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An engine pumps water continuously through a hose. Water leaves the hose with a velocity $\mathrm{v}$ and $\mathrm{m}$ is the mass per unit length of the water jet. What is the rate which kinetic energy is imparted to water ?
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Verified Answer
The correct answer is:
$\frac{1}{2} \mathrm{mv}^3$
$\mathrm{mv}=\frac{\mathrm{dm}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}}=$ Rate of flowing mass
$$
\begin{aligned}
& \mathrm{F}_{\mathrm{av}}=\frac{\mathrm{dm}}{\mathrm{dt}} \times \frac{\mathrm{v}}{2}=\frac{(\mathrm{mv}) \mathrm{v}}{2}=\frac{\mathrm{mv}^2}{2} \\
& \mathrm{p}=\frac{\mathrm{dK}}{\mathrm{dt}}=\frac{\mathrm{mv}^2}{2} \times \mathrm{v}=\frac{\mathrm{mv}^3}{3}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{F}_{\mathrm{av}}=\frac{\mathrm{dm}}{\mathrm{dt}} \times \frac{\mathrm{v}}{2}=\frac{(\mathrm{mv}) \mathrm{v}}{2}=\frac{\mathrm{mv}^2}{2} \\
& \mathrm{p}=\frac{\mathrm{dK}}{\mathrm{dt}}=\frac{\mathrm{mv}^2}{2} \times \mathrm{v}=\frac{\mathrm{mv}^3}{3}
\end{aligned}
$$
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