soluti34649on8_pdf

soluti34649on8_pdf - kinsey(cbk323 – oldhomework 06 –...

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Unformatted text preview: kinsey (cbk323) – oldhomework 06 – Turner – (56725) 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 3 . 19 × 10 7 m / s in the horizontal direction. It enters a uniform electric field of 20900 N / C directed vertically. Ignoring gravitational effects, find the time it takes the proton to travel 0 . 111 m horizon- tally. Correct answer: 3 . 47962 ns. Explanation: Let : v x = 3 . 19 × 10 7 m / s , E = 20900 N / C , and x = 0 . 111 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 = . 111 m 3 . 19 × 10 7 m / s · 10 9 ns s = 3 . 47962 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 . 0121197 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 ) (20900 N / C) 1 . 67262 × 10- 27 kg = 2 . 00198 × 10 12 m / s 2 . The vertical dispacement is Δ y = v t + 1 2 a t 2 = 1 2 a t 2 since v o = 0, so Δ y Δ y = 1 2 ( 2 . 00198 × 10 12 m / s 2 ) × (3 . 47962 × 10- 9 s) 2 × 1000 mm 1 m = . 0121197 mm ....
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This note was uploaded on 11/02/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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soluti34649on8_pdf - kinsey(cbk323 – oldhomework 06 –...

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