The equation frac{dV}{dt} = At - BV describes | vedclass

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(B) Given the equation: $\frac{dV}{dt} = At - BV$According to the principle of homogeneity of dimensions,the dimensions of each term on both sides of

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$LT^{-3}, T^{-1}$

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(B) Given the equation: $\frac{dV}{dt} = At - BV$According to the principle of homogeneity of dimensions,the dimensions of each term on both sides of the equation must be the same.The dimension of $\frac{dV}{dt}$ (acceleration) is $[LT^{-2}]$.For the term $At$:$[At] = [LT^{-2}]$$[A] [T] = [LT^{-2}]$$[A] = [LT^{-2}] / [T] = [LT^{-3}]$For the term $BV$:$[BV] = [LT^{-2}]$$[B] [LT^{-1}] = [LT^{-2}]$$[B] = [LT^{-2}] / [LT^{-1}] = [T^{-1}]$Therefore,the dimensions of $A$ and $B$ are $[LT^{-3}]$ and $[T^{-1}]$ respectively.

The equation $\frac{dV}{dt} = At - BV$ describes the rate of change of velocity of a body falling from rest in a resisting medium. The dimensions of $A$ and $B$ are

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