Answer:
Given, \[m=50JK.E.=625J\] \[v=?\] We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow {{V}^{2}}=\frac{2\times K.E.}{m}\] \[\Rightarrow V=\sqrt{\frac{2\times K.E.}{m}}\] \[\therefore V=\sqrt{\frac{2\times 625}{50}}=\sqrt{25}=5m/s\]
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