Consdier the equation
`(d)/(dt)[intvecF.dvecs]=A[vecF.vecp]` Then dimension of A will be (where `vecF=` force, `dvecs=` small displacement, t=time and `vecp=` linear momentum).

Options

A. `M^(@)L^(@)T^(@)`
B. `M^(1)L^(@)T^(@)`
C. `M^(-1)L^(@)T^(@)`
D. `M^(@)L^(@)T^(-1)`

Explanation

To find the dimension of \( A \) in the equation

\[ \frac{d}{dt} \left( \vec{F} \cdot \vec{s} \right) = A \left( \vec{F} \cdot \vec{p} \right) \]

where \( \vec{F} \) is force, \( \vec{s} \) is small displacement, \( t \) is time, and \( \vec{p} \) is linear momentum, we will proceed step by step.

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