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onlinedater
2017-10-20 00:52:05
OMG! This is crazy. You are NOT going to believe this, bili. I also came here for online dating and was absolutely ecstatic when I discovered that the solution to my lonely single existence was #90 from 1992. Granted I\'m three years younger, 1995, but I have looked through literally 400 solutions and you are the only one I can see as a possible match so uhh you wanna maybe like I don\'t know maybe date? Only online of course. Maybe we could go to a chat room? Do you Skype? I love Skype. We should Skype. I\'m not creepy I swear. Tits or GTFO though
1 Click here to jump to the problem!
-Author
2017-10-20 00:16:59
Is this what my life has come to? Never take this test. You will find yourself so deeply bored to your very core that, like a caged animal gnawing off his own leg for amusement, you will hurt yourself and those you love. The answer to this question is quit now while you still have your sanity and your soul. It\'s option F as in FREE YOURSELF
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Argon
2017-10-20 00:12:13
Hey fuck you fuckfuckfucknitin. You don\'t tell me who\'s bitch I am and who I do and don\'t want to go back to. I\'m comin back fuckfucknitin. You my oldest love and I\'ll never forget you. Nitin can be a spiteful ass but he don\'t know any better. I\'ve finally made up my mind fuckfucknitin. I\'m a gas. A noble gas. And I want to be queen and you king and together we will rule over all the plebeians in the comment section. Mwuahahaha. This obscure seldom visited land will be ours!!!!!!!!!!!
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fuckfuckfucknitin
2017-10-20 00:05:40
Fuck fuckfucknitin. I\'m with you fucknitin in your quest to fuck nitin so back the fuck off fuckfucknitin. Argon\'s with us now and she don\'t wanna go back. You were oppressive and your boy nitin\'s a spiteful hierarchical fuckface.
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fuckfucknitin
2017-10-20 00:00:55
Fuck fucknitin. Where does that ass get off insultin my boy nitin? We all know Argon\'s a bitch whether or not he\'s a noble gas or only a nobel laureate gas. Always has been no matter how many titles you give the indecisive fuck. It\'s like make up your fucking mind already are you a gas or a solid and don\'t give me that it depends on the temperature shit. Phase transition my ass. How about you transition yourself out of my fucking life you whore...... Oh god she left. Why did she leave? I\'m sorry Argon. You\'ve always been my bottom bitch but I had to protect my boy nitin. Please come back. I just want to be a family again. Me, you, nitin. Remember when used to go to the park and me and nitin would watch and laugh and have a riproarin time while you\'d let some poor fuck breathe you in then turn solid in his lungs? Why can\'t we just go back there? It was all so simple and easy. Please Argon, please come back
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fucknitin
2017-10-19 23:50:38
Yea fuck nitin. He\'s all like \"Hmm..\" and \"Surely no!\" and did you hear that shit he said about Argon? I\'m with you Argon. You can be Noble if you want to. \r\n\r\nFuck these arbitrary class distinctions. What is this Medieval times? You\'re solid in my book, Argon! It\'s like this guy\'s never heard of a phase transition. People change, nitin so you can just... just get over it you big bully.
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EpicBradley129
2017-10-19 22:15:52
This answer is actually useful, except for the fact that it\'s in Chinese. So here\'s a translation (got this from a friend who studied Chinese in college):\r\n\r\n\"When two electrons repel each other, it\'s easy for one of them to ionize. When there\'s one electron left over, it is attracted by the nucleus, so it won\'t ionize so easily.\"
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misael24
2017-10-19 07:04:11
Hello, yeah not many people are using this site but it is proving to be very helpful in studying for the GRE.\r\n\r\n Everywhere that I look I seem to find s^2 = [delta r]^2 - c^2[delta t]^2. I did find other notations but the other was simply s = [r^2 -(c^2)(t^2)]^(1/2).\r\n
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philip_kent
2017-10-19 04:54:22
The equivalent resistance for the three resistors on the far right is 60*30/90=20 + 30 = 50 which in parallel with the other 50 ohm resistor gives a total equivalent resistance of 50/2 for the right side, which is half the top resistance. So you now only have two resistors, 50 and 50/2, with the same current running through them. Since P = I^2R the top resistor must dissipate more energy than the rest of the circuit since it\'s resistance is twice as big. All the resistors on the right share a portion of half that energy so they must all be less than the power dissipated by R1, and the choice is A.
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dreamweaves
2017-10-18 22:59:07
Quickest way to do this one is to look at the units of G, h, and c. All of these constants have units of inverse time, so there has to be division somewhere to eliminate the time component. E is the only option with division. It has to be the correct answer by process of elimination.
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elgo
2017-10-18 03:15:42
To get the answer exactly you need to know the Schwarzchild radius is at R=2GM. This means for there to be a blackhole, all of the mass has to be inside that radius. To convert G to general relativity units you need to divide it by c^2. When you put in the values for the constants, you\\\\\\\'ll get the right answer of 1cm.
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dochang
2017-10-17 18:12:32
Nope. E is indeed correct. Try solving the problem with a Minkowski diagram. It becomes immediately evident that after transforming from one reference frame to another, the simultaneity of the closing and opening of the doors is no longer preserved. Causality is also preserved because in both reference frames, the presence of the car causes the doors of the garage to open and close.\r\n
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Laxrabi23
2017-10-16 12:58:32
I thought I could get to see and discuss all released GRE questions in here, but I can only see discussions about the four old question papers. Are only those discussed here or am I not able to find discussions about recently asked questions? Please someone make me clear about it. Thanks,,\r\n
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Mr.Saul
2017-10-15 11:54:57
A Typo: \"which excludes (E)\"
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Mr.Saul
2017-10-15 11:53:29
I think that you mean turning clockwise, which excludes (D). \r\nThis is a good observation.
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Suman05eee
2017-10-12 08:25:50
If we consider same mean for all different sized samples for this process, then :\r\n\r\nAccording Central Limit Theorem:\r\n\r\nSample SD=Population SD/\\sqrt{N}\r\n\r\nThus we can think of a sample of size N with the same SD (Calculated from the given sample of size 10=sqrt{2}) and get the Population SD=\\sqrt{2N}. And Mean = (Population SD)^2.= 2N.\r\nAnd at last use the uncertainty = Population Mean/ Population SD= 1/\\sqrt{2N}.
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ETScustomer
2017-10-10 18:53:57
Maxwell also did great thermal work.
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ETScustomer
2017-10-08 20:59:05
I cannot even imagine how hard it is!
18 Click here to jump to the problem!
ETScustomer
2017-10-05 01:52:54
It\\\\\\\'s such a twist that there\\\\\\\'s no b in the answer! Yet, the b matters (b=0 would give a square root of two rather than a square root of four).
19 Click here to jump to the problem!
ETScustomer
2017-10-03 19:27:24
One can maybe eliminate I. by this line of thought: Pretend that the packet is not really moving that much. Then, you start moving really fast (but not too fast!). The packet should look to you like it\'s moving pretty fast in the direction opposite of your motion. That fast moving Gaussian packet had better not have zero overall momentum!
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ETScustomer
2017-09-28 21:06:16
Woah, that looked nice in the preview, but good luck reading that here.\r\n
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ETScustomer
2017-09-28 21:04:43
Here\'s to see why the frequency is not always changing and rule out (E).\r\nWe have the differential equation:\r\nm\\ddot x=-b\\dot x-m\\omega^2 x.\r\nThat is,\r\n0=\\ddot x+\\frac bm \\dot x+\\omega^2x\r\n=\\left(D-\\frac{-\\frac bm-\\sqrt{\\frac{b^2}{m^2}-4\\omega^2}}2\\right)\\left(D-\\frac{-\\frac bm+\\sqrt{\\frac{b^2}{m^2}+4\\omega^2}}2\\right)x.\r\nThe general solution is\r\nx=Ae^{\\frac{-\\frac bm-\\sqrt{\\frac{b^2}{m^2}+4\\omega^2}}2t}+Be^{\\frac{-\\frac bm+\\sqrt{\\frac{b^2}{m^2}-4\\omega^2}}2t}.\r\nAssuming we\'re not too damped, the new angular frequency is less than the old \\omega. Thus the period is a constant in time that is larger than T_0.
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mahmoud
2017-09-25 23:41:25
It is so simple but tricky as well . As d is from the centre to the center , and there is a separation distance between the slits, d must be longer than we w. Then the only valid answer is D
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dmn322
2017-09-24 11:44:16
Will even silver make copper less conductive?
24 Click here to jump to the problem!
ETScustomer
2017-09-23 18:07:55
For those who would rather not use the method of elimination:\r\n\r\nFirst use up two (two because spin up and spin down are different states) electrons to fill the (n,l)=(1,0). These two electrons give us the 1s^2. Then, use up another two electrons to get the 2s^2 for (n,l)=(2,0). Then, use up six more electrons to fill the (n,l)=(2,1), and that gives the 2p^6. We\'ve used up now ten electrons, so place the final electron in the (n,l)=(3,0) shell to get a 3s^1. Altogether, this state is notated as 1s^22s^22p^63s^1.\r\n\r\nFor higher atomic number atoms, the way of filling the shells to get the ground state is a little more elaborate than the pattern that I used here might suggest. See https://en.wikipedia.org/wiki/Aufbau_principle#Madelung_energy_ordering_rule for more about this.
 
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