The dimensions of frac{alpha}{beta} in the eq | vedclass

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(C) Given the equation $F = \frac{\alpha - t^2}{\beta v^2}$.According to the principle of homogeneity of dimensions,the dimensions of $\alpha$ must be

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$[ML^3T^{-4}]$

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(C) Given the equation $F = \frac{\alpha - t^2}{\beta v^2}$.According to the principle of homogeneity of dimensions,the dimensions of $\alpha$ must be the same as the dimensions of $t^2$.Therefore,$[\alpha] = [T^2]$.Now,the dimensions of the entire expression are $[F] = \frac{[\alpha]}{[\beta][v^2]}$.Substituting the dimensions: $[MLT^{-2}] = \frac{[T^2]}{[\beta][LT^{-1}]^2}$.$[MLT^{-2}] = \frac{[T^2]}{[\beta][L^2T^{-2}]}$.Rearranging for $[\beta]$: $[\beta] = \frac{[T^2]}{[MLT^{-2}][L^2T^{-2}]} = \frac{[T^2]}{[ML^3T^{-4}]} = [M^{-1}L^{-3}T^6]$.Now,calculate the dimensions of $\frac{\alpha}{\beta}$:$\frac{[\alpha]}{[\beta]} = \frac{[T^2]}{[M^{-1}L^{-3}T^6]} = [ML^3T^{-4}]$.

The dimensions of $\frac{\alpha}{\beta}$ in the equation $F = \frac{\alpha - t^2}{\beta v^2}$,where $F$ is force,$v$ is velocity,and $t$ is time,are ..........

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