Numerade

$Number 10 wire has a diameter of $2.59 \mathrm{~mm}$. 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 $R=\rho L / A$ $$ L=\frac{R A}{\rho}=\frac{(1.0 \Omega)(\pi)\left(2.59 \times 10^{-3} \mathrm{~m}\right)^{2} / 4}{2.8 \times 10^{-8} \Omega \cdot \mathrm{m}}=0.19 \mathrm{~km} $$$

Number 10 wire has a diameter of $2.59 \mathrm{~mm}$. 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 $R=\rho L / A$ $$ L=\frac{R A}{\rho}=\frac{(1.0 \Omega)(\pi)\left(2.59 \times 10^{-3} \mathrm{~m}\right)^{2} / 4}{2.8 \times 10^{-8} \Omega \cdot \mathrm{m}}=0.19 \mathrm{~km} $$

svt m.