Question

What should be the power of an engine required to lift 90 metres tons of coal per hour from a minimum of depth 200m.

Answer 1

time interval = 1hour
work = mgh = 90×1000×10×200
$= 1.8 \times {10}^{8}$
power = work/time interval

$= \frac{1.8 \times {10}^{8} }{1} = 1.8 \times {10}^{8}$
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