Answer:
Given, \[V=80\,\,kmph=80\times \frac{5}{18}\] \[=22.22\,\,m/s\] \[m=1000\,\,kgK.E.=?\] We know, \[K.E.=\frac{1}{2}m{{V}^{2}}\] \[=K.E.=\frac{1}{2}\times 1000\times {{(22.22)}^{2}}\] \[=246864.2\,\,J\]
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