Anuclear fission power plant has an actual efficiency of 32%. if 0.18 mw of power are produced by the nuclear fission, how much electric power does the power plant output?

 P₀ = 5.76 x 10⁻² MW

Explanation:

given,

efficiency of the power plant = 32%

Power produced by the nuclear fission = 0.18 MW

the power plant output = ?

using formula of efficiency

where P is the power produced in the power plant

          P₀ is the power output of the power plant

 P₀ = 0.18 x 0.32

 P₀ =  0.0576 MW

 P₀ = 5.76 x 10⁻² MW

Power plant output is equal to  P₀ = 5.76 x 10⁻² MW

adding a(n) to an electric circuit would increase the magnetic field around a wire;

b. solenoid

a solenoid is a type of electromagnet whose purpose is to generate a controlled magnetic field.

if the purpose of the solenoid is instead to impede changes in the electric current, a solenoid can be more specifically classified as an inductor rather than an electromagnet.