A) 9.8 m/s
B) 19.6 m/s
C) 39.2 m/s
D) 98.0 m/s
Correct Answer: B
Solution :
Potential energy of water = kinetic energy at turbine \[mgh=\frac{1}{2}m{{v}^{2}}\]Þ\[v=\sqrt{2gh}=\sqrt{2\times 9.8\times 19.6}=19.6\,m/s\]You need to login to perform this action.
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