question_answer

\[50\text{ }W/{{m}^{2}}\]energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force (in newton) exerted on \[1{{m}^{2}}\] surface area will be\[\left( c=3\times \text{1}{{\text{0}}^{8}}\text{ }m/s \right)\]:

Answer:

\[\left( 20\times {{10}^{-8}} \right)\]