Answer:
326.25 kWh
Explanation:
Efficiency of a machine is defined as the ratio of useful energy to that of the energy consumed by the machine.
Here, efficiency is given as 87% and the energy consumed by the computer is 375 kWh.
Efficiency, [tex]\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}[/tex]
Plug in the values of [tex]\eta=0.87[/tex] and 375 kWh for energy consumed. Solve for useful energy. This gives,
Efficiency, [tex]\eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}\\ 0.87=\frac{\textrm{Useful energy}}{375}\\ \textrm{Useful energy}=0.87\times 375=326.25 \textrm{ kWh}[/tex]
Therefore, the useful energy provided by the computer is 326.25 kWh.
