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A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.1) The intensity of the sun's radiation incident upon the earth is about I=1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?Express your answer numerically in kilowatts to two significant figures.2) What is the total force F on the panels exerted by radiation pressure from the sunlight?Express the total force numerically, to two significant figures, in units of newtons.

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hopemichael95

Answer:

0.000047N

Explanation:

We know that

intensity (I) = P/ A

Where

P= power

A= Area

So lets say that the power absorbed

Will be = Intensity x Area

Which Is = 1.4 x 10^3 x(10)

So

14000 Watt = 14 kWatt

However we know that radiation pressure is equal to

time-averaged intensity all over the speed of light in free space

So

P = (1.4 x 1000)/c

But

F= P x A

So

((1.4 x 1000)/(3 x1 0^8)) x 10

Which is

=0.000046666N

To two SIG figures we have

=0.000047 N

onyebuchinnaji

(a) The total solar power P absorbed by the panels is 14 kW.

(b) The total force F on the panels exerted by radiation pressure from the sunlight is [tex]4.7 \times 10^{-5} \ N[/tex].

The given parameters;

  • area of the panel, A = 10 m²
  • intensity of the sun radiation, I = 1.4 kW/m^2

The total solar power P absorbed by the panels is calculated as follows;

[tex]P = I A\\\\P = (1.4 \ kW /m^2 ) \times 10\ m^2\\\\P = 14 \ kW[/tex]

The total force F on the panels exerted by radiation pressure from the sunlight is calculated as follows;

[tex]Pressure, P = \frac{I}{c} \\\\Force, F = PA\\\\F = (\frac{I}{c} )A\\\\F = \frac{IA}{c} \\\\F = \frac{(1.4 \times 10^3) \times (10)}{3\times 10^8} \\\\F = 4.7 \times 10^{-5} \ N[/tex]

Learn more about power of incident light here: https://brainly.com/question/14258601

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