oldhw6solution_pdf - klarin (sjk772) oldhomework 06 Turner...

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Unformatted text preview: klarin (sjk772) oldhomework 06 Turner (56705) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A proton has an initial velocity of 1 . 33 10 7 m / s in the horizontal direction. It enters a uniform electric field of 7900 N / C directed vertically. Ignoring gravitational effects, find the time it takes the proton to travel 0 . 114 m horizon- tally. Correct answer: 8 . 57143 ns. Explanation: Let : v x = 1 . 33 10 7 m / s , E = 7900 N / C , and x = 0 . 114 m . The electric field vector E is in the vertical ( y ) di- rection, so the electric force vector F elec = q vector E ex- erted by the field on the proton is also in the y-direction, with no component in the x- direction. Hence, the field can exert no force on the proton in the x-direction. This implies a constant speed in the x-direction. Conse- quently, x = v x t t = x v x = . 114 m 1 . 33 10 7 m / s 10 9 ns s = 8 . 57143 ns . 002 (part 2 of 3) 10.0 points What is the vertical displacement of the pro- ton after the electric field acts on it for that time? Correct answer: 0 . 0277982 mm. Explanation: In the vertical direction, the proton experi- ences an electric force with magnitude F elec = q E = ma y a y = q E m = ( 1 . 60218 10- 19 C ) (7900 N / C) 1 . 67262 10- 27 kg = 7 . 56728 10 11 m / s 2 . The vertical dispacement is y = v t + 1 2 a t 2 = 1 2 a t 2 since v o = 0, so y y = 1 2 ( 7 . 56728 10 11 m / s 2 ) (8 . 57143 10- 9 s) 2 1000 mm 1 m = . 0277982 mm ....
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This note was uploaded on 10/10/2010 for the course PHY 56705 taught by Professor Turner during the Fall '10 term at University of Texas at Austin.

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oldhw6solution_pdf - klarin (sjk772) oldhomework 06 Turner...

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