Answer:
Total mass, m =
(volume x density) of water
\[=(1\text{
}{{m}^{2}}\times 1m\times {{10}^{3}}kg/{{m}^{3}})\]
\[={{10}^{3}}\,kg\]
Energy
transferred \[=K.E.=\,\frac{1}{2}\,m{{\upsilon }^{2}}\]
\[=\frac{1}{2}\,\times
{{10}^{3}}\times \,81=40.5\,\times \,{{10}^{3}}\,J\]
\[=4.05\times
{{10}^{4}}J\]
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