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 0 Click here to jump to the problem! huanghandage2019-04-04 20:27:28 We know that the thing is gonna rotate, so eliminate A. Take the derivative of the five solutions, then x\'(0) has to be 2v, so eliminate B, D, and E, and the only choice left is C. Click here to jump to the problem!
 3 Click here to jump to the problem! booyah2019-04-02 05:25:52 I have always thought the \"cyclotron frequency\" referred to the angular frequency, so I left out the 2pi. I feel like ETS should have made which type of frequency they were referring to in the question statement more clearClick here to jump to the problem!
 10 Click here to jump to the problem! andymark2019-02-26 10:01:05 It is the brilliant blog. I like the way you express information to us. It is very useful to me. You are doing a great job and thanks for sharing.\r\nVisit games.lol for Puzzle GamesClick here to jump to the problem!
 14 Click here to jump to the problem! tdl172019-01-05 22:45:55 You can also use dimensional analysis. Whatever that goes into the arc tangent should be purely numerical without any units, and the only answer where this happens is choice A. Vq cancels with mv^2 because they both have units of Joule. L cancels with d because they both have units of length. Click here to jump to the problem!
 15 Click here to jump to the problem! tdl172019-01-03 21:16:15 Since the particles only took one half-life, which is $2.5\\\\\\\\times10^{-8}$ seconds, to travel 15m, it must be traveling close to the speed of light. \\\\r\\\\n\\\\r\\\\nAssume the speed of the particles is some fraction of the speed of light and let this fraction be $a$, we now have $v = ac$. \\\\r\\\\n\\\\r\\\\nUsing $v=\\\\\\\\frac{L\\\\\\\\sqrt{1-\\\\\\\\frac{v^2}{c^2}}}{t_0}$, we can replace v with ac and from there, you can solve for a and that should get you to $a = \\\\\\\\frac{2}{\\\\\\\\sqrt{5}}c$, which is choice C.Click here to jump to the problem!
 23 Click here to jump to the problem! TheBridge2018-10-27 02:12:16 But 1 + i - i is not equal to 0. The eigenvalues are actually 1, $-\\\\\\\\frac{1}{2} \\\\\\\\pm \\\\\\\\frac{\\\\\\\\sqrt{3}}{2}i$.Click here to jump to the problem!
 24 Click here to jump to the problem! TypT2018-10-26 22:10:37 Quick, dumb way I did it without having to figure out what exactly the question was asking:\r\n\r\nSpace time intervals have a Pythagorean relationship, the two numbers we are given are 3 and 5, to complete the Pythagorean triple we need a 4 so we get (C). It even has the trap answer of $4^2 = 16$ (E), to give a little more confidence in (C).Click here to jump to the problem!