Hello all, I have an exam on Monday and am having trouble with this problem, any help would be greatly appreciated! Q: A straight stick of length L' is at rest in the moving S' frame. The stick appears to have length L in the S frame. The S' frame is moving at a velocity √(2/3) c. Calculate the following quantities: gamma = ______ Δx' = ______L Δy' = ______L And calculate the length of the stick L' as observed in the S' frame. ΔL' = ______L And at what angle (with respect to the x'-axis) is the stick L' observed to be in the S' frame? angle' = _____degrees ------------ Relevant equations: 1/Sqrt[1 - (β])^2] Δx' = gamma(Δx - vΔt) ----------- My attempt: the gamma factor is = 1/√[1 - (√[2/3])^2] which is 1.73 then I use Δx' = gamma * x since Δt = 0 so Δx' would be gamma/2 in units of L? - (divided by 2 due to the nature of a 30 60 90 triangle?) so Δx' = .865 L? Then Δy' is equal to √(3)/2 L because there is no contraction of length in the y direction because the motion is along the x-axis, so √(3)/2 is the only factor applied due to the nature of a 30 60 90 triangle. as for the next two im sort of lost. Thanks for any help.