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(B) The power $P$ required to pump water is given by the rate of change of kinetic energy.$P = \frac{d}{dt} (\frac{1}{2} m v^2) = \frac{1}{2} v^2 \fra

Vedclass · Accepted Answer

$400$

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(B) The power $P$ required to pump water is given by the rate of change of kinetic energy.$P = \frac{d}{dt} (\frac{1}{2} m v^2) = \frac{1}{2} v^2 \frac{dm}{dt}$.Here,$\frac{dm}{dt}$ is the mass flow rate,which can be expressed as $\frac{dm}{dt} = \lambda v$,where $\lambda$ is the mass per unit length $(100\, kg/m)$ and $v$ is the velocity $(2\, m/s)$.Thus,$\frac{dm}{dt} = 100\, kg/m 	imes 2\, m/s = 200\, kg/s$.Substituting this into the power formula:$P = \frac{1}{2} 	imes (200\, kg/s) 	imes (2\, m/s)^2$.$P = \frac{1}{2} 	imes 200 	imes 4 = 400\, W$.

An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of $2\, m/s$. The mass per unit length of water in the pipe is $100\, kg/m$. Calculate the power $W$ of the engine. (in $, W$)

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