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check on the my post here h t t p://w w w jiskha . com /display.cgi?id=1389359012
now use conservation of energy "assuming no energy is lost due to this process" hence you should have Potential of m1 = Kinetic m1 + Kinetic m2 or to elaborate m1gR = 1/2(m1)*(v1)^2 + 1/2(m2)*(v2)^2 given all those formula, you should be able to find an
now that gets you to a great relationship between v1 and v2, i suggest get v2 in terms of v1 so v2 = m1v1/m2
Use conservation of momenta you should get 0 = -m2v2 + m1v1 assuming that the block 2 will move to the left
To answer part a) 1. use conservation of energy and obtain how much energy is stored in the block when it reaches the bottom, obviously it is "mgR" since the height is the radius of the ramp. 2. Now this object will transfer energy to the spring and
this will surely hurt more than help but here you go Period = 2*pi sqrt([kl^2+mgl]/[(2mr^2)/3+ ml^2)
This problem is really all about an expression summarizing harmonic motion. So your period for this problem is oddly similar to T = 2pi * sqrt (m/k) but now m = moment of inertia and k = expanded coefficient, that was given above by Mr. Damon Also make
First Solve for the function T(z) T = (-alpha)*z + T(initial) T = -6.5z+293 Now plug that in the hydrostatic pressure equation dP/P = (-M*g/RT)dz integrate that and you will get P(final) = P(initial)*(1-(alpha*z/293))^(Ma*g/(R*alpha)) Part B is easy, get

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