Physics problem using relativistic addition of velocities?
365902
(.8+.9)/(1+.8*.9)=.98837c
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
926868
Solve for momentum.\n\nBefore collision total momentum  = mv\n\nDefine your coordinate system so that the particle was traveling along the x axis in the positive direction, and the collision happens at the origin. Then all of the momentum before the collision is in the x direction. the initial y component of momentum is zero.\n\nparticle 1 goes flying off at v1 at angle a1\n\nparticle 2 goes flying off at v2 at angle a2\n\nThe total momentum afterwards must be equal to the momentum before. So the y component of p1 plus the y component of p2 must equal the y component of the momentum before collision (zero).\n\nSo m*v1*sin(a1) = m*v2*sin(a2)\n\nSimilarly the x components must add up so \n\nm*v1*cos(a1) + m*v2*cos(a2) = mv\n\nNow just solve the first equation for V1 and substitute in the second equation. \n\nThen solve the second equation with substitutions to find the value of v2 in terms of v, and the sin and cos of the two angles. \n\nFinally substitute the value of v2 back into your equation for v1 to find the value of v1.\n\nI leave the simple algebra of solving the equations and the calculation of plugging the numbers into the equations to the student.
901648
put x=2 directly into the given equation to get value of x at t=2...the answer u get is the displacement\n\ndifferentiate the equation wrt time to get velocity equation and then differentiate the velocity equation to get acceleration equation\n\nin both cases put t=2 to get velocity and acceleration\n\nconvert the given equation in terms of sin\ncomparing the new equation with x=asin(wt+@)\nwhere a=amplitude\n          w=angular frequency\n          @=phase\n\nalso use equations w=2 x pi x f and T=1/f\nhere x denotes multiplication \nf=frequency \nT=period
808209
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
62772
1/(1-v^2/c^2)\n\nthis effect causes time dialation, mass expansion, and length contraction.  it is negligible unless the velocity(v) is very very close to the speed of light(c) just try plugging in numbers for v, there wont be any noticeable affect until very very high velocities. the lorentz factor is part of einstein's theory of relativity.
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.
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use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
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The vectorial sum of the two speeds. If the bullet is fired straight foreward, simply add the speeds. If fired straight back, subtract.\n\nThe setup of your question seems to work with ordinary speed bullets and vehicles, so relativistic effects can be ignored along with  friction.
1171207
Let us write coeff. of friction = k     \n                    acceleration due to gravity = g     m/ sq. s\n                    uniform velocity of motion = u      m/s\n                 \n                    Since k=F/R where R = weight of the box acting downward and F = Frictional force acting against the direction of motion.\n                   Hence, F = kR. Again F= ma, where a is the retardation on the box and R=mg.\n                    Therefore, a=kg. (m cancels out on both sides).\n\n                     By using the formula:\n                            v^2=u^2 - 2aS  ( here a is retardation) we have\n                      S= u^2/(2a) = u^2/(2kg).\n\n            Substituting the values:\n                         u=8 m/s\n                         k= 0.16\n                         g=9.8 m/sq. s\n                        The result is S= (8^2)/(2*0.16*9.8) = 20.41 m (approx.).
75568
There are four (at least that I know of) to choose from.  You have to first analyze your problem and find out what you know and then solve for your unknown.  However, these equations assume that the acceleration is constant.\n\nv = v0 + a(t)\nHere:  v=final velocity; v0=initial velocity; a=constant acceleration; t=time\n\ns = s0 + (1/2)(v0 + v)(t)\nHere:  s=final position; s0=initial position; v0=initial velocity; v=final velocity; t=time\n\ns = s0 + v0(t) + (1/2)(a)(t^2)\n\nv^2 = v0^2 + 2*a(s-s0)\n\nI didn't say what each of the variables in the last two equations stood for because they are the same as what you see in the first two equations.  One good way to check if you are on the right track or not is to make sure your units work out right.  For example, if you are looking for velocity, chances are you'll be looking for something like meters per second (m/s) or miles per hour (mi/hr or mph).  Hope this helps!
965485
You can do a theoretical top speed with physics, but it wouldn't account for the two main killers of speed which are aerodynamic drag and internal friction.  Without knowing those, you'd only be getting a false promise from your calculations.
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From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
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)
826674
use conservation of angular momentum in these kind of problems\nNOT energy conservation\nif you dont know how to use the conservation equation, better read up on examples
987748
1.Compute initial velocity V0\nUsing equation S=V0*t-(g*t^2)/2 we find that the initial velocity V0=35/3 + 3g/2\n2.What ever goes up must come down V0*t=-(g*t^2)/2 t=2*V0/g\n3.Total time travel = 2t \nYou to the math\n\nBest\n\nHint:  the arrow will stop midair in about 5.3 seconds