966_Physics ProblemsTechnical Physics

966_Physics ProblemsTechnical Physics - Chapter 34 Emax =...

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Chapter 34 307 P34.51 (a) B E c max max . ==× 667 10 16 T (b) S E c av 2 Wm max . 2 0 17 2 531 10 µ (c) P SA av W 167 10 14 . (d) FP A S c A == F H G I K J av N 556 10 23 . ( the weight of 3 000 H atoms!) FIG. P34.51 P34.52 (a) The power incident on the mirror is: I IA = × 1 340 100 4 21 10 2 7 m W 2 ej af π .. The power reflected through the atmosphere is R = × 0746421 10 314 10 77 . W W . (b) S A R × × = 0625 7 2 . . W 4.00 10 m 3 2 (c) Noon sunshine in Saint Petersburg produces this power-per-area on a horizontal surface: N A = 0 746 1 340 7 00 122 .s i n . 22 . The radiation intensity received from the mirror is 100% 0 513% . . 122 W m 2 2 F H G I K J = of that from the noon Sun in January. P34.53 uE =∈ 1 2 0 2 max E u max . = = 2 95 1 0 mV m P34.54 The area over which we model the antenna as radiating is the lateral surface of a cylinder,
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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