GR 8677927796770177 | # Login | Register

Latest comments
Latest 25 Comments
 0 Click here to jump to the problem! djcluv2019-02-28 09:13:42 scary games for kidsClick here to jump to the problem!
 1 Click here to jump to the problem! djcluv2019-02-28 09:12:02 thanks for having me scary games for kidsClick here to jump to the problem!
 2 Click here to jump to the problem! djcluv2019-02-28 09:10:37 thanks for having me scary games for kidsClick here to jump to the problem!
 3 Click here to jump to the problem! djcluv2019-02-28 09:09:40 thanks for having me scary games for kidsClick here to jump to the problem!
 4 Click here to jump to the problem! andymark2019-02-26 10:06:47 Excellent information on your blog, thank you for taking the time to share with us. Amazing insight you have on this, it’s nice to find a website that details so much information about different artists.\r\nVisit games.lol for Arcade GamesClick here to jump to the problem!
 5 Click here to jump to the problem! andymark2019-02-26 10:04:26 Hi ! Nice post .Points are really good . Great article .Over all nice content , Thanks for posting.\r\nGame Lover - https://games.lol/puzzle/Click here to jump to the problem!
 6 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!
 7 Click here to jump to the problem! andymark2019-02-26 09:59:12 Love to read it,Waiting For More new Update and I Already Read your Recent Post its Great....Thanks\r\n\r\nVisit games.lol for Arcade Games\r\nClick here to jump to the problem!
 8 Click here to jump to the problem! andymark2019-02-26 09:57:41 Love to read it,Waiting For More new Update and I Already Read your Recent Post its Great....Thanks\r\nVisit games.lol for Arcade Games\r\nClick here to jump to the problem!
 9 Click here to jump to the problem! johnbhura12019-01-18 11:29:34 See, to be very true i don\'t know good about Physics but i am glad if someone game answer https://codesuwant.com/ too.Click here to jump to the problem!
 10 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!
 11 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!
 12 Click here to jump to the problem! djcluv2018-12-14 00:44:47 thanks for this blog!! multiplayer racing gamesClick here to jump to the problem!
 13 Click here to jump to the problem! djcluv2018-12-14 00:41:42 thanks for this blog!! multiplayer racing games\r\nClick here to jump to the problem!
 14 Click here to jump to the problem! webtechcoupons2018-11-05 07:04:26 I have perused your blog it is exceptionally useful for me. I need to express profound gratitude to you Godaddy Renewal coupon I have bookmarked your site for future updates. Click here to jump to the problem!
 15 Click here to jump to the problem! webtechcoupons2018-11-05 07:03:03 Aw, I thought this was a significant pleasant post. Godaddy Renewal coupons In idea, I should set up composing like this also – investing energy and real exertion to make an extraordinary article… yet precisely what do I say… I hesitate a lot through no methods seem to go did. Click here to jump to the problem!
 16 Click here to jump to the problem! raytyler2018-11-01 03:14:57 \\lambda_1 = -\\frac{1}{2} + i\\frac{sqrt{3}}{2}\r\n\r\n\\lambda_2 = -\\frac{1}{2} - i\\frac{sqrt{3}}{2}\r\n\r\n\\lambda_3 = 1\r\n192.168.0.1 http://19216801help.com/Click here to jump to the problem!
 17 Click here to jump to the problem! raytyler2018-11-01 03:13:02 1 + i - i is not equal to 0. The eigenvalues are actually 1, -\\\\\\\\\\\\\\\\frac{1}{2} \\\\\\\\\\\\\\\\pm \\\\\\\\\\\\\\\\frac{\\\\\\\\\\\\\\\\sqrt{3}}{2}i.\r\n192.168.0.1 loginClick here to jump to the problem!
 18 Click here to jump to the problem! petyr2018-10-30 06:41:46 dp=v*dm=v*rho*dVolume=v*rho*A*dx=v*rho*A*v*dt\r\n\r\ntherefore dp/dt=F=rho*(v^2)*A.\r\nshowboxClick here to jump to the problem!
 19 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!
 20 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!
 21 Click here to jump to the problem! The Bridge2018-10-26 19:15:17 That won\'t work. The formula on the left is for centripetal acceleration of someone standing on the Equator due to the rotation of the Earth, and works out to about .88 m/s2. This is in fact many times less than $g$, the acceleration due to gravity of something on the Earth\'s surface, which is the right side. If you knew the orbital period of an object at the Earth\'s surface off the top of your head (84 min), then this method would work, but it gives you $g\\approx$10 m/s2 in the table of information at the top of the test anyway, so the left side of the equation is completely unneeded.Click here to jump to the problem!
 22 Click here to jump to the problem! kronotsky2018-10-26 03:27:43 The relative differences in wavelengths is given by the factor (n-1), so this is what you are equating to 2e-4. Other than that, the situation happens to be exactly analogous, since increasing the index of refraction essentially increases the effective optical path length, which is the quantity used to derive the thin film equation.Click here to jump to the problem!
 23 Click here to jump to the problem! Ryry0132018-10-25 12:05:41 My half-BS solution: I tried to figure out which of the answers could reasonably come from 1.5. I thought 1.5 = 3/2, saw 2/3, and then picked it. A guessing strategy for if you know zero Physics.Click here to jump to the problem!
 24 Click here to jump to the problem! kronotsky2018-10-25 02:57:36 I\'m sorry, but you are misinformed about Gauss\'s law. Classically, Maxwell\'s equations hold at any instant in a given frame. Gauss\'s law simply follows from the divergence theorem and the first equation. It may even be more appropriate to say that the first equation is Gauss\'s law. Gauss\'s law certainly applies for radiated fields (and causal propagation fields traveling at the speed of light), though radiation typically breaks the electrostatic symmetry. But in this case, the symmetry argument you use to apply the integral form of Gauss\'s law has exactly the same content as the symmetry argument you use to exclude radiation. Click here to jump to the problem!

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...