The Intensity of solar radiation at the Earth's orbit is 1370 W/m2. However, because ofthe atmosphere, the curvature of the Earth, and rotation (night and day), the actualintensity at the Earth's surface is much lower. At this moment, let us assume theintensity of solar radiation is 575 W/m2. You have installed solar panels on your roof toconvert the sunlight to electricity.If the area of your solar panels is 4 m2, How much power is incident on your array?WattsSubmit AnswerTries 0/2Unfortunately, solar panels are only about 20% efficent...Only 1/5 of the light isconverted to electricity. This being the case, how much electrical power are youactually producing for your home at this moment?Watts

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The Intensity of solar radiation at the Earth's orbit is 1370 W/m2. However, because the atmosphere, the curvature of the Earth, and rotation (night and day), the actual intensity at the Earth's surface is much lower. At this moment, let us assume the intensity of solar radiation is 575 W/m2. You have installed solar panels on your roof convert the sunlight to electricity. If the area of your solar panels is 4 m2, How much power is incident on your array? Watts Submit Answer Tries 0/2 Unfortunately, solar panels are only about 20% efficent...Only 1/5 of the light is

The Intensity of solar radiation at the Earth's orbit is 1370 W/m2. However, because the atmosphere, the curvature of the Earth, and rotation (night and day), the actual intensity at the Earth's surface is much lower. At this moment, let us assume the intensity of solar radiation is 575 W/m2. You have installed solar panels on your roof convert the sunlight to electricity. If the area of your solar panels is 4 m2, How much power is incident on your array? Watts Submit Answer Tries 0/2 Unfortunately, solar panels are only about 20% efficent...Only 1/5 of the light is

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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If a radiation source gives 75 units of radiation at a distance of 2.4 m, at what distance does the source give 25 units of radiation?
(b) The earth has a radius of R = 6.4 x 106 m. It orbits the sun in a nearly circular orbit at an average distance of r = 1.5x 10¹1 m. The solar intensity at the upper atmosphere is 1367 W/m². How much energy does the sun radiate per second?
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An old-fashioned incandescent bulb (100 W) has a tungsten filament inside which glows at a steady temperature of 24000C and radiates electromagnetic waves (part heat, part light). What is the surface area of the filament? Assume emissivity of tungsten is 0.35. a) 9.9*10-4 m2 b) 9.9*10-5 m2 c) 1.5*10-4 m2
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Problem: Radiation Related The energy flux associated with solar radiation incident on the outer surface of the earth's atmosphere has been accurately measured and is known to be 1,368 W/m^2. The diameters of the sun and earth are 1.39 X 10^9 and 1.27 x 10^7 m, respectively, and the distance between the sun and the earth is 1.5 × 10^11 m. (a) What is the emissive power of the sun? (b) Approximating the sun's surface as black, what is its temperature? (c) At what wavelength is the spectral emissive power of the sun a maximum? (d) Assuming the earth's surface to be black and the sun to be the only source of energy for the earth, estimate the earth's surface temperature.
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(a) Calculate the rate in watts at which heat transfer through radiation occurs (almost entirely in the infrared) from 1.0 m2 of the Earth’s surface at night. Assume the emissivity is 0.90, the temperature of the Earth is 15ºC , and that of outer space is 2.7 K. (b) Compare the intensity of this radiation with that coming to the Earth from the Sun during the day, which averages about 800 W/m2, only half of which is absorbed. (c) What is the maximum magnetic field strength in the outgoing radiation, assuming it is a continuous wave?
The energy flux from the sun has a peak value of around 1500 W/m2, and solar cells are around 15% efficient. Estimate the area needed to satisfy US energy needs using solar cells, using the estimates that there are about 350 million people in the US and that we use an average of 90 kW/person. Also assume that there are about 6 hours of "peak" sunlight per day. Answer in square kilometers (do not put units in your answer, however).