Water falls down a 500.0 ,m shaft to reach a | vedclass

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(B) The power output $P$ is given by $50 \%$ of the potential energy per second of the falling water.Power $P = 0.50 	imes \left( \frac{mgh}{t} igh

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(B) The power output $P$ is given by $50 \%$ of the potential energy per second of the falling water.Power $P = 0.50 	imes \left( \frac{mgh}{t} ight)$.Since mass $m = ho V$,where $ho$ is the density of water $(1000 \,kg/m^3)$ and $V$ is the volume,we have:$P = 0.50 	imes \left( \frac{V}{t} ight) ho g h$.Given $P = 1.00 	imes 10^9 \,W$,$g = 10 \,ms^{-2}$,$ho = 1000 \,kg/m^3$,and $h = 500 \,m$.Rearranging for the flow rate $\frac{V}{t}$:$\frac{V}{t} = \frac{P}{0.50 	imes ho 	imes g 	imes h}$.$\frac{V}{t} = \frac{1.00 	imes 10^9}{0.50 	imes 1000 	imes 10 	imes 500}$.$\frac{V}{t} = \frac{1.00 	imes 10^9}{2.5 	imes 10^6} = 400 \,m^3/s$.

Water falls down a $500.0 \,m$ shaft to reach a turbine which generates electricity. How many $m^3$ of water must fall per second in order to generate $1.00 \times 10^9 \,W$ of power? (Assume $50 \%$ efficiency of conversion and $g=10 \,ms^{-2}$)

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