968_Physics ProblemsTechnical Physics

968_Physics - Chapter 34 P34.58 The mirror intercepts power ja e f P = I 1 A1 = 1.00 × 10 3 W m 2 π 0.500 m 2 309 = 785 W In the image 785 W(a I2

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Unformatted text preview: Chapter 34 P34.58 The mirror intercepts power ja e f P = I 1 A1 = 1.00 × 10 3 W m 2 π 0.500 m 2 309 = 785 W . In the image, 785 W (a) I2 = P : A2 I2 = (b) I2 = 2 Emax so 2µ 0 c Emax = 2 µ 0 cI 2 = 2 4π × 10 −7 3.00 × 10 8 6.25 × 10 5 = 21.7 kN C b 0.400P∆t = mc∆T a = 625 kW m 2 2 e Bmax = (c) g π 0.020 0 m fb je je Emax = 72.4 µT c gb ga 0. 400 785 W ∆t = 1.00 kg 4 186 J kg⋅° C 100° C − 20.0° C ∆t = P34.59 f 5 3.35 × 10 J = 1.07 × 10 3 s = 17.8 min 314 W Think of light going up and being absorbed by the bead which presents a face area π rb2 . The light pressure is P = SI =. cc Iπ rb2 4 = mg = ρ π rb3 g 3 c (a) F= (b) P34.60 j P = IA = 8.32 × 10 7 W m 2 π 2.00 × 10 −3 m e and je I= F GH 4ρ gc 3m 3 4π ρ j 2 I JK 13 = 8.32 × 10 7 W m 2 = 1.05 kW Think of light going up and being absorbed by the bead, which presents face area π rb2 . S IF . If we take the bead to be perfectly absorbing, the light pressure is P = av = = c cA (a) F = Fg so I= F c Fg c mgc = = . A A π rb2 From the definition of density, ρ= m m = V 4 3 π rb3 so bg 1 F b 4 3gπ ρ I =G r H m JK mgc F 4π ρ I I= GJ π H 3m K 13 . b Substituting for rb , (b) P = IA 23 F 4ρ I F m I = gc G J G J H 3 K HπK FG IJ HK 4π r 2 ρ gc 3m P= 3 4πρ 23 13 13 F GH 4ρgc 3m = 3 4π ρ I JK 13 . ...
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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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