Question

find the dimension of a/b in relation f=ax+bt^2

Answer 1

hope it will help......

Answer 2

F = ax + bt^2

$f = ax \\ \\ a = \frac{f}{x} = \frac{ {m}^{1} {l}^{1} {t}^{ - 2} }{ {l}^{1} } \\ \\ a = {m}^{1} {t}^{ - 2} \\ \\ f = b {t}^{2} \\ \\ b = \frac{f}{ {t}^{2} } = \frac{ {m}^{1} {l}^{1} {t}^{ - 2} }{ {t}^{2} } = {m}^{ 1} {l}^{1} {t}^{ - 2} {t}^{ - 2} \\ \\ b = {m}^{ 1} {l}^{1} {t}^{ - 4}$

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