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Latest comments
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| Latest 25 Comments | | 0 | Click here to jump to the problem! | Grandpsykick
2010-03-11 15:20:18 | Even more simply put:
1) E fields don't propagate through conductors (if they are normal to the surface).
2) The energy in the field MUST be transmitted through the surface.
3) E is zero (0), thus B cannot be zero (0); thus, the only answer is C.
NOTE: you didn't need any math, or "wave dynamics", or anything but simple knowledge of EM and conservation of energy. |
| 1 | Click here to jump to the problem! | cilginfizikci
2010-03-09 02:36:30 | why the energy can not be "0" ??? dont tell me that this is harmonic oscillator... coz u already change the system... ???????????? |
| 2 | Click here to jump to the problem! | cilginfizikci
2010-03-09 01:06:55 | For the solution at the top: The induced magnetic field direction can not be opposite to what it was before... on the contrary it s the same direction, so that the magnetic flux will be kept constant... Len'z law says that the closed loops doesnt like the change in magnetic flux... there was a magnetic field at the beginning and hen suddenly it s gone so is the magnetic flux... the two cgarhe system will begin rotating to acquire the former magnetic flux value... This approach looks the most formal one, and i tried to solve in this way, however the angular momentum i found is ( mdB(pi^2)(R^2) )/q u0 .... so i dunno what was the guy who prepared this question was thinking... |
| 3 | Click here to jump to the problem! | cilginfizikci
2010-03-08 23:42:31 | you have solved the equation wrongly... V= (gd)^0.5... and according to ur solution the answer is A... check ur calc. pls |
| 4 | Click here to jump to the problem! | chrisiskey
2010-03-08 22:14:25 | No, this makes sense. At the endpoints the velocity of the bob is zero, so there is indeed a need for a nonzero acceleration for the bob to return to the center position; however, the bob has a nonzero velocity at the center position so it is not necessary for a left- or right-pointing acceleration to be present for the bob to continue in the left/right direction. Consider circular motion - a centrally-pointing force is present yet at any instant the bob is moving perpendicular to the direction of the force vector. |
| 6 | Click here to jump to the problem! | fredsp7
2010-03-08 09:27:42 | I tryied to calculate like you said with any angle \theta but my answer came as being: M\leq \frac{2\mu m}{(2\cos \theta \sen \theta - \mu)} and it also provides the correct answer with \theta=45 |
| 7 | Click here to jump to the problem! | fredsp7
2010-03-08 09:09:31 | I tryied to calculate like you said with any angle \theta but my answer came as being: M\leq \frac{2\mu m}{(2\cos \theta \sen \theta - \mu)} and it also provides the correct answer with \theta=45 |
| 8 | Click here to jump to the problem! | Tatyana
2010-03-08 04:41:12 | Actually rotational angular momentum is determined as  only about rotational axis. We cannot calculate total angular momentum about the point P as  . rnThe line of reasoning leading us to the right answer is folowing. The collision is absolutely nonelastic as it is indicated in the formulation of the problem: "two discs stick together". In nonelastic collisions colliding bodies move as a united body after the collision. It means that relative motion of bodies about the center of mass of the system ceased after the collision. So discs will move translationally as a united body. The rotational part will be lost. Consequently the angular moment about the point P - the center of mass of the system in the moment of the collision - will be equal to zero. A is the right answer. |
| 9 | Click here to jump to the problem! | Tatyana
2010-03-08 04:40:28 | Actually rotational angular momentum is determined as only about rotational axis. We cannot calculate total angular momentum about the point P as . rnThe line of reasoning leading us to the right answer is folowing. The collision is absolutely nonelastic as it is indicated in the formulation of the problem: "two discs stick together". In nonelastic collisions colliding bodies move as a united body after the collision. It means that relative motion of bodies about the center of mass of the system ceased after the collision. So discs will move translationally as a united body. The rotational part will be lost. Consequently the angular moment about the point P - the center of mass of the system in the moment of the collision - will be equal to zero. A is the right answer. |
| 11 | Click here to jump to the problem! | Camoph
2010-02-18 08:30:37 | Good solution, but I'm not sure if you have time to do all this in the test, considering that you have 1,7min (average) for every question. Thank you indeed. |
| 13 | Click here to jump to the problem! | Camoph
2010-02-18 08:24:02 | I like your idea about using common, and it works I think, however answers B and c are very close. In the moment of the test, under pressure, is really hard try to deicide between them. There is a way to make a basic caculation to get the right answer between B and C ? |
| 14 | Click here to jump to the problem! | Camoph
2010-02-18 08:22:12 | I like your idea about using common sense, and it works, however, in this case the answers B and C are very close, in the moment of the test, under pressure is very hard try to deicide about one of them. There is an alternative way to do some basic calculation to give the right answer, beteween B and C? |
| 16 | Click here to jump to the problem! | torturedbabycow
2010-02-14 17:50:06 | dogsandfrogs explained it pretty well.
Once you've calculated both lengths in both reference frames, you'll see that whether the car "fits" in the garage also depends on the reference frame. (Which cuts out (A) and (C) immediately.) What this means is that the order of the doors opening/closing ALSO depends on the reference frame, since we're told that they do so automatically as soon as each end of the car passes.
This seems like it might be an issue for causality, but it isn't. (To be pedantic, it's not true that the order of putting-cake-in-mouth and eating-cake is relative - no matter what frame of reference you are in, putting-cake-in-mouth always comes first. Only when the second event is outside the "light cone" of the first can the order of events be reversed based on reference frame, since they're not causally related and it doesn't "matter" which one goes first. :P And that's why there's not really a paradox.) |
| 18 | Click here to jump to the problem! | jchys
2010-02-02 20:38:17 | If you look at the section carefully, it says that
"any particle that has MASS (according to special relativity) phase velocity of matter waves always exceeds c"
photons obviously has no mass. |
| 19 | Click here to jump to the problem! | Nathan_Grey
2010-01-31 22:30:13 | FortranMan: I understand that gamma is not needed for this problem, but gamma is customarily the variable used to define the ratio of specific heats. It is also related to the degrees of freedom by:  =gamma=) |
| 21 | Click here to jump to the problem! | physicsman
2010-01-27 22:23:17 | Q=W= (negative the area inside the loop)
Anyway you want to guess that area is the issue. Its kinda lame without the scale but -300 is the best guess just eyeballing it. |
| 22 | Click here to jump to the problem! | physicsman
2010-01-19 15:38:59 | Qualitatively: 1 proton + 1 electron is needed for every neutron
simply accounting for all the constituents of helium by pulling them from hydrogen will give you the proper number of hydrogen atoms to use. |
| 23 | Click here to jump to the problem! | physicsman
2010-01-18 22:27:50 | It should be noted in the solution that the potential degrees of freedom come form being an harmonic oscillator. If this was a free particle only the translation terms would show up and the energy would be halved |
| 24 | Click here to jump to the problem! | BerkeleyEric
2010-01-12 22:13:16 | C and D have the wrong units, so those can be eliminated immediately. A fails to satisfy the limit of r=0. The expression for B in the limit of theta = pi (with the mass almost hitting the ceiling) gives zero tension, which does not make physical sense. So this leaves E. |
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