The blades of a windmill sweep out a circle of area $A$.
$(a)$ If the wind flows at a velocity $v$ perpendicular to the circle,what is the mass of the air passing through it in time $t?$
$(b)$ What is the kinetic energy of the air?
$(c)$ Assume that the windmill converts $25\%$ of the wind's energy into electrical energy,and that $A=30\;m^{2}, v=36\;km/h$ and the density of air is $1.2\;kg\;m^{-3}.$ What is the electrical power produced?

(C) Area of the circle swept by the windmill $= A$
Velocity of the wind $= v$
Density of air $= \rho$
Volume of the wind flowing through the windmill per second $= A v$
Mass of the wind flowing through the windmill per second $= \rho A v$
Mass $m$ of the wind flowing through the windmill in time $t = \rho A v t$
Kinetic energy of air $= \frac{1}{2} m v^{2} = \frac{1}{2} (\rho A v t) v^{2} = \frac{1}{2} \rho A v^{3} t$
Given: $A = 30\;m^{2}, v = 36\;km/h = 10\;m/s, \rho = 1.2\;kg\;m^{-3}$
Electric energy produced $= 25\%$ of the wind energy $= \frac{25}{100} \times (\frac{1}{2} \rho A v^{3} t) = \frac{1}{8} \rho A v^{3} t$
Electrical power $= \frac{\text{Electrical energy}}{\text{Time}} = \frac{1}{8} \rho A v^{3}$
Power $= \frac{1}{8} \times 1.2 \times 30 \times (10)^{3} = 4.5 \times 10^{3}\;W = 4.5\;kW$