GR 8677927796770177 | # Login | Register

 0 Click here to jump to the problem! onlinedater2017-10-20 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 thoughClick here to jump to the problem!
 2 Click here to jump to the problem! Argon2017-10-20 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!!!!!!!!!!!Click here to jump to the problem!
 3 Click here to jump to the problem! fuckfuckfucknitin2017-10-20 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.Click here to jump to the problem!
 5 Click here to jump to the problem! fucknitin2017-10-19 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.Click here to jump to the problem!
 6 Click here to jump to the problem! EpicBradley1292017-10-19 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.\"Click here to jump to the problem!
 7 Click here to jump to the problem! misael242017-10-19 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 Click here to jump to the problem!
 8 Click here to jump to the problem! philip_kent2017-10-19 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.Click here to jump to the problem!
 9 Click here to jump to the problem! dreamweaves2017-10-18 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.Click here to jump to the problem!
 10 Click here to jump to the problem! elgo2017-10-18 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.Click here to jump to the problem!
 11 Click here to jump to the problem! dochang2017-10-17 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 Click here to jump to the problem!
 14 Click here to jump to the problem! Mr.Saul 2017-10-15 11:53:29 I think that you mean turning clockwise, which excludes (D). \r\nThis is a good observation. Click here to jump to the problem!
 15 Click here to jump to the problem! Suman05eee2017-10-12 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/$\\sqrt{2N}$.Click here to jump to the problem!
 18 Click here to jump to the problem! ETScustomer2017-10-05 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).Click here to jump to the problem!
 21 Click here to jump to the problem! ETScustomer2017-09-28 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\n$m\\ddot x=-b\\dot x-m\\omega^2 x.$\r\nThat is,\r\n$0=\\ddot x+\\frac bm \\dot x+\\omega^2x\r\n=\\left(D-\\frac{-\\frac bm-\\sqrt{\\frac{b^2}{m^2}-4\\omega^2}}2\\right)\\left(D-\\frac{-\\frac bm+\\sqrt{\\frac{b^2}{m^2}+4\\omega^2}}2\\right)x.$\r\nThe general solution is\r\n$x=Ae^{\\frac{-\\frac bm-\\sqrt{\\frac{b^2}{m^2}+4\\omega^2}}2t}+Be^{\\frac{-\\frac bm+\\sqrt{\\frac{b^2}{m^2}-4\\omega^2}}2t}.$\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 $T_0$.Click here to jump to the problem!
 24 Click here to jump to the problem! ETScustomer2017-09-23 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.Click here to jump to the problem!