Number 10 wire has a diameter of 2.59 mm . How many meters of number 10 aluminum wire are needed to give a resistance of $1.0 \Omega$ ? $\rho$ for aluminum is $2.8 \times 10^{-8} \Omega \cdot \mathrm{~m}$.
From $R=\rho L / A$,
$$
L=\frac{R A}{\rho}=\frac{(1.0 \Omega)(\pi)\left(2.59 \times 10^{-3} \mathrm{~m}\right)^2 / 4}{2.8 \times 10^{-8} \Omega \cdot \mathrm{~m}}=0.19 \mathrm{~km}
$$