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energy =alpha×acceleration ×momentum÷plank constant then the dimension of alpha is?​

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Sangamkar786idn

Answer:

The dimension of the physical quantity α in the given equation is [L{-2}T2]1. This result can be obtained by analyzing the dimensions of the other quantities involved in the equation.

Given the equation: [ W = \frac{F}{\alpha \cdot v} ]

We can break down the dimensions of the quantities:

Force (F):

Dimensional formula of force: ([F] = \text{Mass} \cdot \text{Acceleration} = [M][LT^{-2}])

Velocity (v):

Dimensional formula of velocity: ([v] = [LT^{-1}])

Work (W):

Dimensional formula of work: ([W] = [ML2T{-2}])

Now let’s find the dimension of α: [ \alpha = \frac{F}{W \cdot v} ] [ [\alpha] = \frac{[F]}{[W][v]} = \frac{[MLT{-2}]}{[ML2T{-2}][LT{-1}]} = [L{-2}T2] ]

Hence, the dimension of α is [L{-2}T2].

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