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GRE Physics Solutions
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Welcome to GRE Physics Solutions. This is where you'll find complete in-depth solutions to all problems from the ETS released GRE Advanced Physics Examinations.

Free solutions, always and forever. The site aims to provide the most comprehensive and easy-to-access source of solutions to released GRE Physics exams.

 

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Alternate solutions are contributed by users like you in the comments section appending each problem. Your questions pertaining to a particular problem are, in turn, answered by other users. If you find any errors or have an alternate solution you'd like to share---feel free to post a comment!

 

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Top 10 latest comments:

1 Exam GR9277; Problem 68; Page 0
Posted by p3ace, on 2008-05-11 15:32:40.
 
     
So why do we use the formula for diffraction to get the width of a mirror for resolution????    
     
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2 Exam GR9677; Problem 87; Page 0
Posted by p3ace, on 2008-05-05 01:36:02.
 
     
This is an interesting problem. Lenz's Law is easy to brush off as just explaining the minus sign in Faraday's law, but it really should be given more emphasis. One cannot determine the direction without understanding what the minus sign really means. If the change in B is an increase in B then the induced field opposes the original direction of B, but if the change in B is a decrease, then the direction is IN THE SAME DIRECTION AS B.    
     
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3 Exam GR8677; Problem 44; Page 0
Posted by p3ace, on 2008-05-04 14:34:50.
 
     
Only linear momentum for center-of-mass coordinates is needed.    
     
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4 Exam GR8677; Problem 13; Page 0
Posted by glkjap, on 2008-05-04 14:23:09.
 
     
You said: "Interference is produced as long as the sources are coherent, and the sources are coherent as long as there's a constant relation between relative phase in time." But the sources don't have a constant phase. It changes from 0 to 2\pi, right? So how are they coherent?    
     
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5 Exam GR8677; Problem 1; Page 0
Posted by withuman, on 2008-04-30 23:08:04.
 
     
\frac{a}{b}    
     
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6 Exam GR9677; Problem 83; Page 0
Posted by Rus Almighty, on 2008-04-29 08:44:53.
 
     
fix, instead of \frac{\partial\dot{h}}{\partial x} = - k^2 d sin kx dot(x) = - Vx k^2 d cos kx \frac{\partial\dot{h}}{\partial x} = - k^2 d cos kx \dot x = - Vx k^2 d cos kx    
     
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7 Exam GR9677; Problem 83; Page 0
Posted by Rus Almighty, on 2008-04-29 08:35:33.
 
     
A condition for the body to stay on the track: \frac{Vy}{Vx} = \frac{\partial h}{\partial x} let's derive both sides by t and remember that on the top kx= 2 \pi , Vx is V , \dot{Vy} = -g and \dot{Vx} = 0 'cuase there is only gravitational force on the top. we get: \frac{\dot{Vy}}{Vx} = \frac{\partial\dot{h}}{\partial x} now let's calculate \frac{\partial\dot{h}}{\partial x} \frac{\partial h}{\partial x} = - kd sin kx \frac{\partial\dot{h}}{\partial x} = - k^2 d sin kx dot(x) = - Vx k^2 d cos kx And from now on it's only algebra: - \frac{g}{Vx} = - Vx k^2 d cos kx kx= 2 \pi Vx= \sqrt{\frac{g}{k^2 d}} QED    
     
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8 Exam GR0177; Problem 17; Page 0
Posted by judijasa, on 2008-04-25 11:48:14.
 
     
Correction: the order is s,p,d,f,g, not s,p,d,g,f    
     
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9 Exam GR0177; Problem 17; Page 0
Posted by judijasa, on 2008-04-25 11:46:00.
 
     
I think, an easier way to see why the answer is definetly B, instead of calling the "4S" criteria, is that before getting to the 3d subshell you have to fill the 3p just as the s,p,d,g,f order suggests.    
     
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10 Exam GR8677; Problem 92; Page 0
Posted by hanmas, on 2008-04-11 20:48:21.
 
     
s always equals to 1/2 for electron, so there is no such thing as \Delta s = -1.    
     
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