Homework 6-solutions - kumar(vk4433 Homework 6...

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kumar (vk4433) – Homework 6 – markert – (55325) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points The current I = a t 2 - b t + c in a section of a conductor depends on time. What quantity of charge moves across the section of the conductor from t = 0 s to t = t 1 ? 1. q = a t 3 1 - b t 2 + c t 1 2. q = a t 3 1 - b 2 t 2 1 + ct 1 3. q = 2 a t 2 1 - b t 1 4. q = a 3 t 3 1 - b 2 t 2 1 + c 5. q = 2 a t 1 - b 6. q = 2 a t 2 1 - 3 b t 1 + c t 1 7. q = a 3 t 3 1 - b 2 t 2 1 + c t 1 correct 8. q = 3 a t 2 1 - b + c 9. q = 3 a t 2 1 - 2 b t 1 + c 10. q = a t 2 1 - b t 1 + c Explanation: The unit of current is Coulomb per second: I = d q dt or dq = I dt . To find the total charge that passes through the conductor, one must integrate the current over the time interval. q = integraldisplay t 1 0 dq = integraldisplay t 1 0 I dt = integraldisplay t 1 0 ( a t 2 - b t + c ) dt = bracketleftbigg a 3 t 3 1 - b 2 t 2 1 + c t bracketrightbigg vextendsingle vextendsingle vextendsingle vextendsingle t 1 0 = a 3 t 3 1 - b 2 t 2 1 + c t 1 . 002 10.0points Chlorine gas (Cl 2 ) can be produced from a so- lution containing Cl ions ( i.e. , from Na + Cl ) by electrolysis. What must be the current through an electrolytic cell to produce 1.0 kg of Cl 2 gas in 21 hours? The fundamental charge is 1 . 602 × 10 19 C, Avogadro’s number is 6 . 022 × 10 23 , and the atomic mass of chlorine is 35 . 5 g / mol. Correct answer: 35 . 9462 A. Explanation: Let : N A = 6 . 022 × 10 23 / mol , m Cl = 35 . 5 g / mol , q = 1 . 602 × 10 19 C , and t = 21 h . The number of Cl ions in 1.0 kg of Cl 2 is n = 1000 g m Cl N A = 1000 g 35 . 5 g / mol (6 . 022 × 10 23 / mol) = 1 . 69634 × 10 25 . Because one electron is required for each Cl ion to be electrolyzed, the total electric charge passing the solution is Q = n q = (1 . 69634 × 10 25 ) (1 . 602 × 10 19 C) = 2 . 71753 × 10 6 C , so the current is I = Q t = 2 . 71753 × 10 6 C 21 h 1 h 3600 s = 35 . 9462 A . You could also use the mass of the chlorine gas molecule ( m Cl 2 = 2 m Cl ) and twice the number of Cl ions.
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kumar (vk4433) – Homework 6 – markert – (55325) 2 003(part1of2)10.0points The current in a conductor varies over time as shown in the figure below. 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 Current(A) Time(s) How much charge passes through a cross section of the conductor in the time interval t = 0 s to t = 5 s? Correct answer: 18 C. Explanation: Let : Δ t = 5 . 0 s . I = Δ Q Δ t Δ Q = I Δ t = (2 A)(2 s) + (4 A)(1 s) + (6 A)(1 s) + (4 A)(1 s) = 18 C . 004(part2of2)10.0points What constant current would transport the same total charge during the 5 s interval as does the actual current? Correct answer: 3 . 6 A. Explanation: I = Δ Q Δ t = 18 C 5 s = 3 . 6 A . 005 10.0points In a fluorescent tube of diameter 1 . 6 cm , 1 . 6 × 10 18 electrons (with a charge of - e ) and 1 . 3 × 10 18 positive ions (with a charge of + e ) flow in opposite directions through a cross-sectional each second.
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