Question

if a hydroelectric station is 93% efficient. What mass of water must pass through the station each 1.00s to produce 855kW of electricity from a 34.6 meter drop?

Answer 1

$W=mgh\to0.93mgh=Pt\to m=\frac{Pt}{0.93gh}=$

$=\frac{855000\cdot1}{0.93\cdot9.81\cdot34.6}=2709\ (m^3)$