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Latest comments
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| Latest 25 Comments | | 2 | Click here to jump to the problem! | ashestofeonix
2010-08-31 01:54:18 | 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. |
| 5 | Click here to jump to the problem! | his dudeness
2010-08-19 18:37:17 | 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. " |
| 6 | Click here to jump to the problem! | shak
2010-08-19 07:02:42 | 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 |
| 7 | Click here to jump to the problem! | his dudeness
2010-08-18 13:12:00 | 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! |
| 10 | Click here to jump to the problem! | hybridusmanus
2010-08-16 15:33:08 | claire is right, the n=3 state for the potential well is an even state, though 3 is odd. (Think the first state (n=1) has to be the ground state)
Use this mnemonic: For the Quantum Harmonic Oscillator, the value of n in is HARMONY with the state (1 is odd and is an odd state, 2 is even and is an even state, etc...)
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| 13 | Click here to jump to the problem! | shak
2010-08-14 08:57:39 | Most internal conversion electrons come from the K shell (1s state, see electron shell), as these two electrons have the highest probability of being found inside the nucleus. After the electron has been emitted, the atom is left with a vacancy in one of the inner electron shells. This hole will be filled with an electron from one of the higher shells and subsequently a characteristic x-ray or Auger electron will be emitted. |
| 14 | Click here to jump to the problem! | shak
2010-08-13 15:02:04 | The best and easiest way to solve this problem is using Lagrangian equation of motion.. it is one dimensional system.. so small charge -q is moving only along the x-axis.
Lagrangian is
L=E- V (1)
and
E=m*x(dot)^2/2 (2)
V= same potential in problem 3.
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| 15 | Click here to jump to the problem! | his dudeness
2010-08-12 07:17:30 | This question is very poorly worded. Since they don't tell us the thickness of the glass, I guess we are meant to assume very thick glass plates with the light source coming from inside the glass. That is the only way their answer makes sense. |
| 16 | Click here to jump to the problem! | shen
2010-08-11 08:25:04 | Actually u got the wrong tangential vector. It should be found using the gradient dy/dx.
r = [1, x/2]/(1+x^2/4)^1/2
You got the correct answer through a careless mistake that correctly make up for your wrong formulation. Cheers. You can check your answers. |
| 18 | Click here to jump to the problem! | giga17
2010-08-11 02:06:05 | This is one of those problems where dimensional analysis would help a lot. The only way to get a time scale out of the physical quantities from R, L and C is 1/ . |
| 19 | Click here to jump to the problem! | archard
2010-08-06 09:00:12 | The light only undergoes a phase shift when being reflected by a medium with higher index of refraction than the medium it comes from. Glass has a higher index of refraction than air, so the light that reflects from the Glass|Air interface does not undergo a phase shift, while the light that reflects from the Air|Glass interface does. |
| 21 | Click here to jump to the problem! | Phys4
2010-08-03 16:50:03 | If you set up the problem correctly like Yosun does, just do an order of magnitude calculation. You will get -14 and one can plainly see the other solutions have at least two orders of magnitude off. The small coefficients indicate that one does not have to worry about them influencing the order of magnitude. |
| 22 | Click here to jump to the problem! | Phys4
2010-08-02 15:44:18 | Curious, the version of the exam I have (GR9677 obtained from the Ohio State website) defines the specific heat in units of Kelvin and not Celsius. To compute this problem straight through, I would believe that you would need to convert from Celsius to Kelvin. In this case, since you don't, am I safe to assume that the original exam had an error? |
| 23 | Click here to jump to the problem! | signminus
2010-08-02 14:25:18 | Another POE technique here is to realize that the hoop's angular momentum must depend on its radius (think about a hoop with really huge radius versus one with a tiny radius), so only answers (A) and (B) remain. Of course, something more is needed to get the exact answer, but in a pinch you'll be down to a 50-50 random guess at the worst. |
| 24 | Click here to jump to the problem! | signminus
2010-08-02 12:35:10 | Does this actually work? If by the length of the loop you mean its circumference, it actually grows as  rather than  ... |
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