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0 Click here to jump to the problem!
tn
2018-01-23 02:46:39
TYPO! comment above should read:\r\n\r\nIf, E \\propto \\frac{1}{r^2} and S \\propto E^2, then wouldn\'t S \\propto \\frac{1}{r^4}? I think the answer to this problem is contingent upon being able to model the oscillating electron as a Hertzian dipole, which generates a field E \\propto \\frac{1}{r}. Then we can conclude S \\propto \\frac{1}{r^2}.
1 Click here to jump to the problem!
tn
2018-01-23 02:43:54
TYPO! comment above should read:\r\n\r\nIf, E \\propto \\frac{1}{r^2} and S \\propto E^2, then wouldn\'t S \\propto \\frac{1}{r^4}? I think the answer to this problem is contingent upon being able to model the oscillating electron as a Hertzian dipole, which generates a field E \\propto \\frac{1}{r}. Then we can conclude S \\propto \\frac{1}{r^2}.
2 Click here to jump to the problem!
tn
2018-01-23 02:39:21
If, E \\propto \\frac{1}{r^2} and S \\propto E^2, then wouldn\'t S \\propto \\frac{1}{r^4}? I think the answer to this problem is contingent upon being able to model the oscillating electron as a Hertzian dipole, which generates a field E \\propto \\frac{1}{r}. Then we can conclude S \\propto \\frac{1}{r^2}.
3 Click here to jump to the problem!
cairo45
2018-01-09 16:21:17
Earning was never to easy before this.Click here http://www.freepaypal.xyz/ and you can get unlimited dollars into your bank account.
4 Click here to jump to the problem!
Hopeful
2017-12-26 16:23:08
Oops. I see your post is recent. Most of the posts I\'ve seen on here are several years old so I was assuming. \r\n\r\nGo back to school. This test is too difficult to take without the background. I actually did just that myself. Got a BA in visual art and went back and basically started over (except didn\'t have to tak gen Ed again). It\'ll take a few years but 5 years from now 5 years will have passed regardless. \r\n\r\nGood luck!
5 Click here to jump to the problem!
Hopeful
2017-12-26 16:15:28
To the OP, it took 3 to 4 years to learn this material in the first place. I personally don\'t think I could have done it on my own in that time frame. I know your post is several years old but hopefully you went back to school and took the under grad physics sequence otherwise it\'s impossible to take this test.
6 Click here to jump to the problem!
Hopeful
2017-12-26 16:11:59
Hahahaha! I think I\'ll take my chances and continue to study for this test! Hahahahah
7 Click here to jump to the problem!
sean9
2017-11-28 08:41:44
When x = 0, the answer is just that of a point charge Q at distance R.\r\njson formatter\r\n
8 Click here to jump to the problem!
flatearther
2017-11-23 17:10:35
https://www.youtube.com/watch?v=pNe1wWeaHOU&list=PLYI8318YYdkCsZ7dsYV01n6TZhXA6Wf9i
9 Click here to jump to the problem!
flatearther
2017-11-23 17:07:31
The earth is flat! That\\\\\\\'s why is doesn\\\\\\\'t make sense! One thing to know about math, is you can lie in math and no one will know the wiser unless they go check it in real life. Does an empty cup equal a full cup? well in math it does, but in real life you drinking a cup of milk doesn\\\\\\\'t equal to you drinking a cup of air. Youtube search \\\\\\\"Edward Hendrie\\\\\\\" and watch the playlist with 26 videos in it, then you will know about how Einstein is a fraud! also youtube search \\\\\\\"WLC Videos flat earth jesuits\\\\\\\". This problem with newton and physics is that it is a lie, you can lie with math all you want!
10 Click here to jump to the problem!
yummyhat
2017-10-28 00:56:57
yay
11 Click here to jump to the problem!
yummyhat
2017-10-27 22:44:30
thanks rweads (and einstein)
12 Click here to jump to the problem!
capoo0819
2017-10-27 12:50:22
This might be the special case which 2R=d, thus the answer would still be correct.
13 Click here to jump to the problem!
yummyhat
2017-10-27 05:19:08
same\r\n
14 Click here to jump to the problem!
yummyhat
2017-10-27 04:08:14
what if the kid just melts and his goo sticks to the disc
15 Click here to jump to the problem!
yummyhat
2017-10-27 04:04:50
omg physpun
16 Click here to jump to the problem!
yummyhat
2017-10-27 03:46:41
thanks hybrid
17 Click here to jump to the problem!
yummyhat
2017-10-26 20:15:50
feelsbadman
18 Click here to jump to the problem!
yummyhat
2017-10-26 19:56:59
lmao dont think thats the kind of advice he wanted, but yeah plant your corn early
19 Click here to jump to the problem!
yummyhat
2017-10-26 19:41:40
almost every comment is wrong! we are asked for the behavior on the LEFT side of the boundary, i.e. reflection. the official solution is correct, but to explain it qualitatively: conductors cannot support electric fields, so the E field cannot penetrate the plane (and therefore no energy is carried through it). standing waves (no time dependence) do not transfer energy, so this is what is produced.\r\n\r\nby looking at the boundary conditions, we see that E at the surface of the conductor must be zero (because the E field inside the conductor is zero). in other words, the boundary condition fixes a node for E at the surface of the conductor. the B wave is phase shifted by ninety degrees with respect to the E wave, hence B must be at an antinode at the surface of the conductor, and it therefore interferes with itself constructively. \r\n\r\none could predict this by noticing that a phase difference of ninety degrees will zero the poynting vector, which is a description of energy transfer. \r\n\r\nalso, it seems important that if you move away from the surface of the conductor, you leave the node-antinode plane and things are a bit more complicated until the next node-antinode plane. this is why the question specifies that we are discussing the location IMMEDIATELY to the left of the plane. maybe that explains why everyone who took the exam missed it.. xD
20 Click here to jump to the problem!
yummyhat
2017-10-26 16:48:17
best answer
21 Click here to jump to the problem!
capoo0819
2017-10-26 12:54:10
(D) Ar is a nobel gas\\\\r\\\\n->Should be \\\\\\\"noble\\\\\\\"...
22 Click here to jump to the problem!
yummyhat
2017-10-26 12:40:36
use the fact that wavelength*frequency=velocity. notice that the wavelength is determined by the pipe and is constant. use a ratio because the ratio of velocities is already given, and solve for the final frequency.
23 Click here to jump to the problem!
poopterium
2017-10-26 02:05:27
booooooooooooooooooo
24 Click here to jump to the problem!
poopterium
2017-10-26 00:47:57
Just in case anyone comes across this....it comes from integrating over a circular aperture in optics. The 1.22 comes from a zero of a first order Bessel\'s function of the first kind. I believe the zero of the function is about 3.9, then dividing it by pi gives you the 1.22.
 
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