Hw9_pdf - mcconnell (kam2342) homework09 Turner (60230) 1...

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Unformatted text preview: 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 + . 5 m v 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 U = q p V = (1 . 60218 10- 19 C) (- 50000 V) =- 8 . 01089 10- 15 J . Conservation of energy in this case is K + U = K f- K i + U = 0 . So K f- K i =- 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 ....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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Hw9_pdf - mcconnell (kam2342) homework09 Turner (60230) 1...

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