958_Physics ProblemsTechnical Physics

958_Physics ProblemsTechnical Physics - Chapter 34 (a)...

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Unformatted text preview: Chapter 34 (a) efficiency = (b) P34.22 Sav = FG H 299 IJ K useful power output 700 W × 100% = × 100% = 50.0% total power input 1 400 W P 700 W = = 2.69 × 10 5 W m 2 A 0.068 3 m 0.038 1 m b gb g Sav = 269 kW m 2 toward the oven chamber (c) Sav = 2 Emax 2µ 0 c e je je j Emax = 2 4π × 10 −7 T ⋅ m A 3.00 × 10 8 m s 2.69 × 10 5 W m 2 = 1.42 × 10 4 V m = 14.2 kV m Emax : c 7.00 × 10 5 N C (a) Bmax = (b) E2 I = max : 2µ 0 c (c) I= (a) e10.0 × 10 j W I= π e0.800 × 10 mj (b) P34.23 uav = (a) E = cB = 3.00 × 10 8 m s 1.80 × 10 −6 T = 540 V m Bmax = 3.00 × 10 8 m s = 2.33 mT e7.00 × 10 j I= = 650 MW m 2e 4π × 10 je3.00 × 10 j Lπ O P = IA = e6.50 × 10 W m jM e1.00 × 10 mj P = N4 Q 52 2 −7 P : A 8 8 2 −3 2 510 W −3 P34.24 P34.25 −3 2 = 4.97 kW m 2 I 4.97 × 10 3 J m 2 ⋅ s = = 16.6 µJ m3 c 3.00 × 10 8 m s e B2 je e1.80 × 10 j = j −6 2 = 2.58 µJ m3 (b) uav = (c) Sav = cuav = 3.00 × 10 8 2.58 × 10 −6 = 773 W m 2 (d) This is 77.3% of the intensity in Example 34.5 . It may be cloudy, or the Sun may be setting. µ0 4π × 10 e −7 je j ...
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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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