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Unformatted text preview: oldhomewk 13 LEE, BENJAMIN Due: Oct 2 2007, 4:00 am 1 Question 1, chap 27, sect 1. part 1 of 1 10 points How long does it take electrons to get from the car battery to the starting motor? Assume the current is 113 A and the electrons travel through copper wire with cross sectional area 32 mm 2 and length 82 . 3 cm. The mass density of copper is 8960 kg / m 3 and the molar mass is 63 . 5 g / mol. Correct answer: 52 . 8583 min (tolerance 1 %). Explanation: Let : I = 113 A , A = 32 mm 2 = 3 . 2 10 5 m 2 , L = 82 . 3 cm = 0 . 823 m , = 8960 kg / m 3 , and M molecule = 63 . 5 g / mol = 0 . 0635 kg / mol . Current is equal to the charge passing a cross section per unit time. In this problem, the current I comes from the motion of the electrons inside the copper wire. If the time that the electrons move from the battery to the starting motor is t , and the charge inside this region is Q , we have I = Q t . The volume of this copper wire segment is V = AL, so the mass is M = V = AL = (8960 kg / m 3 ) (3 . 2 10 5 m 2 ) (0 . 823 m) = 0 . 235971 kg . This mass corresponds to . 235971 kg . 0635 kg / mol = 3 . 71607 mol , so the number of electrons is N = (3 . 71607 mol) (6 . 02 10 23 / mol) = 2 . 23708 10 24 , and the total charge is Q = N e = (2 . 23708 10 24 ) (1 . 602 10 19 C) = 358379 C . Thus the time it takes to move this amount of charge from the battery to the starting motor is t = Q I = 358379 C 113 A parenleftbigg 1 min 60 s parenrightbigg = 52 . 8583 min . Question 2, chap 1, sect 1. part 1 of 1 10 points How much does it cost to watch a com plete 19 h long World Series on a 276 W television set? Assume that electricity costs $0 . 07 / kW h. Correct answer: 36 . 708 cents (tolerance 1 %). Explanation: Let : P = 276 W = 0 . 276 kW , t = 19 h , and R = $0 . 07 / kW h . The cost is c = ( P t ) R = (0 . 276 kW) (19 h) ($0 . 07 / kW h) 100 cents dollar = 36 . 708 cents . Question 3, chap 27, sect 4. part 1 of 1 10 points oldhomewk 13 LEE, BENJAMIN Due: Oct 2 2007, 4:00 am 2 Consider the circuit shown in the figure. In this circuit, r i is the internal resistance of the battery (it cannot be removed and must be considered as part of the battery); R o is an external resistance; a and b are the battery terminals; S is a switch....
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 Fall '07
 Turner
 Charge, Current

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