1110_Physics ProblemsTechnical Physics

# 1110_Physics ProblemsTechnical Physics - Chapter 39 P39.53...

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Chapter 39 451 P39.53 P == = = × dE dt dmc dt c dm dt 2 22 6 377 10 ej . W Thus, dm dt = × × 300 10 419 10 26 8 2 9 . . . Js ms kg s P39.54 21 0 2 2 mc e = M e V E γ 102 M e V Section 39.10 The General Theory of Relativity *P39.55 (a) For the satellite Fm a = : GM m r mv r m r r T E 2 2 2 2 F H G I K J π GM T r r r E 3 11 24 2 2 13 7 4 6 67 10 5 98 10 43 080 4 266 10 = = ×⋅× F H G G I K J J .. . N m kg s kg m 2 2 bg (b) v r T × 2 2 2 66 10 43 080 387 10 7 3 . . m s (c) The small fractional decrease in frequency received is equal in magnitude to the fractional increase in period of the moving oscillator due to time dilation: fractional change in f =− − =− −× × F H G G G I K J J J =− − − × × F H G I K J L N M M O Q P P F H G G I K J J × 1 1 1 3 87 10 3 10 1 11 1 2 31 0 834 10 38 2 3 8 2 11 . . . (d) The orbit altitude is large compared to the radius of the Earth, so we must use
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