Question on: WAEC Physics - 2014
a ca travelling at 30m-1 overcomes a frictional resistance of 100N while moving. Calculate the power developed by the engine. [1 hp = 0.75 kW]
A
0.23hp
B
0.40hp
C
4.00hp
D
4.40hp
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Correct Option: C
P = FV = \(\frac{100 \times 30}{1000} = 3 kW\)
\(\frac{3}{0.75} = 4hp\)
\(\frac{3}{0.75} = 4hp\)
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