An engine develops 10 text{ kW} of power. How | vedclass

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(C) Power $(P)$ is defined as the rate of doing work,given by $P = \frac{W}{t}$.Here,the work done $(W)$ to lift a mass $(m)$ to a height $(h)$ is equ

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$8$

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(C) Power $(P)$ is defined as the rate of doing work,given by $P = \frac{W}{t}$.Here,the work done $(W)$ to lift a mass $(m)$ to a height $(h)$ is equal to the change in potential energy: $W = mgh$.Therefore,$P = \frac{mgh}{t}$.Given values: $P = 10 	ext{ kW} = 10 	imes 10^3 	ext{ W}$,$m = 200 	ext{ kg}$,$h = 40 	ext{ m}$,and $g = 10 	ext{ m/s}^2$.Rearranging the formula for time $(t)$: $t = \frac{mgh}{P}$.Substituting the values: $t = \frac{200 	imes 10 	imes 40}{10 	imes 10^3}$.$t = \frac{80000}{10000} = 8 	ext{ s}$.Thus,the time taken is $8 	ext{ seconds}$.

An engine develops $10 \text{ kW}$ of power. How much time will it take to lift a mass of $200 \text{ kg}$ to a height of $40 \text{ m}$? $(g = 10 \text{ m/s}^2)$

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