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Hw9_pdf - mcconnell(kam2342 – homework09 – Turner...

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mcconnell (kam2342) – 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 × 10 5 V / m di- rected along the positive x axis. The proton undergoes a displacement of 0 . 5 m in the di- rection of the electric field as shown in the figure. + + + + + + + + + + + + + - - - - - - - - - - - - - v A = 0 + 0 . 5 m v 0 1 × 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 . 5 m. Correct answer: 8 . 01089 × 10 - 15 J. Explanation: The change in the potential energy of the proton is Δ U = q p Δ V = (1 . 60218 × 10 - 19 C) ( - 50000 V) = - 8 . 01089 × 10 - 15 J . Conservation of energy in this case is Δ K + Δ U = K f - K i + Δ U = 0 . So K f - K i = - Δ U = 8 . 01089 × 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 . 37 × 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?
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