The conversion of 1 ;MW power in a new system | vedclass

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Question

(A) The dimensional formula for power is $[P] = [M^1 L^2 T^{-3}]$.We use the formula for conversion between two systems: $n_2 = n_1 [M_1/M_2]^a [L_1/L

Vedclass · Accepted Answer

$2.16 	imes 10^{12} \; 	ext{unit}$

Vedclass · Answer

(A) The dimensional formula for power is $[P] = [M^1 L^2 T^{-3}]$.We use the formula for conversion between two systems: $n_2 = n_1 [M_1/M_2]^a [L_1/L_2]^b [T_1/T_2]^c$.Here,$n_1 = 10^6 \; W$,$M_1 = 1 \; kg$,$L_1 = 1 \; m$,$T_1 = 1 \; s$.In the new system,$M_2 = 10 \; kg$,$L_2 = 1 \; dm = 0.1 \; m$,$T_2 = 1 \; 	ext{minute} = 60 \; s$.Substituting the values: $n_2 = 10^6 	imes [1 \; kg / 10 \; kg]^1 	imes [1 \; m / 0.1 \; m]^2 	imes [1 \; s / 60 \; s]^{-3}$.$n_2 = 10^6 	imes [1/10] 	imes [10]^2 	imes [1/60]^{-3}$.$n_2 = 10^6 	imes 0.1 	imes 100 	imes 60^3$.$n_2 = 10^7 	imes 216000 = 2.16 	imes 10^{12} \; 	ext{units}$.

The conversion of $1 \;MW$ power in a new system having basic units of mass,length,and time as $10 \;kg$,$1 \;dm$,and $1 \;minute$ respectively is:

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