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