In a system of units if force (F), accelerati | vedclass

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Question

(b) $E = K{F^a}{A^b}{T^c}$

$\left[ {M{L^2}{T^{ - 2}}} \right] = {\left[ {ML{T^{ - 2}}} \right]^a}{\left[ {L{T^{ - 2}}} \right]^b}{\left[ T \right]^

Vedclass · Accepted Answer

$FA{T^2}$

Vedclass · Answer

(b) $E = K{F^a}{A^b}{T^c}$

$\left[ {M{L^2}{T^{ - 2}}} ight] = {\left[ {ML{T^{ - 2}}} ight]^a}{\left[ {L{T^{ - 2}}} ight]^b}{\left[ T ight]^c}$

$\left[ {M{L^2}{T^{ - 2}}} ight] = \left[ {{M^a}{L^{a + b}}{T^{ - 2a - 2b + c}}} ight]$

$	herefore $ $a = 1$, $a + b = 2$ $&rArr;$ $b = 1$

and $ - 2a - 2b + c = - 2\;\; \Rightarrow \;c = 2$

$	herefore $ $E = KFA{T^2}$.

In a system of units if force $(F)$, acceleration $(A) $ and time $(T)$ are taken as fundamental units then the dimensional formula of energy is

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