oldhw13_pdf

# oldhw13_pdf - mcconnell(kam2342 – oldhomework 13 –...

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Unformatted text preview: mcconnell (kam2342) – oldhomework 13 – Turner – (60230) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points How long does it take electrons to get from the car battery to the starting motor? Assume the current is 141 A and the electrons travel through copper wire with cross sectional area 37 . 2 mm 2 and length 96 . 9 cm. The mass density of copper is 8960 kg / m 3 and the molar mass is 63 . 5 g / mol. Correct answer: 57 . 9815 min. Explanation: Let : I = 141 A , A = 37 . 2 mm 2 = 3 . 72 × 10- 5 m 2 , L = 96 . 9 cm = 0 . 969 m , ρ = 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 = ρV = ρAL = (8960 kg / m 3 ) (3 . 72 × 10- 5 m 2 ) × (0 . 969 m) = 0 . 322979 kg . This mass corresponds to . 322979 kg . 0635 kg / mol = 5 . 08629 mol , so the number of electrons is N = (5 . 08629 mol) (6 . 02 × 10 23 / mol) = 3 . 06195 × 10 24 , and the total charge is Q = N e = (3 . 06195 × 10 24 ) (1 . 602 × 10- 19 C) = 4 . 90524 × 10 5 C . Thus the time it takes to move this amount of charge from the battery to the starting motor is t = Q I = 4 . 90524 × 10 5 C 141 A parenleftbigg 1 min 60 s parenrightbigg = 57 . 9815 min . 002 10.0 points How much does it cost to watch a com- plete 22 h long World Series on a 516 W television set? Assume that electricity costs \$0 . 07 / kW · h. Correct answer: 79 . 464 cents. Explanation: Let : P = 516 W = 0 . 516 kW , Δ t = 22 h , and R = \$0 . 07 / kW · h . The cost is c = ( P Δ t ) R = (0 . 516 kW) (22 h) (\$0 . 07 / kW · h) × 100 cents dollar = 79 . 464 cents . 003 10.0 points 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 mcconnell (kam2342) – oldhomework 13 – Turner – (60230)...
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oldhw13_pdf - mcconnell(kam2342 – oldhomework 13 –...

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