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Unformatted text preview: fierro (jmf2547) – HW 09 – berk – (60290) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A positron is accelerated from rest between two points due to a fixed electrostatic po tential difference, and acquires a speed of 28 % percent of the speed of light, which is 3 × 10 8 m / s. Denote m 1 as the mass of a positron, e the charge of a positron, and v 1 the velocity of the positron. Find the magnitude of the potential differ ence  Δ V  between the two points. 1.  Δ V  = m 1 v 2 1 e 2.  Δ V  = 1 2 m 1 v 2 1 e correct 3.  Δ V  = em 2 v 2 1 4.  Δ V  = 1 2 m 1 v 2 1 5.  Δ V  = m 2 v 2 1 e 6.  Δ V  = 1 2 m 2 v 2 1 e 7.  Δ V  = 1 2 em 1 v 2 1 8.  Δ V  = 1 2 m 2 v 2 1 9.  Δ V  = 1 2 em 2 v 2 1 10.  Δ V  = em 1 v 2 1 Explanation: According to energy conservation, the po tential energy lost Δ U by the positron be tween two points is equal to the kinetic energy Δ K gained by the positron, Δ U = Δ K Since the positron starts from rest, Δ K = 1 2 m 1 v 2 1 , Δ U = e Δ V Therefore e Δ V = 1 2 m 1 v 2 1 , where Δ V is the potential difference between two points. Thus the potential difference is  Δ V  = 1 2 m 1 v 2 1 e 002 (part 2 of 2) 10.0 points A proton is also accelerated from rest between the same two points. Use 9 . 11 × 10 − 31 kg for the mass of the positron, with the mass of the proton given as 1 . 67 × 10 − 27 kg. What final speed will be reached by this proton? Correct answer: 1 . 96192 × 10 6 m / s. Explanation: Let : p = 0 . 28 , c = 3 × 10 8 m / s , m 1 = 9 . 11 × 10 − 31 kg , and m 2 = 1 . 67 × 10 − 27 kg ....
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 Electric Potential, Potential Energy, Potential difference

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