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Construct Your Own ProblemConsider a space sail such as mentioned in Example 29.5. Construct a problem in which you calculate the light pressure on the sail in $\mathrm{N} / \mathrm{m}^{2}$ produced by reflecting sunlight. Also calculate the force that could be produced and how much effect that would have on a spacecraft. Among the things to be considered are the intensity of sunlight, its average wavelength, the number of photons per square meter this implies, the area of the space sail, and the mass of the system being accelerated.

$P_{\text {sail }}=4.56 \times 10^{-6} \mathrm{N} / \mathrm{m}^{2}$

Physics 103

Chapter 29

Introduction to Quantum Physics

Quantum Physics

University of Michigan - Ann Arbor

University of Washington

University of Sheffield

McMaster University

Lectures

02:51

In physics, wave optics is…

10:02

Interference is a phenomen…

04:14

One possible means of achi…

02:20

Radiation PressureIt h…

05:49

Some scientists have sugge…

04:05

A possible means of space …

05:55

Construct Your Own Problem…

07:12

Solar wind (radiation) tha…

05:41

03:07

The intensity of the Sun&#…

03:24

02:49

03:45

06:49

The radiation pressure (Se…

06:21

Solar Sail 1. During $2004…

03:25

06:27

Solar Sail 2. NASA is givi…

02:13

04:41

$\bullet$$\bullet$ Solar s…

0:00

NASA is giving serious con…

04:21

04:43

Refer to Conceptual Exampl…

So here we have the intensity of radiation at a distance are from a point source and that point sources radiating a total power p. This would be the intensity for party. This would be the intensity equaling need power divided by the effective area. Essentially, this would be divided by this would be equaling rather than the power divided by four pi r squared and so at a distance. Here we have our equaling 2.0 inches from a cellphone radiating a total power of P equaling 2.0 wants. Or better yet, we can say 2.0 times 10 to the third. Miller Watts. We can say that the intensity then I is equaling 2.0 times 10 to the third Mila Watts, and this will be divided by four pi multiplied by 2.0 inches, multiplied by 2.54 centimeters for every one inch, and this would be squared. And this is giving us 6.2 Mila watts for every centimeter squared. So this would be our final answer for part A. And from here you can see that the intensity, this intensity I is 24% higher then the maximum allowed leakage from a microwave again at this distance of two inches. So continuing on for Part B, we have a Bluetooth headset, and here we can say that we have a Bluetooth headset. The way you say blue tooth is emitting 2.5 million watts of power, and it's going in the air at a distance. Our sub H. This would be equaling 2.0 inches, or simply 5.1 centimeters from the brain. And so the cell phone, which is emitting to Watts of power. We have a cell phone where we have to watts, and this is located in the pocket, where the distance from the brain to the pocket would be equaling. Tend to the second centimeters, essentially 11 meter or 10 to the second centimeters, so the total radiation intensity at the brain would be. You could say I totally at the brain. This would be willing, the intensity from the phone, plus the intensity of the headset. We can say of the Bluetooth headset, and so this would be equaling 2.0 times, 10 to the third. Miller Watts, divided by for pie, multiplied by 10 to the second centimeters. Quantity squared plus 2.5 million watts, divided by four pi multiplied by 5.1 centimeters. Quantity squared. And so the total is found to be equaling 2.4 times, 10 to the negative second Miller Watts For every centimeter squared, this would be our final answer for Part B. That is the end of the solution. Thank you for watching.

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