Do you mean this relationship?
dS = \frac{\delta Q}{T}
\Delta S = \int \frac{\delta Q_{reversible}}{T}
If yes, then this would be a good starting points I guess:
http://en.wikipedia.org/wiki/Clausius_theorem
And then the explanation under the "definition" section here...
Oh ok, fair enough.
Yeah. I was looking at beams with point loads but it was too difficult to come up with an expression that described the bending moment (because it was described piecewise with discontinuity in shear forces) so I gave up.
I think I see the problem with my expressions...
Hello, I have a small question about moments and bending moments.
So, if I have a beam with a loading given by q (N/m) which is given as a function of x then what do these calculations get me?
\int xq(x) dx
\int (\int q(x) dx) dx
The first integral gives me the moment about a point because I...
Ok, never mind.
I tried solving the M=EI \frac{d^2y}{dx^2}
equation for the scenario 1 on the link, using boundary conditions y(0)=0 and y'(0)=0 and I got the given result so I guess it is how they got all the results.
I used the result I got on my second post and it gave the right answer.
The...
So, I have found some equations for deflection of beams that I'm assuming come from solving the above differential equation for basic scenarios:
http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf
So, my question now is, can the principle of superposition be...
Homework Statement
Homework Equations
For small deflections:
M=EI \frac{d^2y}{dx^2}
The Attempt at a Solution
To solve a problem like this, I think I was told I need to study the deflection and displacement of the beam.
If I said that the deflection at points x=0 and x=3L have to be 0...
I think the charge is going to equalise. So you can just add the charge of A and C and divide it by 2 to get the total charge on each one.
So, lets say A has 0 and B and C have Q.
A touches B so they both have Q/2.
A touches C. The charge distributes equally. So, total charge is 3Q/2. So, there...
Homework Statement
Homework Equations
dI = \frac{\partial I}{\partial V_{GS}} dV_{GS} + \frac{\partial I}{\partial V_{DS}}dV_{DS}
The Attempt at a Solution
I was just wondering why \small V_{G} is at 0. I have solved the entire question (there are more parts :P), but I could only do it...
I was thinking maybe use the F = E q = Vq/d = ma ?
But that means any non zero potential will be able to bring it to rest (lower potential will simply take long to do it and the nucleus will travel a larger distance while decelerating).
Thanks man.
I looked at your question but I don't know how to do it.
I was thinking maybe use the F = E q = Vq/d = ma ?
But that means any non zero potential will be able to bring it to rest (lower potential will simply take long to do it and the nucleus will travel a larger distance while...
How will you get a function for phi in terms of t so you can differentiate?
Look at this link:
http://en.wikipedia.org/wiki/Faraday's_law_of_induction#Faraday.27s_law_as_two_different_phenomena
It talks about circuit moving and field changing as 2 different phenomena and I think it will be...
I found another thread that went through some of the derivation:
https://www.physicsforums.com/showthread.php?t=267367
They talk about phasor representations and so on, which is a little beyond my mathematical knowledge so I think I'll just have to learn the relationship you gave me.
How would...
I think you are right.
How did you derive the
Z = R + j(XL - XC)
I think you are using the final general solution (sort of the solution you get after you plough through all the maths).
I think you get Z = R + jXL for inductor and Z = R - jXC for a capacitor. But am not too sure.
I wanted to...
Yo dawg, consider each side of the rectangular loop.
The top horizontal side "cuts" no flux cause it ain't got no balls.
The right side cuts flux so there IS an EMF cause it's one badass playa.
The left side cuts flux too and there IS an EMF generated but it's like ya know consider the direction...