HW 8 solutions (Ch 28)

HW 8 solutions (Ch 28) - Physics 6C HW#8 solution 28.2 a P...

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Unformatted text preview: Physics 6C HW#8 solution May 29, 2008 28.2 a) P = eAT 4 T = ( P eA ) 1 / 4 = 3 . 85 10 26 W 1(4 (6 . 96 10 8 m ) 2 )(5 . 67 10- 8 W/m 2 K 4 ) 1 / 4 = 5 . 78 10 3 K b) max = 2 . 898 10- 3 mK T = 2 . 898 10- 3 mK 5 . 78 10 3 K = 5 . 01 10- 7 m 28.4 a) E = hf = (6 . 626 10- 34 J s )(620 10 12 s- 1 )( 1 . 00 eV 1 . 60 10- 19 J ) = 2 . 57 eV b) E = hf = (6 . 626 10- 34 J s )(3 . 10 10 9 s- 1 )( 1 . 00 eV 1 . 60 10- 19 J ) = 1 . 28 10- 5 eV c) E = hf = (6 . 626 10- 34 J s )(46 . 10 6 s- 1 )( 1 . 00 eV 1 . 60 10- 19 J ) = 1 . 91 10- 7 eV d) = c f = 3 . 00 10 8 m/s 620 10 12 Hz = 484 nm, visible light(blue) = c f = 9.68 cm, radio wave = c f = 6.52 m, radio wave 1 28.7 We take sin = 0 . 030 radians. Then, the pendu- lums total energy is E = mgh = mg ( L- L cos ) E = (1 Kg )(9 . 80 m/s 2 )(1 . 00- . 9995) = 4 . 41 10- 3 J The frequency of oscillation is f = 1 2 p g L = 0 . 498 Hz....
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This note was uploaded on 07/02/2008 for the course PHYS 6C taught by Professor Staff during the Spring '01 term at UCLA.

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HW 8 solutions (Ch 28) - Physics 6C HW#8 solution 28.2 a P...

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