Answer:
(B)
[tex]\eta=7.5kW[/tex]
Explanation:
It is actually an easy problem, the energy efficiency is a dimensionless number, which is the report that indicates what can be recovered profitably from the machine from what has been spent to make it work. It is defined as:
[tex]\eta=\frac{P_o_u_t}{P_i_n}[/tex]
Where:
[tex]P_o_u_t=[/tex] Output power.
[tex]P_i_n=[/tex] Input power.
Replacing the data given by the problem:
[tex]0.8=\frac{6000}{P_i_n}[/tex]
Isolating [tex]P_i_n[/tex]
[tex]P_i_n=\frac{6000}{0.8} =7500W=7.5kW[/tex]
