## Physics (10th Edition)

Length $l=2.80\,m$ Area $A=\pi r^{2}=\pi (1.03\times10^{-3}\,m)^{2}=3.33\times10^{-6}\,m^{2}$ $R=\frac{V}{I}=\frac{0.0320\,V}{1.35\,A}=0.0237\,\Omega$ Another equation for resistance is $R=\rho \frac{l}{A}$ Rearranging this equation, we get $\rho=\frac{RA}{l}=\frac{0.0237\,\Omega\times3.33\times10^{-6}\,m^{2}}{2.80\,m}=2.82\times10^{-8}\,\Omega\cdot m$ The material with resistivity equal to $2.82\times10^{-8}\,\Omega \cdot m$ is aluminum.