A force is represented by F = ax^2 + bt^{1/2} | vedclass

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Question

(A) According to the principle of homogeneity of dimensions,each term in the equation must have the same dimensions as force $F$.$1$. For the term $ax

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

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(A) According to the principle of homogeneity of dimensions,each term in the equation must have the same dimensions as force $F$.$1$. For the term $ax^2$:$[ax^2] = [F] = [MLT^{-2}]$$[a] = [F] / [x^2] = [MLT^{-2}] / [L^2] = [ML^{-1}T^{-2}]$$2$. For the term $bt^{1/2}$:$[bt^{1/2}] = [F] = [MLT^{-2}]$$[b] = [F] / [t^{1/2}] = [MLT^{-2}] / [T^{1/2}] = [MLT^{-5/2}]$$3$. Calculating the dimensions of $b^2/a$:$[b^2/a] = [b]^2 / [a] = ([MLT^{-5/2}])^2 / [ML^{-1}T^{-2}]$$[b^2/a] = [M^2 L^2 T^{-5}] / [ML^{-1}T^{-2}] = [M^{2-1} L^{2-(-1)} T^{-5-(-2)}] = [ML^3 T^{-3}]$

$A$ force is represented by $F = ax^2 + bt^{1/2}$,where $x$ is distance and $t$ is time. The dimensions of $b^2/a$ are:

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