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Q. The dimensions of kinetic energy is:

KEAMKEAM 2004Physical World, Units and Measurements

Solution:

$ KE=\frac{1}{2}m{{v}^{2}} $
$ \therefore $ $ [KE]=[M]{{[L{{T}^{-1}}]}^{2}}[M{{L}^{2}}{{T}^{-2}}] $