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# HW 12 - yang(ey942 – ohw 12 – turner –(56705 This...

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yang (ey942) – ohw 12 – turner – (56705) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A total charge of 11 mC passes through a cross-sectional area of a nichrome wire in 3 . 6 s. a) What is the current in the wire? Correct answer: 0 . 00305556 A. Explanation: Let : Δ Q 1 = 11 mC and Δ t 1 = 3 . 6 s . I = Δ Q Δ t = 0 . 011 C 3 . 6 s = 0 . 00305556 A . 002 (part 2 of 3) 10.0 points b) How many electrons pass through the cross-sectional area in 6.50 s? Correct answer: 1 . 24132 × 10 17 . Explanation: Let : q e = 1 . 6 × 10 19 C and Δ t 2 = 6 . 50 s . Δ Q = N q e N = Δ Q q e = I Δ t 2 q e = (0 . 00305556 A)(6 . 5 s) 1 . 6 × 10 19 C = 1 . 24132 × 10 17 . 003 (part 3 of 3) 10.0 points c) If the number of charges that pass through the cross-sectional area during the given time interval doubles, what is the resulting cur- rent? Correct answer: 0 . 00611111 A. Explanation: Let : Δ Q 2 = 2 Δ Q 1 . I = 2 Δ Q 1 Δ t 1 = 2 (0 . 011 C) 3 . 6 s = 0 . 00611111 A . 004 10.0 points An electrician finds that a 0 . 6 m length of a certain type of wire has a resistance of 0 . 12 Ω. What is the total resistance of the 115 m of this wire he plans to use to wire a house? Correct answer: 23 Ω. Explanation: Let : 1 = 0 . 6 m , 2 = 115 m , and R = 0 . 12 Ω . A 0 . 6 m length of the wire has a resistance of λ = R 1 , so the total resistance of the wire is R tot = λ ℓ = R 1 = parenleftbigg 0 . 12 Ω 0 . 6 m parenrightbigg (115 m) = 23 Ω . 005 (part 1 of 2) 10.0 points Suppose that a uniform wire of resistance R is stretched uniformly to N times its original length. Assume the total volume of the wire re- mains the same, and the volume is given by the product of the cross sectional area and the length. Denote the original area by A and the area after the stretch by A .

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yang (ey942) – ohw 12 – turner – (56705) 2 The ratio A A is 1. A A = 1 2. A A = 1 N 2 3. A A = N 2 4. A A = 1 N 2 / 3 5. A A = 0 6. A A = N 7. A A = N 8. A A = 1 N correct 9. A A = N 3 / 2 10. A A = 1 N Explanation: Let : = N ℓ . The volume before the stretch is V = A ℓ where is the length. The volume after the stretch is V = A = A N ℓ.
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