This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Kapoor (mk9499) – oldhomework 13 – Turner – (60230) 1 This printout should have 15 questions. Multiplechoice 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 31 . 5 mm 2 and length 86 . 1 cm. The mass density of copper is 8960 kg / m 3 and the molar mass is 63 . 5 g / mol. Correct answer: 43 . 6251 min. Explanation: Let : I = 141 A , A = 31 . 5 mm 2 = 3 . 15 × 10 5 m 2 , L = 86 . 1 cm = 0 . 861 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 . 15 × 10 5 m 2 ) × (0 . 861 m) = 0 . 243009 kg . This mass corresponds to . 243009 kg . 0635 kg / mol = 3 . 82691 mol , so the number of electrons is N = (3 . 82691 mol) (6 . 02 × 10 23 / mol) = 2 . 3038 × 10 24 , and the total charge is Q = N e = (2 . 3038 × 10 24 ) (1 . 602 × 10 19 C) = 3 . 69069 × 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 = 3 . 69069 × 10 5 C 141 A parenleftbigg 1 min 60 s parenrightbigg = 43 . 6251 min . 002 10.0 points How much does it cost to watch a com plete 19 h long World Series on a 491 W television set? Assume that electricity costs $0 . 07 / kW · h. Correct answer: 65 . 303 cents. Explanation: Let : P = 491 W = 0 . 491 kW , Δ t = 19 h , and R = $0 . 07 / kW · h . The cost is c = ( P Δ t ) R = (0 . 491 kW) (19 h) ($0 . 07 / kW · h) × 100 cents dollar = 65 . 303 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 Kapoor (mk9499) – oldhomework 13 – Turner – (60230)...
View
Full
Document
This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

Click to edit the document details