# hw05 - = V P i i = 0.833 A Q27.02 a Series b Parallel c...

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PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 5. 26.08, 26.39, 26.43; Q27.02, Q27.07 26.08. a) J = n p ev p = (8.70 cm -3 )(10 6 cm 3 /m 3 )(1.60 ×10 -19 C)(470 ×10 3 m/s) J = 6.54 ×10 -7 A/m 2 b) i = JA , where A is the area that intercepts the protons. Since Earth looks like a disk from the Sun, A = π R 2 , where R is the radius of Earth. i = J R 2 = (6.54 ×10 -7 A/m 2 ) (6.37 ×10 6 m) 2 i = 8.34 ×10 7 A 26.39. A 9 . 10 V 115 W 1250 so , a) = = = = i V P i iV P = = = = 6 . 10 W 1250 V) 115 ( so , b) 2 2 2 R P V R R V P c) W = Pt = (1250 W)(1.0 hr)(3600 s/hr) W = 4.5 × 10 6 J = 4.5 MJ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 26.43. a) Cost = ( Number of kW·h)( Price/ kW·h). For a 30 day month, this becomes: Cost = (0.10 kW)(24 hr/day)(30 days)(\$0.06/kW·h) Cost = \$4.32 b) R V P 2 = , so W 100 ) V 120 ( 2 2 = = P V R R = 144 c) P = iV , so V 120 W 100 =

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Unformatted text preview: = V P i i = 0.833 A Q27.02. a) Series b) Parallel c) Parallel Q27.07. The smallest resistance will draw the largest current. Therefore, we must order the resistances. Since R 1 > R 2 , R 2 will draw more current than R 1 . When in series, the resistance will be R 1 + R 2 > R 1 . Therefore, when in series, the combination will draw less current than R 1 . When in parallel, the resistance will be, R R R R R R R R 1 2 1 2 2 2 1 2 1 + = + < ( / ) , so the parallel combination will draw more current than R 2 . Therefore the order according to current drawn, from highest to lowest will be: parallel , R 2 , R 1 , series ....
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## This note was uploaded on 04/07/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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hw05 - = V P i i = 0.833 A Q27.02 a Series b Parallel c...

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