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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 . = = × 6 67 10 16 T (b) S E c av 2 W m = = × max . 2 0 17 2 5 31 10 µ (c) P = = × S A av W 1 67 10 14 . (d) F PA S c A = = F H G I K J = × av N 5 56 10 23 . ( the weight of 3 000 H atoms!) FIG. P34.51 P34.52 (a) The power incident on the mirror is: P I IA = = = × 1 340 100 4 21 10 2 7 W m m W 2 e j a f π . . The power reflected through the atmosphere is P R = × = × 0 746 4 21 10 3 14 10 7 7 . . . W W e j . (b) S A R = = × × = P 3 14 10 0 625 7 2 . . W 4.00 10 m W m 3 2 π e j (c) Noon sunshine in Saint Petersburg produces this power-per-area on a horizontal surface: P N A = °= 0 746 1 340 7 00 122 . sin . W m W m 2 2 e j . The radiation intensity received from the mirror is 0 625 100% 0 513% . . W m 122 W m 2 2 F H G I K J = of that from the noon Sun in January. P34.53 u E = 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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