Answer
5.85 m
Work Step by Step
$R=\frac{V^{2}}{P}=\frac{(75.0V)^{2}}{5000W}= 1.125 \Omega$
Rearranging the equation $R=\rho\frac{l}{A}$, we have
$l=\frac{RA}{\rho}= \frac{1.125\Omega\times2.60\times10^{-6}m^{2}}{5.00\times10^{-7}\Omega m}= 5.85 m$
