An engine pumps water continuously through a hose. Water leaves the hose with a velocity $y$ and $m$ is the…
- $\frac{1}{2} \mathrm{mv}^3$
- $\mathrm{mv}^3$
- $\frac{1}{2} \mathrm{mv}^2$
- $\frac{1}{2} m^2 v^2$
Solution
Mass flowing per second \(=\mathrm{mv}\)
\(\therefore\) Rate of kinetic energy or K.E. per second
\(=1 / 2(m v) v^2=1 / 2 m v^3\)
Asked in: NEET 2009 (Screening)
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