A force F is given by F = at + bt^2,where t i | vedclass

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(B) According to the principle of dimensional homogeneity,the dimensions of each term in an equation must be the same.Since $F = at + bt^2$,the dimens

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$MLT^{-3}$ and $MLT^{-4}$

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(B) According to the principle of dimensional homogeneity,the dimensions of each term in an equation must be the same.Since $F = at + bt^2$,the dimensions of $F$ must be equal to the dimensions of $at$ and $bt^2$.The dimension of force $F$ is $[MLT^{-2}]$.For term $at$: $[a][t] = [MLT^{-2}] \implies [a] = [MLT^{-2}] / [T] = [MLT^{-3}]$.For term $bt^2$: $[b][t^2] = [MLT^{-2}] \implies [b] = [MLT^{-2}] / [T^2] = [MLT^{-4}]$.Thus,the dimensions of $a$ and $b$ are $[MLT^{-3}]$ and $[MLT^{-4}]$ respectively.

$A$ force $F$ is given by $F = at + bt^2$,where $t$ is time. What are the dimensions of $a$ and $b$?

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