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jedymark
2018-06-07 07:20:21
they are not necessarily each others anti particle: https://happywheelsnew.com
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aprilrussell
2018-05-30 03:51:45
run 3
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aprilrussell
2018-05-30 03:50:43
This seems the best solution. Yosun\'s is rather ambiguous. run 3
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aprilrussell
2018-05-30 03:48:38
At the very least we can eliminate choice (C) by units. \r\nhttp://fivenightsat-freddys.com
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aprilrussell
2018-05-30 03:45:18
A quick way I did this is similar to chemicalsoul\'s solution. https://run2.online/run-3 Thanks
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aprilrussell
2018-05-30 03:43:14
Relativity is so glamorous! I love it. \r\nhttps://run2.online/run-3
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WoosterGrad
2018-05-29 21:51:47
For the sake of other people studying for the GRE:\r\n\r\nC would be correct if the question asked for the expectation value of the energy-- the number you would get if you took many, many measurements and averaged them.\r\n\r\nHowever, the question asks for a possible value of ONE measurement. A single measurement of the energy will collapse the wave function to a single energy eigenstate.\r\n\r\nThe energy of a state is proportional to the square of the principle quantum number: E_n=n^2E_1.\r\n\r\nThe principle quantum numbers are 1, 2, and 3 in this superposition, so a single measurement would yield E_1, 4E_1, or 9E_1. \r\n\r\nThis reveals choice D as the correct answer.
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jonny77
2018-05-25 12:47:30
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2018-05-25 12:47:16
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jonny77
2018-05-25 12:47:07
I couldn\'t resist commenting. Perfectly written!\r\n\r\n http://medikapedia.com/ selengkapnya \r\n http://servicewikawaterheater.com Telepon Service Wika
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2018-05-25 12:46:44
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11 Click here to jump to the problem!
BillNyeTheRussianSpy
2018-05-22 03:06:49
I think, notationally, it is easier to have E_{n}=n^2 E_{1}. But it still comes out the same.\r\n\r\nE_{2}=2eV=4E_{1} implies E_{1}=1/2eV
12 Click here to jump to the problem!
BillNyeTheRussianSpy
2018-05-22 03:06:49
I think, notationally, it is easier to have E_{n}=n^2 E_{1}. But it still comes out the same.\r\n\r\nE_{2}=2eV=4E_{1} implies E_{1}=1/2eV
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BillNyeTheRussianSpy
2018-05-22 02:28:36
Don\'t look a crap mouth in the horse, as they say.
14 Click here to jump to the problem!
BillNyeTheRussianSpy
2018-05-22 02:25:21
I just remembered E=-\\mu \\cdot B from a class where \\mu is the electron spin magnetic moment, not the nuclear magnetic moment
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BillNyeTheRussianSpy
2018-05-20 03:49:57
I come for the physics, but stay for the suicide.
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timfinnigan
2018-05-20 03:19:45
This E value will allow you to eliminate choices D&E
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timfinnigan
2018-05-20 03:18:29
Plug \\psi into the TDSE and then set x=0. You will end up with E=(hb)^2 / 2M (should be h-bar). Put that into the TDSE and then cancel like terms. You will end up with V(x)=choice B.
18 Click here to jump to the problem!
timfinnigan
2018-05-19 03:40:44
Simple Work-KE Thm:\r\n\r\nW=\\Delta KE=KE_{f}-KE_{i}=0-(1/2)(5)(10)^2 =-250\r\n\r\nW=Fd=F(.025)\r\n\r\nF=W/.025=-250/.025=-10^4\r\n\r\nThe magnitude of the force, 10^4, is the answer (the negative could be thought of as direction).
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ellascott
2018-05-16 04:32:49
http://rollingskygame.com/run-3\r\n
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ellascott
2018-05-16 04:32:07
This topic is so much helpful topic.Thanks admin to giving this topic.And also thanks other users http://rollingskygame.com/run-3\r\n\r\n
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dwight5
2018-05-14 04:08:42
Agree!\r\ngmail sign up
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jeffgames
2018-04-14 08:04:48
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TimToolMan
2018-04-12 16:28:04
Disregard this one, formatting mistake. See my other comment.
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TimToolMan
2018-04-12 16:27:04
We know that \\beta=v/c=pc/E (using relativistic momentum and energy). Since v=c we have 1=pc/E which implies E=pc. And, from the relation E^2 = (pc)^2+(mc^2)^2, we can say that E=pc implies m=0.
 
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