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(A) The dimensional formula for power is $[P] = [M L^2 T^{-3}]$.Let the new units be $M' = a \ kg$,$L' = b \ m$,and $T' = c \ s$.Therefore,$1 \ kg = \

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$\frac{6 c^3}{a b^2}$

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(A) The dimensional formula for power is $[P] = [M L^2 T^{-3}]$.Let the new units be $M' = a \ kg$,$L' = b \ m$,and $T' = c \ s$.Therefore,$1 \ kg = \frac{1}{a} M'$,$1 \ m = \frac{1}{b} L'$,and $1 \ s = \frac{1}{c} T'$.Given power $P = 6 \ W = 6 \ kg \cdot m^2 \cdot s^{-3}$.Substituting the new units into the expression for power:$P = 6 	imes (\frac{1}{a} M') 	imes (\frac{1}{b} L')^2 	imes (\frac{1}{c} T')^{-3}$$P = 6 	imes \frac{1}{a} 	imes \frac{1}{b^2} 	imes c^3 	imes (M' L'^2 T'^{-3})$$P = \frac{6 c^3}{a b^2}$ new units.Thus,the magnitude of $6 \ W$ in the new system is $\frac{6 c^3}{a b^2}$.

In a new system of units,the unit of mass is $a \ kg$,and the units of length and time are $b \ m$ and $c \ s$,respectively. What is the magnitude of $6 \ W$ of power in this system?

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