Please Help me solve this relativity problem.?
144170
Space would have be the absolute reference. Since according to Relativity that reference does not exist than there is no way of determining any parameter.\nTo Answer that you ahave to go outside relativity theory where an absolute frame of reference is found.
367853
F = m*a \n\nm = 2 kg (two masses; two strings)\na = gravitational constant which happens to be 9.81 m/s^2 on the surface of the earth (or 32.2 ft/s^2 in EE units)\n\nsum the forces in the y-direction:\n\nT = F = 2 * m * a = 2 N\n\n(assuming massless strings)
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
981582
Check Wikipedia: http://en.wikipedia.org/wiki/Theory_of_relativity
1150360
The acceleration due to gravity is 32.2 ft. per second per second, which is pretty near 32.  Now we need two fundamental equations of physics:\n\nv = a / t\n\nwhich we can re-write\n\nt = a / v\n\nSince a (acceleration) is 32 ft/sec^2, and v (velocity) is 32 ft/sec, time (t) is 32/32 = 1 second until the ball's velocity equals zero.\n\nSo how much distance in 1 second? The second equation gives us the answer:\n\nd = 1/2 a t^2\n\nwe know a = 32, and since we now know t = 1, we can solve the equation. \n\nt^2 = t * t = 1 * 1 = 1\n\nso\nd = 1/2 * 32 * 1\nd = 16 feet
610467
Go to your local physics professor. If you are correct, (s)he will no doubt help you to get it published. If you are incorrect, (s)he should be willing to show you where your math is off.
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
997326
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
1161934
Your understanding of relativity is wrong.\n\nYou cannot even accelerate to the speed of light, let alone beyond it. Relativity predicts that infinite enrgy would be required to accelerate a mass to the speed of light. IT makes no predictions that would make travel backwards in time possible.
776431
Integrate the acceleration equation with respect to time, twice.  Use, Xo = 500 and Vo = 10...solve for t.
18392
According to the lorentz transformation the space time coordinate of two inertial reference frame moving at a relative speed v along x axis is\nx'=(x-v*t)/L\ny'=y \nz'=z\nt'=(t-v*x/c^2)\nwhere L is lorentz factor=sqrt(1/(1-v^2/c^2));\n\nThe tricky thhing is this if any object moves faster than light the L of above equation will become imaginary.\nSo for an observer the proper time t' and the relativistic mass m=Lm0 becomes imaginary...\n Thus it cannot be faster than light.\nHowever theoritically there exist tachyon particles which moves in superluminal(faster than light)  velocity with imaginary proper time.\n But those crazy physix!!!
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
1014031
I totally agree with you. Relativity is a big factor in life.
835037
Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)
308200
Neither Observer EXPERIENCES a change. The ONLY way any comparison can be made between the two unmentioned but essential clocks is to bring them together in the same rest frame again: then whichever clock ticktocked the least would be shown to be the one which accelerated and slowed down in their relative travel.
876226
Visit this link for a good explanation:\n\nhttp://en.wikipedia.org/wiki/Theory_of_relativity
932886
the speed of light = 299 792 458 m / s or 1079252848800 M/H\ncircumference = 2*Pi*R \nso 2* 3.14159*R= 107925848800M/H\n1079252848800/(2*3.14159)=R\nRadius= 1.71768571 × 1011(<--- times 10 to the 11th power) meters