## Physics (10th Edition)

$1.0\times10^{2}\Omega$
V= 120 V P= 140 W We know that P= $\frac{V^{2}}{R}$ Thus, we find: R= $\frac{V^{2}}{P}=\frac{(120V)^{2}}{140 W}\approx 1.0\times10^{2} \Omega$