Answer
(a) less than
(b) $0.583c$
(c) $2.84cm$
Work Step by Step
(a) We know that due to length contraction, the lab traveled distance is less than $3.50cm$.
(b) We can determine the electron's speed as follows:
$v=\frac{L_{\circ}}{\Delta t}$
We plug in the known values to obtain:
$v=\frac{3.50\times 10^{-2}m}{0.200\times 10^{-9}m/s}$
$v=1.75\times 10^8m/s=0.583c$
(c) We know that
$L=L_{\circ}\sqrt{1-\frac{v^2}{c^2}}$
We plug in the known values to obtain:
$L=(3.50\times 10^{-2}m )\sqrt{1-\frac{(1.75\times 10^8m/s)^2}{(3\times 10^8m/s)^2}}$
$L=2.84cm$
