Physics: Asunny day has a light intensity of 1200 w/m^2. the efficiency of the cells is 9.1%. what total area must the cells be to generate 6.7 kw of power while the light is shining? a. 61.4 m^2 b. 4.71 m^2 c. 047 m^2 d. 061 m^2 e. none of the above and the answer is

Asunny day has a light intensity of 1200 w/m^2.

divanov

31.12.2021

Option a.

Total area of the cells to generate 6.7 kW, A = 61.4 $m^{2}$

Given:

The efficiency of the cell, $\eta _{cell}$ = 9.1%

Power generated by the cells, P = 6.7 kW

Intensity of light, I = 1200 W/ $m^{2}$

Solution:

Using the formula for intensity of light, I = $\frac{P}{\eta _{cell}A}$

rearranging the above formula for Area, A:

Area, A = $\frac{P}{I\eta _{cell}}$

Now, putting values in the above formula:

Area, A = $\frac{6.7\times 10^{3}}{1200\times 0.091}}$

Area, A = 61.355 ≈ 61.4 $m^{2}$

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marissalwilliams3

21.01.2021

BP! SIZZLE! BOP! S! And people talking i guess

Explanation:

hope this helps!

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