Answer
$0.54V$
Work Step by Step
We know that
$V=IR$
$\implies V=I\rho \frac{L}{A}$
We plug in the known values to obtain:
$V=\frac{(5.0A)(1.72\times 10^{-8}\Omega.m)(12ft)(\frac{0.3048m}{1ft})}{1.17\times 10^{-6}m^2}$
$V=0.54V$
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