Question

Can somebody tell me why we assume V=0 in module 1 problem 1? I understand why it is so with gravity...

Answer 1

by continuing equation 1.3 we have:\[[T]=[M]^{-3V}[L]^{-3V+W+X+2Y+2Z}[T]^{-2X}\]now we must equalize powers from two sides of equation for each dimension (L,M,T). so we have 3 equations to solve:\[-3V=0\rightarrow V=0\]\[-2X=1\rightarrow X=-1/2\]\[-3V+W+X+2Y+2Z=0\]now we solve the third equation in continue

Answer 2

ParsaN, should the first factor on the right side of the equation be M^v since the exponential property of L^-3 dictates it's not distributable to M? We would be left with M^v, in which case v still = 0. Thanks or your post.