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968_Physics ProblemsTechnical Physics

968_Physics ProblemsTechnical Physics - Chapter 34 P34.58...

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Chapter 34 309 P34.58 The mirror intercepts power P = = × = I A 1 1 3 2 1 00 10 0 500 785 . . W m m W 2 e j a f π . In the image, (a) I A 2 2 = P : I 2 2 785 625 = = W 0.020 0 m kW m 2 π b g (b) I E c 2 2 0 2 = max µ so E cI max . . . = = × × × = 2 2 4 10 3 00 10 6 25 10 21 7 0 2 7 8 5 µ π e je je j kN C B E c max max . = = 72 4 T µ (c) 0 400 . P t mc T = 0 400 785 1 00 4 186 100 20 0 3 35 10 1 07 10 17 8 5 3 . . . . . . W kg J kg C C C J 314 W s min a f b gb ga f t t = ⋅° ° ° = × = × = P34.59 Think of light going up and being absorbed by the bead which presents a face area π r b 2 . The light pressure is P S c I c = = . (a) F I r c mg r g b b A = = = π ρ π 2 3 4 3 and I gc m = F H G I K J = × 4 3 3 4 8 32 10 1 3 7 ρ π ρ . W m 2 (b) P = = × × = IA 8 32 10 2 00 10 1 05 7 3 2 . . . W m m kW 2 e j e j π P34.60 Think of light going up and being absorbed by the bead, which presents face area π r b 2 . If we take the bead to be perfectly absorbing, the light pressure is
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