## Essential University Physics: Volume 1 (4th Edition)

We first find the wind output per unit area: $P=\frac{1}{2}\rho v^3$ $P = \frac{1}{2}(.3)(1.2)(12^2)=25.92 \ W$ We multiply this by the overall area: $P_t=25.92\times \pi \times (\frac{95}{2})^2 \times 800=147 \ GW$ Thus, it could replace the power plant.