Question

check the dimensional correctness of physical equation ... s=ut+1/2×at^2​

Answer 1

Explanation:

Writing the dimensions of either side of the given equation.

LHS=8=displacement=[M0LT0]

RHS=ut=velocity×time=[M0LT−1][T]=[M0LT0]

and 21at2=(acceleration)×(time)2=[M0LT−2][T]2=[M0LT0]

As LHS=RHS, formula is dimensionally correct.

Answer 2

Answer:

We know that L.H.S = s and R.H.S = ut + 1/2at2

The dimensional formula for the L.H.S can be written as s = [L1M0T0] ………..(1)

We know that R.H.S is ut + ½ at2 , simplifying we can write R.H.S as [u][t] + [a] [t]2

[L1M0T-1][L0M0T-1] +[L1M0T-2][L0M0T0]

=[L1M0T0]………..(2)

From (1) and (2), we have [L.H.S] = [R.H.S]

Explanation:

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