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(C) The power $P$ is defined as the energy $E$ produced per unit time $\Delta t$,where $E = \Delta m c^2$.Thus,$P = \frac{\Delta m c^2}{\Delta t}$,whi

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$4 	imes 10^{-2} \, g$

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(C) The power $P$ is defined as the energy $E$ produced per unit time $\Delta t$,where $E = \Delta m c^2$.Thus,$P = \frac{\Delta m c^2}{\Delta t}$,which implies $\frac{\Delta m}{\Delta t} = \frac{P}{c^2}$.Given $P = 10^9\, W$ and $c = 3 	imes 10^8\, m/s$,the mass consumed per second is:$\frac{\Delta m}{\Delta t} = \frac{10^9}{(3 	imes 10^8)^2} = \frac{10^9}{9 	imes 10^{16}} = \frac{1}{9} 	imes 10^{-7} \, kg/s$.To find the mass consumed per hour,we multiply by the number of seconds in an hour $(3600\, s)$:$\Delta m = (\frac{1}{9} 	imes 10^{-7} \, kg/s) 	imes 3600\, s = 400 	imes 10^{-7} \, kg = 4 	imes 10^{-5} \, kg$.Converting to grams $(1\, kg = 1000\, g)$:$\Delta m = 4 	imes 10^{-5} 	imes 10^3 \, g = 4 	imes 10^{-2} \, g$.

Imagine that a reactor converts all given mass into energy and that it operates at a power level of $10^9\, W$. The mass of the fuel consumed per hour in the reactor will be: (velocity of light,$c = 3 \times 10^8\, m/s$)

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