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(B) The power $P$ is defined as the rate of doing work,which is equal to the change in kinetic energy divided by time.$P = \frac{\Delta K.E.}{t} = \fr

Vedclass · Accepted Answer

$2.05$

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(B) The power $P$ is defined as the rate of doing work,which is equal to the change in kinetic energy divided by time.$P = \frac{\Delta K.E.}{t} = \frac{\frac{1}{2}m(v^2 - u^2)}{t}$Given:Mass $m = 2.05 	imes 10^6 \; kg$Initial velocity $u = 5 \; m/s$Final velocity $v = 25 \; m/s$Time $t = 5 \; min = 5 	imes 60 \; s = 300 \; s$Substituting the values:$P = \frac{0.5 	imes 2.05 	imes 10^6 	imes (25^2 - 5^2)}{300}$$P = \frac{0.5 	imes 2.05 	imes 10^6 	imes (625 - 25)}{300}$$P = \frac{0.5 	imes 2.05 	imes 10^6 	imes 600}{300}$$P = 0.5 	imes 2.05 	imes 10^6 	imes 2$$P = 2.05 	imes 10^6 \; W = 2.05 \; MW$Thus,the power of the engine is $2.05 \; MW$.

$A$ force applied by an engine of a train of mass $2.05 \times 10^6 \; kg$ changes its velocity from $5 \; m/s$ to $25 \; m/s$ in $5$ minutes. The power of the engine is ........... $MW$.

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