p34_084 - A , we note that the exposed area of the small...

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84. Since intensity is power divided by area (and the area is spherical in the isotropic case), then the intensity at a distance of r = 20 m from the source is I = P 4 πr 2 =0 . 040 W / m 2 . as illustrated in Sample Problem 34-2. Now, in Eq. 34-32 for a totally absorbing area
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Unformatted text preview: A , we note that the exposed area of the small sphere is that on a at circle A = (0 . 020 m) 2 = 0 . 0013 m 2 . Therefore, F = IA c = (0 . 040)(0 . 0013) 3 10 8 = 1 . 7 10 13 N ....
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