p34_018 - 18(a The power received is Pr =(1.0 1012 W(1000...

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18. (a) The power received is P r =(1 . 0 × 10 12 W) π [(1000 ft)(0 . 3048 m / ft)] 2 / 4 4 π (6 . 37 × 10 6 m) 2 =1 . 4 × 10 22 W . (b) The power of the source would be P =4 πr 2 I =4 π [(2 . 2 ×
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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