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NOW: WITH FREE PRINTABLE SOLUTIONS. See Gold Linkbar above.
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.
A Brief Guide to Basic Site Navigation:
1-liner: Jump directly to a particular problem via the QuickLoad bar. (On the grey bar on top of the page, you can quickly jump to a problem by selecting GRxxxx, Problem y.)
In the Meta-Search below, you can search for problem(s) by type and subtype---or search * in solutions or user-comments.
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:
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Exam GR8677; Problem 18; Page 0
Posted by Kabuto Yakushi, on 2010-09-02 09:49:11. |
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I just plugged the wave function into the schrodingers equation.
is obviously zero so V(x)
equals the second time derivative of psi time the
. Solving one gets B).
Thanks for the great site Yosun. |
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Exam GR8677; Problem 8; Page 0
Posted by Kabuto Yakushi, on 2010-09-02 09:25:45. |
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Just use the Impulse-Momentum theorem.
F t = M V
the answer falls out. |
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Exam GR9277; Problem 19; Page 0
Posted by ashestofeonix, on 2010-08-31 01:54:18. |
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| uhh... I hate to be a stickler about something as silly as correct units but.... the density of water is 1 g/cm, that is 1kg/dm^3... And most of the earth is not water... most of the SURFACE of the earth is covered by water, but most of the volume is iron & nickel. There really is a big difference. |
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Exam GR9277; Problem 29; Page 0
Posted by shak, on 2010-08-29 15:55:08. |
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Exam GR9677; Problem 91; Page 0
Posted by shak, on 2010-08-20 17:19:23. |
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Exam GR8677; Problem 93; Page 0
Posted by his dudeness, on 2010-08-19 18:37:17. |
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| Yeah I definitely fell for it too... Here's Wikipedia to the rescue: "Occasionally it is maintained that µ is always < 1, but this is not true. While in most relevant applications µ < 1, a value above 1 merely implies that the force required to slide an object along the surface is greater than the normal force of the surface on the object. " |
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Exam GR9677; Problem 81; Page 0
Posted by shak, on 2010-08-19 07:02:42. |
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| Can u please explain me, why are u dividing frequency of A4 to the frequency of D2 to find harmonics?
even , if then answer is close to 6 , but it is smaller than 6, and therefore answer should be 5,i.e lowest? thank you |
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Exam GR8677; Problem 31; Page 0
Posted by his dudeness, on 2010-08-18 13:12:00. |
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| kroner, the situation is the exact opposite of what you said. If point B has a negative voltage with respect to point A, then the electric field points from A-->B. This means that positive charges want to drop in voltage and electrons want to go up in voltage. So we need the detector to be at a negative value that's sufficiently high to stop the electron current which arises naturally (even at zero voltage). Tricky stuff! |
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Exam GR8677; Problem 1; Page 0
Posted by atp_ch, on 2010-08-17 05:43:03. |
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Exam GR9677; Problem 66; Page 0
Posted by shak, on 2010-08-16 21:09:25. |
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