I don't know if you meant this
1/(8zxt^2) or not. Substitute the units into the eqwuation and solve for z.
Consider the equation
v = 1/8zxt2. The dimensions of the variables v, x, and t are [L]/[T], [L], and [T], respectively. The numerical factor 8 is dimensionless. What must be the dimensions of the variable z, such that both sides of the equation have the same dimensions? (Use the following as necessary: [L], [T].)
4 answers
I'm sorry, yes that is what I meant, but the difficulty I'm having is solving for z. I get it to a certain point and then I get stuck.
z=1/(8xt^2)
looking at 8xt^2, units must be L*T^2
so units for z must be 1/(LT^2)
looking at 8xt^2, units must be L*T^2
so units for z must be 1/(LT^2)
I think you forgot the v:
v = 1/8zxt^2
z = 1/8vxt^2
= 1/(L/T * L * T^2)
= 1/L^2T
v = 1/8zxt^2
z = 1/8vxt^2
= 1/(L/T * L * T^2)
= 1/L^2T
3 answers