Question

Test dimensionally. If the
equation v²=
$u^{2}$
+2as may be
correct​

Answer 1

LHS =[v^2]

=L^2T^-2

RHS =[u^2]+[2][a][s]

=L2T^-2+M^0L^0T^0xLT^-2xL

=L^2T^-2+L^2T^-2

=L^2T^-2

=LHS

Therefore, v^2=u^2+2as is dimensionally correct.

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Answer 2

Answer:

There are three terms in this equation v^2, u^2 and 2ax. The equation may be correct if the dimensions of these terms are equal.

${ {v}^{2} } = ({ \frac{l}{t} })^{2} = {l}^{2} {t}^{ - 2}$

${u}^{2} = ( {\frac{l}{t} })^{2} = {l}^{2} {t}^{ - 2}$

and,

$2ax = {a} \: x \: = \frac{l}{ { t }^{2} } = l{t^{ - 2} }$

Thus, the equation is correct!