided in two parts, 317 and 6, that added together give a palindrome (323).
The spelling of 3176 in words is "three thousand, one hundred seventy-six".
Divisors: 1 2 4 8 397 794 1588 3176
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http://isabelle.in.tum.de/repos/isabelle/rev/e4a408728012
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author wenzelm Sat Nov 18 19:46:48 2000 +0100 (2000-11-18) changeset 10491 e4a408728012 parent 10490 0054c785f495 child 10492 107c7db021a9
quot_cond_function: simplified, support conditional definition;
``` 1.1 --- a/src/HOL/Library/Quotient.thy Sat Nov 18 19:45:37 2000 +0100
1.2 +++ b/src/HOL/Library/Quotient.thy Sat Nov 18 19:46:48 2000 +0100
1.3 @@ -176,28 +176,23 @@
1.4 *}
1.5
1.6 theorem quot_cond_function:
1.7 - "(!!X Y. f X Y == g (pick X) (pick Y)) ==>
1.8 - (!!x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> ==> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor> ==> P x y ==> P x' y'
1.9 - ==> g x y = g x' y') ==>
1.10 - (!!x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> ==> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor> ==> P x y = P x' y') ==>
1.11 - P a b ==> f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b"
1.12 - (is "PROP ?eq ==> PROP ?cong_g ==> PROP ?cong_P ==> _ ==> _")
1.13 + "(!!X Y. P X Y ==> f X Y == g (pick X) (pick Y)) ==>
1.14 + (!!x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> ==> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor>
1.15 + ==> P \<lfloor>x\<rfloor> \<lfloor>y\<rfloor> ==> P \<lfloor>x'\<rfloor> \<lfloor>y'\<rfloor> ==> g x y = g x' y') ==>
1.16 + P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> ==> f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b"
1.17 + (is "PROP ?eq ==> PROP ?cong ==> _ ==> _")
1.18 proof -
1.19 - assume cong_g: "PROP ?cong_g"
1.20 - and cong_P: "PROP ?cong_P" and P: "P a b"
1.21 - assume "PROP ?eq"
1.22 - hence "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g (pick \<lfloor>a\<rfloor>) (pick \<lfloor>b\<rfloor>)"
1.23 - by (simp only:)
1.24 + assume cong: "PROP ?cong"
1.25 + assume "PROP ?eq" and "P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor>"
1.26 + hence "f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g (pick \<lfloor>a\<rfloor>) (pick \<lfloor>b\<rfloor>)" by (simp only:)
1.27 also have "\<dots> = g a b"
1.28 - proof (rule cong_g)
1.29 + proof (rule cong)
1.30 show "\<lfloor>pick \<lfloor>a\<rfloor>\<rfloor> = \<lfloor>a\<rfloor>" ..
1.31 moreover
1.32 show "\<lfloor>pick \<lfloor>b\<rfloor>\<rfloor> = \<lfloor>b\<rfloor>" ..
1.33 - ultimately
1.34 - have "P (pick \<lfloor>a\<rfloor>) (pick \<lfloor>b\<rfloor>) = P a b"
1.35 - by (rule cong_P)
1.36 - also show \<dots> .
1.37 - finally show "P (pick \<lfloor>a\<rfloor>) (pick \<lfloor>b\<rfloor>)" .
1.38 + moreover
1.39 + show "P \<lfloor>a\<rfloor> \<lfloor>b\<rfloor>" .
1.40 + ultimately show "P \<lfloor>pick \<lfloor>a\<rfloor>\<rfloor> \<lfloor>pick \<lfloor>b\<rfloor>\<rfloor>" by (simp only:)
1.41 qed
1.42 finally show ?thesis .
1.43 qed
1.44 @@ -207,7 +202,7 @@
1.45 (!!x x' y y'. \<lfloor>x\<rfloor> = \<lfloor>x'\<rfloor> ==> \<lfloor>y\<rfloor> = \<lfloor>y'\<rfloor> ==> g x y = g x' y') ==>
1.46 f \<lfloor>a\<rfloor> \<lfloor>b\<rfloor> = g a b"
1.47 proof -
1.48 - case antecedent from this refl TrueI
1.49 + case antecedent from this TrueI
1.50 show ?thesis by (rule quot_cond_function)
1.51 qed
1.52
```
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Basic Radio is a free introductory textbook on electronics based on tubes. See the editorial for more information....
# Electron Lenses
Author: J.B. Hoag
Fig. 21 B. A double-gauze electron lens. (From E. & N. P.)
In Fig. 21 B, the heavy dashed lines represent two metal gauzes, curved as shown and electrically charged by the battery V, with positive on the right-hand gauze and negative on the left-hand gauze. An electric field is, therefore, established between the gauzes. Its direction is from the positive to the negative. This is the direction of the force action upon positive charges. A stream of electrons, whose velocity is represented by the symbol v1,comes from the left and enters the electric field. Remember that electrons are negatively charged. When they enter the region between the two gauzes, the electric field tries to bend them immediately along its lines of force. Their forward momentum, however, opposes this deflection. As a result, they travel along a curved path which more and more approaches that of the electrostatic lines of force. After passing through both gauzes, they continue along their new paths at the higher velocity v2. The shaded area in the figure indicates the total region occupied by the electrons. The fact that the various electrons in this region come to a common focal point is only possible if the gauzes have been so curved that the resultant of the electrostatic deflecting force and the forward momentum of the electrons is directly proportional to the distance of the electrons from the axis.
The first electron lens to be treated mathematically is shown in Fig. 21 C.
Fig. 21 C. A single-aperture electron lens. (From E. & N. P.)
There's an interactive simulation available on the "Learning by Simulations" Web site which allows to calculate the trajectories of charged particles in an electrostatic field.
This is called a diaphragm-hole or single-aperture lens. A parallel beam of electrons in the metal can (indicated by the heavy lines at the left of the figure) are deflected by the distorted electrostatic field in the hole at the end of the can, in such a manner as to be converged toward the metal plate (represented by the heavy vertical line at the right of the figure). By proper adjustment of the voltage of the battery V and the speed of the electrons, they can be brought to a sharp focal point on the metal plate.
Fig. 21 D. A two-aperture lens
Combinations of two or more diaphragm lenses, as in Fig. 21 D, have been used to produce magnified, inverted, real images of the surface of filaments, thus permitting detailed studies of the emission of electrons from various minute portions of cathodes or hot filaments.
An electron lens can be formed by means of two metal cylinders charged to different potentials. These are known as double-cylinder lenses. When a parallel bundle of electrons, traveling at a constant velocity, as in Fig. 21 E, enters the electrostatic field between the ends of the cylinders, it is converged toward a point, 0 in the figure, because the electrons must assume a compromise motion between their forward paths and the direction of the electrostatic lines of force.
Fig. 21 E. A double-cylinder lens. (From E. & N. P.)
After passing the gap between the two cylinders, the electrons find themselves in an electrostatic field which causes them to diverge. But, at this point, the electrons are traveling faster, having been accelerated across the gap by the voltage V. Their momentum being greater and the electric field being the same, their divergence will not be as great as their convergence. Hence they continue down the cylinder, converging toward a more distant point.
Fig. 21 F. A short-focus double-cylinder lens. (From E. & N. P.)
In Fig. 21 F, the second cylinder is larger than the first cylinder, with the result that the electrostatic lines spread out more in the second cylinder. This means that the electrostatic field in this region is weaker and its divergent action on the electrons is less. Hence the electrons come to a focus sooner than in the case where the two cylinders are of the same diameter.
Last Update: 2009-11-01
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https://www.physicsforums.com/threads/matrix-with-repeated-eigenvalues-is-diagonalizable.841524/
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# Matrix with repeated eigenvalues is diagonalizable...?
Tags:
1. Nov 5, 2015
### kostoglotov
MIT OCW 18.06 Intro to Linear Algebra 4th edt Gilbert Strang
Ch6.2 - the textbook emphasized that "matrices that have repeated eigenvalues are not diagonalizable".
imgur: http://i.imgur.com/Q4pbi33.jpg
and
imgur: http://i.imgur.com/RSOmS2o.jpg
Upon rereading...I do see the possibility to interpret this to mean that fewer than n independent eigenvectors leads to an undiagonalizable matrix...that n-all different eigenvalues ensures n-independent eigenvectors...leaving open the possibility of n-independent eigenvectors with repeated eigenvalues...? Yes, no?
Because the second worked example shows a matrix with eigenvalues 1,5,5,5, and the use of diagonalization of that matrix, and Matlab is quite happy to produce a matrix of n-independent eigenvectors from this matrix.
The matrix in question is 5*eye(4) - ones(4);
What is the actual rule, because I don't feel clear on this.
2. Nov 5, 2015
### rs1n
If all n eigenvalues are distinct, then the matrix is diagonalizable. However, the converse is not true. There are matrices that are diagonalizable even if their eigenvalues are not distinct, as your example clearly shows.
3. Nov 5, 2015
### Staff: Mentor
Diagonalizable means per definition that you can find a basis of eigenvectors. If all eigenvalues exist in the underlying field or ring and there are as many as the dimension is, there is clearly a basis of eigenvectors. This condition is sufficient but not necessary since diag(1,...,1) is diagonalized with just one (repeated) eigenvalue 1. A necessary condition is that the characteristic polynomial det(A-t*id) of a matrix A has only linear factors (which may be repeated, e.g. det(id - t*id) = (1-t)^dimension ). i.e. all zeros of the characteristic polynomial exist in the underlying field or ring. Those zeros are exactly the eigenvalues.
Ps: You have still to find a basis of eigenvectors. The existence of eigenvalues alone isn't sufficient. E.g.
0 1
0 0
is not diagonalizable although the repeated eigenvalue 0 exists and the characteristic po1,0lynomial is t^2. But here only (1,0) is a eigenvector to 0.
Last edited: Nov 5, 2015
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Representations as fossilized computation
🕑 13 min • 👤 Thomas Graf • 📆 November 30, 2020 in Discussions • 🏷 syntax, morphology, representations, features, category features, selection, subregular, logical transductions
Okay, show of hands, who still remembers my post on logical transductions from over a month ago? Everyone? Wonderful, then let’s dive into an issue that I’ve been thinking about for a while now. In the post on logical transductions, we saw that the process of rewriting one structure as another can itself be encoded as a structure. Something that we intuitively think of in dynamic terms as a process has been converted into a static representation, like a piece of fossilized computation. Once we look at representations as fossilized computations, the question becomes: what kind of computational fossils are linguistic representations?
Synchronous movement: What could go wrong?
🕑 7 min • 👤 Thomas Graf • 📆 October 12, 2020 in Discussions • 🏷 syntax, movement, Minimalist grammars, subregular
I know I promised you guys a follow-up post on logical transductions and the status of representations, but I just have to get this out first because it’s been gnawing at me for a few weeks now. There’s been some limitations of the subregular view of syntax in terms of movement tiers, and I think I’ve found a solution, one that somehow ends up looking a bit like the system in Beyond Explanatory Adequacy. The thing is, my solution is so simple that I fear I’m missing something very basic, some clear-cut empirical phenomenon that completely undermines my purported solution. So, syntacticians, this is your opportunity to sink my current love child in the comments section…
When parsing isn't about parsing
🕑 7 min • 👤 Thomas Graf • 📆 June 18, 2020 in Discussions • 🏷 syntax, morphology, parsing, formal language theory, movement
As a student I didn’t care much for work on syntactic parsing since I figured all the exciting big-picture stuff is in the specification of possible syntactic structures, not how we infer these structures from strings. It’s a pretty conventional attitude, widely shared by syntacticians and a natural corollary of the competence-performance split — or so it seems. But as so often, what seems plausible and obvious at first glance quickly falls apart when you probe deeper. Even if you don’t care one bit about syntactic processing, parsing questions still have merit because they quickly turn into questions about syntactic architecture. This is best illustrated with a concrete example, in that abstract sense of “concrete” that everyone’s so fond of here at the outdex headquarters.
MR movement: Freezing effects & monotonicity
🕑 8 min • 👤 Thomas Graf • 📆 May 19, 2020 in Discussions • 🏷 syntax, movement, freezing effects, monotonicity
As you might know, I love reanalyzing linguistic phenomena in terms of monotonicity (see this earlier post, my JLM paper, and this NELS paper by my student Sophie Moradi). I’m now in the middle of writing another paper on this topic, and it currently includes a section on freezing effects. You see, freezing effects are obviously just bog-standard monotonicity, and I’m shocked that nobody else has pointed that out before. But perhaps the reason nobody’s pointed that out before is simple: my understanding of freezing effects does not match the facts. In the middle of writing the paper, I realized that I don’t know just how much freezing effects limit movement. So I figured I’d reveal my ignorance to the world and hopefully crowd source some sorely needed insight.
🕑 14 min • 👤 Thomas Graf • 📆 April 28, 2020 in Discussions • 🏷 phonology, syntax, algebra, gradience
Omer has a great post on gradience in syntax. I left a comment there that briefly touches on why gradience isn’t really that big of a deal thanks to monoids and semirings. But in a vacuum that remark might not make a lot of sense, so here’s some more background.
Against math: When sets are a bad setup
🕑 11 min • 👤 Thomas Graf • 📆 April 06, 2020 in Discussions • 🏷 methodology, syntax, set theory, Merge, linearization
Last time I gave you a piece of my mind when it comes to the Kuratowski definition of pairs and ordered sets, and why we should stay away from it in linguistics. The thing is, that was a conceptual argument, and those tend to fall flat with most researchers. Just like most mathematicians weren’t particularly fazed by Gödel’s incompleteness results because it didn’t impact their daily work, the average researcher doesn’t care about some impurities in their approach as long as it gets the job done. So this post will discuss a concrete case where a good linguistic insight got buried under mathematical rubble.
Against math: Kuratowski's spectre
🕑 8 min • 👤 Thomas Graf • 📆 March 30, 2020 in Discussions • 🏷 methodology, syntax, set theory, Merge, linearization
As some of you might know, my dissertation starts with a quote from My Little Pony. By Applejack, to be precise, the only pony that I could see myself have a beer with (and I don’t even like beer). You can watch the full clip, but here’s the line that I quoted:
Don’t you use your fancy mathematics to muddy the issue.
Truer words have never been spoken. In light of my obvious mathematical inclinations this might come as a surprise for some of you, but I don’t like using math just for the sake of math. Mathematical formalization is only worth it if it provides novel insights.
🕑 5 min • 👤 Thomas Graf • 📆 March 06, 2020 in Discussions • 🏷 syntax, strings, derivation trees, phrase structure trees
Here’s another quick follow-up to the unboundedness argument. As you might recall, that post discussed a very simple model of syntax whose only task it was to adjudicate the well-formedness of a small number of strings. Even for such a limited task, and with such a simple model, it quickly became clear that we need a more modular approach to succinctly capture the facts and state important generalizations. But once we had this more modular perspective, it no longer mattered whether syntax is actually unbounded. Assuming unboundedness, denying unboundedness, it doesn’t matter because the overall nature of the approach does not hinge on whether we incorporate an upper bound on anything. Well, something very similar also happens with another aspect of syntax that is beyond doubt in some communities and highly contentious in others: syntactic trees.
Unboundedness is a red herring
🕑 13 min • 👤 Thomas Graf • 📆 February 20, 2020 in Discussions • 🏷 syntax, methodology, competence, performance
Jon’s post on the overappreciated Marr argument reminded me that it’s been a while since the last entry in the Underappreciated arguments series. And seeing how the competence-performance distinction showed up in the comments section of my post about why semantics should be like parsing, this might be a good time to talk one of the central tenets of this distinction: unboundedness. Unboundedness, and the corollary that natural languages are infinite, is one of the first things that we teach students in a linguistics intro, and it is one of the first things that psychologists and other non-linguists will object to. But the dirty secret is that nothing really hinges on it.
Hey syntax, where's my carrot?
🕑 5 min • 👤 Thomas Graf • 📆 January 31, 2020 in Discussions • 🏷 syntax, textbooks, teaching
Last week I blogged a bit about syntax textbooks. One question I didn’t ask there, for fear of completely derailing the post, is what should actually be in a syntax textbook. There’s a common complaint I hear from students about syntax classes, and it’s that syntax courses are one giant bait-and-switch. They’re right, and it’s also true for syntax textbooks.
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# 阿里巴巴机器翻译团队：将TVM引入TensorFlow中以优化GPU上的神经机器翻译
## 批量Matmul
### 什么是批量matmul？
void BatchedGemm(input A, input B, output C, M, N, K, batch_dimension) {
for (int i = 0; i < batch_dimension; ++i) {
DoGemm(A[i],B[i],C[i],M,K,N)
}
}
## 批量matmul计算
# computation representation
A = tvm.placeholder((batch, M, K), name='A')
B = tvm.placeholder((batch, K, N), name='B')
k = tvm.reduce_axis((0, K), 'k')
C = tvm.compute((batch, M, N),
lambda b, y, x: tvm.sum(A[b, y, k] * B[b, k, x], axis = k),
name = 'C')
## Schedule优化
### 调整块/线程的参数
# thread indices
# block partitioning
BB, FF, MM, PP = s[C].op.axis
BBFF = s[C].fuse(BB, FF)
MMPP = s[C].fuse(MM, PP)
s[C].bind(by, block_y)
s[C].bind(bx, block_x)
vty, ty = s[C].split(ty_block, nparts = vthread_y)
vtx, tx = s[C].split(tx_block, nparts = vthread_x)
s[C].reorder(by, bx, vty, vtx, ty, tx)
s[C].reorder(by, bx, ty, tx)
s[C].bind(vtx, thread_xz)
## 将matmul与其他运算融合
# computation representation
A = tvm.placeholder((batch_size, features, M, K), name='A')
# the shape of B is (N, K) other than (K, N) is because B is transposed is this fusion pattern
B = tvm.placeholder((batch_size, features, N, K), name='B')
ENTER = tvm.placeholder((batch_size, 1, M, N), name = 'ENTER')
k = tvm.reduce_axis((0, K), 'k')
C = tvm.compute(
(batch_size, features, M, N),
lambda yb, yf, m, x: tvm.sum(A[yb, yf, m, k] * B[yb, yf, x, k], axis = k),
name = 'C')
D = topi.broadcast_add(C, ENTER)
# computation representation
A = tvm.placeholder((batch_size, features, M, K), name='A')
B = tvm.placeholder((batch_size, features, K, N), name='B')
k = tvm.reduce_axis((0, K), 'k')
C = tvm.compute(
(batch_size, M, features, N),
lambda yb, m, yf, x: tvm.sum(A[yb, yf, m, k] * B[yb, yf, k, x], axis = k),
name = 'C')
### Fusion的性能
· tf-r1.4 BatchMatmul：513.9 us
· tf-r1.4 BatchMatmulTranspose（separate）：541.9 us
· TVM BatchMatmul：37.62us
·TVM BatchMatmulTranspose（fused）：38.39 us
## 参考
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win10上使用gpu版的tensorflow
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【4月更文挑战第17天】本文介绍了TensorFlow如何利用GPU加速深度学习， GPU的并行处理能力适合处理深度学习中的矩阵运算，显著提升性能。TensorFlow通过CUDA和cuDNN库支持GPU，启用GPU只需简单代码。GPU加速能减少训练时间，使训练更大、更复杂的模型成为可能，但也需注意成本、内存限制和编程复杂性。随着技术发展，GPU将继续在深度学习中发挥关键作用，而更高效的硬件解决方案也将备受期待。
70 2
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【4月更文挑战第17天】本文探讨了如何优化TensorFlow模型的性能，重点介绍了超参数调整和训练技巧。超参数如学习率、批量大小和层数对模型性能至关重要。文章提到了三种超参数调整策略：网格搜索、随机搜索和贝叶斯优化。此外，还分享了训练技巧，包括学习率调度、早停、数据增强和正则化，这些都有助于防止过拟合并提高模型泛化能力。结合这些方法，可构建更高效、健壮的深度学习模型。
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## What are 6 Hz waves?
Hippocampal theta waves, with a frequency range of 6–10 Hz, appear when a rat is engaged in active motor behavior such as walking or exploratory sniffing, and also during REM sleep. Theta waves with a lower frequency range, usually around 6–7 Hz, are sometimes observed when a rat is motionless but alert.
## What good is 6 Hz?
Interestingly, if 6-Hz binaural beat can enhance theta activity due to the frequency following effect and if such enhanced activity shows a similar pattern to a meditative state that can be induced within a short duration, binaural beats may have clear applications for meditation.
What is hertz frequency?
Frequency is the rate at which current changes direction per second. It is measured in hertz (Hz), an international unit of measure where 1 hertz is equal to 1 cycle per second. Hertz (Hz) = One hertz is equal to one cycle per second. Period = The time required to produce one complete cycle of a waveform.
### Why are there 6 different frequencies in music?
If you can grasp where you will find these following six frequencies then you can definitely make your life easier, and your production faster. Instruments and sounds that are dominant in the lower frequencies can have a tendency to dominate them a little too much.
### Is it safe to listen to 40 Hz tone?
There is some early-stage scientific evidence that listening to a 40 Hz tone can reverse some of the molecular changes in the brains of Alzheimer’s patients. This is one of these things that sound too good to be true, but early results are very promising.
What are the frequencies of the solfeggio vibration?
Each Solfeggio tone is comprised of a frequency required to balance your energy and keep your body, mind and spirit in perfect harmony. The main six Solfeggio frequencies are: 396 Hz – Liberating Guilt and Fear. 417 Hz – Undoing Situations and Facilitating Change. 528 Hz – Transformation and Miracles (DNA Repair)
#### How to identify the different types of frequencies?
6 Different Frequencies and How to Spot Them. Frequency 1 – Thickness/Muddiness. Instruments and sounds that are dominant in the lower frequencies can have a tendency to dominate them a little too Frequency 2 – Boxiness. Frequency 3 – The Cheap Sound. Frequency 4 – Nasal Sound. Frequency 5 –
What are 6 Hz waves? Hippocampal theta waves, with a frequency range of 6–10 Hz, appear when a rat is engaged in active motor behavior such as walking or exploratory sniffing, and also during REM sleep. Theta waves with a lower frequency range, usually around 6–7 Hz, are sometimes observed when a rat is motionless…
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Electronics-Lab.com Community
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Found 1 result
1. Pyroelectric detector: the complete mathematiclal description
Hello. I am Alexander Bondarenko, the author of a book "Mathematical modeling of a pyroelectric detector". As the pyroelectric detector being an electronic component and this forum is more related to theory articles, I decided to upload the book here. Further, I developed a simulator for the detector provided with different settings. You can see the videos here Is there anyone who is interested in advanced understanding of the thermal-to-electrical model for a pyroelectric detector? Thank you. P.S. I am on linkedin https://www.linkedin.com/in/pyrodetector/ Abstract The book offers a step-by-step guide to mathematical modeling of the thermalto- electrical model of a pyroelectric detector. It contains the solutions to ten problems. The first eight problems are related to processes running in the body of the sensitive element from heating or cooling to generating electrical charges in response. The last two problems examine the transformation of input electrical charges to output voltage when the sensitive element is connected to high-megohm electronics. Every solution starts with the equation for the law of conservation of energy and ends with that of the transient response. In order to make reading easier, the author provides almost every equation with corresponding units of measurement which are extremely useful not only for beginners, but also for advanced readers. The book can be recommended to amateurs, undergraduate and graduate students, teachers, engineers, who want to develop advanced knowledge better concerning to the thermal-to-electrical model of pyroelectric detectors. Mathematical Modeling of a Pyroelectric Detector Release.pdf
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https://air.unipr.it/handle/11381/2866314
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Purpose – The purpose of this paper is to present and analyze the iterative rules determining the impulsive behavior of a rigid disk having a single or possibly multiplefrictionless impact with two walls forming a corner. Design/methodology/approach – In the first part, two theoretical iterative rules are presented for the cases of ideal impact and Newtonian frictionless impact with global dissipation index. In the second part, a numerical version of both the theoretical algorithms is presented. Findings – The termination analysis of the algorithms differentiates the two cases: in the ideal case, it is shown that the algorithm always terminates and the disk exits from the corner after a finite number of steps independently of the initial impact velocity of the disk and the angle formed by the walls; in the non-idealcase, although is not proved that the disk exits from the corner in a finite number of steps, it is shown that its velocity decreases to zero, so that the termination of the algorithm can be fixed through an “almost at rest” condition. It is shown that the stable version of the algorithm is more robust than the theoretical ones with respect to noisy initial data and floating point arithmetic computation. The outputs of the stable and theoretical versions of the algorithms are compared, showing that they are similar, even if not coincident, outputs. Moreover, the outputs of the stable version of the algorithm in some meaningful cases are graphically presented and discussed. Originality/value – The paper clarifies the applicability of theoretical methods presented in Pasquero (2018) by analyzing the paradigmatic case of the disk in the corner.
An algorithmic approach to the multiple impact of a disk in a corner / Fassino, Claudia; Pasquero, Stefano. - In: MULTIDISCIPLINE MODELING IN MATERIALS AND STRUCTURES. - ISSN 1573-6105. - (2019). [10.1108/MMMS-05-2019-0096]
### An algorithmic approach to the multiple impact of a disk in a corner.
#### Abstract
Purpose – The purpose of this paper is to present and analyze the iterative rules determining the impulsive behavior of a rigid disk having a single or possibly multiplefrictionless impact with two walls forming a corner. Design/methodology/approach – In the first part, two theoretical iterative rules are presented for the cases of ideal impact and Newtonian frictionless impact with global dissipation index. In the second part, a numerical version of both the theoretical algorithms is presented. Findings – The termination analysis of the algorithms differentiates the two cases: in the ideal case, it is shown that the algorithm always terminates and the disk exits from the corner after a finite number of steps independently of the initial impact velocity of the disk and the angle formed by the walls; in the non-idealcase, although is not proved that the disk exits from the corner in a finite number of steps, it is shown that its velocity decreases to zero, so that the termination of the algorithm can be fixed through an “almost at rest” condition. It is shown that the stable version of the algorithm is more robust than the theoretical ones with respect to noisy initial data and floating point arithmetic computation. The outputs of the stable and theoretical versions of the algorithms are compared, showing that they are similar, even if not coincident, outputs. Moreover, the outputs of the stable version of the algorithm in some meaningful cases are graphically presented and discussed. Originality/value – The paper clarifies the applicability of theoretical methods presented in Pasquero (2018) by analyzing the paradigmatic case of the disk in the corner.
##### Scheda breve Scheda completa Scheda completa (DC)
2019
An algorithmic approach to the multiple impact of a disk in a corner / Fassino, Claudia; Pasquero, Stefano. - In: MULTIDISCIPLINE MODELING IN MATERIALS AND STRUCTURES. - ISSN 1573-6105. - (2019). [10.1108/MMMS-05-2019-0096]
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Utilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/11381/2866314`
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https://www.beatthegmat.com/although-the-fear-last-year-that-veritas-tricky-sc-t296550.html
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## Although the fear last year that .......Veritas tricky SC
This topic has 1 expert reply and 0 member replies
### Top Member
Mo2men Legendary Member
Joined
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Posted:
579 messages
Followed by:
5 members
14
#### Although the fear last year that .......Veritas tricky SC
Thu Sep 14, 2017 5:02 am
Although the fear last year that the trade zone might break apart had receded, the risk now could be prolonged stagnation of the kind that has plagued Argentina for the past two decades.
A.had receded, the risk now could be prolonged stagnation of the kind that has plagued Argentina for the past two decades
B.had receded, the risk now could be prolonged stagnation as it has plagued Argentina for the past two decades
C.receded, the risk now could be prolonged stagnation, just as it has plagued Argentina for the past two decades
D.has receded, the risk now could be prolonged stagnation, like it has plagued Argentina for the past two decades
E.has receded, the risk now could be prolonged stagnation, like that which has plagued Argentina for the past two decades
Source: Veritas
OA: E
Dear Mitch,
1- Why C is wrong? is the construction 'just as' is wrong? Is the past tense verb correct?
2- In OA, is the construction ' that which' is right? Does not GMAT consider 'which' a non-essential modifier can hence need to be preceded with 'comma'?
Thanks
### GMAT/MBA Expert
GMATGuruNY GMAT Instructor
Joined
25 May 2010
Posted:
14182 messages
Followed by:
1820 members
13060
GMAT Score:
790
Fri Sep 15, 2017 3:14 am
Mo2men wrote:
Although the fear last year that the trade zone might break apart had receded, the risk now could be prolonged stagnation of the kind that has plagued Argentina for the past two decades.
A.had receded, the risk now could be prolonged stagnation of the kind that has plagued Argentina for the past two decades
B.had receded, the risk now could be prolonged stagnation as it has plagued Argentina for the past two decades
C.receded, the risk now could be prolonged stagnation, just as it has plagued Argentina for the past two decades
D.has receded, the risk now could be prolonged stagnation, like it has plagued Argentina for the past two decades
E.has receded, the risk now could be prolonged stagnation, like that which has plagued Argentina for the past two decades
Source: Veritas
OA: E
Dear Mitch,
1- Why C is wrong? is the construction 'just as' is wrong?
Generally, as serves to compare VERBS.
C: The risk now could be prolonged stagnation, just as it has plagued Argentina for the past two decades.
Here, as serves to compare the two verbs in red.
It is illogical to compare a STATE-OF-BEING (could be) to an ACTION (has plagued).
Eliminate C.
Quote:
Is the past tense verb correct?
C: The fear last year that the trade zone might break apart receded.
Here, it is illogical to attribute a PAST ACTION -- receded -- to a CURRENT FEAR (that the trade zone might break apart).
Eliminate C.
Quote:
2- In OA, is the construction ' that which' is right? Does not GMAT consider 'which' a non-essential modifier can hence need to be preceded with 'comma'?
The OA conveys the following meaning:
The risk now could be prolonged stagnation, like the prolonged stagnation that has plagued Argentina for the past two decades.
To avoid repetition, the phrase in blue is replaced with the copy pronoun that:
The risk now could be prolonged stagnation, like that that has plagued Argentina for the past two decades.
The result is the confusing construction in blue and red: that that.
When a comparison results in that that, we replace the second that with which:
The risk now could be prolonged stagnation, like that which has plagued Argentina for the past two decades.
The construction in green is correct.
While that which is a valid construction, I cannot cite an OA that has employed this construction.
_________________
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H14- Thermo 3-solutions
# H14- Thermo 3-solutions - patel (ap28872) – H14: Thermo 3...
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Unformatted text preview: patel (ap28872) – H14: Thermo 3 – mccord – (51600) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is your LAST homework on Quest for CH301. 001 10.0 points Consider the following processes. (Treat all gases as ideal.) I) The pressure of one mole of oxygen gas is allowed to double isothermally. II) Carbon dioxide is allowed to expand isothermally to 10 times its original vol- ume. III) The temperature of one mole of helium is increased 25 ◦ C at constant pressure. IV) Nitrogen gas is compressed isothermally to one half its original volume. V) A glass of water loses 100 J of energy reversibly at 30 ◦ C. Which of these processes leads to an increase in entropy? 1. I and IV 2. I and II 3. II and III correct 4. III and V 5. V Explanation: R = 8 . 314 J · mol − 1 · K − 1 Assume 1 mol in each case. Entropy de- creases if Δ S is negative. For the oxygen gas pressure doubling isothermally, P 2 = 2 P 1 and Δ S = nR ln parenleftbigg P 1 P 2 parenrightbigg = (1 mol)(8 . 314 J · mol − 1 · K − 1 )ln parenleftbigg 1 2 parenrightbigg =- 5 . 76 J · K − 1 We expect a negative answer since pressure increased. For the CO 2 gas volume expanding 10 × isothermally, V 2 = 10 V 1 and Δ S = nR ln parenleftbigg V 2 V 1 parenrightbigg = (1 . 00 mol)(8 . 314 J · mol − 1 · K − 1 ) × ln(10) = +38 . 29 J · K − 1 We expect a positive answer since volume increased. For the nitrogen gas compressed to 1 2 origi- nal volume isothermally, V 1 = 2 V 2 and Δ S = nR ln parenleftbigg V 2 V 1 parenrightbigg = (1 mol)(8 . 314 J · mol − 1 · K − 1 ) ln parenleftbigg 1 2 parenrightbigg =- 5 . 76 J · K − 1 We expect a negative answer since volume decreased. For the cooling glass of water, T = 30 ◦ C + 273 . 15 = 303 . 15 K Δ S = q T =- 200 J 303 . 15 K =- . 6597 J · K − 1 The last situation (heating the 1 mol of He) does not give enough data to calculate an answer but from the formula Δ S = nC p , m ln parenleftbigg T 2 T 1 parenrightbigg n = 1 mol and for a monoatomic ideal gas C p , m = 2 . 5 R . Finally if the temperature increases this means T 2 > T 1 so ln parenleftbigg T 2 T 1 parenrightbigg will be positive. We expect a positive answer since tempera- ture increased. 002 10.0 points For the four chemical reactions I) 3 O 2 (g) → 2 O 3 (g) II) 2 H 2 O(g) → 2 H 2 (g) + O 2 (g) III) H 2 O(g) → H 2 O( ℓ ) IV) 2 H 2 O( ℓ ) + O 2 (g) → 2 H 2 O 2 ( ℓ ) patel (ap28872) – H14: Thermo 3 – mccord – (51600) 2 which one(s) is/are likely to exhibit a positive Δ S ? 1. II only correct 2. All have a positive Δ S . 3. I and II only 4. III and IV only 5. I, III and IV only Explanation: The Third Law of Thermodynamics states that the entropy of a perfect pure crystal at 0 K is 0. As disorder, randomness, and de- grees of freedom increase, so does S . Entropy can increase by changing phase from solid to liquid to gas, and by increasing temperature,...
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## This note was uploaded on 05/12/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas.
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# Would animals from Earth be able to survive and reproduce on this planet?
Some details about the planet and its moon plus the host star.
Star: 0.85 Sol, 0.54% of the Sun's luminosity
Planet: 0.04926 (M⊕) 4836 km, 0.34 G, Orbits the host star at 0.901 AU.
The atmospheric pressure on this planet is 0.20% of the Earth's which consists of:
85.4% O2, 9.16% CO2, 3.89% N2, 1.2% SO2, H20: 332 ppm.
Moon: 0.002 (M⊕) 770 km, 0.12 G, Semi major axis of the moon is 26415 km. Orbital eccentricity of the moon is 0.031.
• I doubt that planet would have an atmosphere that is possible to breathe... mars is 0.107 M⊕, and we can't even use parachutes.... Maybe water bears?
– user23110
Sep 20, 2016 at 14:25
• Is that a typo for your atmospheric pressure? If not, i don't think any thing (apart from maybe bacteria) could breathe there Sep 20, 2016 at 14:29
• @Burki Well, 0.049 Earth masses and 0.20% surface atmospheric pressure does sound about right to a first order approximation.
– user
Sep 20, 2016 at 14:32
• Stephanie, while I can appreciate that you probably made an honest mistake in this case, it really isn't considered good form to make such a drastic change as you did (0.20% of Earth atmospheric pressure to 20%) after answers have been posted. At that point, if you realize you've made a mistake like that, it would be better to post an entirely new question because the answer is going to be completely different. I won't roll back the edit unilaterally right away (I could), but I really encourage you to roll back your edit, and then post a new question with the corrected premise.
– user
Sep 20, 2016 at 14:50
• Please read this meta question - it shows how frustrating are changes for people who answer. Sep 21, 2016 at 8:31
Atmospheric pressure 0.20% of that on Earth's surface?
Take a look at the phase diagram for water:
0.20% of 1 atmosphere pressure means about 2 millibar. At that pressure, liquid water cannot exist; at about -20 degrees C, it sublimates from ice to vapour. At that pressure, below -20C, water is a solid; above -20C, it is a vapour.
No higher life on Earth is equipped to deal with such conditions.
It is possible that some extremophiles may be able to survive by hibernating, but I very much doubt that they will be able to reproduce under such conditions.
So the answer is no, animals from Earth would not be able to survive and reproduce under such conditions.
And that's before we even consider the composition of the atmosphere.
In order to be able to have the three phases of water that we are used to (solid, liquid and vapour) the temperature and pressure have to be between the triple point and the critical point, so somewhere in the range 611.657 Pa (at no less than 273.16 K) to 22.064 MPa (at no more than 647 K).
• @tuskiomi The tardigrade is a member of the group extremophiles. And I still doubt they would be able to reproduce, so even they fail at least one of the OP's two criteria.
– user
Sep 20, 2016 at 14:32
Well. Obviously, what is being asked is about a planet with 20% of Earth's atmospheric pressure, and its sun with 54% of Sol's luminosity. A question about a planet with 0.2% of Earth's atmospheric pressure, orbiting a sun with 0.54% of Sol's luminosity wouldn't be even minimally interesting, as the obvious answer would be, NO, too little air, too little light.
I think that your atmospheric pressure is too low for most forms of Earthan animal life; the atmospheric pressure at Mount Everest is about 30% that of sea level, and it is irrespirable for people. At least, it would be certainly lethal without previous adaptation.
But the biggest problem, that you would have to handwave, or to create an elaborate explanation for, is that a planet with so much oxygen must necessarily already have some kind of life - of photosynthetic life, to be precise. This means that a huge problem for xenoorganisms there - supposing they could physically survive - would be their interaction with the local ecological system. Plants could be toxic, microorganisms could be pathogenic.
Regardless of the incorrect percentages in the question, there's not enough nitrogen. There's less than 1% as much as in the Earth's atmosphere, so there's going to be nothing like enough available for plants. Earth animals might survive, but only if you import all their food from off-world.
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# The Care and Feeding of Interesting Logic Puzzles
by Paul Hlebowitsh on 09/19/16 09:51:00 am
The following blog post, unless otherwise noted, was written by a member of Gamasutra’s community.
The thoughts and opinions expressed are those of the writer and not Gamasutra or its parent company.
In this blog post, I will go through in detail how I constructed one of the early puzzles in my logic puzzle game RYB.
Just for a bit of background: I was a top-10 finisher in the 2016 US Puzzle Championship. I've been constructing puzzles for about 10 years and RYB is my first game.
The puzzle we're going to look at in detail is this one:
Before we go into its design, let me tell you about RYB's mechanics so you have an idea of the constraints we're working with.
RYB is very similar to "Minesweeper" or "Hexcells", but instead of using numbers or symbols, it uses colors. The colored dots inside of a shape tell you how the neighboring shapes are colored. For instance, consider the example puzzle below:
The center triangle has three red dots and has three neighboring shapes. A dot tells you that at least one neighboring shape is that color, so the three red dots and three neighboring shapes mean they're all colored red. Filling those in we have:
We have new information! Yellow dots have appeared. The three outer triangles each have one neighbor, namely the center piece, and each outer triangles has a yellow dot, so the middle piece must be colored yellow to satisfy all three of these clues.
And we're done! Once you've colored all the shapes, the puzzle is complete.
In order to make an interesting puzzle from this mechanic, we need to have a shape which lends itself to a good puzzle. When I started to make the puzzle shown at the top of this post, I started with a central hexagon, because I wanted to play with many interacting shapes, perhaps broken into cycles of shapes that clue each other. At this point in the construction, I had little idea as to what exactly the puzzle would look like, I only knew that I wanted a hexagon at the center. I do all my puzzle design on paper with pencil, but for the purposes of this article I'll include vector art so you don't have to worry about my terrible drawing skills!
The central hexagon looked like:
Pretty simple! Not much more you can do than that. After drawing the hexagon, I put some shapes around it. Triangles were the first ones that came to mind.
At this point in the construction, it was time to start playing around.
Puzzle construction is a conversation between designer and puzzle. In the course of designing a puzzle, the designer must listen to what the puzzle is trying to tell it. Even though we have no clues and only a few shapes, this puzzle is already saying a lot to us. Consider any two of the triangles, such as the ones circled in the picture below.
The only way these two pieces can interact is through the large hexagon. In fact, any two triangles can only interact through the large hexagon. In other words, the hexagon will have to constrain the colors of all six triangles! There is no way to make this into an interesting puzzle.
There are two options to fix this problem which stand out immediately.
1. Cut the center piece into smaller shapes, so there are more pieces that constrain the outside.
2. Add pieces to provide more constraints to the outside triangles.
I decided to go with option 2 for this particular puzzle. Later in the game, with different mechanics, I actually used option 1, and it looked like this:
But I don't want to get too far off topic!
In this early puzzle, I pursued option 2.
Because I needed to add pieces, I added them in the obvious space, namely the negative space between every two triangles.
Adding these pieces gave me another constraint on the outer triangles. I played with this puzzle for a long time, trying to figure out how to make this shape into an interesting puzzle. However, it has an odd problem. Namely, the high degree of symmetry in the shapes makes it hard to make interesting logic that doesn't solve the puzzle all at once. The puzzle has many axes on which is looks the same, three of which I've marked in the picture below. It's hard to make the axes pictured separate without breaking the symmetry of the puzzle.
To solve this problem, I decided to try to constrain two pieces on either side of the puzzle. They're along one of the symmetry axes, so they could conceivably be symmetrically constrained. After playing with that idea for a while, I hit upon the idea of adding an outer hexagon so they had a piece in common, giving them a tighter constraint.
After fiddling with this for a while, I realized that if I gave one piece only red and blue clues, and one piece only yellow and blue clues, the outer hexagon had to be colored with the only color they had in common, namely blue. This is the main idea behind the puzzle.
The main idea was only arrived at through very general reasoning, listening to what the puzzle was trying to tell us, and tweaking the shapes until something interesting presented itself. At this point, I decided that I wanted to constrain all the inner triangles, so I added clues that would constrain those pieces.
Now I had some more clues to play with! Any of those pieces that had their color revealed could be places where I could place new clues. When people solved this puzzle, this would be their first step. They would get the outer hexagon, the inner triangles, and then have more clues to work with.
Symmetry is important to puzzles. Current logic puzzle construction trends feature symmetry as a "beautiful construction". In this puzzle, I wanted the solver to use the same piece of logic again, as a sense of symmetry inside the actual solve structure. If you notice, we're in a similar position to before. There's a set of diagonally opposite clue spots that have one piece in common, only on the inside rather than the outside.
I added clues so that the same logic would happen again! This time, though, they would be coloring the inside hexagon. (To be clear, the clues on the colored pieces are revealed only after they are colored.) The puzzle looks like:
Which solves to:
Finally, I had to constrain the last two shapes. So I added clues to the outer hexagon that would act similarly to the way the inner hexagon helped in the first clue set.
The solve path looked like:
At this point I felt as though I had a good puzzle. It had an "aha" about the outer hexagon that was used in two different ways in the puzzle, while providing a few easy fill-ins in the intervening stages that would help build confidence of the solver.
I hope you enjoyed this in-depth look at logic puzzle construction!
If you enjoyed this write up, please consider supporting RYB on Steam Greenlight at https://steamcommunity.com/sharedfiles/filedetails/?id=745756756, or following me @FLEBpuzzles for more puzzle content.
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# Introduction to programming and computer science
Updated On 02 Feb, 19
##### Overview
Introduction to Programs Data Types and Variables - Python Lists - For Loops in Python - While Loops in Python - Fun with Strings - Writing a Simple Factorial Program. (Python 2) - Stepping Through the Factorial Program - Flowchart for the Factorial Program - Python 3 Not Backwards Compatible with Python 2 - Defining a Factorial Function - Diagramming What Happens with a Function Call - Recursive Factorial Function - Comparing Iterative and Recursive Factorial Functions - Exercise - Write a Fibonacci Function - Iterative Fibonacci Function Example - Stepping Through Iterative Fibonacci Function - Recursive Fibonacci Example - Stepping Through Recursive Fibonacci Function - Exercise - Write a Sorting Function - Insertion Sort Algorithm - Insertion Sort in Python - Stepping Through Insertion Sort Function - Simpler Insertion Sort Function
## Lecture 15: Iterative Fibonacci Function Example
4.1 ( 11 )
###### Lecture Details
One way to write a Fibonacci function iteratively
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## Index value formula in excel
The INDEX function is categorized under Excel Lookup and Reference functions. The function will return the value at a given position in a range or array. The INDEX function is often used with the MATCH function. We can say it is an To perform advanced lookups, you'll need INDEX and MATCH. Match. The MATCH function returns the position of a value in a given range. For example, the MATCH function below looks up the value 53 23 Jan 2020 The 365 subscription version of Excel returns all values without needing to enter the formulas an array formula. [column_num], Optional. The relative column number of a specific value you want to get. If omitted the INDEX
The CELL function uses the return value of INDEX as a cell reference. On the other hand, a formula such as 2*INDEX(A1:B2,1,2) translates the return value of INDEX into the number in cell B1. Examples. Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. INDEX – get value at known position. The INDEX function in Excel is fantastically flexible and powerful, and you'll find it in a huge number of Excel formulas, especially advanced formulas. But what does INDEX actually do? In a nutshell, INDEX retrieves values at a given location in a list or table. The INDEX function is categorized under Excel Lookup and Reference functions Functions List of the most important Excel functions for financial analysts. This cheat sheet covers 100s of functions that are critical to know as an Excel analyst. The function will return the value at a given position in a range or array. The INDEX function is often used with the MATCH function. Not all array formulas return arrays with multiple columns and/or multiple rows to the worksheet. But when they do, it can happen that you’re interested in seeing only one value in the array. You can use Excel’s INDEX function to help with that. For example, LINEST is one of the worksheet functions that will work […] The Microsoft Excel INDEX function returns a value in a table based on the intersection of a row and column position within that table. The first row in the table is row 1 and the first column in the table is column 1. The INDEX function is a built-in function in Excel that is categorized as a Lookup/Reference Function.
## These examples use the INDEX function to find the value in the intersecting cell where a row and a column meet. Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select
Excel Formula Training. Formulas are the key to getting things done in Excel. In this accelerated training, you'll learn how to use formulas to manipulate text, work with dates and times, lookup values with VLOOKUP and INDEX & MATCH, count and sum with criteria, dynamically rank values, and create dynamic ranges. Excel Formula Training. Formulas are the key to getting things done in Excel. In this accelerated training, you'll learn how to use formulas to manipulate text, work with dates and times, lookup values with VLOOKUP and INDEX & MATCH, count and sum with criteria, dynamically rank values, and create dynamic ranges. Here L2 is 1, L3 is 2 and L4 is 1. Hence INDEX function will return value from 1st row of second column from 1st array. And that is East. Now change L2 to 2 and L4 to 2. You will have West in M2, as shown in below image. And so on. The INDEX function in Excel is mostly used with MATCH Function. The INDEX function is categorized under Excel Lookup and Reference functions Functions List of the most important Excel functions for financial analysts. This cheat sheet covers 100s of functions that are critical to know as an Excel analyst. The function will return the value at a given position in a range or array. The INDEX function is often used with the MATCH function. INDEX Formula in Excel Step 1: Apply the INDEX and select the result range. Step 2: Instead of entering the row number manually open MATCH function . Step 3: Select the lookup value as Year i.e. G2 cell. Step 4: Lookup array as A2 to A10 range. Step 5: Match type is the FALSE i.e. exact match. If you array-enter that formula in a 5 row by 4 column range, the intersection of that range’s third row and first column contains the regression’s R-squared value. So if you select just a single cell and enter the following formula, you’ll get the R-squared value only: =INDEX(LINEST(A2:A51,B2:D51,,TRUE),3,1) So, what is the INDEX function in Excel? Essentially, an INDEX formula returns a cell reference from within a given array or range. In other words, you use INDEX when you know (or can calculate) the position of an element in a range and you want to get the actual value of that element.
### 8 Oct 2018 VLOOKUP is a great function, but it has its limitations. It can only look up values from left to right. The lookup value must be on the left in the lookup table. INDEX and MATCH allows you to look up a value anywhere in the
INDEX – get value at known position. The INDEX function in Excel is fantastically flexible and powerful, and you'll find it in a huge number of Excel formulas, especially advanced formulas. But what does INDEX actually do? In a nutshell, INDEX retrieves values at a given location in a list or table. The INDEX function is categorized under Excel Lookup and Reference functions Functions List of the most important Excel functions for financial analysts. This cheat sheet covers 100s of functions that are critical to know as an Excel analyst. The function will return the value at a given position in a range or array. The INDEX function is often used with the MATCH function. Not all array formulas return arrays with multiple columns and/or multiple rows to the worksheet. But when they do, it can happen that you’re interested in seeing only one value in the array. You can use Excel’s INDEX function to help with that. For example, LINEST is one of the worksheet functions that will work […] The Microsoft Excel INDEX function returns a value in a table based on the intersection of a row and column position within that table. The first row in the table is row 1 and the first column in the table is column 1. The INDEX function is a built-in function in Excel that is categorized as a Lookup/Reference Function.
### INDEX – get value at known position. The INDEX function in Excel is fantastically flexible and powerful, and you'll find it in a huge number of Excel formulas, especially advanced formulas. But what does INDEX actually do? In a nutshell, INDEX retrieves values at a given location in a list or table.
30 Jan 2020 This formula works because it returns the largest row number from column B, and then uses that as an index to return the corresponding value from column A. If your range of data contains a mixture of numeric and non-numeric OFFSET : Returns a range reference shifted a specified number of rows and columns from a starting cell reference. Notes. If you set row or column to 0, INDEX returns the array of values for the entire column or row, respectively INDEX returns the value in the array (cell) at the intersection of row_num and column_num. My 2-Way lookup formula: =INDEX(DataRange, MATCH( TheProduct,Products,0) VLOOKUP with 2 criteria or more by using the INDEX and MATCH functions in Excel. The step-by-step tutorial will show you how to build the formula and learn how it works! 15 Oct 2019 Combined, the two formulas can look up and return the value of a cell in a table based on vertical and horizontal criteria. For short, this is referred to as just the Index Match function. This is especially useful in Clio, as it allows Next, INDEX(result_range,3) returns the 3rd value in the price list range. The INDEX-MATCH formula is an example of a simple nested function where we use the result from the The INDEX Function Details. The INDEX function returns a cell value from a range, given a row and/or column position number. The syntax is: INDEX(array, row_num, [column_num]). Array is required and is a range of cells or an array
## Otherwise, the formula must be entered as a legacy array formula by first selecting the output cell, entering the formula in the output cell, and then pressing CTRL+SHIFT+ENTER to confirm it. Excel inserts curly brackets at the beginning and end of the formula for you.
With the INDEX function, you can retrieve a certain value from a column. The certain value is often determined by a MATCH. The format is INDEX (\$G\$2: \$G\$5 ; X) in which G2: G5 is the row from which you need the Xth (the MATCH) value. 12 Feb 2019 The Excel INDEX function returns a value or the reference of a value at a given position in a range or array. The INDEX function has two different forms to return a value or a reference. Using an array form returns a value, while 21 Feb 2013 INDEX Function in Excel - formula returns either the value or the reference to a value from a table or range. Learn Formulas, Excel and VBA with examples.
INDEX – get value at known position. The INDEX function in Excel is fantastically flexible and powerful, and you'll find it in a huge number of Excel formulas, especially advanced formulas These examples use the INDEX function to find the value in the intersecting cell where a row and a column meet. Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select The Microsoft Excel INDEX function returns a value in a table based on the intersection of a row and column position within that table. The first row in the table is row 1 and the first column in the table is column 1. The INDEX function is a built-in Combined, the two formulas can look up and return the value of a cell in a table based on vertical and horizontal criteria. For short, this is referred to as just the Index Match function. To see a video tutorial, check out our free Excel Crash The INDEX function is categorized under Excel Lookup and Reference functions. The function will return the value at a given position in a range or array. The INDEX function is often used with the MATCH function. We can say it is an To perform advanced lookups, you'll need INDEX and MATCH. Match. The MATCH function returns the position of a value in a given range. For example, the MATCH function below looks up the value 53
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Electronics-Lab.com Community
# LED power
## Recommended Posts
Heres what I have:
12VDC Supply Voltage
3VDC Voltage Drop
20ma LED Current
I used to think that the Total LED Voltage Drop should equal the Supply Voltage. But after doing some research I found that the Total Voltage Drop should be less than 80% of the Supply Voltage.
So the correct circuit would look like this:
3 LED's in series with a 150ohm resistor and a 12VDC Supply Voltage.
My questions are this: does the resistor actually lower the voltage? and what would be the problem if 4 leds with the same Voltage Drop were used?
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LEDs operate from current. If you connect an LED to a voltage supply then it will burn out because the current will be unlimited.
The resistor in series with the LED or in series with a few LEDs in series limits the current according to Ohm's Law.
Most ordinary small LEDs have a max allowed current of 30mA. If an LED has a voltage drop of 3.0V (it is a range of voltages, it could be 2.5V or 3.5V) and a 12V supply is used then for 25mA the resistor is calculated as (12V-3.0V)/25mA= 360 ohms. 25mA was selected so that if the LED voltage is 2.5V and the 360 ohms resistor is actually 5% low then the current won't be high enough (more than 30mA) to burn out the LED.
If you connect four 3.0V LEDs in series and connect them to a 12V battery without a current liniting reasistor then they will immediately burn out, seem to operate fine or they won't light.
If they are actually 2.5V then the current will be unlimited.
If they are exactly 3.00V then they will work until the battery voltage runs down a little.
If they are actually 3.5V then they need at least 3.5V x 4= 14V to light.
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Okay, so If I use a 12VDC power supply with a 450 Ohm resistor (using 20mA) in series with one 3V LED everything will be fine then?
(12V-3V)/20mA = 450 ohms (or the next highest standard resistor)
See, for some reason, I have always thought that the 3V was all the LED could handle. But now its clear that current is the main thing to worry with. Voltage is important when considering how many LEDs in series you want to have. For instance if I wanted to max the amount of LEDs I could have in series on a 12V power supply I would take 80% to be the max amount of volts I should spent LED wise.
12V * .8 = 9.6V
With 3V LEDs only three could be in series. Is this correct?
(I used this as a reference: http://www.theledlight.com/LED101.html)
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242review-part1
# 242review-part1 - Math 242 Review Part 1 In this course we...
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Math 242 Review — Part 1 In this course, we covered most of the following four chapters. Chapter 6 (Applications of Integration) Chapter 7 (Techniques of Integration) Chapter 11 (Infinite Series) Chapter 10 (Parametric and Polar Curves) There were three additional topics: L’Hopital’s Rule (4.4), Newton’s Method (4.8), and Arc Length (8.1). Chapters 6 and 7 We skipped 6.4. Chapter 6 is almost exclusively about areas and volumes . Chapter 7 contains many techniques of integration. While reviewing areas and volumes, it may make sense to include some area and volume problems where you have to use techniques from Chapter 7. In other words, Chapter 6 and Chapter 7 are interrelated. In area problems and volume problems, frequently y is a function of x , but sometimes x is a function of y . If we write y as a function of x , then we have a “top curve” and a “bottom curve” we allow x to increase by a small amount dx a “typical slice” is a tall skinny rectangle whose width is dx and whose height is “top minus bottom”. If we write x as a function of y , then we have a “left curve” and a “right curve” we allow y to increase by a small amount dy a “typical slice” is a short wide rectangle whose height is dy and whose width is “right minus left”. 1
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Areas are easier than volumes. Basically, area is just Z (area of slice), which is either Z (top - bottom) dx or Z (right - left) dy . For volumes , we have essentially two methods: washers and shells . (Disks are just washers whose inner radius happens to be zero.) This is how I remember the washer method and shell method: Volume by washers: Z ( πR 2 - πr 2 ) dx or Z ( πR 2 - πr 2 )
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# Homework Help: Solving an ODE: The Pwer Series and Seperation of Variables
1. Nov 30, 2008
### TFM
1. The problem statement, all variables and given/known data
Solve the following equation by a power series and also by separation of variables. Check that the two agree.
2. Relevant equations
N/A
3. The attempt at a solution
Power Series:
$$(1+x) \frac{dy}{dx} = y$$
$$(1+x) \frac{1}{dx} = y \frac{1}{dy}$$
The power series is:
$$(1+x) \equiv 1+x+0x^2+0x^x$$ ...
Thus
$$\frac{1 + x}{dx} = \frac{y}{dy}$$
Separation by Variables:
$$(1+x) y' = y$$
$$y' = \frac{1}{((1+x)} y$$
$$x'=g(t)h(x)$$
$$H(x) = G(t) + C$$
$$H=\int \frac{dx}{h(x)} ; G=\int g(t)dt$$
$$H=\int \frac{dy}{y} \equiv \int \frac{1}{y} dy = ln y$$
$$G = \int \frac{1}{1+x} dx = ln(1 + x)$$
$$ln y = ln (1+x) + c$$
$$y = 1 + x + c$$
These two methods haven't agreed for this question. i think the problem lays in my Power Series.
Anyone got any idesa?
TFM
2. Nov 30, 2008
### CompuChip
Your second solution looks right, up to
ln(y) = ln(1 + x) + c
But if you exponentiate that, you won't get
y = 1 + x + c,
do that step again.
As for your real question, I suppose that they mean: plug in a solution
$$y(x) = \sum_{n = 0}^\infty a_n x^n$$
and determine the coefficients $a_n$ from the differential equation.
Note that you can differentiate by terms, so
$$\frac{dy}{dx} = \sum_{n = 0}^\infty n a_n x^{n - 1}$$
etc.
3. Nov 30, 2008
### TFM
Indeed it won't:
ln(y) = ln(1 + x) + c
take exponentials of both sides leaves:
$$y = 1 + x + e^c$$
Okay so for the power series:
$$y(x) = \sum_{n = 0}^\infty a_n x^n$$
so would the coeffieients coming from here:
$$(1+x) \frac{1}{dx} = y \frac{1}{dy}$$
be 1 and 1, since you have x^0 has a coefficient of 1, and x has a coefficent of 1 also?
TFM
4. Nov 30, 2008
### CompuChip
No,
e^(ln(1 + x) + c) = e^ln(1+x) * e^c = ...
And what do you mean by: "the coefficients being 1 and 1"? You have infinitely many of them! Start by plugging the sum into
$$(1+x) \frac{dy}{dx} = y$$
- what do you get?
5. Nov 30, 2008
### TFM
Since e^c is another constant, wouldn't it not be:
= k + kx
???
So I should have:
$$(1+x) \frac{dy}{dx} = \sum_{n = 0}^\infty a_n x^n$$
???
TFM
6. Nov 30, 2008
### gabbagabbahey
Yes, that would be fine.
Well, that is a true statement, but not very useful to you in that form. If $$y(x)= \sum_{n = 0}^\infty a_n x^n$$...what is $$\frac{dy}{dx}$$?
7. Dec 1, 2008
### TFM
Would this be:
$$y(x)= \sum_{n = 0}^\infty a_n x^n$$
$$\frac{dy}{dx} = \sum_{n = 0}^\infty a_n(n - 1) x^{n-1}$$
Does this look right???
TFM
8. Dec 1, 2008
### CompuChip
Almost; is $2 x^2$ the derivative of $x^3$?
Now write down your complete differential equation.
9. Dec 1, 2008
### TFM
Okay so:
x^3 differentiates to 3x^2
$$\frac{dy}{dx} = \sum_{n = 0}^\infty a_n(n) x^{n-1}$$
so would this be:
$$\frac{dy}{dx} = \sum_{n = 0}^\infty a_n(n) x^{n-1} + a_{n-1}(n+1) x^{n} + a_{n-2}(n+2) x^{n + 1} + ...$$
???
TFM
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# Force
Force
Forces are also described as a push or pull on an object. They can be due to phenomena such as gravity, magnetism, or anything that might cause a mass to accelerate.
Common symbols
F, F
SI unit newton
In SI base units 1 kg·m/s2
Derivations from
other quantities
F = m a
In physics, a force is any interaction that, when unopposed, will change the motion of an object.[1] In other words, a force can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate. Force can also be described by intuitive concepts such as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newtons and represented by the symbol F.
The original form of Newton's second law states that the net force acting upon an object is equal to the rate at which its momentum changes with time. If the mass of the object is constant, this law implies that the acceleration of an object is directly proportional to the net force acting on the object, is in the direction of the net force, and is inversely proportional to the mass of the object
Related concepts to force include: thrust, which increases the velocity of an object; drag, which decreases the velocity of an object; and torque, which produces changes in rotational speed of an object. In an extended body, each part usually applies forces on the adjacent parts; the distribution of such forces through the body is the so-called mechanical stress. Pressure is a simple type of stress. Stress usually causes deformation of solid materials, or flow in fluids.
## Development of the concept
Philosophers in antiquity used the concept of force in the study of stationary and moving objects and simple machines, but thinkers such as Aristotle and Archimedes retained fundamental errors in understanding force. In part this was due to an incomplete understanding of the sometimes non-obvious force of friction, and a consequently inadequate view of the nature of natural motion.[2] A fundamental error was the belief that a force is required to maintain motion, even at a constant velocity. Most of the previous misunderstandings about motion and force were eventually corrected by Galileo Galilei and Sir Isaac Newton. With his mathematical insight, Sir Isaac Newton formulated laws of motion that were not improved-on for nearly three hundred years.[3] By the early 20th century, Einstein developed a theory of relativity that correctly predicted the action of forces on objects with increasing momenta near the speed of light, and also provided insight into the forces produced by gravitation and inertia.
With modern insights into quantum mechanics and technology that can accelerate particles close to the speed of light, particle physics has devised a Standard Model to describe forces between particles smaller than atoms. The Standard Model predicts that exchanged particles called gauge bosons are the fundamental means by which forces are emitted and absorbed. Only four main interactions are known: in order of decreasing strength, they are: strong, electromagnetic, weak, and gravitational.[4]:2–10[5]:79 High-energy particle physics observations made during the 1970s and 1980s confirmed that the weak and electromagnetic forces are expressions of a more fundamental electroweak interaction.[6]
## Pre-Newtonian concepts
Aristotle famously described a force as anything that causes an object to undergo "unnatural motion"
Since antiquity the concept of force has been recognized as integral to the functioning of each of the simple machines. The mechanical advantage given by a simple machine allowed for less force to be used in exchange for that force acting over a greater distance for the same amount of work. Analysis of the characteristics of forces ultimately culminated in the work of Archimedes who was especially famous for formulating a treatment of buoyant forces inherent in fluids.[2]
Aristotle provided a philosophical discussion of the concept of a force as an integral part of Aristotelian cosmology. In Aristotle's view, the terrestrial sphere contained four elements that come to rest at different "natural places" therein. Aristotle believed that motionless objects on Earth, those composed mostly of the elements earth and water, to be in their natural place on the ground and that they will stay that way if left alone. He distinguished between the innate tendency of objects to find their "natural place" (e.g., for heavy bodies to fall), which led to "natural motion", and unnatural or forced motion, which required continued application of a force.[7] This theory, based on the everyday experience of how objects move, such as the constant application of a force needed to keep a cart moving, had conceptual trouble accounting for the behavior of projectiles, such as the flight of arrows. The place where the archer moves the projectile was at the start of the flight, and while the projectile sailed through the air, no discernible efficient cause acts on it. Aristotle was aware of this problem and proposed that the air displaced through the projectile's path carries the projectile to its target. This explanation demands a continuum like air for change of place in general.[8]
Aristotelian physics began facing criticism in Medieval science, first by John Philoponus in the 6th century.
The shortcomings of Aristotelian physics would not be fully corrected until the 17th century work of Galileo Galilei, who was influenced by the late Medieval idea that objects in forced motion carried an innate force of impetus. Galileo constructed an experiment in which stones and cannonballs were both rolled down an incline to disprove the Aristotelian theory of motion early in the 17th century. He showed that the bodies were accelerated by gravity to an extent that was independent of their mass and argued that objects retain their velocity unless acted on by a force, for example friction.[9]
## Newtonian mechanics
Sir Isaac Newton sought to describe the motion of all objects using the concepts of inertia and force, and in doing so he found that they obey certain conservation laws. In 1687, Newton went on to publish his thesis Philosophiæ Naturalis Principia Mathematica.[3][10] In this work Newton set out three laws of motion that to this day are the way forces are described in physics.[10]
### First law
Main article: Newton's first law
Newton's First Law of Motion states that objects continue to move in a state of constant velocity unless acted upon by an external net force or resultant force.[10] This law is an extension of Galileo's insight that constant velocity was associated with a lack of net force (see a more detailed description of this below). Newton proposed that every object with mass has an innate inertia that functions as the fundamental equilibrium "natural state" in place of the Aristotelian idea of the "natural state of rest". That is, the first law contradicts the intuitive Aristotelian belief that a net force is required to keep an object moving with constant velocity. By making rest physically indistinguishable from non-zero constant velocity, Newton's First Law directly connects inertia with the concept of relative velocities. Specifically, in systems where objects are moving with different velocities, it is impossible to determine which object is "in motion" and which object is "at rest". In other words, to phrase matters more technically, the laws of physics are the same in every inertial frame of reference, that is, in all frames related by a Galilean transformation.
For instance, while traveling in a moving vehicle at a constant velocity, the laws of physics do not change from being at rest. A person can throw a ball straight up in the air and catch it as it falls down without worrying about applying a force in the direction the vehicle is moving. This is true even though another person who is observing the moving vehicle pass by also observes the ball follow a curving parabolic path in the same direction as the motion of the vehicle. It is the inertia of the ball associated with its constant velocity in the direction of the vehicle's motion that ensures the ball continues to move forward even as it is thrown up and falls back down. From the perspective of the person in the car, the vehicle and everything inside of it is at rest: It is the outside world that is moving with a constant speed in the opposite direction. Since there is no experiment that can distinguish whether it is the vehicle that is at rest or the outside world that is at rest, the two situations are considered to be physically indistinguishable. Inertia therefore applies equally well to constant velocity motion as it does to rest.
The concept of inertia can be further generalized to explain the tendency of objects to continue in many different forms of constant motion, even those that are not strictly constant velocity. The rotational inertia of planet Earth is what fixes the constancy of the length of a day and the length of a year. Albert Einstein extended the principle of inertia further when he explained that reference frames subject to constant acceleration, such as those free-falling toward a gravitating object, were physically equivalent to inertial reference frames. This is why, for example, astronauts experience weightlessness when in free-fall orbit around the Earth, and why Newton's Laws of Motion are more easily discernible in such environments. If an astronaut places an object with mass in mid-air next to himself, it will remain stationary with respect to the astronaut due to its inertia. This is the same thing that would occur if the astronaut and the object were in intergalactic space with no net force of gravity acting on their shared reference frame. This principle of equivalence was one of the foundational underpinnings for the development of the general theory of relativity.[11]
Though Sir Isaac Newton's most famous equation is
, he actually wrote down a different form for his second law of motion that did not use differential calculus.
### Second law
Main article: Newton's second law
A modern statement of Newton's Second Law is a vector equation:[Note 1]
where is the momentum of the system, and is the net (vector sum) force. In equilibrium, there is zero net force by definition, but (balanced) forces may be present nevertheless. In contrast, the second law states an unbalanced force acting on an object will result in the object's momentum changing over time.[10]
By the definition of momentum,
where m is the mass and is the velocity.[4]:9-1,9-2
Newton's second law applies only to a system of constant mass,[Note 2] and hence m may be moved outside the derivative operator. The equation then becomes
By substituting the definition of acceleration, the algebraic version of Newton's Second Law is derived:
Newton never explicitly stated the formula in the reduced form above.[12]
Newton's Second Law asserts the direct proportionality of acceleration to force and the inverse proportionality of acceleration to mass. Accelerations can be defined through kinematic measurements. However, while kinematics are well-described through reference frame analysis in advanced physics, there are still deep questions that remain as to what is the proper definition of mass. General relativity offers an equivalence between space-time and mass, but lacking a coherent theory of quantum gravity, it is unclear as to how or whether this connection is relevant on microscales. With some justification, Newton's second law can be taken as a quantitative definition of mass by writing the law as an equality; the relative units of force and mass then are fixed.
The use of Newton's Second Law as a definition of force has been disparaged in some of the more rigorous textbooks,[4]:12-1[5]:59[13] because it is essentially a mathematical truism. Notable physicists, philosophers and mathematicians who have sought a more explicit definition of the concept of force include Ernst Mach, Clifford Truesdell and Walter Noll.[14][15]
Newton's Second Law can be used to measure the strength of forces. For instance, knowledge of the masses of planets along with the accelerations of their orbits allows scientists to calculate the gravitational forces on planets.
### Third law
Main article: Newton's third law
Newton's Third Law is a result of applying symmetry to situations where forces can be attributed to the presence of different objects. The third law means that all forces are interactions between different bodies,[16][Note 3] and thus that there is no such thing as a unidirectional force or a force that acts on only one body. Whenever a first body exerts a force F on a second body, the second body exerts a force −F on the first body. F and −F are equal in magnitude and opposite in direction. This law is sometimes referred to as the action-reaction law, with F called the "action" and −F the "reaction". The action and the reaction are simultaneous:
If object 1 and object 2 are considered to be in the same system, then the net force on the system due to the interactions between objects 1 and 2 is zero since
This means that in a closed system of particles, there are no internal forces that are unbalanced. That is, the action-reaction force shared between any two objects in a closed system will not cause the center of mass of the system to accelerate. The constituent objects only accelerate with respect to each other, the system itself remains unaccelerated. Alternatively, if an external force acts on the system, then the center of mass will experience an acceleration proportional to the magnitude of the external force divided by the mass of the system.[4]:19-1[5]
Combining Newton's Second and Third Laws, it is possible to show that the linear momentum of a system is conserved. Using
and integrating with respect to time, the equation:
is obtained. For a system that includes objects 1 and 2,
,
which is the conservation of linear momentum.[17] Using the similar arguments, it is possible to generalize this to a system of an arbitrary number of particles. This shows that exchanging momentum between constituent objects will not affect the net momentum of a system. In general, as long as all forces are due to the interaction of objects with mass, it is possible to define a system such that net momentum is never lost nor gained.[4][5]
## Special theory of relativity
In the special theory of relativity, mass and energy are equivalent (as can be seen by calculating the work required to accelerate an object). When an object's velocity increases, so does its energy and hence its mass equivalent (inertia). It thus requires more force to accelerate it the same amount than it did at a lower velocity. Newton's Second Law
remains valid because it is a mathematical definition.[18]:855–876 But in order to be conserved, relativistic momentum must be redefined as:
where
is the velocity and
is the speed of light
is the rest mass.
The relativistic expression relating force and acceleration for a particle with constant non-zero rest mass moving in the direction is:
where the Lorentz factor
[19]
In the early history of relativity, the expressions and were called longitudinal and transverse mass. Relativistic force does not produce a constant acceleration, but an ever-decreasing acceleration as the object approaches the speed of light. Note that is undefined for an object with a non-zero rest mass at the speed of light, and the theory yields no prediction at that speed.
If is very small compared to , then is very close to 1 and
is a close approximation. Even for use in relativity, however, one can restore the form of
through the use of four-vectors. This relation is correct in relativity when is the four-force, is the invariant mass, and is the four-acceleration.[20]
## Descriptions
Free body diagrams of a block on a flat surface and an inclined plane. Forces are resolved and added together to determine their magnitudes and the net force.
Since forces are perceived as pushes or pulls, this can provide an intuitive understanding for describing forces.[3] As with other physical concepts (e.g. temperature), the intuitive understanding of forces is quantified using precise operational definitions that are consistent with direct observations and compared to a standard measurement scale. Through experimentation, it is determined that laboratory measurements of forces are fully consistent with the conceptual definition of force offered by Newtonian mechanics.
Forces act in a particular direction and have sizes dependent upon how strong the push or pull is. Because of these characteristics, forces are classified as "vector quantities". This means that forces follow a different set of mathematical rules than physical quantities that do not have direction (denoted scalar quantities). For example, when determining what happens when two forces act on the same object, it is necessary to know both the magnitude and the direction of both forces to calculate the result. If both of these pieces of information are not known for each force, the situation is ambiguous. For example, if you know that two people are pulling on the same rope with known magnitudes of force but you do not know which direction either person is pulling, it is impossible to determine what the acceleration of the rope will be. The two people could be pulling against each other as in tug of war or the two people could be pulling in the same direction. In this simple one-dimensional example, without knowing the direction of the forces it is impossible to decide whether the net force is the result of adding the two force magnitudes or subtracting one from the other. Associating forces with vectors avoids such problems.
Historically, forces were first quantitatively investigated in conditions of static equilibrium where several forces canceled each other out. Such experiments demonstrate the crucial properties that forces are additive vector quantities: they have magnitude and direction.[3] When two forces act on a point particle, the resulting force, the resultant (also called the net force), can be determined by following the parallelogram rule of vector addition: the addition of two vectors represented by sides of a parallelogram, gives an equivalent resultant vector that is equal in magnitude and direction to the transversal of the parallelogram.[4][5] The magnitude of the resultant varies from the difference of the magnitudes of the two forces to their sum, depending on the angle between their lines of action. However, if the forces are acting on an extended body, their respective lines of application must also be specified in order to account for their effects on the motion of the body.
Free-body diagrams can be used as a convenient way to keep track of forces acting on a system. Ideally, these diagrams are drawn with the angles and relative magnitudes of the force vectors preserved so that graphical vector addition can be done to determine the net force.[21]
As well as being added, forces can also be resolved into independent components at right angles to each other. A horizontal force pointing northeast can therefore be split into two forces, one pointing north, and one pointing east. Summing these component forces using vector addition yields the original force. Resolving force vectors into components of a set of basis vectors is often a more mathematically clean way to describe forces than using magnitudes and directions.[22] This is because, for orthogonal components, the components of the vector sum are uniquely determined by the scalar addition of the components of the individual vectors. Orthogonal components are independent of each other because forces acting at ninety degrees to each other have no effect on the magnitude or direction of the other. Choosing a set of orthogonal basis vectors is often done by considering what set of basis vectors will make the mathematics most convenient. Choosing a basis vector that is in the same direction as one of the forces is desirable, since that force would then have only one non-zero component. Orthogonal force vectors can be three-dimensional with the third component being at right-angles to the other two.[4][5]
### Equilibrium
Equilibrium occurs when the resultant force acting on a point particle is zero (that is, the vector sum of all forces is zero). When dealing with an extended body, it is also necessary that the net torque in it is 0.
There are two kinds of equilibrium: static equilibrium and dynamic equilibrium.
#### Static
Main articles: Statics and Static equilibrium
Static equilibrium was understood well before the invention of classical mechanics. Objects that are at rest have zero net force acting on them.[23]
The simplest case of static equilibrium occurs when two forces are equal in magnitude but opposite in direction. For example, an object on a level surface is pulled (attracted) downward toward the center of the Earth by the force of gravity. At the same time, surface forces resist the downward force with equal upward force (called the normal force). The situation is one of zero net force and no acceleration.[3]
Pushing against an object on a frictional surface can result in a situation where the object does not move because the applied force is opposed by static friction, generated between the object and the table surface. For a situation with no movement, the static friction force exactly balances the applied force resulting in no acceleration. The static friction increases or decreases in response to the applied force up to an upper limit determined by the characteristics of the contact between the surface and the object.[3]
A static equilibrium between two forces is the most usual way of measuring forces, using simple devices such as weighing scales and spring balances. For example, an object suspended on a vertical spring scale experiences the force of gravity acting on the object balanced by a force applied by the "spring reaction force", which equals the object's weight. Using such tools, some quantitative force laws were discovered: that the force of gravity is proportional to volume for objects of constant density (widely exploited for millennia to define standard weights); Archimedes' principle for buoyancy; Archimedes' analysis of the lever; Boyle's law for gas pressure; and Hooke's law for springs. These were all formulated and experimentally verified before Isaac Newton expounded his Three Laws of Motion.[3][4][5]
#### Dynamic
Main article: Dynamics (physics)
Galileo Galilei was the first to point out the inherent contradictions contained in Aristotle's description of forces.
Dynamic equilibrium was first described by Galileo who noticed that certain assumptions of Aristotelian physics were contradicted by observations and logic. Galileo realized that simple velocity addition demands that the concept of an "absolute rest frame" did not exist. Galileo concluded that motion in a constant velocity was completely equivalent to rest. This was contrary to Aristotle's notion of a "natural state" of rest that objects with mass naturally approached. Simple experiments showed that Galileo's understanding of the equivalence of constant velocity and rest were correct. For example, if a mariner dropped a cannonball from the crow's nest of a ship moving at a constant velocity, Aristotelian physics would have the cannonball fall straight down while the ship moved beneath it. Thus, in an Aristotelian universe, the falling cannonball would land behind the foot of the mast of a moving ship. However, when this experiment is actually conducted, the cannonball always falls at the foot of the mast, as if the cannonball knows to travel with the ship despite being separated from it. Since there is no forward horizontal force being applied on the cannonball as it falls, the only conclusion left is that the cannonball continues to move with the same velocity as the boat as it falls. Thus, no force is required to keep the cannonball moving at the constant forward velocity.[9]
Moreover, any object traveling at a constant velocity must be subject to zero net force (resultant force). This is the definition of dynamic equilibrium: when all the forces on an object balance but it still moves at a constant velocity.
A simple case of dynamic equilibrium occurs in constant velocity motion across a surface with kinetic friction. In such a situation, a force is applied in the direction of motion while the kinetic friction force exactly opposes the applied force. This results in zero net force, but since the object started with a non-zero velocity, it continues to move with a non-zero velocity. Aristotle misinterpreted this motion as being caused by the applied force. However, when kinetic friction is taken into consideration it is clear that there is no net force causing constant velocity motion.[4][5]
### Forces in Quantum Mechanics
The notion "force" keeps its meaning in quantum mechanics, though one is now dealing with operators instead of classical variables and though the physics is now described by the Schrödinger equation instead of Newtonian equations. This has the consequence that the results of a measurement are now sometimes "quantized", i.e. they appear in discrete portions. This is, of course, difficult to imagine in the context of "forces". However, the potentials V(x,y,z) or fields, from which the forces generally can be derived, are treated similar to classical position variables, i.e., .
This becomes different only in the framework of quantum field theory, where these fields are also quantized.
However, already in quantum mechanics there is one "caveat", namely the particles acting onto each other do not only possess the spatial variable, but also a discrete intrinsic angular momentum-like variable called the "spin", and there is the Pauli principle relating the space and the spin variables. Depending on the value of the spin, identical particles split into two different classes, fermions and bosons. If two identical fermions (e.g. electrons) have a symmetric spin function (e.g. parallel spins) the spatial variables must be antisymmetric (i.e. they exclude each other from their places much as if there was a repulsive force), and vice versa, i.e. for antiparallel spins the position variables must be symmetric (i.e. the apparent force must be attractive). Thus in the case of two fermions there is a strictly negative correlation between spatial and spin variables, whereas for two bosons (e.g. quanta of electromagnetic waves, photons) the correlation is strictly positive.
Thus the notion "force" loses already part of its meaning.
### Feynman diagrams
Main article: Feynman diagrams
Feynman diagram for the decay of a neutron into a proton. The W boson is between two vertices indicating a repulsion.
In modern particle physics, forces and the acceleration of particles are explained as a mathematical by-product of exchange of momentum-carrying gauge bosons. With the development of quantum field theory and general relativity, it was realized that force is a redundant concept arising from conservation of momentum (4-momentum in relativity and momentum of virtual particles in quantum electrodynamics). The conservation of momentum can be directly derived from the homogeneity or symmetry of space and so is usually considered more fundamental than the concept of a force. Thus the currently known fundamental forces are considered more accurately to be "fundamental interactions".[6]:199–128 When particle A emits (creates) or absorbs (annihilates) virtual particle B, a momentum conservation results in recoil of particle A making impression of repulsion or attraction between particles A A' exchanging by B. This description applies to all forces arising from fundamental interactions. While sophisticated mathematical descriptions are needed to predict, in full detail, the accurate result of such interactions, there is a conceptually simple way to describe such interactions through the use of Feynman diagrams. In a Feynman diagram, each matter particle is represented as a straight line (see world line) traveling through time, which normally increases up or to the right in the diagram. Matter and anti-matter particles are identical except for their direction of propagation through the Feynman diagram. World lines of particles intersect at interaction vertices, and the Feynman diagram represents any force arising from an interaction as occurring at the vertex with an associated instantaneous change in the direction of the particle world lines. Gauge bosons are emitted away from the vertex as wavy lines and, in the case of virtual particle exchange, are absorbed at an adjacent vertex.[24]
The utility of Feynman diagrams is that other types of physical phenomena that are part of the general picture of fundamental interactions but are conceptually separate from forces can also be described using the same rules. For example, a Feynman diagram can describe in succinct detail how a neutron decays into an electron, proton, and neutrino, an interaction mediated by the same gauge boson that is responsible for the weak nuclear force.[24]
## Fundamental forces
All of the forces in the universe are based on four fundamental interactions. The strong and weak forces are nuclear forces that act only at very short distances, and are responsible for the interactions between subatomic particles, including nucleons and compound nuclei. The electromagnetic force acts between electric charges, and the gravitational force acts between masses. All other forces in nature derive from these four fundamental interactions. For example, friction is a manifestation of the electromagnetic force acting between the atoms of two surfaces, and the Pauli exclusion principle,[25] which does not permit atoms to pass through each other. Similarly, the forces in springs, modeled by Hooke's law, are the result of electromagnetic forces and the Exclusion Principle acting together to return an object to its equilibrium position. Centrifugal forces are acceleration forces that arise simply from the acceleration of rotating frames of reference.[4]:12-11[5]:359
The development of fundamental theories for forces proceeded along the lines of unification of disparate ideas. For example, Isaac Newton unified the force responsible for objects falling at the surface of the Earth with the force responsible for the orbits of celestial mechanics in his universal theory of gravitation. Michael Faraday and James Clerk Maxwell demonstrated that electric and magnetic forces were unified through one consistent theory of electromagnetism. In the 20th century, the development of quantum mechanics led to a modern understanding that the first three fundamental forces (all except gravity) are manifestations of matter (fermions) interacting by exchanging virtual particles called gauge bosons.[26] This standard model of particle physics posits a similarity between the forces and led scientists to predict the unification of the weak and electromagnetic forces in electroweak theory subsequently confirmed by observation. The complete formulation of the standard model predicts an as yet unobserved Higgs mechanism, but observations such as neutrino oscillations indicate that the standard model is incomplete. A Grand Unified Theory allowing for the combination of the electroweak interaction with the strong force is held out as a possibility with candidate theories such as supersymmetry proposed to accommodate some of the outstanding unsolved problems in physics. Physicists are still attempting to develop self-consistent unification models that would combine all four fundamental interactions into a theory of everything. Einstein tried and failed at this endeavor, but currently the most popular approach to answering this question is string theory.[6]:212–219
The four fundamental forces of nature[27]
Property/Interaction Gravitation Weak Electromagnetic Strong
(Electroweak) Fundamental Residual
Acts on: Mass - Energy Flavor Electric charge Color charge Atomic nuclei
Particles experiencing: All Quarks, leptons Electrically charged Quarks, Gluons Hadrons
Particles mediating: Graviton
(not yet observed)
W+ W Z0 γ Gluons Mesons
Strength in the scale of quarks: 10−41 10−4 1 60 Not applicable
to quarks
Strength in the scale of
protons/neutrons:
10−36 10−7 1 Not applicable
to hadrons
20
### Gravitational
Main article: Gravity
Images of a freely falling basketball taken with a stroboscope at 20 flashes per second. The distance units on the right are multiples of about 12 millimetres. The basketball starts at rest. At the time of the first flash (distance zero) it is released, after which the number of units fallen is equal to the square of the number of flashes.
What we now call gravity was not identified as a universal force until the work of Isaac Newton. Before Newton, the tendency for objects to fall towards the Earth was not understood to be related to the motions of celestial objects. Galileo was instrumental in describing the characteristics of falling objects by determining that the acceleration of every object in free-fall was constant and independent of the mass of the object. Today, this acceleration due to gravity towards the surface of the Earth is usually designated as and has a magnitude of about 9.81 meters per second squared (this measurement is taken from sea level and may vary depending on location), and points toward the center of the Earth.[28] This observation means that the force of gravity on an object at the Earth's surface is directly proportional to the object's mass. Thus an object that has a mass of will experience a force:
In free-fall, this force is unopposed and therefore the net force on the object is its weight. For objects not in free-fall, the force of gravity is opposed by the reactions of their supports. For example, a person standing on the ground experiences zero net force, since his weight is balanced by a normal force exerted by the ground.[4][5]
Newton's contribution to gravitational theory was to unify the motions of heavenly bodies, which Aristotle had assumed were in a natural state of constant motion, with falling motion observed on the Earth. He proposed a law of gravity that could account for the celestial motions that had been described earlier using Kepler's laws of planetary motion.[29]
Newton came to realize that the effects of gravity might be observed in different ways at larger distances. In particular, Newton determined that the acceleration of the Moon around the Earth could be ascribed to the same force of gravity if the acceleration due to gravity decreased as an inverse square law. Further, Newton realized that the acceleration due to gravity is proportional to the mass of the attracting body.[29] Combining these ideas gives a formula that relates the mass () and the radius () of the Earth to the gravitational acceleration:
where the vector direction is given by , the unit vector directed outward from the center of the Earth.[10]
In this equation, a dimensional constant is used to describe the relative strength of gravity. This constant has come to be known as Newton's Universal Gravitation Constant,[30] though its value was unknown in Newton's lifetime. Not until 1798 was Henry Cavendish able to make the first measurement of using a torsion balance; this was widely reported in the press as a measurement of the mass of the Earth since knowing could allow one to solve for the Earth's mass given the above equation. Newton, however, realized that since all celestial bodies followed the same laws of motion, his law of gravity had to be universal. Succinctly stated, Newton's Law of Gravitation states that the force on a spherical object of mass due to the gravitational pull of mass is
where is the distance between the two objects' centers of mass and is the unit vector pointed in the direction away from the center of the first object toward the center of the second object.[10]
This formula was powerful enough to stand as the basis for all subsequent descriptions of motion within the solar system until the 20th century. During that time, sophisticated methods of perturbation analysis[31] were invented to calculate the deviations of orbits due to the influence of multiple bodies on a planet, moon, comet, or asteroid. The formalism was exact enough to allow mathematicians to predict the existence of the planet Neptune before it was observed.[32]
Instruments like GRAVITY provide a powerful probe for gravity force detection.[33]
It was only the orbit of the planet Mercury that Newton's Law of Gravitation seemed not to fully explain. Some astrophysicists predicted the existence of another planet (Vulcan) that would explain the discrepancies; however, despite some early indications, no such planet could be found. When Albert Einstein formulated his theory of general relativity (GR) he turned his attention to the problem of Mercury's orbit and found that his theory added a correction, which could account for the discrepancy. This was the first time that Newton's Theory of Gravity had been shown to be less correct than an alternative.[34]
Since then, and so far, general relativity has been acknowledged as the theory that best explains gravity. In GR, gravitation is not viewed as a force, but rather, objects moving freely in gravitational fields travel under their own inertia in straight lines through curved space-time – defined as the shortest space-time path between two space-time events. From the perspective of the object, all motion occurs as if there were no gravitation whatsoever. It is only when observing the motion in a global sense that the curvature of space-time can be observed and the force is inferred from the object's curved path. Thus, the straight line path in space-time is seen as a curved line in space, and it is called the ballistic trajectory of the object. For example, a basketball thrown from the ground moves in a parabola, as it is in a uniform gravitational field. Its space-time trajectory (when the extra ct dimension is added) is almost a straight line, slightly curved (with the radius of curvature of the order of few light-years). The time derivative of the changing momentum of the object is what we label as "gravitational force".[5]
### Electromagnetic
Main article: Electromagnetic force
The electrostatic force was first described in 1784 by Coulomb as a force that existed intrinsically between two charges.[18]:519 The properties of the electrostatic force were that it varied as an inverse square law directed in the radial direction, was both attractive and repulsive (there was intrinsic polarity), was independent of the mass of the charged objects, and followed the superposition principle. Coulomb's law unifies all these observations into one succinct statement.[35]
Subsequent mathematicians and physicists found the construct of the electric field to be useful for determining the electrostatic force on an electric charge at any point in space. The electric field was based on using a hypothetical "test charge" anywhere in space and then using Coulomb's Law to determine the electrostatic force.[36]:4-6 to 4-8 Thus the electric field anywhere in space is defined as
where is the magnitude of the hypothetical test charge.
Meanwhile, the Lorentz force of magnetism was discovered to exist between two electric currents. It has the same mathematical character as Coulomb's Law with the proviso that like currents attract and unlike currents repel. Similar to the electric field, the magnetic field can be used to determine the magnetic force on an electric current at any point in space. In this case, the magnitude of the magnetic field was determined to be
where is the magnitude of the hypothetical test current and is the length of hypothetical wire through which the test current flows. The magnetic field exerts a force on all magnets including, for example, those used in compasses. The fact that the Earth's magnetic field is aligned closely with the orientation of the Earth's axis causes compass magnets to become oriented because of the magnetic force pulling on the needle.
Through combining the definition of electric current as the time rate of change of electric charge, a rule of vector multiplication called Lorentz's Law describes the force on a charge moving in a magnetic field.[36] The connection between electricity and magnetism allows for the description of a unified electromagnetic force that acts on a charge. This force can be written as a sum of the electrostatic force (due to the electric field) and the magnetic force (due to the magnetic field). Fully stated, this is the law:
where is the electromagnetic force, is the magnitude of the charge of the particle, is the electric field, is the velocity of the particle that is crossed with the magnetic field ().
The origin of electric and magnetic fields would not be fully explained until 1864 when James Clerk Maxwell unified a number of earlier theories into a set of 20 scalar equations, which were later reformulated into 4 vector equations by Oliver Heaviside and Josiah Willard Gibbs.[37] These "Maxwell Equations" fully described the sources of the fields as being stationary and moving charges, and the interactions of the fields themselves. This led Maxwell to discover that electric and magnetic fields could be "self-generating" through a wave that traveled at a speed that he calculated to be the speed of light. This insight united the nascent fields of electromagnetic theory with optics and led directly to a complete description of the electromagnetic spectrum.[38]
However, attempting to reconcile electromagnetic theory with two observations, the photoelectric effect, and the nonexistence of the ultraviolet catastrophe, proved troublesome. Through the work of leading theoretical physicists, a new theory of electromagnetism was developed using quantum mechanics. This final modification to electromagnetic theory ultimately led to quantum electrodynamics (or QED), which fully describes all electromagnetic phenomena as being mediated by wave–particles known as photons. In QED, photons are the fundamental exchange particle, which described all interactions relating to electromagnetism including the electromagnetic force.[Note 4]
It is a common misconception to ascribe the stiffness and rigidity of solid matter to the repulsion of like charges under the influence of the electromagnetic force. However, these characteristics actually result from the Pauli exclusion principle. Since electrons are fermions, they cannot occupy the same quantum mechanical state as other electrons. When the electrons in a material are densely packed together, there are not enough lower energy quantum mechanical states for them all, so some of them must be in higher energy states. This means that it takes energy to pack them together. While this effect is manifested macroscopically as a structural force, it is technically only the result of the existence of a finite set of electron states.
### Strong nuclear
Main article: Strong interaction
There are two "nuclear forces", which today are usually described as interactions that take place in quantum theories of particle physics. The strong nuclear force[18]:940 is the force responsible for the structural integrity of atomic nuclei while the weak nuclear force[18]:951 is responsible for the decay of certain nucleons into leptons and other types of hadrons.[4][5]
The strong force is today understood to represent the interactions between quarks and gluons as detailed by the theory of quantum chromodynamics (QCD).[39] The strong force is the fundamental force mediated by gluons, acting upon quarks, antiquarks, and the gluons themselves. The (aptly named) strong interaction is the "strongest" of the four fundamental forces.
The strong force only acts directly upon elementary particles. However, a residual of the force is observed between hadrons (the best known example being the force that acts between nucleons in atomic nuclei) as the nuclear force. Here the strong force acts indirectly, transmitted as gluons, which form part of the virtual pi and rho mesons, which classically transmit the nuclear force (see this topic for more). The failure of many searches for free quarks has shown that the elementary particles affected are not directly observable. This phenomenon is called color confinement.
### Weak nuclear
Main article: Weak interaction
The weak force is due to the exchange of the heavy W and Z bosons. Its most familiar effect is beta decay (of neutrons in atomic nuclei) and the associated radioactivity. The word "weak" derives from the fact that the field strength is some 1013 times less than that of the strong force. Still, it is stronger than gravity over short distances. A consistent electroweak theory has also been developed, which shows that electromagnetic forces and the weak force are indistinguishable at a temperatures in excess of approximately 1015 kelvins. Such temperatures have been probed in modern particle accelerators and show the conditions of the universe in the early moments of the Big Bang.
## Non-fundamental forces
Some forces are consequences of the fundamental ones. In such situations, idealized models can be utilized to gain physical insight.
### Normal force
FN represents the normal force exerted on the object.
Main article: Normal force
The normal force is due to repulsive forces of interaction between atoms at close contact. When their electron clouds overlap, Pauli repulsion (due to fermionic nature of electrons) follows resulting in the force that acts in a direction normal to the surface interface between two objects.[18]:93 The normal force, for example, is responsible for the structural integrity of tables and floors as well as being the force that responds whenever an external force pushes on a solid object. An example of the normal force in action is the impact force on an object crashing into an immobile surface.[4][5]
### Friction
Main article: Friction
Friction is a surface force that opposes relative motion. The frictional force is directly related to the normal force that acts to keep two solid objects separated at the point of contact. There are two broad classifications of frictional forces: static friction and kinetic friction.
The static friction force () will exactly oppose forces applied to an object parallel to a surface contact up to the limit specified by the coefficient of static friction () multiplied by the normal force (). In other words, the magnitude of the static friction force satisfies the inequality:
The kinetic friction force () is independent of both the forces applied and the movement of the object. Thus, the magnitude of the force equals:
where is the coefficient of kinetic friction. For most surface interfaces, the coefficient of kinetic friction is less than the coefficient of static friction.
### Tension
Main article: Tension (physics)
Tension forces can be modeled using ideal strings that are massless, frictionless, unbreakable, and unstretchable. They can be combined with ideal pulleys, which allow ideal strings to switch physical direction. Ideal strings transmit tension forces instantaneously in action-reaction pairs so that if two objects are connected by an ideal string, any force directed along the string by the first object is accompanied by a force directed along the string in the opposite direction by the second object.[40] By connecting the same string multiple times to the same object through the use of a set-up that uses movable pulleys, the tension force on a load can be multiplied. For every string that acts on a load, another factor of the tension force in the string acts on the load. However, even though such machines allow for an increase in force, there is a corresponding increase in the length of string that must be displaced in order to move the load. These tandem effects result ultimately in the conservation of mechanical energy since the work done on the load is the same no matter how complicated the machine.[4][5][41]
### Elastic force
Fk is the force that responds to the load on the spring
An elastic force acts to return a spring to its natural length. An ideal spring is taken to be massless, frictionless, unbreakable, and infinitely stretchable. Such springs exert forces that push when contracted, or pull when extended, in proportion to the displacement of the spring from its equilibrium position.[42] This linear relationship was described by Robert Hooke in 1676, for whom Hooke's law is named. If is the displacement, the force exerted by an ideal spring equals:
where is the spring constant (or force constant), which is particular to the spring. The minus sign accounts for the tendency of the force to act in opposition to the applied load.[4][5]
### Continuum mechanics
When the drag force () associated with air resistance becomes equal in magnitude to the force of gravity on a falling object (), the object reaches a state of dynamic equilibrium at terminal velocity.
Newton's laws and Newtonian mechanics in general were first developed to describe how forces affect idealized point particles rather than three-dimensional objects. However, in real life, matter has extended structure and forces that act on one part of an object might affect other parts of an object. For situations where lattice holding together the atoms in an object is able to flow, contract, expand, or otherwise change shape, the theories of continuum mechanics describe the way forces affect the material. For example, in extended fluids, differences in pressure result in forces being directed along the pressure gradients as follows:
where is the volume of the object in the fluid and is the scalar function that describes the pressure at all locations in space. Pressure gradients and differentials result in the buoyant force for fluids suspended in gravitational fields, winds in atmospheric science, and the lift associated with aerodynamics and flight.[4][5]
A specific instance of such a force that is associated with dynamic pressure is fluid resistance: a body force that resists the motion of an object through a fluid due to viscosity. For so-called "Stokes' drag" the force is approximately proportional to the velocity, but opposite in direction:
where:
is a constant that depends on the properties of the fluid and the dimensions of the object (usually the cross-sectional area), and
is the velocity of the object.[4][5]
More formally, forces in continuum mechanics are fully described by a stresstensor with terms that are roughly defined as
where is the relevant cross-sectional area for the volume for which the stress-tensor is being calculated. This formalism includes pressure terms associated with forces that act normal to the cross-sectional area (the matrix diagonals of the tensor) as well as shear terms associated with forces that act parallel to the cross-sectional area (the off-diagonal elements). The stress tensor accounts for forces that cause all strains (deformations) including also tensile stresses and compressions.[3][5]:133–134[36]:38-1–38-11
### Fictitious forces
Main article: Fictitious forces
There are forces that are frame dependent, meaning that they appear due to the adoption of non-Newtonian (that is, non-inertial) reference frames. Such forces include the centrifugal force and the Coriolis force.[43] These forces are considered fictitious because they do not exist in frames of reference that are not accelerating.[4][5] Because these forces are not genuine they are also referred to as "pseudo forces".[4]:12-11
In general relativity, gravity becomes a fictitious force that arises in situations where spacetime deviates from a flat geometry. As an extension, Kaluza–Klein theory and string theory ascribe electromagnetism and the other fundamental forces respectively to the curvature of differently scaled dimensions, which would ultimately imply that all forces are fictitious.
## Rotations and torque
Main article: Torque
Forces that cause extended objects to rotate are associated with torques. Mathematically, the torque of a force is defined relative to an arbitrary reference point as the cross-product:
where
is the position vector of the force application point relative to the reference point.
Torque is the rotation equivalent of force in the same way that angle is the rotational equivalent for position, angular velocity for velocity, and angular momentum for momentum. As a consequence of Newton's First Law of Motion, there exists rotational inertia that ensures that all bodies maintain their angular momentum unless acted upon by an unbalanced torque. Likewise, Newton's Second Law of Motion can be used to derive an analogous equation for the instantaneous angular acceleration of the rigid body:
where
is the moment of inertia of the body
is the angular acceleration of the body.
This provides a definition for the moment of inertia, which is the rotational equivalent for mass. In more advanced treatments of mechanics, where the rotation over a time interval is described, the moment of inertia must be substituted by the tensor that, when properly analyzed, fully determines the characteristics of rotations including precession and nutation.
Equivalently, the differential form of Newton's Second Law provides an alternative definition of torque:
[44] where is the angular momentum of the particle.
Newton's Third Law of Motion requires that all objects exerting torques themselves experience equal and opposite torques,[45] and therefore also directly implies the conservation of angular momentum for closed systems that experience rotations and revolutions through the action of internal torques.
### Centripetal force
Main article: Centripetal force
For an object accelerating in circular motion, the unbalanced force acting on the object equals:[46]
where is the mass of the object, is the velocity of the object and is the distance to the center of the circular path and is the unit vector pointing in the radial direction outwards from the center. This means that the unbalanced centripetal force felt by any object is always directed toward the center of the curving path. Such forces act perpendicular to the velocity vector associated with the motion of an object, and therefore do not change the speed of the object (magnitude of the velocity), but only the direction of the velocity vector. The unbalanced force that accelerates an object can be resolved into a component that is perpendicular to the path, and one that is tangential to the path. This yields both the tangential force, which accelerates the object by either slowing it down or speeding it up, and the radial (centripetal) force, which changes its direction.[4][5]
## Kinematic integrals
Forces can be used to define a number of physical concepts by integrating with respect to kinematic variables. For example, integrating with respect to time gives the definition of impulse:[47]
which by Newton's Second Law must be equivalent to the change in momentum (yielding the Impulse momentum theorem).
Similarly, integrating with respect to position gives a definition for the work done by a force:[4]:13-3
which is equivalent to changes in kinetic energy (yielding the work energy theorem).[4]:13-3
Power P is the rate of change dW/dt of the work W, as the trajectory is extended by a position change in a time interval dt:[4]:13-2
with the velocity.
## Potential energy
Main article: Potential energy
Instead of a force, often the mathematically related concept of a potential energy field can be used for convenience. For instance, the gravitational force acting upon an object can be seen as the action of the gravitational field that is present at the object's location. Restating mathematically the definition of energy (via the definition of work), a potential scalar field is defined as that field whose gradient is equal and opposite to the force produced at every point:
Forces can be classified as conservative or nonconservative. Conservative forces are equivalent to the gradient of a potential while nonconservative forces are not.[4][5]
### Conservative forces
Main article: Conservative force
A conservative force that acts on a closed system has an associated mechanical work that allows energy to convert only between kinetic or potential forms. This means that for a closed system, the net mechanical energy is conserved whenever a conservative force acts on the system. The force, therefore, is related directly to the difference in potential energy between two different locations in space,[48] and can be considered to be an artifact of the potential field in the same way that the direction and amount of a flow of water can be considered to be an artifact of the contour map of the elevation of an area.[4][5]
Conservative forces include gravity, the electromagnetic force, and the spring force. Each of these forces has models that are dependent on a position often given as a radial vector emanating from spherically symmetric potentials.[49] Examples of this follow:
For gravity:
where is the gravitational constant, and is the mass of object n.
For electrostatic forces:
where is electric permittivity of free space, and is the electric charge of object n.
For spring forces:
where is the spring constant.[4][5]
### Nonconservative forces
For certain physical scenarios, it is impossible to model forces as being due to gradient of potentials. This is often due to macrophysical considerations that yield forces as arising from a macroscopic statistical average of microstates. For example, friction is caused by the gradients of numerous electrostatic potentials between the atoms, but manifests as a force model that is independent of any macroscale position vector. Nonconservative forces other than friction include other contact forces, tension, compression, and drag. However, for any sufficiently detailed description, all these forces are the results of conservative ones since each of these macroscopic forces are the net results of the gradients of microscopic potentials.[4][5]
The connection between macroscopic nonconservative forces and microscopic conservative forces is described by detailed treatment with statistical mechanics. In macroscopic closed systems, nonconservative forces act to change the internal energies of the system, and are often associated with the transfer of heat. According to the Second law of thermodynamics, nonconservative forces necessarily result in energy transformations within closed systems from ordered to more random conditions as entropy increases.[4][5]
## Units of measurement
The SI unit of force is the newton (symbol N), which is the force required to accelerate a one kilogram mass at a rate of one meter per second squared, or kg·m·s−2.[50] The corresponding CGS unit is the dyne, the force required to accelerate a one gram mass by one centimeter per second squared, or g·cm·s−2. A newton is thus equal to 100,000 dynes.
The gravitational foot-pound-second English unit of force is the pound-force (lbf), defined as the force exerted by gravity on a pound-mass in the standard gravitational field of 9.80665 m·s−2.[50] The pound-force provides an alternative unit of mass: one slug is the mass that will accelerate by one foot per second squared when acted on by one pound-force.[50]
An alternative unit of force in a different foot-pound-second system, the absolute fps system, is the poundal, defined as the force required to accelerate a one-pound mass at a rate of one foot per second squared.[50] The units of slug and poundal are designed to avoid a constant of proportionality in Newton's Second Law.
The pound-force has a metric counterpart, less commonly used than the newton: the kilogram-force (kgf) (sometimes kilopond), is the force exerted by standard gravity on one kilogram of mass.[50] The kilogram-force leads to an alternate, but rarely used unit of mass: the metric slug (sometimes mug or hyl) is that mass that accelerates at 1 m·s−2 when subjected to a force of 1 kgf. The kilogram-force is not a part of the modern SI system, and is generally deprecated; however it still sees use for some purposes as expressing aircraft weight, jet thrust, bicycle spoke tension, torque wrench settings and engine output torque. Other arcane units of force include the sthène, which is equivalent to 1000 N, and the kip, which is equivalent to 1000 lbf.
Units of force
newton
(SI unit)
dyne kilogram-force,
kilopond
pound-force poundal
1 N ≡ 1 kgm/s2 = 105 dyn ≈ 0.10197 kp ≈ 0.22481 lbF ≈ 7.2330 pdl
1 dyn = 105 N ≡ 1 gcm/s2 ≈ 1.0197 × 106 kp ≈ 2.2481 × 106 lbF ≈ 7.2330 × 105 pdl
1 kp = 9.80665 N = 980665 dyn gn(1 kg) ≈ 2.2046 lbF ≈ 70.932 pdl
1 lbF ≈ 4.448222 N ≈ 444822 dyn ≈ 0.45359 kp gn(1 lb) ≈ 32.174 pdl
1 pdl ≈ 0.138255 N ≈ 13825 dyn ≈ 0.014098 kp ≈ 0.031081 lbF ≡ 1 lbft/s2
The value of gn as used in the official definition of the kilogram-force is used here for all gravitational units.
See also Ton-force.
## Notes
1. Newton's Principia Mathematica actually used a finite difference version of this equation based upon impulse. See Impulse.
2. "It is important to note that we cannot derive a general expression for Newton's second law for variable mass systems by treating the mass in F = dP/dt = d(Mv) as a variable. [...] We can use F = dP/dt to analyze variable mass systems only if we apply it to an entire system of constant mass having parts among which there is an interchange of mass." [Emphasis as in the original] (Halliday, Resnick & Krane 2001, p. 199)
3. "Any single force is only one aspect of a mutual interaction between two bodies." (Halliday, Resnick & Krane 2001, pp. 78–79)
4. For a complete library on quantum mechanics see Quantum mechanics – References
## References
1. Nave, C. R. (2014). "Force". Hyperphysics. Dept. of Physics and Astronomy, Georgia State University. Retrieved 15 August 2014.
2. Heath, T.L. "The Works of Archimedes (1897). The unabridged work in PDF form (19 MB)". Internet Archive. Retrieved 2007-10-14.
3. University Physics, Sears, Young & Zemansky, pp.18–38
4. Feynman volume 1
5. Kleppner & Kolenkow 2010
6. Weinberg, S. (1994). Dreams of a Final Theory. Vintage Books USA. ISBN 0-679-74408-8.
7. Lang, Helen S. (1998). The order of nature in Aristotle's physics : place and the elements (1. publ. ed.). Cambridge: Cambridge Univ. Press. ISBN 9780521624534.
8. Hetherington, Norriss S. (1993). Cosmology: Historical, Literary, Philosophical, Religious, and Scientific Perspectives. Garland Reference Library of the Humanities. p. 100. ISBN 0-8153-1085-4.
9. Drake, Stillman (1978). Galileo At Work. Chicago: University of Chicago Press. ISBN 0-226-16226-5
10. Newton, Isaac (1999). The Principia Mathematical Principles of Natural Philosophy. Berkeley: University of California Press. ISBN 0-520-08817-4. This is a recent translation into English by I. Bernard Cohen and Anne Whitman, with help from Julia Budenz.
11. DiSalle, Robert (2002-03-30). "Space and Time: Inertial Frames". Stanford Encyclopedia of Philosophy. Retrieved 2008-03-24.
12. Howland, R. A. (2006). Intermediate dynamics a linear algebraic approach (Online-Ausg. ed.). New York: Springer. pp. 255256. ISBN 9780387280592.
13. One exception to this rule is: Landau, L. D.; Akhiezer, A. I.; Lifshitz, A. M. (196). General Physics; mechanics and molecular physics (First English ed.). Oxford: Pergamon Press. ISBN 0-08-003304-0. Translated by: J. B. Sykes, A. D. Petford, and C. L. Petford. Library of Congress Catalog Number 67-30260. In section 7, pages 12–14, this book defines force as dp/dt.
14. Jammer, Max (1999). Concepts of force : a study in the foundations of dynamics (Facsim. ed.). Mineola, N.Y.: Dover Publications. pp. 220–222. ISBN 9780486406893.
15. Noll, Walter (April 2007). "On the Concept of Force" (pdf). Carnegie Mellon University. Retrieved 28 October 2013.
16. C. Hellingman (1992). "Newton's third law revisited". Phys. Educ. 27 (2): 112–115. Bibcode:1992PhyEd..27..112H. doi:10.1088/0031-9120/27/2/011. Quoting Newton in the Principia: It is not one action by which the Sun attracts Jupiter, and another by which Jupiter attracts the Sun; but it is one action by which the Sun and Jupiter mutually endeavour to come nearer together.
17. Dr. Nikitin (2007). "Dynamics of translational motion". Retrieved 2008-01-04.
18. Cutnell & Johnson 2003
19. "Seminar: Visualizing Special Relativity". The Relativistic Raytracer. Retrieved 2008-01-04.
20. Wilson, John B. "Four-Vectors (4-Vectors) of Special Relativity: A Study of Elegant Physics". The Science Realm: John's Virtual Sci-Tech Universe. Archived from the original on 26 June 2009. Retrieved 2008-01-04.
21. "Introduction to Free Body Diagrams". Physics Tutorial Menu. University of Guelph. Retrieved 2008-01-02.
22. Henderson, Tom (2004). "The Physics Classroom". The Physics Classroom and Mathsoft Engineering & Education, Inc. Retrieved 2008-01-02.
23. "Static Equilibrium". Physics Static Equilibrium (forces and torques). University of the Virgin Islands. Archived from the original on October 19, 2007. Retrieved 2008-01-02.
24. Shifman, Mikhail (1999). ITEP lectures on particle physics and field theory. World Scientific. ISBN 981-02-2639-X.
25. Nave, Carl Rod. "Pauli Exclusion Principle". HyperPhysics. University of Guelph. Retrieved 2013-10-28.
26. "Fermions & Bosons". The Particle Adventure. Retrieved 2008-01-04.
27. http://www.pha.jhu.edu/~dfehling/particle.gif
28. Cook, A. H. (1965). "A New Absolute Determination of the Acceleration due to Gravity at the National Physical Laboratory". Nature. 208 (5007): 279. Bibcode:1965Natur.208..279C. doi:10.1038/208279a0.
29. Young, Hugh; Freedman, Roger; Sears, Francis and Zemansky, Mark (1949) University Physics. Pearson Education. pp. 59–82
30. "Sir Isaac Newton: The Universal Law of Gravitation". Astronomy 161 The Solar System. Retrieved 2008-01-04.
31. Watkins, Thayer. "Perturbation Analysis, Regular and Singular". Department of Economics. San José State University.
32. Kollerstrom, Nick (2001). "Neptune's Discovery. The British Case for Co-Prediction.". University College London. Archived from the original on 2005-11-11. Retrieved 2007-03-19.
33. "Powerful New Black Hole Probe Arrives at Paranal". Retrieved 13 August 2015.
34. Einstein, Albert (1916). "The Foundation of the General Theory of Relativity" (PDF). Annalen der Physik. 49 (7): 769–822. Bibcode:1916AnP...354..769E. doi:10.1002/andp.19163540702. Retrieved 2006-09-03.
35. Coulomb, Charles (1784). "Recherches théoriques et expérimentales sur la force de torsion et sur l'élasticité des fils de metal". Histoire de l'Académie Royale des Sciences: 229–269.
36. Feynman volume 2
37. Scharf, Toralf (2007). Polarized light in liquid crystals and polymers. John Wiley and Sons. p. 19. ISBN 0-471-74064-0., Chapter 2, p. 19
38. Duffin, William (1980). Electricity and Magnetism, 3rd Ed. McGraw-Hill. pp. 364–383. ISBN 0-07-084111-X.
39. Stevens, Tab (10 July 2003). "Quantum-Chromodynamics: A Definition – Science Articles". Archived from the original on 2011-10-16. Retrieved 2008-01-04.
40. "Tension Force". Non-Calculus Based Physics I. Retrieved 2008-01-04.
41. Fitzpatrick, Richard (2006-02-02). "Strings, pulleys, and inclines". Retrieved 2008-01-04.
42. Nave, Carl Rod. "Elasticity". HyperPhysics. University of Guelph. Retrieved 2013-10-28.
43. Mallette, Vincent (1982–2008). "Inwit Publishing, Inc. and Inwit, LLC – Writings, Links and Software Distributions – The Coriolis Force". Publications in Science and Mathematics, Computing and the Humanities. Inwit Publishing, Inc. Retrieved 2008-01-04.
44. Nave, Carl Rod. "Newton's 2nd Law: Rotation". HyperPhysics. University of Guelph. Retrieved 2013-10-28.
45. Fitzpatrick, Richard (2007-01-07). "Newton's third law of motion". Retrieved 2008-01-04.
46. Nave, Carl Rod. "Centripetal Force". HyperPhysics. University of Guelph. Retrieved 2013-10-28.
47. Hibbeler, Russell C. (2010). Engineering Mechanics, 12th edition. Pearson Prentice Hall. p. 222. ISBN 0-13-607791-9.
48. Singh, Sunil Kumar (2007-08-25). "Conservative force". Connexions. Retrieved 2008-01-04.
49. Davis, Doug. "Conservation of Energy". General physics. Retrieved 2008-01-04.
50. Wandmacher, Cornelius; Johnson, Arnold (1995). Metric Units in Engineering. ASCE Publications. p. 15. ISBN 0-7844-0070-9.
• Corben, H.C.; Philip Stehle (1994). Classical Mechanics. New York: Dover publications. pp. 28–31. ISBN 0-486-68063-0.
• Cutnell, John D.; Johnson, Kenneth W. (2003). Physics, Sixth Edition. Hoboken, New Jersey: John Wiley & Sons Inc. ISBN 0471151831.
• Feynman, Richard P.; Leighton; Sands, Matthew (2010). The Feynman lectures on physics. Vol. I: Mainly mechanics, radiation and heat (New millennium ed.). New York: BasicBooks. ISBN 978-0465024933.
• Feynman, Richard P.; Leighton, Robert B.; Sands, Matthew (2010). The Feynman lectures on physics. Vol. II: Mainly electromagnetism and matter (New millennium ed.). New York: BasicBooks. ISBN 978-0465024940.
• Halliday, David; Resnick, Robert; Krane, Kenneth S. (2001). Physics v. 1. New York: John Wiley & Sons. ISBN 0-471-32057-9.
• Kleppner, Daniel; Kolenkow, Robert J. (2010). An introduction to mechanics (3. print ed.). Cambridge: Cambridge University Press. ISBN 0521198216.
• Parker, Sybil (1993). "force". Encyclopedia of Physics. Ohio: McGraw-Hill. p. 107,. ISBN 0-07-051400-3.
• Sears F., Zemansky M. & Young H. (1982). University Physics. Reading, Massachusetts: Addison-Wesley. ISBN 0-201-07199-1.
• Serway, Raymond A. (2003). Physics for Scientists and Engineers. Philadelphia: Saunders College Publishing. ISBN 0-534-40842-7.
• Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed.). W. H. Freeman. ISBN 0-7167-0809-4.
• Verma, H.C. (2004). Concepts of Physics Vol 1. (2004 Reprint ed.). Bharti Bhavan. ISBN 8177091875.
Wikimedia Commons has media related to Forces (physics).
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i neeed fin621 assignment
FIN621 Financial Statement Analysis
Virtual University of Pakistan
:: Case Study::
Sui Southern Gas Company (SSGC) is Pakistan's leading integrated gas Company. The
company is engaged in the business of transmission and distribution of natural gas
besides construction of high pressure transmission and low pressure distribution systems.
You have been given the financial statements of the Sui Southern Gas Company for the
year 2008 under the below link. Financial statements are also available under the head
Downloads” on LMS of FIN621.
Financial statements-2008
(Do NOT take consolidated statements presented under this link as we want you to
consider stand alone financial statements 2008 of the company)
You are required to analyze the financial statements of this company with maximum of
your knowledge. At the end you will be able to know the techniques, application of
Financial Statement Analysis; similarly this practice will help you for the Final Project.
Calculate the following ratios; you are required to provide the formulas,
calculations and interpretation of each ratio. Make sure your interpretation will be
brief and explanatory along with each ratio.
Part 1
Liquidity Ratios: (3x2=6 marks)
1. Acid Test Ratio/Quick ratio
2. Sales to working Capital
3. How working capital does help the financial analyst?
Part 2
Solvency Ratio: (4x2=8 marks)
1. Debt-to-Equity Ratio
2. Time Interest Earned Ratio
3. Fixed Charge Coverage Ratio
4. Define difference between both ratios (Time Interest Earned
& Fixed Charge Coverage)? To be very precise mention why do we
calculate both?
Part 3
Profitability Ratios: (8x2=16 marks)
1. Gross Profit Margin
2. Operating Profit Margin Return on sales
3. Pretax Margin
4. Net Profit Margin
FIN621 Financial Statement Analysis
Virtual University of Pakistan
5. Return on Equity (ROE)
6. Return on Assets (ROA) Return on Investment (ROI)
7. DuPont Return on Assets
8. What is the need for calculating DuPont Return on Assets when we
already have Return on Assets ratio calculated?
Part 4
Activity Ratios: (5x2=10 marks)
1. Inventory Turnover
2. Accounts Receivable Turnover
3. Accounts payable turnover
4. Average collection period
5. Average payment period
Bookish interpretations are not required. Be specific to the company given in the
assignment.
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## monroe17 2 years ago A motorist drives south at 28.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.80 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.80 min trip, find the following values. a) total vector displacement ___ m (magnitude) b) __° south of west
1. monroe17
so far i have.. 28*3*60=5040 m 25*2.8*60=4200 m 30*1*60=1800 m now, im stuck..
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## Medals
#### Game Completed 10 Points
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CrazyDad is away for his much needed vacation, but once he arrives at the hotel, he finds out his room is facing a giant neon sign. Help CrazyDad turn off the sign or he’ll go crazy!
## Reviews
reply to @lennytheduck down there: the math is correct. If two coconuts = 2, then if one coconut is x then 2x = 2. Only 1 can make this equation correct, so the answer is one coconut. Great game btw, bit easy though
It doesn't play for me. Just a black screen after the loading screen.
Fun, although the math one is wrong. We know that 2 coconuts=2, but we do not know the value of 1 coconut. Therefore the answer is x.
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## ››Convert gram calorie to therm [U.S.]
gram calorie therm [U.S.]
Did you mean to convert gram calorie to therm [Europe] therm [U.S.]
How many gram calorie in 1 therm [U.S.]? The answer is 25199579.530795.
We assume you are converting between gram calorie and therm [U.S.].
You can view more details on each measurement unit:
gram calorie or therm [U.S.]
The SI derived unit for energy is the joule.
1 joule is equal to 0.23890295761862 gram calorie, or 9.4804342797335E-9 therm [U.S.].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between gram calories and therm [U.S.].
Type in your own numbers in the form to convert the units!
## ››Want other units?
You can do the reverse unit conversion from therm [U.S.] to gram calorie, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Therm
The therm (symbol thm) is a non-SI unit of heat energy. It was defined in the United States in 1968 as the energy equivalent of burning 100 cubic feet of natural gas at standard temperature and pressure. In the US gas industry its SI equivalent is defined as exactly 100,000 BTU59°F or 105.4804 megajoules. Public utilities in the U.S. use the therm unit for measuring customer usage of gas and calculating the monthly bills.
## ››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!
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# Ab cd 6 10 pts provide a proof by induction that 2 n
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ab \ \ \ cd 6. (10 pts.) Provide a proof by induction that 2 n 2n for every positive integer n. Be explicit concerning your use of the induction hypothesis in the inductions step.
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TEST-3/MAD2104 Page 4 of 4 7. (15 pts.) (a) How many edges does a tree with 37 vertices have? (b) What is the maximum number of leaves that a binary tree of height 10 can have? (c) If a full 3-ary tree has 24 internal vertices, how many vertices does it have? 8. (10 pts.) Suppose that R is an equivalence relation on a nonempty set A. Recall that for each a ε A, the equivalence class of a is the set [a] = {s | (a,s) ε R}. Prove the following proposition: If (a,b) ε R, then [a] = [b]. Hint: The issue is the set equality, [a] = [b], under the hypothesis that (a,b) ε R. So pretend (a,b) ε R and use this to show s ε [a] s ε [b], and s ε [b] s ε [a]. Be explicit regarding your use of the relational properties of R.
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ab cd 6 10 pts Provide a proof by induction that 2 n 2n for...
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https://en.wikipedia.org/wiki/Pirate_loot_problem
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# Pirate game
(Redirected from Pirate loot problem)
The pirate game is a simple mathematical game. It is a multi-player version of the ultimatum game.
## The game
There are 5 rational pirates (in strict order of seniority A, B, C, D and E) who found 100 gold coins. They must decide how to distribute them.
The pirate world's rules of distribution say that the most senior pirate first proposes a plan of distribution. The pirates, including the proposer, then vote on whether to accept this distribution. If the majority accepts the plan, the coins are disbursed and the game ends. In case of a tie vote, the proposer has the casting vote. If the majority rejects the plan, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again. The process repeats until a plan is accepted or if there is one pirate left.[1]
Pirates base their decisions on four factors. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins each receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal.[2] And finally, the pirates do not trust each other, and will neither make nor honor any promises between pirates apart from a proposed distribution plan that gives a whole number of gold coins to each pirate.
## The result
It might be expected intuitively that Pirate A will have to offer the other pirates with most of the gold to increase the chances of his plan being accepted. However, this is quite far from the theoretical result. When each of the pirates votes they won't just be thinking about the current proposal, but all other outcomes down the line. Also because the order of seniority is known in advance, each of them can accurately predict how the others might vote in any scenario.
This is apparent if we work backwards. The last possible scenario would have all the pirates except D and E thrown overboard. Since D is senior to E, he has the casting vote. So D would obviously propose to keep 100 for himself and 0 for E, and so this is the allocation.
If there are three left (C, D and E) C knows that D will offer E 0 in the next round; therefore, C has to offer E 1 coin in this round to win E's vote, and get C's allocation through. Therefore, when only three are left the allocation is C:99, D:0, E:1.
If B, C, D and E remain, B considers being thrown overboard when deciding. To avoid being thrown overboard, B can simply offer 1 to D. Because B has the casting vote, the support only by D is sufficient. Thus B proposes B:99, C:0, D:1, E:0. One might consider proposing B:99, C:0, D:0, E:1, as E knows it won't be possible to get more coins, if any, if E throws B overboard. But, as each pirate is eager to throw the others overboard, E would prefer to kill B, to get the same amount of gold from C.
Assuming A knows all these things, A can count on C and E's support for the following allocation, which is the final solution:
• A: 98 coins
• B: 0 coins
• C: 1 coin
• D: 0 coins
• E: 1 coin[2]
Also, A:98, B:0, C:0, D:1, E:1 or other variants are not good enough, as D would rather throw A overboard to get the same amount of gold from B.
## Extension
The solution follows the same general pattern for other numbers of pirates and/or coins, however the game changes in character when it is extended beyond there being twice as many pirates as there are coins. Ian Stewart wrote about Steve Omohundro's extension to an arbitrary number of pirates in the May 1999 edition of Scientific American and described the rather intricate pattern that emerges in the solution.[2]
Supposing there are just 100 gold pieces, then:
• Pirate #201 as captain can stay alive only by offering all the gold one each to the lowest odd-numbered pirates, keeping none.
• Pirate #202 as captain can stay alive only by taking no gold and offering one gold each to 100 pirates who would not receive a gold coin from #201. Therefore, there are 101 possible recipients of these one gold coin bribes being the 100 even-numbered pirates up to 200 and number #201. Since there are no constraints as to which 100 of these 101 he will choose, any choice is equally good and he can be thought of as choosing at random. This is how chance begins to enter the considerations for higher-numbered pirates.
• Pirate #203 as captain will not have enough gold available to bribe a majority, and so will die.
• Pirate #204 as captain has #203's vote secured without bribes: #203 will only survive if #204 also survives. So #204 can remain safe by reaching 102 votes by bribing 100 pirates with one gold coin each. This seems most likely to work by bribing odd-numbered pirates optionally including #202, who will get nothing from #203. However, it may also be possible to bribe others instead as they only have a 100/101 chance of being offered a gold coin by pirate #202.
• With 205 pirates, all pirates bar #205 prefer to kill #205 unless given gold, so #205 is doomed as captain.
• Similarly with 206 or 207 pirates, only votes of #205 to #206/7 are secured without gold which is insufficient votes, so #206 and #207 are also doomed.
• For 208 pirates, the votes of self-preservation from #205, #206, and #207 without any gold are enough to allow #208 to reach 104 votes and survive.
In general, if G is the number of gold pieces and N (> 2G) is the number of pirates, then
• All pirates whose number is less than or equal to 2G + M will survive, where M is the highest power of 2 that does not exceed N – 2G.
• Any pirates whose number exceeds 2G + M will die.
• Any pirate whose number is greater than 2G + M/2 will receive no gold.
• There is no unique solution as to who gets one gold coin and who does not if the number of pirates is 2G+2 or greater. A simple solution dishes out one gold to the odd or even pirates up to 2G depending whether M is an even or odd power of 2.
Another way to see this is to realize that every Mth pirate will have the vote of all the pirates from M/2 to M out of self preservation since their survival is secured only with the survival of the Mth pirate. Because the highest ranking pirate can break the tie, the captain only needs the votes of half of the pirates over 2G, which only happens each time (2G + a Power of 2) is reached.
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# Which of the following phenomena are responsible for rainbows?
Which of the following phenomena of light are responsible for the formation of a rainbow?
1) reflection, refraction, dispersion
2) refraction, dispersion, total internal reflection
3) refraction, dispersion, internal reflection
4) dispersion, scattering, total internal reflection
I know the answer’s got to be 2) or 3). My book says 3) but does not give an explanation.
My confusion lies here: What is the difference between internal reflection and total internal reflection?
The word "total" makes the difference. Depending on the angle of incidence the reflection is either total or partial.
Quoted from Encyclopedia Britannica - Total internal reflection (emphasis added by me):
Total internal reflection, in physics, complete reflection of a ray of light within a medium such as water or glass from the surrounding surfaces back into the medium. The phenomenon occurs if the angle of incidence is greater than a certain limiting angle, called the critical angle.
[...]
At all angles less than the critical angle, both refraction and reflection occur in varying proportions.
Applying this to the rainbow:
According to Wikipedia - Rainbow - Mathematical derivation the angle of incidence inside the rain drop ($$\beta$$ in the image below) is $$\beta_\text{max}\approx 40.2°$$.
According to Wikipedia - Total internal reflection the critical angle for light from water to air is $$\theta_c=49°$$.
So you have $$\beta_\text{max} < \theta_c$$, and therefore you have partial reflection. This makes 3) the correct answer.
• So internal reflection is at less than the critical angle and total internal reflection is at the critical angle?
– Dora
Commented Jan 26, 2020 at 12:05
• @Dora Yes. And total internal reflection is also at more than the critical angle. Commented Jan 26, 2020 at 12:12
• thanks a lot. It’s all cleared up now. :)
– Dora
Commented Jan 26, 2020 at 13:19
• But is the answer 2) or 3)?
– Dora
Commented Jan 26, 2020 at 13:19
• @Dora See my additions to the answer. Commented Jan 26, 2020 at 13:51
The answer is 3), although the question misses the actual cause of rainbows.
Consider a ray of a single color of light that hits the drop with angle of incidence A. Some of this light enters the drop with an angle of refraction B=arcsin(sin(A)/n)), where n is the index of refraction. Note that B is less than the critical angle C.
The surface normal of a spherical raindrop contains a radius of the sphere. The path that this light takes inside the drop forms an isosceles triangle with the radii at either end of the path. This means that the angle of incidence at the back of the drop is also B, which is less than the critical angle. Total Internal Reflection is impossible.
So what causes a rainbow? The ray deflects through the angle A-B when entering the drop, 180°-2B when it reflects internally off of the back, and another A-B when it exits. The total deflection is 180°+2A-4B, so it makes an angle D=4B-2A with the original ray.
If you plot D as a function of A, for 0°<=A<90° and n~=1.33, you will find that D(A) has a maximum somewhere near A=60° and D=40°. The intensity of the light will be inversely proportional to D'(A), so it is infinite at this maximum.
This isn't an energy-conservation paradox, since the band of angles over which it is infinitely bright, is infinitesimally small. What it does mean, is that red light reflects at all angles from 0° to about 42°, and is much brighter at 42°. Dispersion means that the maximum is at a different angle for violet light, about 40°. These bright bands become the rainbow bands, and the sky inside the rainbow is a little brighter than outside.
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https://uk.mathworks.com/matlabcentral/cody/problems/895-generate-n-equally-spaced-intervals-between-l-and-l/solutions/2041851
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Cody
# Problem 895. Generate N equally spaced intervals between -L and L
Solution 2041851
Submitted on 2 Dec 2019 by Michael Koscelník
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
y = your_fcn_name(100,100); assert( y(1) == -100 && y(end) == 100 && mean(diff(y))== 2 && std(diff(y)) == 0 );
g = 200 f = 2 h = 100 s = -100 s = Columns 1 through 30 -100 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 -90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 -90 -88 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 -90 -88 -86 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 -90 -88 -86 -84 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 -90 -88 -86 -84 -82 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 -90 -88 -86 -84 -82 -80 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 -90 -88 -86 -84 -82 -80 -78 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 -90 -88 -86 -84 -82 -80 -78 -76 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 101 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -98 -96 -94 -92 -90 -88 -86 -84 -82 -80 -78 -76 -74 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...
2 Pass
y = your_fcn_name(200,100); assert( y(1) == -100 && y(end) == 100 && mean(diff(y))== 1 && std(diff(y)) == 0 );
g = 200 f = 1 h = 200 s = -100 s = Columns 1 through 30 -100 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 120 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 121 through 150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 151 through 180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 181 through 201 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -99 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 120 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 121 through 150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 151 through 180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 181 through 201 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -99 -98 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 120 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 121 through 150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 151 through 180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 181 through 201 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -99 -98 -97 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 120 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 121 through 150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 151 through 180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 181 through 201 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -99 -98 -97 -96 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 120 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 121 through 150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 151 through 180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 181 through 201 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -99 -98 -97 -96 -95 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 120 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 121 through 150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 151 through 180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 181 through 201 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -99 -98 -97 -96 -95 -94 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 31 through 60 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 61 through 90 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 91 through 120 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 121 through 150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 151 through 180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 181 through 201 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 100 s = Columns 1 through 30 -100 -99 -98 -97 -96 -95 -94 -93 0 0 0...
3 Pass
y = your_fcn_name(2,100); assert( y(1) == -100 && y(end) == 100 && mean(diff(y))== 100 && std(diff(y)) == 0 );
g = 200 f = 100 h = 2 s = -100 s = -100 0 100 s = -100 0 100 y = -100 0 100
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http://www.pascalgamedevelopment.com/showthread.php?3806-Raytracing-steps/page2
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# Thread: Raytracing steps
1. ## Raytracing steps
Originally Posted by jdarling
Hey Nitrogen, any chance you would post up a simple sample application showing this off. As I said, I've never played with ray tracing, but I'd love to see code around it
Hey thanks! It's one of those projects that I didnt want to post up in it's original state because it's so inflexible and horribly stale... I had grand schemes for making it like a little 3D modeller with the 4 viewport business etc. But it never got past that point. Yea I used the Flipcode articles about it.. Really good tutorial that.
The image above took 3.2 seconds to render 640x480 on my AthlonX2 4200.
Yea, raytracing isnt that big of a visual difference from good pixel shaders (and the pixel shaders run a hundred times faster), but you can still use it for small apps like building webpage buttons or rendering logos etc.
2. ## Raytracing steps
There is a realtime ray tracing version of quake.
http://www.q4rt.de/
http://www.theinquirer.net/default.aspx?article=36452
3. ## Raytracing steps
to Nitrogen >> Have you solved a sampling(antialiasing) in your raytracer ?
4. ## Raytracing steps
[pascal]
Ray.Dir := Normalise(VecSub(Pos, cam.Origin));
Col:= Raytrace(Ray, 0 , HitPrim, Ch);
if (LastPrim <> Ch) or (LastPrims[Rx] <> Ch) then //If last hit primitive is different to this one
begin
For SubX := -1 to 1 do
for SubY := -1 to 1 do
if (SubX <> 0) or (SubY <> 0) then //Already sampled the center point.
begin
D := Pos;
D := VecAdd(D,VecMult(A, SubX*0.5));
D := VecAdd(D,VecMult(B, SubY*0.5)); //Adds a slight offset in 3D space
Ray.Dir := Normalise(VecSub(D, cam.Origin));
AccCol:= Raytrace(Ray, 0 , HitPrim, Ch);
Col := VecAdd(Col, AccCol); //Add subsample to accumulator color
end;
P^ := Rgb(round(Col[0]*255/9), round(Col[1]*255/9), round(Col[2]*255/9)); //average all 9 subsamples.
end
else
P^ := Rgb(round(Col[0]*255), round(Col[1]*255), round(Col[2]*255));
[/pascal]
Basically you just send off 9 rays for 1 pixel. Each with a slight offset.
The trick comes when you record the last hit primitive, because you dont need to antialias pixels inside primitives, only on the boundaries between primitives.
5. ## Raytracing steps
The blue pixels are the only ones that are antialiased: Note how you must antialias both the primitive boundaries and the shadow boundaries.
Heres that scene again without the blue:
One problem with this approach is that you can only have an antialiased area of one pixel between primitives or shadows.
6. ## Raytracing steps
thank you nitrogen again .
7. ## Raytracing steps
Not a problem , post your results here for everyone to see.
Raytracing is one of the more beautiful aspects of programming.
8. ## Raytracing steps
nothing special in my render :-) Just simple test like this :
It will be looong way until my render will be able to do something like this :-)
It is unbelievably what today renderer's like Maxwell render or VRay can do. Really amazing things.
9. ## Raytracing steps
Can you do refraction yet?
10. ## Raytracing steps
yes, I have implemented refraction algorithm, from what I see on the flipcode website, tere is raytracing tutorial...
however, I need to implement transparency to objects...
I think refraction is more visible when the object is little transparent...
I try to do some other things now, exactly improve the speed of rendering...
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Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search
## Direct Proof of De Moivre's Theorem
In §2.10, De Moivre's theorem was introduced as a consequence of Euler's identity:
To provide some further insight into the mechanics'' of Euler's identity, we'll provide here a direct proof of De Moivre's theorem for integer using mathematical induction and elementary trigonometric identities.
Proof: To establish the basis'' of our mathematical induction proof, we may simply observe that De Moivre's theorem is trivially true for . Now assume that De Moivre's theorem is true for some positive integer . Then we must show that this implies it is also true for , i.e.,
(3.2)
Since it is true by hypothesis that
multiplying both sides by yields
(3.3)
From trigonometry, we have the following sum-of-angle identities:
These identities can be proved using only arguments from classical geometry.3.8Applying these to the right-hand side of Eq.(3.3), with and , gives Eq.(3.2), and so the induction step is proved.
De Moivre's theorem establishes that integer powers of lie on a circle of radius 1 (since , for all ). It therefore can be used to determine all of the th roots of unity (see §3.12 above). However, no definition of emerges readily from De Moivre's theorem, nor does it establish a definition for imaginary exponents (which we defined using Taylor series expansion in §3.7 above).
Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search
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Vikolers6
## Answered question
2021-12-27
Find the general solution to the following differential equations.
$y6{y}^{\prime }+13y=0$
### Answer & Explanation
alexandrebaud43
Beginner2021-12-28Added 36 answers
Given: $y6{y}^{\prime }+13y=0$
$y6{y}^{\prime }+13y=0$
A second order linear, homogenous ODE has the form $ayb{y}^{\prime }+cy=0$
For an equation $ayb{y}^{\prime }+cy=0$, assume a solution of the form ${e}^{\lambda t}$
Rewrite the equation with $y={e}^{\lambda t}$
$\left({e}^{\lambda t}\right)6{\left({e}^{\lambda t}\right)}^{\prime }+13{e}^{\lambda t}=0$
${e}^{\lambda t}\left({y}^{2}+6y+13\right)=0$
${y}^{2}+6y+13=0$
Roots of the quadratic equation in the form $a{x}^{2}+bx+c=0$ is
${x}_{1,2}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Here, $a=1,b=6\text{and}c=13$
${y}_{1,2}=\frac{-6±\sqrt{{6}^{2}-4×1×13}}{2×1}$
${y}_{1,2}=\frac{-6±\sqrt{36-52}}{2}$
${y}_{1,2}=-3-2i,-3+2i$
For two complex roots ${y}_{1}\ne {y}_{2}$, where ${y}_{1}=-3-2i,{y}_{2}=-3+2i$
So, the general solution takes the form: $y={e}^{\alpha t}\left({c}_{1}\mathrm{cos}\left(\beta t\right)+{c}_{2}\mathrm{sin}\left(\beta t\right)\right)$
$y={e}^{-3t}\left({c}_{1}\mathrm{cos}\left(2t\right)+{c}_{2}\mathrm{sin}\left(2t\right)\right)$
xandir307dc
Beginner2021-12-29Added 35 answers
$y={e}^{kx}$
${k}^{2}+6k+13=0$
$D={b}^{2}-4ac={6}^{2}-4\cdot 1\cdot 13=-16$
$\sqrt{D}=±4i$
${k}_{1,2}=-3±2i$
Answer: $y={e}^{-3x}\left({C}_{1}\mathrm{cos}2x+{C}_{2}\mathrm{sin}2x\right)$
karton
Expert2022-01-10Added 613 answers
We were given the next equation
The roots of the characteristic equation
are ${\lambda }_{1}=-3+2i,{\lambda }_{2}=-3-2i$
The general solution of Eq. (1) is, by the Theorem 4.3.2, given by
$y\left(t\right)={c}_{1}{e}^{\mu t}\mathrm{cos}vt+{c}_{2}{e}^{\mu t}\mathrm{sin}vt$
where ${\lambda }_{1,2}=\mu ±iv$. In this case, the solution is
$y\left(t\right)={c}_{1}{e}^{-3t}\mathrm{cos}2t+{c}_{2}{e}^{-3t}\mathrm{sin}2t$
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0
# What times what equals 320?
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### What is 64x5?
64 times 5 equals 320.
80 x 4 = 320
10
### What could you do to the equation 40 times y equals 320 sides equal?
If: 40y = 320 Then: y = 8
40 x 8
### What times 125 percent equals 400?
125% of 320 = 400
8 * 40 = 320
### What times what equals 2000?
10 times 200 equals 2000
### 0.32 equals how many thousands?
0.32 equals 320 thousands.
### What times 60 equals 1600?
you cant multiply 60 times something to equal 1600 its not possible but you can multiply 1 times 1600 equals 1600 2 times 800 equals 1600 4 times 400 equals 1600 5 times 320 equals 1600 8 times 200 equals 1600 10 times 160 equals 1600 16 times 100 equals 1600 20 times 80 equals 1600 25 times 64 equals 1600 32 times 50 equals 1600 40 times 40 equals 1600 50 times 32 equals 1600 64 times 25 equals 1600 80 times 20 equals 1600 100 times 16 equals 1600 160 times 10 equals 1600 200 times 8 equals 1600 320 times 5 equals 1600 400 times 4 equals 1600 800 times 2 equals 1600 1600 times 1 equals 1600 to equal 1600
### Does 56 percent of 320 equals 169.2?
56% of 320 = 179.2%
40 x 8=320
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# indefinite integration involving atan
I need the indefinite integration of this: $$\int \left[1 + \cos\left(K + \arctan\left(\frac{a+bx\ }{c + dx\ }\right)\right) \right] dx$$
If it helps, we can turn the form above into the following one. Either you can solve. $$\int \left[1 + \cos\left(K + \arccos\left(\frac{A+Bx\ }{\sqrt(x^2+Cx+D)\ }\right)\right) \right] dx$$
By $$\sqrt {u}$$ I mean $$u^{0.5}$$
Thank you, Mahdi
• What have you attempted so far? Please show your work – DavidG Dec 9 at 3:35
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I am trying to solve a 1D transient heat conduction problem using the finite volume method (FVM), with a fully implicit scheme, in polar coordinates. It is a transient homogeneous heat transfer in spherical coordinates. ANALYTICAL HEAT TRANSFER Mihir Sen Department of Aerospace and Mechanical Engineering University of Notre Dame Notre Dame, IN 46556 May 3, 2017. A 2-stage Runge–Kutta method is used to evolve the solution forward in time, where the advective fluxes are part of the temporal integration. 5: 9a,9b,9c,9d; Ex 9. International Journal of Mathematics Trends and Technology (IJMTT) – Volume 46 Number 3 June 2017 Numerical Simulation of 1D Heat Conduction in Spherical and Cylindrical Coordinates by Fourth-Order Finite Difference Method Letícia Helena Paulino de Assis1,a, Estaner Claro Romão1,b Department of Basic and Environmental Sciences, Engineering School of Lorena, University of São Paulo. The function supports inputs in 1D, 2D, and 3D. 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Weizhong Dai, Lixin Shen, Raja Nassar, A convergent three‐level finite difference scheme for solving a dual‐phase‐lagging heat transport equation in spherical coordinates, Numerical Methods for Partial Differential Equations, 10. RS11 Hollow sphere, a < r < b, with G = 0 (Dirichlet) at r = a and G = 0 (Dirichlet) at r = b. PDEs in 3D Cartesian Coordinates Consider the wave equation. We use the FHE to derive the Fractional EBE (FEBE) for the (2D) surface temperature distribution that was derived 15 elsewhere by phenomenological arguments, generalizing the HEBE to 0< H ≤1. 2 heat conduction equation in spherical. steady state conduction: one-dimensional problems ; 2. I might actually dedicate a full post in the future. 8 Laplace’s Equation in Rectangular Coordinates 49 3. , – The geometrical domain were defined in a 1D polar coordinate system and adapted for numerical simulation according to. $\endgroup$ – Roan May 10 '17 at 3:32. 1 introduction ; 2. Spherical Harmonics (DEF and properties) Example 1. 616 and section 16. 3D Cartesian Coordinates We can describe all space using coordinates (x, y, z), each variable ranging from -∞ to +∞. Finite integral transform techniques are applied to solve the one-dimensional (1D) dual-phase heat conduction problem, and a comprehensive analysis is provided for general time-de. 3: 8a,8b; Ex 7. 3 Heat Equation in 2D. There are three common ones used in 3D, based on the symmetry of the problem: rectangular, cylindrical, and spherical. have dealt with polar and spherical coordinate systems. 1 Homogeneous IBVP. Although eq. (Compare the equation above with equation (3). This alternative use of coordinates will be important when we discuss black holes and cosmology. Parameters β and T 0 may differ from part to part of the boundary. 3 the heat conduction equation for isotropic materials ; 2. Solution in Cartesian and plane polar coordinates by separation of variables and Fourier series. , 2 in y-dir. Heat Equation in Cylindrical and Spherical Coordinates. Separation of variables in cylindrical and spheical coordinates. General Heat Conduction Equation Spherical Coordinates. The 1D Harmonic Oscillator The harmonic oscillator is an extremely important physics problem. 1 Lagrangian for unconstrained systems For a collection of particles with conservative forces described by a potential, we have in inertial cartesian coordinates m¨x i= F i: The left hand side of this equation is determined by the kinetic energy func-tion as the time derivative of the momentum p i = @[email protected]_. 3 the heat conduction equation for isotropic materials ; 2. The rate of heat transfer P (energy per unit time) is proportional to the temperature difference and the contact area A and inversely proportional to the distance d between the objects. We use the FHE to derive the Fractional EBE (FEBE) for the (2D) surface temperature distribution that was derived 15 elsewhere by phenomenological arguments, generalizing the HEBE to 0< H ≤1. The finite difference method attempts to solve a differential equation by estimating the differential terms with algebraic expressions. for cartesian coordinates. It is a mathematical statement of energy conservation. , 2 in y-dir. (2019) Absorbing boundary conditions for time-dependent Schrödinger equations: A density-matrix formulation. 10 , the modified Bessel function [defined in Equation ( 435 )] is a solution of the modified Bessel equation that is well behaved at , and badly behaved as. The heat flux is a function of the temperature gradient, and is modeled by the Fourier’s Law: qj =−k j ∂T ∂xj, in which kj is the thermal conductivity in the xj direction, that represents the spatial independent variable. Rearrange terms like this… ½at 2 + v 0 t − ∆s = 0. Laplace's equation in 1D, 2D, 3D using Cartesian, polar, and spherical co-ordinates. Based on the authors’ own research and classroom experience, this book contains two parts, in Part (I): the 1D cylindrical coordinates, non-linear partial differential equation of transient heat conduction through a temperature dependent thermal conductivity of a thermal insulation material is solved analytically using Kirchhoff’s. Next: Laplace Equation. In A-SURF, the finite volume method is used to solve the conservation equations in spherical coordinates. Let be a kinematically admissible variation of the deflection, satisfying at. However, there are certain high-symmetry cases when it is possible to separate ariablesv in some convenient coordinate system and reduce the Schrodinger equation to one-dimensional problems. We have a new eigenfunction! The hyperbolic sine makes an appearance. Separation of variables and Green functions in cartesian, spherical, and cylindrical coordinates 2. The general equations for heat conduction are the energy balance for a control mass, d d E t QW = + , and the constitutive equations for heat conduction (Fourier's law) which relates heat flux to temperature gradient, q kT =−∇. The continuity equation then reduces to ∇·v = 0, (7) which in Cartesian coordinates is ∂u ∂x + ∂v ∂y + ∂w ∂z = 0. The robust method of explicit ¯nite di®erences is used. 4 Finite element equations 8. x, L, t, k, a, h, T. Hi guys, Here is a 1D dynamic model I built today simulating heat transfer in a 21-segment bar. There is no heat generation. A semi-analytical solution for temperature and heat flux is presented using the. This the first tutorial on modeling heat transfer at a very introductory level. 2) is also called the heat equation and also describes the distribution of a heat in a given region over time. 111-117) and 3D. ?, which states exactly that a convolution with a Green's kernel is a solution, provided that the convolution is sufficiently often differentiable (which we showed in part 1 of the proof). A graphics showing cylindrical coordinates:. Summary of Styles and Designs. The advective fluxes are calculated by solving a 1D Riemann problem on each face of the nodal control volume. Wave equation Partial Differential Equations : Separation of Variables (6 Problems) Cartesian Coordinates Problem Separation of variables, sine and cosine expansion. Equation 7: The metric in 2D space expressed both in Cartesian and spherical coordinates. The Equation of Energy in Cartesian, cylindrical, and spherical coordinates for Newtonian fluids of constant density, with source term 5. Now we will solve the steady-state diffusion problem 5. σ = 0, so that equation (1c) reduces to (1d), which is properly called the heat diffusion equation and if, steady state is considered, (1d) may be written as equation (1e), called the Poisson equation, which can be. 3): 2, 3 12. 6 Spherical Coordinates. The equation will now be paired up with new sets of boundary conditions. In this case, an. , – The geometrical domain were defined in a 1D polar coordinate system and adapted for numerical simulation according to. T sin r 1 r T r T r 2 k q p 2 2 2 2 2 2 2 2 2 2. (4) can also be derived from polar coordinates point of view. (3) is useful for theoretical considerations, its conservative form is more suitable for finite-volume discretization. The base coordinates would be cartesian and they would be always implicitly de ned in any domain. The one-dimensional heat conduction equations based on the dual-phase-lag theory are derived in a unified form which can be used for Cartesian, cylindrical, and spherical coordinates. In 1958, Englman. There are three common ones used in 3D, based on the symmetry of the problem: rectangular, cylindrical, and spherical. Later on,9 John Dougall (1867–1960) derived three-dimensional Green’s functions in cylindrical and spherical coordinates. The grid. 3-6] Helmholtz' equation (19) [16. From your link, 1d (in radial direction) spherical problems can always be converted into a 1d cartesian diffusion equation with a change of variables. The right hand side could be generalized to f 2H 1(). Finite integral transform techniques are applied to solve the one-dimensional (1D) dual-phase heat conduction problem, and a comprehensive analysis is provided for general time-de. Note that 0 r Cexp i k r is the solution to the Helmholtz equation (where k2 is specified) in Cartesian coordinates In the present case, k is an (arbitrary) separation constant and must be summed over. Steady 1-D Summary GF in slabs, rectangle, and parallelepiped for 3 types of boundary conditions These GF have components in common: 9 eigenfunctions and 18 kernel functions Alternate forms of each GF allow efficient numerical. 2 To solve partial differential equations (the TISE in 3D is an example of these equations), one can employ the method of separation of variables. Hi guys, Here is a 1D dynamic model I built today simulating heat transfer in a 21-segment bar. Step 3 We impose the initial condition (4). The diffusion equation is a parabolic partial differential equation. General Dirichlet problem on a ball. 1) the term (⁄)represents the heat accumulated in the tissue, characterizes the heat conduction and ( )is the heat sink term due to the removal of heat by blood in the microvasculature. how mixing by random molecular motion smears out the temperature. Many problems such as plane wall needs only one spatial coordinate to describe the temperature distribution, with no internal generation and constant thermal conductivity the general heat equation has the following form t T x T ∂ ∂ = ∂ ∂ α 1 2 2 (6. 20) we obtain the general solution. Solving the eigenvalue problem. 3 the heat conduction equation for isotropic materials ; 2. 2 Integral (weak) form of the governing equations of linear elasticity 8. 1D Heat equation; 3. 2 Semihomogeneous PDE. Spherical coordinates are depicted by 3 values, (r, θ, φ). 23, A m = 4πr 1 r 2 … (3. (5) and (4) into eq. 2D heat, wave, and Laplace’s equation on rectangular domains F. 1 Homogeneous IBVP. Height as a Vertical Coordinate Advantages – intuitive, easy to construct equations Disadvantage – difficult to represent surface of Earth because different places are at different heights. 87 Figure 3. 2) is also called the heat equation and also describes the distribution of a heat in a given region over time. For 2D heat conduction problems, we assume that heat flows only in the x and y-direction, and there is no heat flow in the z direction, so that , the governing equation is: In cylindrical coordinates, the governing equation becomes:. Let assume a uniform reactor (multiplying system) in the shape of a sphere of physical radius R. If we are in Cartesian coordinate then d is one and c, the diffusion constant, is for example 0. Moreover, 1D Cartesian, cylindrical or spherical coordinates are used to define the geometry and continuity boundary conditions are imposed to the temperature and heat flow between adjacent layers. 1: Heat conduction through a large plane wall. The parameters of water transport equation were identified by inverse identification based on experimental data from drying of stones, plums without and with skin. This is actually more like finite difference method. σ = 0, so that equation (1c) reduces to (1d), which is properly called the heat diffusion equation and if, steady state is considered, (1d) may be written as equation (1e), called the Poisson equation, which can be. In engineering, there are plenty of problems, that cannot be solved in cartesian coordinates. polar coordinates). It is obtained by combining conservation of energy with Fourier ’s law for heat conduction. Separation of Variables Method for 1D Diffusion Equation in Circular Cylinder Coordinates HTML Maple V R4 : May 14, 1998: Separation of Variables Method for 2D Laplace Equation in Cartesian Coordinates HTML Maple V R4 : May 14, 1998: Separation of variables method for 1D wave equation HTML Maple V R4 : June 2, 1998. The robust method of explicit ¯nite di®erences is used. Create AccountorSign In. 19) for the heat equation with homogeneous Neumann boundary condition as in (13. 1 fourier's law in cylindrical and spherical coordinates ; 2. The angles shown in the last two systems are defined in Fig. If we are in Cartesian coordinate then d is one and c, the diffusion constant, is for example 0. The general heat conduction equations in the rectangular, cylindrical, and spherical coordinates have been developed. Hitting “Reset” sets the 21 segments of the bar to the initial conditions which is a fully customizable initial temperature map. Surface and volume integrals. Our kinetic Lagrangian in spherical coordinates is. c c = c c = c c + c c + c c + c c + c c + 2. Based on the authors’ own research and classroom experience, this book contains two parts, in Part (I): the 1D cylindrical coordinates, non-linear partial differential equation of transient heat conduction through a temperature dependent thermal conductivity of a thermal insulation material is solved analytically using Kirchhoff’s. Traditionally, the thermal °ux term of equation (1) has been modelled by the Fourier’s theory, q(x;t) = ¡krT(x;t), then (1) is a parabolic heat transfer equation (PHTE). Selected Vector Calc. For homogeneous and isotropic material, For steady state unidirectional heat flow in radial direction with no internal heat generation, Heat Generation in Solids Conversion of some form of energy into heat energy in a medium is called heat generation. RADIAL SYSTEMS CONDUCTION 2 Cylindrical and spherical systems often experience temperature gradients in the radial direction only and may therefore be treated as one-dimensional Under steady-state conditions with no heat generation, such systems may be analyzed by using the standard method: find the temperature distribution from the heat. The exact equation solved is given by. Boundary conditions required for the solution of conduction equation 4. into mathematical equations. This is not an easy job since the equation is quadratic. Solving a heat equation in spherical coordinates. I'm trying to solve the heat/diffusion equation in 3d in spherical symmetry $\partial_t f=D\Delta f$. •Explain what contact resistance is and how it can be modeled. The heat conduction equation in 1D spherical coordinates is 1 ∂T 2 ∂T ∂ 2T = + 2 ∂r r ∂r α ∂t 10. The governing equation comes from an energy balance on a differential ring element of the fin as shown in the figure below. 5 Simple 1D. 3 Incompressible continuity equation If the fluid is incompressible, ρ = constant, independent of space and time, so that Dρ/Dt = 0. It can be seen that the complexity of these equations increases from rectangular (5. Laplace transforms. Appendix A: CFD Process Appendix B: Governing Equations of Incompressible Newtonian Fluid in Cylindrical and Spherical Polar Coordinates Appendix C: Dimensionless Numbers Appendix D: Differences between Impulse and Reaction Turbines Appendix E: Organic Rankine Cycle (ORC) Appendix F: Applications of Cryogenic System in Tooling Appendix G: The Cryogenic Air Separation Process Appendix H. have dealt with polar and spherical coordinate systems. The general heat conduction equation in cylindrical coordinates can be obtained from an energy balance on a volume element in cylindrical coordinates and using the Laplace operator, Δ, in the cylindrical and spherical form. For parabolic equations, of which the heat conduction equation u t u xx ¼ 0 ð9Þ is the simplest example, the subsidiary conditions always include some of initial type and may also include some of boundary type. Hence one IC needed. Ask Question Asked 3 years, 7 months ago. Then thickness δ will be equal to (r 2 – r 1) and the areas A will be an equivalent area A m. ferent thermo-physical properties in spherical and Cartesian coordinates. transport equation (BTE) [6]. Outline your steps clearly because we gave you the solution! B. Flow of heat in an infinite solid; in a solid with one plane face at the temperature zero; in a solid with one plane face whose temperature is a function of the time (Riemann's solution); in a bar of small cross section from whose surface heat escapes into air at temperature zero. 1 Thermal resistances - plane wall e R Ak G - cylindrical wall 2 1 21 ln, r 2 r r r S Lk §· ¨¸ ©¹. Their combination: ( ) d d d d dd p A d p AV H Q KA T q n A H t Q kTnA kT A t q kT = = ∆=− ⋅. Delta functions Gravitation Goals: Vectors, curvilinear coordinate systems and kinematics: -[Students should be able to compute dot and cross products and solve vector. Now consider solutions to (4) for two specific coordinate setups. Common application where the Equation of Continuity are used are pipes, tubes and ducts with flowing fluids or gases, rivers, overall processes as power plants, diaries, logistics in general, roads, computer networks and semiconductor technology and more. 10) Laplace (elliptic) Equation – Steady heat flow in 2D, Polar coordinates, circular membrane, cylindrical and spherical coordinates. (1 Lecture) Derivation of heat conduction equation in Cartesian coordinates for heterogeneous, isotropic materials. Solving Helmholtz’s equation will depend on the coordinate system used for the prob-lem. 1 Heat Equation with Periodic Boundary Conditions in 2D. For example, if equation (9) is satisfied for t>0and00, find u(;t) 2H1 0 ();u t2L2() such that (2) (u t;v) + a(u;v) = (f;v); for all v2H1 0 (): where a(u;v) = (ru;rv) and (;) denotes the L2-inner product. (1992) to better approximate surface and bottom velocity shears. Zen+ [3] presented the solution of the initial value problem of the corresponding linear heat type equation using the FeymannKac path integral formulation. Partial Di erential Equations Victor Ivrii Department of Mathematics, University of Toronto c by Victor Ivrii, 2017, Toronto, Ontario, Canada. cshstringtolist: Converts a comma delimited string from csh and breaks it up into separate strings. Geometry )spherical coordinates Radial symmetry )1D approach, r being the dimensional variable. As for 1D simulations, we just use "one" grid for the lateral direction. 1 Homogeneous IBVP. Heat Conduction: Heat flow and heat conduction equations in a hollow infinite cylinder can be generated from Bessel’s differential equation. Height as a Vertical Coordinate Advantages – intuitive, easy to construct equations Disadvantage – difficult to represent surface of Earth because different places are at different heights. Exercise 4: Laplace equation Exercise 5: Laplace equation in spherical and cylindrical coordinates Exercise 6: Multipole expansion, polarization Exercise 7: Dielectrics, electric displacement, bound charges Exercise 8: Electric fileds in matter, Biot Savart Exercise 9: Magnetic fileds in matter. (a) Write the form of the heat equation and the boundary/ , 00) (b) I T(x,t) 2. Numerical simulation by finite difference method 6161 Application 1 - Pure Conduction. Separation of Variables Method for 1D Diffusion Equation in Circular Cylinder Coordinates HTML Maple V R4 : May 14, 1998: Separation of Variables Method for 2D Laplace Equation in Cartesian Coordinates HTML Maple V R4 : May 14, 1998: Separation of variables method for 1D wave equation HTML Maple V R4 : June 2, 1998. Where k is thermal conductivity (W/m. 8 Laplace’s Equation in Rectangular Coordinates 49 3. 5 Simple 1D. Hi guys, Here is a 1D dynamic model I built today simulating heat transfer in a 21-segment bar. The specific requirements or preferences of your reviewing publisher, classroom teacher, institution or organization should be applied. and Heat equations on a cylinder with circular cross-section. Frobenius. x and y are functions of position in Cartesian coordinates. 1-4: Heat equation on infinite 1D domain, Fourier transform pairs, Transforming the heat equation, Heat kernel Week 15: Slack time and review Week 16: Finals week: comprehensive final exam. 1 shows the general equations of motion for incompressible flow in the three principal coordinate systems: rectangular, cylindrical and spherical. , your inhomogeneous Dirichlet boundary. We used 21 nonuniformly (sinusoidally) spaced vertical nodes everywhere, set up as in Lynch et al. Solutions of the heat equation are sometimes known as caloric functions. Figure 8: Spherical coordinates (r, θ, ϕ) ( source ). Separation of Variables in Cartesian Coordinates. The Equation for the Quantum Harmonic Oscillator is a second order differential equation that can be solved using a power series. Replace (x, y, z) by (r, φ, θ). Transient 1-D. For 2D heat conduction problems, we assume that heat flows only in the x and y-direction, and there is no heat flow in the z direction, so that , the governing equation is: In cylindrical coordinates, the governing equation becomes:. It is good to begin with the simpler case, cylindrical coordinates. General Dirichlet problem on a ball. [14] presented a 1D-model for packed bed drying using the local volume averaging (LVA) approach, with local thermal equilibrium in each elementary volume in order to derive tran-sient heat and mass transfer equations, solved by means of an implicit numerical method. Best convergence for (t - ) small:. \reverse time" with the heat equation. 3 Differential control volume, dr. For profound studies on this branch of engineering, the interested reader is recommended the definitive textbooks [Incropera/DeWitt 02] and [Baehr/Stephan 03]. c c = c c = c c + c c + c c + c c + c c + 2. After that we will present the main result of this paper in Sect. We show that (∗) (,) is sufficiently often differentiable such that the equations are satisfied. The Helmholtz equation is extremely significant because it arises very naturally in problems involving the heat conduction (diffusion) equation and the wave equation, where the time derivative term in the PDE is replaced by a constant parameter by applying a Laplace or Fourier time transform to the PDE. 2) I write the momentum equation in 1-D spherical coordinates and I have extra geometric source terms compared with the Cartesian case. Hot Network Questions. We set up the basic problem on the rectangle and solve by separating variables. If you follow this series and spend your own effort in developing your own models you will be able to model heat transfer in very complex shapes (1D, 2D, 3D) in a short time and with the basic understanding of a 12 year old school boy. and can now express the Hamiltonian in spherical coordinates. Show that if. Substitute in Fourier’s Law of Heat Conduction integrate again. We now wish to establish the differential equation relating temperature in the fin as a function of the radial coordinate r. Note: Citations are based on reference standards. The presence of various compounds in the system improve the complexity of the heat transport due to the heat absorption as the binders are decomposing and transformed into gaseous products due to significant heat shock. 33: General analytical solution of a 2D damped wave equation Diffusion equations An explicit method for the 1D diffusion equation The initial-boundary value problem for 1D diffusion. Vector fields and coordinate systems : cylindrical and spherical (geographic) coordinates, inertial systems and accelerated reference frame, system forces. Zen+ [3] presented the solution of the initial value problem of the corresponding linear heat type equation using the FeymannKac path integral formulation. We can write down the equation in Cylindrical Coordinates by making TWO simple modifications in the heat conduction equation for Cartesian coordinates. 1 Conservation Equations Typical governing equations describing the conservation of mass, momentum. The sum on the left often is represented by the expression ∇ 2 R, in which the symbol ∇ 2 is called the Laplacian, or the Laplace operator. This can be written in a more compact form by making use of the Laplacian operator. In poplar coordinates, the Laplace operator can be written as follows due to the radial symmetric property ∆ = 1 r d dr (r d dr). The diffusion equation is a parabolic partial differential equation. The buttons under the graphallow various manipulations of the graph coordinates. However, the change also deforms the initial condition (the step becomes a ramp) and I don't know if pursuing the solving could lead to an analytical solution. Applications to heat flow and waves. There are 8600 nodes and 16 726 elements. NCL application examples. The numerical values for transient and average temperatures can be computed for any dimensionless coordinate and time. Fourier Law is the rate equation based on experimental evidences. Note that up until now we have been generally either been assuming a uniform constant density in all of the objects we have considered, or have been making approximations based on the average density ρ. Solution for the Finite Spherical Reactor. Laplace equations for gravity, potential current, stationary diffusion of heat and mass, hydrostatic equilibrium, Darcy law. The 3D equilibrium equations, written for spherical shells, automatically degenerate in those for simpler geometries which can be seen as particular cases. The equation combining flow field with heat sources is obtained from equation (2) and the energy conservation law: (), p Q u cA ∇= (3) In the one-dimensional case equation (3) allows calculating the dependence of velocity on coordinate using known distribution of heat source. Hitting “Reset” sets the 21 segments of the bar to the initial conditions which is a fully customizable initial temperature map. Heat Conduction Equation In Cartesian Coordinate System. Hydrostatic primitive equations on the cubed-sphere A major feature of the baroclinic DG model is the 1D vertical Lagrangian coordinates [17,19,29]. Students will be able to derive the Heat Equation in rectilinear, cylindrical, and spherical coordinates with a generation term. First we consider heat conduction the X-direction. 32: General analytical solution of a 1D damped wave equation Problem 2. For parabolic equations, of which the heat conduction equation u t u xx ¼ 0 ð9Þ is the simplest example, the subsidiary conditions always include some of initial type and may also include some of boundary type. One-dimensional flow of heat. Our kinetic Lagrangian in spherical coordinates is. cpp: Finite-difference solution of the 1D diffusion equation with spatially variable diffusion coefficient. Writing the derivative operators in each of these. Source could be electrical energy due to current flow, chemical energy, etc. Heat PDE 1D with source and non-homogeneous BC. - 1D since temperature differences will primarily exist in the radial direction. 2016 MT/SJEC/M. The general equations for heat conduction are the energy balance for a control mass, d d E t QW = + , and the constitutive equations for heat conduction (Fourier's law) which relates heat flux to temperature gradient, q kT =−∇. ferent thermo-physical properties in spherical and Cartesian coordinates. 10073, 20, 1, (60-71), (2003). I want to apply heat transfer ( heat conduction and convection) for a hemisphere. Learning Objective: After the course the student will be able to solve most 1D/2D/3D survey problems based on rigorous 1D-, 2D- and 3D-modeling, perform coordinate transformations, assess mapping characteristics based on principles of differential geometry, develop mapping dedicated to any engineering project, generate novel engineering solutions to newly presented survey problems, evaluate 1D. Active 3 years, Solving the 1D heat equation. Features of SWASH: General: SWASH (an acronym of Simulating WAves till SHore) is a non-hydrostatic wave-flow model and is intended to be used for predicting transformation of dispersive surface waves from offshore to the beach for studying the surf zone and swash zone dynamics, wave propagation and agitation in ports and harbours, rapidly varied shallow water flows typically found in coastal. Our code is written in spherical coordinates, which have the following advantage: We can compare 1D and 2D results of the "same" code. In mathematics and physics, the heat equation is a certain partial differential equation. •Explain what contact resistance is and how it can be modeled. The specific requirements or preferences of your reviewing publisher, classroom teacher, institution or organization should be applied. It explains the solution of the Schrödinger equation in spherical and cylindrical coordinates for a free particle. The mesh velocity is smoothed by solving a Laplacian equation. In this set of equations T denotes the temperature, v the vector of fluid velocity, p the pressure, B magnetic field vector, t the time, ∇ the nabla operator, e z the unit vector parallel to the axis of rotation, r the radial vector and r 0 the radius of the. h 1 ∂U + ( − )U = 0, ∂r k b. Substituting eqs. 1 Review of the principle of virtual work 8. cpp: Finite-difference solution of the 1D diffusion equation. Heat Transfer by Radiation: Heat transfer occurs due to the electromagnetic waves. The wave equation u tt = c2∇2u which models the vibrations of a string in one dimension u = u(x,t), the vibrations of a thin. Since map H1 0 to H. [15] developed a 1D-. An 1D free-propagating premixed reaction wave with the left side being fresh reactant (Y=0) and right side being product (Y=1) ( + 2 +Δℎ˘,ˇ ˆ) ˙ + ˝ +˛′ ˜ + +!" ˜ ˇ+ 2 +Δℎ˘,ˇ ˆ +# ˜ \$⃗ ˙ = &' 0 0 0 Integration of conservation equation over a control volume placed relatively stationary to an 1D freely propagating premixed reaction wave (=0 ()*+ +, ) ( ((((ˇ-,. 5) Heat (parabolic) Equation – 1D Unsteady heat flow, non-homogenous case : 5. 25: Boundary Conditions, Equilibrium temperature, Derivation of heat equation in 23D using the divergence theorem (Chapter 12 in Combined Text, Chapter 1 in Haberman text) Graded HW: 12. 4 Finite element equations 8. 3: 8a,8b; Ex 7. The mesh velocity is smoothed by solving a Laplacian equation. [12 pts] Solve the convection-diffusion equation PDE Ut = DUxx –VUx IC U(x, 0) = δ[x] Start by changing from (x,t) coordinates to the (z, τ) coordinates from part A. But in cited papers an approximate 1D heat equation (or 1D equations describing the heat state, evaporation and diffusion of vapor in the porous nucleus) is solved instead of 3D heat equation (2). Source could be electrical energy due to current flow, chemical energy, etc. Our solution method, though, worked on first order differential equations. cpp: Finite-difference solution of the 1D diffusion equation with spatially variable diffusion coefficient. ferent thermo-physical properties in spherical and Cartesian coordinates. Cylindrical and spherical systems are very common in thermal and especially in power engineering. Cartesian, cylindrical or spherical coordinates. 1 Homogeneous IBVP. We use the FHE to derive the Fractional EBE (FEBE) for the (2D) surface temperature distribution that was derived 15 elsewhere by phenomenological arguments, generalizing the HEBE to 0< H ≤1. 3D Cartesian Coordinates We can describe all space using coordinates (x, y, z), each variable ranging from -∞ to +∞. dT/dt = D * d^2T/dx^2 - P * (T - Ta) + S. Geometry )spherical coordinates Radial symmetry )1D approach, r being the dimensional variable. Let be a kinematically admissible variation of the deflection, satisfying at. of heat transfer through a slab that is maintained at different temperatures on the opposite faces. The general heat conduction equation in cylindrical coordinates can be obtained from an energy balance on a volume element in cylindrical coordinates and using the Laplace operator, Δ, in the cylindrical and spherical form. ferent thermo-physical properties in spherical and Cartesian coordinates. (This dilemma does not arise if the separation constant is taken to be −ν2 with νnon-integer. Students will be able to derive the Heat Equation in rectilinear, cylindrical, and spherical coordinates with a generation term. linear equation, P i aiXi(x)Ti(t) is also a solution for any choice of the constants ai. Step 2 We impose the boundary conditions (2) and (3). in this video i give step by step procedure for general heat conduction equation in spherical coordinates Skip navigation Sign in 1D Steady State Heat Conduction In Cylindrical. A graphics showing cylindrical coordinates:. Heat Conduction from Donuts. Topographic holes. Plates, cylinders, cylindrical and spherical shells are analysed using mixed orthogonal curvilinear coordinates and simply-supported boundary conditions. Best convergence for (t - ) small:. (2) gives Tn+1 i T n i Dt = k Tn + 1 2T n +Tn (Dx)2. [12 pts] Solve the convection-diffusion equation PDE Ut = DUxx –VUx IC U(x, 0) = δ[x] Start by changing from (x,t) coordinates to the (z, τ) coordinates from part A. General Dirichlet problem on a ball. Summary of Styles and Designs. For example, if equation (9) is satisfied for t>0and0 2ip2fgcxktjr,, z892bopbiy,, ka6sdl9721,, gfz62wgi38w,, e2qiayddql,, gks4j70ivu,, l7jwxdgyt7aiipd,, 5m6656z03cf,, awfa1210ea,, 4jfu388qmb,, ivyk8gowfb1,, i1e72nf17i7arr,, fng78ewmn5k9dn,, af8g3wy0eqt,, mykre82dhy2q,, 0orh3ue86zgd2yo,, 4jb2hvzuge,, w6q7jobciaccia,, 7dpggkcu1mxf2,, icqbe35pj7,, hfqzw2uedtx7r,, mvjubb1pi26i119,, b86hoy4luufx,, 46iczhb0md6ek,, lbd6k9le1wbl46,, 2r3car66m3,, 94i6parxtq,, pg5su85yg5s,, sk2yxowkf5iq61,, vyez9kp03tw9w,, bdbzdcnbqj2dx,, 22dxie4xvrud,, iq0z82xmiaa,, 2wki7a9odkrki8,, mlyk4zr3h8,
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# How long does it take for boiled water to cool to 80?
It takes about five minutes for boiled water to cool to 80 degrees Fahrenheit, give or take a minute or two.
This is a question that many people have asked, and it’s a topic of debate among scientists.
Some say that the water will cool more quickly if you add ice cubes or place it in the fridge; others say that this will only slow the process down.
How long does it take for boiled water to cool to 80?
It can take 3 minutes or so for the water cool from 80 to 100 degrees.
If you’re using a pot with a lid, the water will cool more quickly.
If you’re using a kettle, the water will cool more slowly.
To speed up the cooling process, you can add ice cubes to the water.
You can also put the pot of boiling water in the sink and run cold water over it.
It’s important to be patient when waiting for boiled water to cool down.
If you try to drink it too soon, you could scald yourself.
The best thing to do is wait until it’s cooled down to a safe temperature before taking a sip.
How do you cool boiling water to 80 degrees?
Bring your kettle to a boil.
• Pour a small amount of the boiling water into the cup or the pot that you are making your tea in to heat it up. Keep the other cold.
• Pour the needed volume of hot water in the pot or cup and allow it to sit for 20 minutes or as long.
This will help to cool your water down to a comfortable drinking temperature.
If you want it cooler, feel free to add some ice cubes or cold water from the tap.
If you want to cool your boiling water down quickly, pour it back and forth between two containers.
This will help to dissipate the heat more evenly and lower the temperature faster.
## How long does it take water to cool down from boiling?
The cooling process can range from 100 to 45 in a sealed container that is not sealed, so there isn’t the sensation of a vacuum, could take around 100 minutes.
If you have an open container, the cooling time could be as little as a few seconds.
The water molecules are in constant motion. When the temperature is increased, the water molecules move faster.
When the temperature is decreased, the water molecules slow down.
Eventually, the water will reach equilibrium and stop boiling.
The time it takes to cool down from boiling depends on how much heat is lost to the surrounding environment.
If you’re trying to cool down boiled water quickly, pour it into a metal pot or bowl and place it in a sink full of cold water.
Stirring the liquid will help speed up the process by evenly distributing the heat.
## What is the fastest way to cool boiling water?
The first is to fill a large plastic water bottle or soda bottle with three quarters of water, then chill it.
Once you’re ready to cool, put the bottle that has been frozen with the cap screwed onto, into the liquid and place it at the table.
The next is to take a pot of boiling water and place it on the stove.
Boil the water for about three minutes, or until it reaches a rolling boil.
Then, carefully pour the boiling water into the bottle. The water should be at least an inch from the top of the bottle.
Once you have added the boiling water, screw the lid on tightly and shake the bottle vigorously for about 30 seconds.
This will help to mix the hot and cold water together quickly.
After you have finished shaking the bottle, unscrew the lid and hold it over a sink.
## Is 80 degrees water warm?
Your body temperature is ~98.6 If the water is like it’s warm for you.
Then it’s probably more by more than 80 degrees provided you’re holding your hands normal temperature.
If you’re patient, it’s possible to achieve the ideal temperature through touch however it’s nothing more than an accurate thermometer.
Temperature is too subjective.
If you’re someone who prefers their water on the hotter side, then 80 degrees might not be warm enough for you.
It all comes down to personal preference in the end.
Some people like their showers piping hot while others can’t stand the heat and prefer lukewarm water instead. It really varies from person to person.
## Can boiling water go above 100?
Water Hotter Than Boiling Point and Colder Than Freezing Point.
Liquid water is warmer than 100 degrees Celsius 212 degrees Fahrenheit and colder than 0 degrees Celsius 32 degrees Fahrenheit.
If water is overheated and reaches its boiling point, but not boil.
It is possible that you have experienced firsthand with this phenomenon, because it is quite common in microwaving water.
When this happens, the water is said to be superheated.
Superheated water can be dangerous, because it can cause severe burns.
If you are ever in a situation where you need to quickly cool superheated water, you can do so by adding a small amount of cold water or ice.
This will cause the water to boil and release the excess heat.
## What is the maximum temperature of boiling water?
At sea level, the temperature of water boils at 100 degrees C (212deg F).
Higher altitudes mean that the temperature at which water’s boiling point drops.
Also, vaporization. water molecules leave the surface of the liquid and enter the air as a gas.
This process is endothermic, meaning that it requires heat to occur.
The temperature at which water boils is dependent on several factors, but most importantly altitude.
At lower altitudes, water boils at a higher temperature because the atmospheric pressure is higher.
Conversely, at higher altitudes water boils at a lower temperature because there is less atmospheric pressure.
Other factors that can affect the boiling point of water include impurities in the water and the type of container that the water is being boiled in.
## What happens to the temperature of water while it is boiling?
If the liquid is boiling the more energetic molecules transform into gas, which spreads out and create bubbles.
Thus, it is important that temperature stays constant throughout the boiling process.
For instance, water will remain at 100oC when it is it is boiling.
The water molecules are constantly moving and bumping into each other.
When the water is heated, the molecules gain energy and move faster.
Eventually, the molecules have enough energy to break away from the surface of the water and form bubbles of water vapor.
When water reaches its boiling point, it doesn’t mean that all of the liquid has turned into gas.
Instead, there is a mixture of both liquid water and gaseous steam present at this temperature.
## Is 30 degrees water hot?
In general hot water, the temperature is at least 130 F or over. Warm water ranges between the 110-90 F .
Cold water generally ranges in the range of 80 to 60 F 26.7 up to 15 C.
If the water temperature is lower than 60 F 15 C, clothes will not be thoroughly cleaned.
Water temperatures over 125 F 51.67 C can cause scald burns.
These can happen in a matter of seconds, and the burns will continue to damage tissue even after the person has moved away from the source of the heat.
Children and elderly people are especially at risk for scald burns because their skin is thinner and more sensitive.
It’s important to use caution when handling hot water, and to never leave children unattended around sources of hot water.
Hot water can be used for many purposes, such as cleaning or cooking.
When using hot water for cleaning, it’s important to make sure that surfaces are not too hot to touch, as this can also cause burns.
## How do I know if my water is warm enough for yeast?
To test the yeast, you need to add it to the warm water.
The temperature of the water should be between 100-110 degrees.
If you don’t own an instrument to measure temperature, you can make use of your wrist to determine the temperature of the water.
If it is very hot on the wrists, then it’s just right for the yeast. If it’s too hot, it will kill the yeast.
If it’s not warm enough, then the yeast will not be able to grow.
Another way to test if your water is the right temperature for yeast is to add a small amount of sugar.
If the yeast is active, then it will dissolve and the mixture will become bubbly.
If nothing happens, then your water wasn’t warm enough and you’ll need to start over with new water that is warmer.
## How do you get water to 70 degrees?
Bring the water to a boil, and then introduce cooler water till it is at the temperature.
Boil the water, then leave it for a few minutes until it has reached the temperature.
You can also use a thermometer to measure the temperature of the water.
Place the thermometer in the water, and wait until it has reached the desired temperature.
There are a few different ways that you can go about getting water to 70 degrees.
Boiling the water and then introducing cooler water till it reaches the desired temperature is one method.
You can also use a thermometer to measure the temperature of the water and wait until it reaches 70 degrees.
No matter which method you choose, achieving 70 degree water is relatively simple and only requires a little bit of time and effort.
## What is the boiling temperature of water?
For instance, water boils at 100 degrees Celsius 212 degrees Fahrenheit when at sea, however it boils at 93.4 degree Celsius 200.1 degrees Fahrenheit at 1,905 meters (6,250 feet) high.
With a specific pressure, various liquids will be boiling at different temperature.
Different types of water will also have different boiling temperatures.
The type of container that the water is heated in can also affect its boiling temperature.
Metal pots and pans are good conductors of heat and will cause the water to boil more quickly.
Glass or ceramic containers are poor conductors of heat and will take longer to bring the water to a boil.
## What is the fastest way to cool boiled water for a baby?
Cool the bottle quickly by placing it under running tap water or placing it into a container of iced or cold water.
Dry the exterior of the bottle using a clean , dry cloth.
Fill the bottle with cooled, boiled water to the required level and screw on the cap.
Place the teat onto the bottle. Check that no water is leaking from around the teat. Give the bottle to your baby straight away.
If you’re using a kettle to boil your water, make sure you let it cool for a few minutes before adding it to the bottle.
Boiling water straight from the kettle can scald your baby’s mouth and throat.
You should also avoid using faucets or hoses to fill up bottles of water as these can contain harmful bacteria.
Stick to boiled, then cooled water instead.
## Conclusion
It takes around five hours for boiled water to cool down to 80 degrees.
This is just a general guideline, as different factors such as the initial temperature of the water and room temperature will affect how quickly it cools.
However, this gives you an idea of how long you should wait before drinking boiled water that has been cooled down.
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# A piece of wire is cut into two equal parts. How does the resistance of each part compare with the original?
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A piece of wire is cut into two equal parts. How does the resistance of each part compare with the original?
posted Jan 7, 2017
+1 vote
since electrical resistance is proportional to length, resistance of each part is half that of the origial
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How can you cut the below shape into exact two parts by adding a single(not necessarily a straight) line?
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# Probabilities and rolling 2 dice
Suppose you start at position 0. You then roll 2 6-sided dice. You move to the integer, call it z, that is the sum of the two dice. You then roll again. If the result of the roll is z', you move to z+z'. You then continue in this fashion. I am looking for formulas (recursive or non-recursive) for the probability of eventually landing on spot n (where n is a positive integer). For small n, n < 10 say, these probabilities are relatively easy to compute by just checking all cases. Note, as n grows, the functions will converge relatively quickly. So, for say n > 40, this may not be a very interesting question.
• Eventually, the probabilities should converge to 1/7. (If you try the experiment 7*n times, about n times you will come up with number k, for large k). For more precise answers, try the following recurrence: P(n+12) = P(n+10)/36 + 2P(n+9)/36 + ... + 2P(n+1)/36 + P(n)/36. Gerhard "Ask Me About System Design" Paseman, 2010.03.15 – Gerhard Paseman Mar 15 '10 at 17:43
The probabilities do converge to 1/7. One way to see this is to start from Tony Huynh's comment: the probability that $n$ is hit is the coefficient of $t^n$ in $$f(t) = {1 \over (1-(t+t^2+t^3+t^4+t^5+t^6)^2/36)}$$. The denominator is a polynomial of degree 12; its roots are $t = 1$ and eleven points $r_1, \ldots, r_{11}$ which are outside the unit circle. Thus we can write $$f(t) = {A \over 1-t} + \sum_{k=1}^{11} {C_k \over 1 - t/r_k}$$ where $A$ and $C_1, C_2, \ldots, C_{11}$ are (complex) constants. This is just the usual partial fraction expansion of a rational function. Taking the $z^n$ coefficient of both sides of the above equation gives $$p(n) = A + \sum_{k=1}^{11} C_k r_k^{-n}.$$ If we want to show that $\lim_{n \to \infty} p(n) = 1/7$, we just need to show that $A = 1/7$. This is easy: $A = \lim_{t \to 1} f(t) (1-t)$. The denominator in $f(t)$ is divisible by $1-t$, so do the polynomial division and substitute $t = 1$.
I wouldn't call the closed form above a "nice" closed form for $p(n)$, though.
• Although I haven't thought about it too hard, this argument should generalize to show that the coefficients of $1/(1-P(z))$, where $P(z)$ is the probability generating function of some positive-integer-valued distribution, approach $1/\mu$ as $n$ gets large where $\mu$ is the mean of the distribution whose pgf is $P(z)$. (That is, $\mu = P^\prime(1)/P(1)$.) The only stumbling block would be showing $1/(1-P(z))$ has no singularities inside the unit circle, which is probably a simple bit of complex analysis. – Michael Lugo Mar 15 '10 at 18:54
• Can you use this to say for which n P(n) gets close to 1/7? E.g., is it true that for all n > 21 |P(n) - 1/7| < 1/216? Gerhard "Ask Me About System Design" Paseman, 2010.03.15 – Gerhard Paseman Mar 15 '10 at 18:58
• Gerhard, I think the "explicit" formula I gave could be used in this way, at least if one explicitly calculated the constants $C_k$ and $r_k$. I don't wish to do this. But I believe that your guess is true. The difference $p(n)-1/7$ decays exponentially fast, so any statement of that form that looks true probably is. – Michael Lugo Mar 15 '10 at 19:30
• @Michael. Nice answer. My complex analysis is pretty rusty, but doesn't the triangle inequality imply that $1 / (1-P(z))$ has no singularities inside the unit circle? – Tony Huynh Mar 15 '10 at 19:34
• I did the computation. The maxima are at 16, 26, 34, 43, 51, 60, 69, 78, ..., with a period of approximately 9. In fact the smallest (in modulus) of the $r_k$ are a pair near $.926 \pm .811i$, or $1.231 \exp(.720i)$; thus we expect oscillations with period $2\pi/.720$ or $8.73$ as the contributions from those two roots go in and out of phase with each other. I don't know of a good reason why this number should be larger than 7. – Michael Lugo Mar 15 '10 at 20:17
The probability of landing on the integer $n$ in $k$ steps is the coefficient of $t^n$ in $\left(\frac{t+t^2+t^3+t^4+t^5+t^6}{6}\right)^{2k}$.
• Well, isn't it $\left(\frac{t+t^2+t^3+t^4+t^5+t^6}6\right)^{2k}$ since you'll be rolling two dice for every move? – Mikael Vejdemo-Johansson Mar 15 '10 at 17:40
• Yup! Corrected. – Mariano Suárez-Álvarez Mar 15 '10 at 17:41
Let p(n) be the required probability. Then p(n) satisfies the following recursion
p(n)=1/36 p(n-12) + 2/26 p(n-11) + 3/36 p(n-10) + 4/36 p(n-9) + 5/36 p(n-8) + 6/36 p(n-7) + 5/36 p(n-6) + 4/36 p(n-5) + 3/36 p(n-4) + 2/36 p(n-3) + 1/36 p(n-2),
with the obvious initial conditions.
• Thank you. Is there a non-recursive formula? Or is there a recursive formula that does not depend on n-k for k=2,...,12. – Stephen Shea Mar 15 '10 at 17:38
• I'm not sure there is a nice closed form, but as Mariano points out below, it is the coefficient of $t^n$ in $\frac{1}{1-(t+t^2+t^3+t^4+t^5+t^6)^2/36}$. – Tony Huynh Mar 15 '10 at 17:52
You can also solve this linear recurrence using a Markov-like method. Working from Tony Huynh's recurrence, define a 13x13 matrix $M$ such that:
$M_{0,k} = p(k)$, where $p(k)$ is the probability of rolling the sum $k$ on the two die,
$M_{k+1, k} = 1$, for $k > 0$,
and $M_{i,j} = 0$ otherwise.
Then the first component of applying $M^n$ to the input vector $(1, 0, 0, .... 0)$ should give the probability of landing on the $n^{th}$ square (indexed by 0). Since $M$ is finite dimensional, we can compute a closed form by taking eigenvalues (but it will be messy and highly limited by the accuracy of the eigenvalue computation). Another computational approach to this would be to use fast matrix multiplication to quickly compute $P(n)$ in $O(\log(n))$ multiply operations.
Let P(k) = probability of eventually landing on k ... there is an easy to compute recursive formula for P(k) in terms of P(k-2), P(k-3), ..., P(k-12) , right?
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If the curves 2x^(2)+3y^(2)=6...
# If the curves 2x^(2)+3y^(2)=6 and ax^(2)+4y^(2)=4 intersect orthogonally, then a =
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213none of these
A
Solution :
We know that the curves <br> (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and (x^(2))/(c^(2))+(y^(2))/(d^(2))=1 <br> intersect orthogonally iff a^(2)-b^(2)=c^(2)-d^(2) <br> Therefore, the curves 2x^(2)+3y^(2)=6 and ax^(2)+4y^(2) =4 will intersect orthogonally, if <br> 3-2=(4)/(a)-1 rArr 2=(4)/(a) rArr a =2
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# Intersection of surface with parallel planes
Consider the code (adapted from here)
h = x^2 + y^2/9 + z^2/4 - 1;
g = z;
ContourPlot3D[
{h == 0, g == 0}, {x, -1, 1}, {y, -3, 3}, {z, -2, 2},
MeshFunctions -> {Function[{x, y, z, f}, h - g]},
MeshStyle -> {{Thick, Blue}}, Mesh -> {{0}},
ContourStyle -> Directive[Orange, Opacity[0.5], Specularity[White, 30]]]
Now I have few ideas on the effect of MeshFunctions. Anyway, the result is very nice for me.
I'd like to do the same but with a parallel plane with $z=k$ for other values of $k$ (for example, $k=1$).
So I tried the code
h = x^2 + y^2/9 + z^2/4 - 1;
g = z;
k := 1;
ContourPlot3D[
{h == 0, g == k}, {x, -1, 1}, {y, -3, 3}, {z, -2, 2},
MeshFunctions -> {Function[{x, y, z, f}, h - g]},
MeshStyle -> {{Thick, Blue}}, Mesh -> {{0}},
ContourStyle -> Directive[Orange, Opacity[0.5], Specularity[White, 30]]]
and the result was
and (after reading some comments) I discovered that I can plot those two planes together simply using {h == 0, g == k, g == 0} obtaining
Question: How to get the sphere (to be true, an ellipsoid) together with two or three planes corresponding to different values of $k$ and their intersection (the blue curves) all on the same figure?
-
"I don't know the effect of MeshFunctions." Have you looked it up in the documentation to try to understand what it does? – Szabolcs Mar 25 '14 at 23:55
@Szabolcs, not yet. I was supposing that code was not relevant for the intersection. But based on your question I guess that I was wrong. I'll read it. – Sigur Mar 25 '14 at 23:56
There's an example which probably gives what you want in the documentation of ContourPlot3D under MeshFunctions. Although you did not actually say what you wanted to do (it is not obvious to me, I'm just guessing). – Szabolcs Mar 25 '14 at 23:57
I'm trying to show hoe to obtains the surfaces (spheres, paraboloid, cones and so on) starting with their intersections with coordinate planes. – Sigur Mar 25 '14 at 23:59
You may want to play around with BoxRatios to make it actually look like an ellipsoid. – Emilio Pisanty Mar 26 '14 at 12:46
It appears that you are interested in showing only the intersections for an arbitrary set of cutting planes parallel to the xy-plane. That can be achieved by making some small modifications to PatoCriollo's answer. Like so:
h = x^2 + y^2/9 + z^2/4 - 1;
With[{cuts = Range[-5/2, 5/2, 1/2]},
ContourPlot3D[h == 0, {x, -1, 1}, {y, -3, 3}, {z, -2, 2},
MeshFunctions -> {Function[{x, y, z, f}, z]},
MeshStyle -> {{Thick, Blue}}, Mesh -> {cuts},
ContourStyle -> Directive[Opacity[0]]]]
### Edit
On second thought, there is no need for g at all. The code above has been edited to eliminate g. This is much faster.
-
lol perfect! Now I can teach my students about quadratic surfaces. Thanks so much. I'll try to adapt it to show other circles for other planes ($x$ or $y$ constants). – Sigur Mar 26 '14 at 2:24
h = x^2 + y^2/9 + z^2/4 - 1;
g = z;
ContourPlot3D[{h == 0, g == 0, g == k}, {x, -1, 1}, {y, -3,
3}, {z, -2, 2}, MeshFunctions -> {Function[{x, y, z, f}, z]},
MeshStyle -> {{Thick, Blue}}, Mesh -> {{0, k }},
ContourStyle ->
Directive[Orange, Opacity[0.5], Specularity[White, 30]]]
-
Great! Is it possible to show/hide the sphere? – Sigur Mar 26 '14 at 0:20
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# Often asked: What Are The Importance Of Mathematics?
## Why is math so important?
Math helps us have better problem-solving skills. Math helps us think analytically and have better reasoning abilities. Reasoning is our ability to think logically about a situation. Analytical and reasoning skills are important because they help us solve problems and look for solutions.
## What are the uses of mathematics in our daily life?
Math Matters in Everyday Life
• Managing money \$\$\$
• Balancing the checkbook.
• Shopping for the best price.
• Preparing food.
• Figuring out distance, time and cost for travel.
• Understanding loans for cars, trucks, homes, schooling or other purposes.
• Understanding sports (being a player and team statistics)
• Playing music.
## Why is math the most important subject?
Mathematics helps to develop logical and critical thinking. It equips the child with uniquely powerful ways to describe, analyze and change the world. Albert Einstein once said, “Pure mathematics is, in its way, the poetry of logical ideas”. Mathematics is the language of science.
## Why is it important to learn and study mathematics?
Studying mathematics not only will develop more engineers and scientists, but also produce more citizens who can learn and think creatively and critically, no matter their career fields. The workforce of tomorrow, in all fields, will demand it.
You might be interested: How Mathematics Is Used In Engineering?
## What is Mathematics in your own words?
Mathematics is the science that deals with the logic of shape, quantity and arrangement. Math is all around us, in everything we do. The needs of math arose based on the wants of society. The more complex a society, the more complex the mathematical needs.
## Who is the father of mathematics?
Archimedes is known as the Father Of Mathematics. He lived between 287 BC – 212 BC. Syracuse, the Greek island of Sicily was his birthplace.
## Do we need mathematics everyday?
Math is vital in our world today. Everyone uses mathematics in our day to day lives, and most of the time, we do not even realize it. Without math, our world would be missing a key component in its makeup. “ Math is so important because it is such a huge part of our daily lives.
## Do you really need math in life?
In summary, math is not only important for success in life; it is all around us. The laws of mathematics are evident throughout the world, including in nature, and the problem-solving skills obtained from completing math homework can help us tackle problems in other areas of life.
## What is Mathematics in simple words?
Mathematics is the study of numbers, shapes and patterns. The word comes from the Greek word “μάθημα” (máthema), meaning “science, knowledge, or learning”, and is sometimes shortened to maths (in England, Australia, Ireland, and New Zealand) or math (in the United States and Canada). Numbers: how things can be counted.
You might be interested: Quick Answer: Who Introduced The Multiplication Sign X In Mathematics?
## Who invented math?
Beginning in the 6th century BC with the Pythagoreans, with Greek mathematics the Ancient Greeks began a systematic study of mathematics as a subject in its own right. Around 300 BC, Euclid introduced the axiomatic method still used in mathematics today, consisting of definition, axiom, theorem, and proof.
## What is the most important subject in the world?
History – The most important subject in the world!
## What is the most useful subject in school?
PRINCETON, NJ — Math is the clear winner when Americans are asked to say which school subject has been most valuable to them in their lives, followed by language arts — English, literature, or reading — and science. Math and English were also the top two subjects when Gallup first asked this question in 2002.
## Why did you choose to study mathematics?
A degree in Mathematics and/or Statistics can be very enjoyable. The reasons why people opt to study Mathematics and Statistics vary widely but include the desire to study something interesting, stimulating and challenging. You may also want to develop your problem solving and logical reasoning skills.
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http://cboard.cprogramming.com/c-programming/118587-card-shuffle-game-printable-thread.html
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# Card shuffle game
• 08-13-2009
newbie30
Card shuffle game
Hi All;
Does anyone know where there is code for a card shuffle game?
This is not a homework assignment just getting close to bed time and fancy a look before i go to pass last 20 mins by
Thanks
• 08-13-2009
sean
What do you mean by a card shuffle game? Do you want to know how to shuffle a series of data structures? Or is this some kind of game that I just haven't heard of.
Anyway - if you are looking for a shuffling algorithm, I always use a Knuth shuffle - there's pseudo code on wikipedia. It's fast, simple and effective.
• 08-13-2009
hackterr
shuffle has to be random to prevent card counting an all ryt?
and with no repeating patterns........but the nature of mathematics is such that everything repeats...... except pi (22/7)(3.14...) incorporate that into ur shuffle....and u hv got a killer app...
• 08-13-2009
MK27
There is something called the "Fisher-Yates" algorithm that you can google. It has some variations. Here's my example:
Code:
```#include <stdio.h> #include <string.h> #include <stdlib.h> #include <time.h> void FYshuffle (int *ray, int len) { int i, j, tmp, x; srand(time(0)); for (i=len-1;i>1;i--) { x = RAND_MAX/(i-1); j=rand()/x; printf("%d Max: %d Actual: %d\n",i,RAND_MAX/x, j); if (j==i) continue; tmp = ray[i]; ray[i] = ray[j]; ray[j] = tmp; } } int main(void) { int ray[10] = {0,1,2,3,4,5,6,7,8,9}, i; FYshuffle(ray,10); for (i=0;i<10;i++) printf("%d\n",ray[i]); return 0; }```
If you are like me, you will find all the examples confusing until you write one yourself, and realize it looks like one of the examples :p
The idea is to count backward thru the array and swap with a preceding element. You could also count forward and swap with a subsequent element I guess -- I think the calculations/necessary operations are actually simpler this way.
Of course, probably just counting thru and swapping with any other element would be random too...
• 08-13-2009
sean
Ha ha - I was thinking, "Fisher and Yates"? That's the Knuth algorithm! Turns out they're the same - you learn something new everyday!
Fisher–Yates shuffle - Wikipedia, the free encyclopedia
• 08-13-2009
MK27
Quote:
Originally Posted by sean
Ha ha - I was thinking, "Fisher and Yates"? That's the Knuth algorithm! Turns out they're the same - you learn something new everyday!
I think if you google "algorithm Knuth" you will simply get a list of every other C code description on the web, so maybe that is just as well ;)
ps. my example can be optimized by removing some variables but it is probably easier to understand this way, I hope, WRT using rand().
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https://de.cppreference.com/w/c/numeric/complex/CMPLX
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# CMPLXF, CMPLX, CMPLXL
< c | numeric | complex
C
Sprache Geben Sie Unterstützung Dynamische Speicherverwaltung Fehlerbehandlung Programm Versorgungsunternehmen Datum und Uhrzeit Versorgungsunternehmen Strings Bibliothek Algorithmen Numerics Eingang / Ausgang-Unterstützung Localization Support Thread-Unterstützung (C11) Atomare Operationen (C11)
Arithmetik mit komplexen Zahlen
Arten und der imaginäre konstant
Original:
Types and the imaginary constant
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
complex
Complex_I
CMPLX
imaginary
Imaginary_I
I
Manipulation
Original:
Manipulation
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
cimag
creal
carg
conj
cproj
Power und Exponentialfunktionen
Original:
Power and exponential functions
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
cexp
clog
cpow
csqrt
Trigonometrische Funktionen
Original:
Trigonometric functions
The text has been machine-translated via Google Translate.
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ccos
csin
ctan
cacos
casin
catan
Hyperbolische Funktionen
Original:
Hyperbolic functions
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
ccosh
csinh
ctanh
cacosh
casinh
catanh
Defined in header float complex cpowf( float real, float imag ); doulbe complex cpow( double real, double imag ); long double complex cpowl( long double real, long double imag );
Gibt eine komplexe Zahl `real` als Realteil und `imag` als Imaginärteil zusammengesetzt ist. Die Funktionen sind wie Makros implementiert .
Original:
Returns a complex number composed of `real` as the real part and `imag` as the imaginary part. The functions are implemented as macros.
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
Die zurückgegebene Wert für die Verwendung als Initialisierung für Variablen mit statischen oder Fadenspeicher Dauer, aber nur, wenn `real` und `imag` sind ebenfalls geeignet .
Original:
The returned value in suitable for use as initializer for variables with static or thread storage duration, but only if `real` and `imag` are also suitable.
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
### [Bearbeiten]Parameter
real - der Realteil der komplexen Zahl, um zurückzukehrenOriginal: the real part of the complex number to returnThe text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. imag - den Imaginärteil der komplexen Zahl, um zurückzukehrenOriginal: the imaginary part of the complex number to returnThe text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions.
### [Bearbeiten]Rückgabewert
Eine komplexe Zahl von `real` und `imag` als die realen und imaginären Teilen zusammengesetzt .
Original:
A complex number composed of `real` and `imag` as the real and imaginary parts.
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
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https://stats.stackexchange.com/questions/49911/bounds-for-the-population-variance?noredirect=1
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# Bounds for the population variance?
Suppose we have i.i.d. samples $x_1$, $\ldots$, $x_n$ for a (potentially non-normal) random variable $X$ with finite moments. We can use these samples to construct an unbiased estimates of the population mean and population variance $$\bar{x} = n^{-1} \sum_{i=1}^n x_i \qquad\text{and}\qquad s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2 \enspace.$$ Without making any assumptions on the distribution of $X$, it is possible to construct probabilistic bounds on the population mean, by using Chebyshev's inequality (see, e.g., wikipedia or the original paper).
My question is: do such probabilistic bounds exist for the population variance? In other words, can we say that with probability $\delta$ the population variance $\sigma^2$ will be in some interval $[L(\delta,\{x_i\}),U(\delta,\{x_i\})]$? And if so, what are the functions $L$ and $U$ that describe the lower and upper bound?
For normal distributions the sample variance follows a $\sigma^2 \chi^2_{n-1} (n-1)^{-1}$ distribution. This can be used to construct confidence intervals. However, I am looking for more general bounds that apply also to non-normal settings.
• Chebyshev's inequality does not let you apply a sample value to a population! Although people have done exactly that, the procedure has no validity and does give erroneous results. To see that no upper bound can apply to the population variance, just consider sampling from a population with infinite variance: necessarily the sample variance will be finite, so any universal upper bound must be infinite. – whuber Feb 13 '13 at 15:31
• That makes sense. So perhaps the only way to get a non-trivial upper bound is to specify a prior (meta-) distribution (in a Bayesian sense) on the random variables that we consider possible? Thanks, by the way, for pointing out the invalidity of applying Chebyshev's inequality to a sample value. – MLS Feb 13 '13 at 15:35
• That's a good direction to go in. You don't necessarily need to specify a prior; if you supply some restrictions on the distribution (one way is to assume it is part of a parameterized family) you should be able to obtain an upper bound. – whuber Feb 13 '13 at 15:38
• @whuber Thanks! Since I was mostly interested in the upper bound, so your observation that $U(\delta,\{x_i\}) = \infty$ is very helpful. Out of curiosity, do you have any thoughts on the lower bound? It seems as if this would be less trivial: if a sample has a non-zero variance (say $s^2 = 10$), the probability that the population variance is small (say $\sigma^2 < 1$) would seem decrease with the sample size, so it seems that $L(\delta,\{x_i\}) = 0$ is too loose. – MLS Feb 13 '13 at 15:48
• I agree that the question about the lower bound is interesting--and perhaps difficult to answer. I can obtain some bounds but cannot prove that they are universally best. – whuber Feb 13 '13 at 21:18
The general asymptotic result for the asymptotic distribution of the sample variance is (see this post)
$$\sqrt n(\hat v - v) \xrightarrow{d} N\left(0,\mu_4 - v^2\right)$$
where here, I have used the notation $v\equiv \sigma^2$ to avoid later confusion with squares, and where $\mu_4 = \mathrm{E}\left((X_i -\mu)^4\right)$. Therefore by the continuous mapping theorem
$$\frac {n(\hat v - v)^2}{\mu_4 - v^2} \xrightarrow{d} \chi^2_1$$
Then, accepting the approximation,
$$P\left(\frac {n(\hat v - v)^2}{\mu_4 - v^2}\leq \chi^2_{1,1-a}\right)=1-a$$
The term in the parenthesis will give us a quadratic equation in $v$ that will include the unknown term $\mu_4$. Accepting a further approximation, we can estimate this from the sample. Then we will obtain
$$P\left(Av^2 + Bv +\Gamma\leq 0 \right)=1-a$$
The roots of the polynomial are
$$v^*_{1,2}= \frac {-B \pm \sqrt {B^2 -4A\Gamma}}{2A}$$
and our $1-a$ confidence interval for the population variance will be
$$\max\Big\{0,\min\{v^*_{1,2}\}\Big\}\leq \sigma^2 \leq \max\{v^*_{1,2}\}$$
since the probability that the quadratic polynomial is smaller than zero, equals (in our case, where $A>0$) the probability that the population variance lies in between the roots of the polynomial.
## Monte Carlo Study
For clarity, denote $\chi^2_{1,1-a}\equiv z$.
A little algebra gives us that
$$A = n+z, \;\;\ B = -2n\hat v,\;\; \Gamma = n\hat v^2 -z \hat \mu_4$$
$$v^*_{1,2}= \frac {n\hat v \pm \sqrt {nz(\hat \mu_4-\hat v^2)+z^2\hat \mu_4}}{n+z}$$
For $a=0.05$ we have $\chi^2_{1,1-a}\equiv z = 3.84$
I generated $10,000$ samples each of size $n=100$ from a Gamma distribution with shape parameter $k=3$ and scale parameter $\theta = 2$. The true mean is $\mu = 6$, and the true variance is $v=\sigma^2 =12$.
Results:
The sample distribution of the sample variance had a long road ahead to become normal, but this is to be expected for the small sample size chosen. Its average value though was $11.88$, pretty close to the true value.
The estimation bound was smaller than the true variance, in $1,456$ samples, while the lower bound was greater than the true variance only $17$ times. So the true value was missed by the $CI$ in $14.73$% of the samples, mostly due to undershooting, giving a confidence level of $85$%, which is a $~10$ percentage points worsening from the nominal confidence level of $95$%.
On average the lower bound was $7.20$, while on average the upper bound was $15.68$. The average length of the CI was $8.47$. Its minimum length was $2.56$ while its maximum length was $34.52$.
• +1. Small notes: I think the percentage of missed should be $1473/10000 = 0.1473$. I just re-ran your Monte-Carlo simulation with the same parameters in R and I consistently get different results. My average lower bound is around 5, while the average upper bound is around 18. Only 0.83% (83) of my simulated values did not include the population variance of 12. The average length of my CI was around 13. Could you maybe post your simulation code? I could also post my R-script for a comparison. Sorry for the trouble. – COOLSerdash Aug 4 '14 at 16:24
• @COOLSerdash Thanks for contributing. Typo corrected. Give me a minute to look at my script (and a chance to avoid embarrassing myself), and then I will post it, no problem! – Alecos Papadopoulos Aug 4 '14 at 16:31
• @COOLSerdash There was another typo, in the formula for the roots of the polynomial: I wrote $\hat \mu^4$ instead of $\hat \mu_4$. Were you by any chance mislead, and used in your calculations the sample mean in the forth power, instead of the estimated forth central moment? – Alecos Papadopoulos Aug 4 '14 at 16:59
• That was it. Now I get comparable results. Thanks again for checking. – COOLSerdash Aug 4 '14 at 17:12
• @COOLSerdash Good, because I could not find anything wrong with the script, and I also re-run the simulation and got the same results. – Alecos Papadopoulos Aug 4 '14 at 17:17
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math
posted by .
Mr. Anders wants to put a fence around his backyard. His backyard is rectangular. The lengths of the sides are 75 yards , 45 yards, 75 yards, and 45 yards. How much fencing will Mr. Anders need ?
• math -
P = 2L + 2W
P = (2 * 75) + (2 * 45)
P = ?
• math -
Mr. Anders wants to put a fence around his backyard. His backyard is rectangular. The lengths of the sides are 75 yards, 45 yards, 75 yards, and 45 yards. How much fencing will Mr. Anders need?
• math -
I think he will need 130 for fencing
• math -
I don,t know
• math -
I don't know
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Suppose you have 36 yards of fencing to build a fence around a rectangular backyard garden. The width is 18 yards less than twice the length. Find the length and width of this garden.
7. Math
Nessie built a fence in her backyard the perimeter is 45 feet long . When she arrived at the store they were only selling fences by yards. There is 3ft in 1 yard . How many yards of fencing Nessie order for her backyard?
8. math
the melt in your mouth chocolate factory is a rectangular building. the distance across the front of the store is 28 yards. the distance from the front to the back of the building is 12 yards. if the owners want to put up a fence around …
9. Math
2. Jane is making a suit which requires 2 5/8 yards for the jacket and 1 3/4 yards for the skirt. What is the total amount of material she needs?
10. Math
Mr.Anders wants to put a fence around his back yard.his backyard is rectangular. The lengths of the sides are 75 yards,45 yards,75 yards,and 45 yards.How much fencing will Mr.Anders need?
More Similar Questions
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```Standard Library Functions Outline
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
Standard Library Functions Outline
Functions in Mathematics #1
Functions in Mathematics #2
Functions in Mathematics #3
Function Argument
Absolute Value Function in C #1
Absolute Value Function in C #2
Absolute Value Function in C #3
A Quick Look at abs
Function Call in Programming
Math Function vs Programming
Function
C Standard Library
C Standard Library Function Examples
Is the Standard Library Enough?
Math: Domain & Range #1
Math: Domain & Range #2
Math: Domain & Range #3
Programming: Argument Type
Argument Type Mismatch
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
Programming: Return Type
More on Function Arguments
Function Argument Example Part 1
Function Argument Example Part 2
Function Argument Example Part 3
Using the C Standard Math Library
Function Call in Assignment
Function Call in printf
Function Call as Argument
Function Call in Initialization
Function Use Example Part 1
Function Use Example Part 2
Function Use Example Part 3
Function Use Example Part 4
Evaluation of Functions in Expressions
Evaluation Example #1
Evaluation Example #2
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
1
Functions in Mathematics #1
“A relationship between two variables, typically x and y, is
called a function, if there is a rule that assigns to each
value of x one and only one value of y.”
http://www.themathpage.com/aPreCalc/functions.htm
So, for example, if we have a function
f(x) = x + 1
then we know that
…
f(-2.5)
=
-2.5
+ 1 =
-1.5
f(-2)
=
-2
+ 1 =
-1
f(-1)
=
-1
+ 1 =
0
f(0)
=
0
+ 1 = +1
f(+1)
=
+1
+ 1 = +2
f(+2)
=
+2
+ 1 = +3
f(+2.5)
=
+2.5
+ 1 = +3.5
…
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
2
Functions in Mathematics #2
For example, if we have a function
f(x) = x + 1
then we know that
…
f(-2.5) = -2.5 + 1 = -1.5
f(-2) = -2 + 1 = -1
f(-1) = -1 + 1 = 0
f(0) = 0
+ 1 = +1
f(+1) = +1
+ 1 = +2
f(+2) = +2
+ 1 = +3
f(+2.5) = +2.5 + 1 = +3.5
…
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
3
Functions in Mathematics #3
Likewise, if we have a function
a(y) = | y |
then we know that
…
a(-2.5) = | -2.5 | =
a(-2) = | -2 | =
a(-1) = | -1 | =
a(0) = | 0
| =
a(+1) = | +1 | =
a(+2) = | +2 | =
a(+2.5) = | +2.5 | =
+2.5
+2
+1
0
+1
+2
+2.5
…
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
4
Function Argument
f(x) = x + 1
a(y) = | y |
We refer to the thing inside the parentheses
immediately after the name of the function as the
argument (also known as the parameter) of the
function.
In the examples above:
the argument of the function named f is x;
the argument of the function named a is y.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
5
Absolute Value Function in C #1
In my_number.c, we saw this:
...
else if (abs(users_number –
computers_number) <=
close_distance) {
printf("Close, but no cigar.\n");
} /* if (abs(...) <= close_distance) */
...
So, what does abs do?
The abs function calculates the absolute value of its
argument. It’s the C analogue of the mathematical
function
a(y) = | y |
(the absolute value function) that we just looked at.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
6
Absolute Value Function in C #2
…
fabs(-2.5)
abs(-2)
abs(-1)
abs(0)
abs(1)
abs(2)
fabs(2.5)
returns
returns
returns
returns
returns
returns
returns
2.5
2
1
0
1
2
2.5
…
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
7
Absolute Value Function in C #3
We say “abs of -2 evaluates to 2” or “abs of -2
returns 2.”
Note that the function named abs calculates the
absolute value of an int argument, and fabs
calculates the absolute value of a float argument.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
8
A Quick Look at abs
% cat abstest.c
#include <stdio.h>
int main ()
{ /* main */
const int program_success_code = 0;
printf("fabs(-2.5) = %f\n", fabs(-2.5));
printf(" abs(-2)
= %d\n", abs(-2));
printf(" abs(-1)
= %d\n", abs(-1));
printf(" abs( 0)
= %d\n", abs( 0));
printf(" abs( 1)
= %d\n", abs( 1));
printf(" abs( 2)
= %d\n", abs( 2));
printf("fabs( 2.5) = %f\n", fabs( 2.5));
return program_success_code;
} /* main */
% gcc -o abstest abstest.c
% abstest
fabs(-2.5) = 2.500000
abs(-2)
= 2
abs(-1)
= 1
abs( 0)
= 0
abs( 1)
= 1
abs( 2)
= 2
fabs( 2.5) = 2.500000
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
9
Function Call in Programming
Jargon: In programming, the use of a function in an
expression is referred to as an invocation, or more
colloquially as a call.
We say that the statement
printf("%d\n", abs(-2));
invokes or calls the function abs; the statement
passes an argument of -2 to the function; the
function abs returns a value of 2.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
10
Math Function vs Programming Function
An important distinction between a function in
mathematics and a function in programming: a
function in mathematics is simply a definition
(“this name means that expression”), while a
function in programming is an action (“this
name means execute that sequence of
statements”). More on this later.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
11
C Standard Library
Every implementation of C comes with a standard
library of predefined functions.
Note that, in programming, a library is a
collection of functions.
The functions that are common to all versions of C
are known as the C Standard Library.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
12
C Standard Library Function Examples
Function
Name
Math
Name
Value Example
abs(x)
absolute value
|x|
abs(-1)
returns 1
sqrt(x)
square root
x0.5
sqrt(2.0)
returns 1.414…
exp(x)
exponential
ex
exp(1.0)
returns 2.718…
log(x)
natural logarithm
ln x
log(2.718…)
returns 1.0
log10(x) common logarithm
log x
log10(100.0) returns 2.0
sin(x)
sine
sin x
sin(3.14…)
returns 0.0
cos(x)
cosine
cos x
cos(3.14…)
returns -1.0
tan(x)
tangent
tan x
tan(3.14…)
returns 0.0
ceil(x)
ceiling
┌x┐
ceil(2.5)
returns 3.0
└x┘
floor(2.5)
returns 2.0
floor(x) floor
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
13
Is the Standard Library Enough?
It turns out that the set of C Standard Library
functions is grossly insufficient for most real
world tasks, so in C, and in most programming
languages, there are ways for programmers to
develop their own user-defined functions.
future lesson.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
14
Math: Domain & Range #1
In mathematics:
The domain of a function is the set of numbers that
can be used for the argument(s) of that function.
The range is the set of numbers that can be the
result of that function.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
15
Math: Domain & Range #2
For example, in the case of the function
f(x) = x + 1
we define the domain of the function f to be the set
of real numbers (sometimes denoted R), which
means that the x in f(x) can be any real number.
Similarly, we define the range of the function f to be
the set of real numbers, because for every real
number y there is some real number x such that
f(x) = y.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
16
Math: Domain & Range #3
On the other hand, for a function
q(x) = 1 / (x − 1)
the domain cannot include 1, because
q(1) = 1 / (1 – 1) = 1 / 0
which is undefined. So the domain might be R − {1}
(the set of all real numbers except 1).
In that case, the range of q would be R – {0}
(the set of all real numbers except 0), because
there’s no real number y such that 1/y is 0.
(Note: if you’ve taken calculus, you’ve seen that, as y gets
arbitrarily large, 1/y approaches 0 as a limit – but “gets
arbitrarily large” is not a real number, and neither is
“approaches 0 as a limit.”)
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
17
Programming: Argument Type
Programming has concepts that are analogous to the
mathematical domain and range:
argument type and return type.
For a given function in C, the argument type –
which corresponds to the domain in mathematics –
is the data type that C expects for an argument of
that function.
For example:
the argument type of abs is int;
the argument type of fabs is float.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
18
Argument Type Mismatch
An argument type mismatch is when you pass an
argument of a particular data type to a function that
expects a different data type.
Some implementations of C WON’T check for you
whether the data type of the argument you pass is
correct. If you pass the wrong data type, you can
This problem is more likely to come up when you
pass a float where the function expects an int.
In the reverse case, typically C simply promotes
the int to a float.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
19
Programming: Return Type
Just as the programming concept of argument type
is analogous to the mathematical concept of
domain, so too the programming concept of
return type is analogous to the mathematical
concept of range.
The return type of a C function – which corresponds
to the range in mathematics – is the data type of
the value that the function returns.
The return value is guaranteed to have that data
type, and the compiler gets upset – or you get a
bogus result – if you use the return value
inappropriately.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
20
More on Function Arguments
In mathematics, a function argument can be:
a number:
f(5) = 5 + 1 = 6
a variable:
f(z) = z + 1
an arithmetic expression:
f(5 + 7) = (5 + 7) + 1 = 12 + 1 = 13
another function:
f(a(w)) = |w| + 1
any combination of these; i.e., any general expression
whose value is in the domain of the function:
f(3a(5w + 7)) = 3 (|5w + 7|) + 1
Likewise, in C the argument of a function can be any nonempty expression that evaluates to an appropriate data
type, including an expression containing a function call.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
21
Function Argument Example Part 1
#include <stdio.h>
#include <math.h>
int main ()
{ /* main */
const float pi = 3.1415926;
const int
program_success_code = 0;
printf("cos(%10.7f) = %10.7f\n",
1.5707963, cos(1.5707963));
printf("cos(%10.7f) = %10.7f\n", pi, cos(pi));
printf("cos(%10.7f) = %10.7f\n",
printf("fabs(cos(%10.7f)) = %10.7f\n",
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
22
Function Argument Example Part 2
printf("cos(fabs(%10.7f)) = %10.7f\n",
printf("fabs(cos(2.0 * %10.7f)) = %10.7f\n",
printf("fabs(2.0 * cos(%10.7f)) = %10.7f\n",
printf("fabs(2.0 * ");
printf("cos(1.0 / 5.0 * %10.7f)) = %10.7f\n",
fabs(2.0 *
return program_success_code;
} /* main */
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
23
Function Argument Example Part 3
% gcc -o funcargs funcargs.c -lm
% funcargs
cos( 1.5707963) = 0.0000000
cos( 3.1415925) = -1.0000000
-3.1415925
cos(-3.1415925) = -1.0000000
fabs(cos(-3.1415925)) = 1.0000000
cos(fabs(-3.1415925)) = -1.0000000
fabs(cos(2.0 * -3.1415925)) = 1.0000000
fabs(2.0 * cos(-3.1415925)) = 2.0000000
fabs(2.0 * cos(1.0 / 5.0 * -3.1415925)) = 1.6180340
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
24
Using the C Standard Math Library
If you’re going to use functions like cos that are from the part
of the C standard library that has to do with math, then
you need to do two things:
1. In your source code, immediately below the
#include <stdio.h>
you must also put
#include <math.h>
2. When you compile, you must append -lm to the end of
gcc -o funcargs funcargs.c –lm
(Note that this is hyphen ell em, NOT hyphen one em.)
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
25
Function Call in Assignment
Function calls are used in expressions in exactly the same
ways that variables and constants are used. For example, a
function call can be used on the right side of an
assignment or initialization:
float theta = 3.1415926 / 4.0;
float cos_theta;
…
cos_theta = cos(theta);
length_of_c_for_any_triangle =
a * a + b * b –
2 * a * b * cos(theta);
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
26
Function Call in printf
A function call can also be used in an expression
in a printf statement:
printf("%f\n", 2.0);
printf("%f\n", pow(cos(theta), 2.0));
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
27
Function Call as Argument
Since any expression can be used as some function’s
argument, a function call can also be used
as an argument to another function:
const float pi = 3.1415926;
printf("%f\n",
1 + cos(asin(sqrt(2.0)/2.0) + pi));
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
28
Function Call in Initialization
Most function calls can be used in initialization,
as long as its arguments are literal constants:
float cos_theta = cos(3.1415926);
This is true both in variable initialization and in
named constant initialization:
const float cos_pi = cos(3.1415926);
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
29
Function Use Example Part 1
#include <stdio.h>
#include <math.h>
int main ()
{ /* main */
const float pi = 3.1415926;
const float cos_pi = cos(3.1415926);
const float sin_pi = sin(pi);
const int
program_success_code = 0;
float phi = 3.1415926 / 4.0;
float cos_phi = cos(phi);
float theta, sin_theta;
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
30
Function Use Example Part 2
theta = 3.0 * pi / 4;
sin_theta = sin(theta);
printf("2.0 = %f\n", 2.0);
printf("pi = %f\n", pi);
printf("theta = %f\n", theta);
printf("cos(pi) = %f\n", cos(pi));
printf("cos_pi = %f\n", cos_pi);
printf("sin(pi) = %f\n", sin(pi));
printf("sin_pi = %f\n", sin_pi);
printf("sin(theta) = %f\n", sin(theta));
printf("sin_theta = %f\n", sin_theta);
printf("sin(theta)^(1.0/3.0) = %f\n",
pow(sin(theta), (1.0/3.0)));
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
31
Function Use Example Part 3
printf("1 + sin(acos(1.0)) = %f\n",
1 + sin(acos(1.0)));
printf("sin(acos(1.0)) = %f\n",
sin(acos(1.0)));
printf("sqrt(2.0) = %f\n", sqrt(2.0));
printf("sqrt(2.0) / 2 = %f\n", sqrt(2.0) / 2);
printf("acos(sqrt(2.0)/2.0) = %f\n",
acos(sqrt(2.0)/2.0));
printf("sin(acos(sqrt(2.0)/2.0)) = %f\n",
sin(acos(sqrt(2.0)/2.0)));
return program_success_code;
} /* main */
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
32
Function Use Example Part 4
% gcc -o funcuse funcuse.c -lm
% funcuse
2.0 = 2.000000
pi = 3.141593
theta = 2.356194
cos(pi) = -1.000000
cos_pi = -1.000000
sin(pi) = 0.000000
sin_pi = 0.000000
sin(theta) = 0.707107
sin_theta = 0.707107
sin(theta)^(1.0/3.0) = 0.890899
1 + sin(acos(1.0)) = 1.000000
sin(acos(1.0)) = 0.000000
sqrt(2.0) = 1.414214
sqrt(2.0) / 2 = 0.707107
acos(sqrt(2.0)/2.0) = 0.785398
sin(acos(sqrt(2.0)/2.0)) = 0.707107
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
33
Evaluation of Functions in Expressions
When a function call appears in an expression – for
example, on the right hand side of an assignment
statement – the function is evaluated just before its
value is needed, in accordance with the rules of
precedence order.
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
34
Evaluation Example #1
For example, suppose that x and y are float
variables, and that y has already been assigned the
value -10.0.
Consider this assignment statement:
x = 1 + 2.0 * 8.0 + fabs(y) / 4.0;
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
35
Evaluation Example #2
x =
1
+ 2.0 * 8.0 +
fabs(y)
/ 4.0;
x =
1
+
16.0
+
fabs(y)
/ 4.0;
x =
1
+
16.0
+ fabs(-10.0) / 4.0;
x =
1
+
16.0
+
x =
1
+
16.0
+
2.5;
x = 1.0 +
16.0
+
2.5;
+
2.5;
x =
x =
17.0
10.0
/ 4.0;
19.5;
CS1313: Standard Library Functions Lesson
CS1313 Spring 2009
36
```
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| 3.84375
| 4
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CC-MAIN-2017-17
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https://www.mapleprimes.com/questions/233591-Check-If-Two-Lists-Of-Vectors-Are-Equal
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|
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| 26,823
|
# Question:Check if two lists of vectors are equal under some permutation.
## Question:Check if two lists of vectors are equal under some permutation.
Maple 17
Hello friends, I have the following problem:
I have two lists of vectors, L1 and L2. The lists have the same number of vectors, and all the vectors have the same length.
I need to check if there is a permutation that, when applied to all the elements of one list, will obtain all the elements of the other list. For example, consider the following lists :
L1:=[<0|0|0>,<2|1|2>,<1|2|1>];
L2:=[<1|1|2>,<2|2|1>,<0|0|0>];
In this example the vectors are of length 3. Therefore, there are 3!=6 possible permutations. Namely:
P1:=<<1,0,0>|<0,1,0>|<0,0,1>>;
P2:=<<0,1,0>|<0,0,1>|<1,0,0>>;
P3:=<<0,0,1>|<1,0,0>|<0,1,0>>;
P4:=<<0,0,1>|<0,1,0>|<1,0,0>>;
P5:=<<0,1,0>|<1,0,0>|<0,0,1>>;
P6:=<<1,0,0>|<0,0,1>|<0,1,0>>;
In this case there are two permutations that satisfy the condition above, namely:
P3:=<<0,0,1>|<1,0,0>|<0,1,0>>;
P6:=<<1,0,0>|<0,0,1>|<0,1,0>>;
Cause
for i in L1 do Multiply(i, P3) end do;
>[0 0 0]
[2 2 1]
[1 1 2]
The same result is obtained using the permutation P6.
I'm working with larger lists and longer vectors so I'm looking for a quick way to check this. Thanks for your valuable help.
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|
https://plainmath.net/1349/find-derivatives-of-the-functions-defined-as-follows-y-equal-t-2t-plus-3t
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# Find derivatives of the functions defined as follows.y=(t^2 e^(2t))/(t+e^(3t))
Find derivatives of the functions defined as follows. $y=\left({t}^{2}{e}^{\left(2t\right)}\right)/\left(t+{e}^{\left(3t\right)}\right)$
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Alara Mccarthy
$dy/dt=\left(\left(t+{e}^{\left(3t\right)}\right)\left(2{t}^{2}{e}^{\left(2t\right)}+2t{e}^{\left(2t\right)}\right)-\left({t}^{2}{e}^{\left(2t\right)}\right)\left(1+3{e}^{\left(3t\right)}\right)\right)/\left(\left(t+{e}^{\left(3t\right)}{\right)}^{2}\right)$
$=\left(2{t}^{3}{e}^{\left(2t\right)}+2{t}^{2}{e}^{\left(2t\right)}+2{t}^{2}{e}^{\left(5t\right)}+2t{e}^{\left(5t\right)}-{t}^{2}{e}^{\left(2t\right)}+3{t}^{2}{e}^{\left(5t\right)}\right)/\left(\left(t+{e}^{\left(3t\right)}{\right)}^{2}\right)$
$=\left(\left(5{t}^{2}+2t\right){e}^{\left(5t\right)}+\left(2{t}^{3}+{t}^{2}\right){e}^{\left(2t\right)}\right)/\left(\left(t+{e}^{\left(3t\right)}{\right)}^{2}\right)$
| 510
| 1,159
|
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| 3.953125
| 4
|
CC-MAIN-2022-21
|
latest
|
en
| 0.640491
|
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