Question

50 W/m² 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 exerted on 1 m² surface area will be close to (c = 3 × 10⁸ m/s)
(A) 15 × 10⁻⁸ N
(B) 35 × 10⁻⁸ N
(C) 10 × 10⁻⁸ N
(D) 20 × 10⁻⁸ N

Answer 1

Heya Mate!! Plzz refer to the above attach,ent

Answer 2

The force exerted on 1 m² surface area

= 20 × $10^{-8}$ N

Given:

50 W/m² energy density of sunlight is normally incident on the surface of a solar panel.

25% part of incident energy is reflected from the surface.

c = Speed of light =3 × 10⁸ m/s

To find:

The force exerted on 1 m² surface area.

Formula used:

Force on the surface (25% reflecting and rest absorbing)

F=$\frac{25}{100}$($\frac{2I}{C}$)+$\frac{75}{100}$($\frac{I}{C}$) = $\frac{125}{100}$($\frac{I}{C}$)

Where, F = force exerted in 1 m² surface area.

            c = Speed of light =3 × 10⁸ m/s

            I =  Incident energy.

Explanation:

Given that,       I =  Incident energy = 50 W/m²

       F = $\frac{125}{100}$($\frac{I}{C}$)  

      F =  $\frac{125}{100}$($\frac{50}{3\times 10^{8} }$)  

      F = 20.83 × $10^{-8}$ N ≅  20 × $10^{-8}$ N

The force exerted on 1 m² surface area = 20 × $10^{-8}$ N

To learn more...

1) Incident energy is equal to binding energy + kinetic energy

https://brainly.in/question/9601287

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