Homwork Combined - Kapoor (mk9499) homework09 Turner...

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Unformatted text preview: Kapoor (mk9499) homework09 Turner (60230) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A proton is released from rest in a uniform electric field of magnitude 1 . 5 10 5 V / m di- rected along the positive x axis. The proton undergoes a displacement of 0 . 4 m in the di- rection of the electric field as shown in the figure. + + + + + + + + + + + + +------------- v A = 0 + . 4 m v 1 . 5 10 5 V / m A B Apply the principle of energy conserva- tion to find the amount of the kinetic energy gained after it has moved 0 . 4 m. Correct answer: 9 . 61306 10- 15 J. Explanation: The change in the potential energy of the proton is U = q p V = (1 . 60218 10- 19 C) (- 60000 V) =- 9 . 61306 10- 15 J . Conservation of energy in this case is K + U = K f- K i + U = 0 . So K f- K i =- U = 9 . 61306 10- 15 J . 002 10.0 points Two alpha particles (helium nuclei), each consisting of two protons and two neu- trons, have an electrical potential energy of 6 . 41 10- 19 J . Given: k e = 8 . 98755 10 9 N m 2 / C 2 , q p = 1 . 6021 10- 19 C , and g = 9 . 8 m / s 2 . What is the distance between these parti- cles at this time? Correct answer: 1 . 43954 10- 9 m. Explanation: Let: U electric = 6 . 41 10- 19 J , k e = 8 . 98755 10 9 N m 2 / C 2 , q p = 1 . 6021 10- 19 C , q = 2 q p = 3 . 2042 10- 19 C , and q n = 0 C . q 1 = q 2 = 2 q p + 2 q n = 2 (1 . 6021 10- 19 C) + 2 (0 C) = 3 . 2042 10- 19 C U electric = k e q 1 q 2 r r = k e q 1 q 2 U electric = k e q 2 1 U electric = (8 . 99 10 9 N m 2 / C 2 ) (3 . 2042 10- 19 C) 2 6 . 41 10- 19 J = 1 . 43954 10- 9 m . 003 (part 1 of 2) 10.0 points A proton is accelerated through a potential difference of 4 . 10 6 V. a) How much kinetic energy has the proton acquired? Correct answer: 6 . 4 10- 13 J. Explanation: Let : V = 4 . 10 6 V and q = 1 . 60 10- 19 C . K = U = q V = (1 . 60 10- 19 C) (4 . 10 6 V) = 6 . 4 10- 13 J . Kapoor (mk9499) homework09 Turner (60230) 2 004 (part 2 of 2) 10.0 points b) If the proton started at rest, how fast is it moving? Correct answer: 2 . 76603 10 7 m / s. Explanation: Let : m = 1 . 673 10- 27 kg . Since K i = 0 J , K = K f = 1 2 mv 2 f v f = radicalbigg 2 K f m = radicalBigg 2 (6 . 4 10- 13 J) 1 . 673 10- 27 kg = 2 . 76603 10 7 m / s . 005 10.0 points It takes 145 J of work to move 2 . 7 C of charge from a positive plate to a negative plate. What voltage difference exists between the plates? Correct answer: 53 . 7037 V. Explanation: Let : W = 145 J and q = 2 . 7 C ....
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Homwork Combined - Kapoor (mk9499) homework09 Turner...

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