(10%) Problem 10: The power output of the Sun is about 4.00 x 1026 W 20% Part (a

Question

(10%) Problem 10: The power output of the Sun is about 4.00 x 1026 W 20% Part (a) Calculate the power per unit area (intensity, in kilowatts per square meter reaching Earth's upper atmosphere from the Sun . The radius of the Earth's orbit is 1.5 x 1011 m. 20% Part (b) Part of this power is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m2 reaches Earth's surface Calculate the area, in square kilometers, of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves and the fact that the sun is directly overhead only briefly.) 20% Part (c) With the same assumptions, what area (again in square kilometers) would be needed to meet the United States' energy needs (which is 1.05 x 1020 J/yr)? 20% Part (d) What would this area be (in square kilometers) for Australia's energy needs (5.4 x 1018 J/yr)? 20% Part (e) what about China's energy needs (6.3 x 1019 J/yr), in square kilometers? Grade Summary Deductions Potential ACN 0% 100% sin Submissions Attempts remaining: 5 (190 per attempt) detailed view 78 9 tanO t ( acos0 coS cotan) asin) atan) acotan)sinh(0 cosh0tanh cotanh0 0 END Degrees O Radians CLEAR Submit Hint Hints: 1 % deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback

Explanation / Answer

e] Since intensity is same. in all these countries,

For answer to part e, multiply the answer of part d by (6.3e19/5.4e18)

i.e. multiply the answer of part d by 11.6667

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