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HW 6 - Kapoor(mk9499 oldhomework 06 Turner(60230 This...

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Kapoor (mk9499) – oldhomework 06 – Turner – (60230) 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 2 . 9 × 10 7 m / s in the horizontal direction. It enters a uniform electric field of 6700 N / C directed vertically. Ignoring gravitational effects, find the time it takes the proton to travel 0 . 027 m horizon- tally. Correct answer: 0 . 931034 ns. Explanation: Let : v x = 2 . 9 × 10 7 m / s , E = 6700 N / C , and x = 0 . 027 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 = 0 . 027 m 2 . 9 × 10 7 m / s · 10 9 ns s = 0 . 931034 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 . 000278156 mm. Explanation: In the vertical direction, the proton experi- ences an electric force with magnitude F elec = q E = m a y a y = q E m = ( 1 . 60218 × 10 - 19 C ) (6700 N / C) 1 . 67262 × 10 - 27 kg = 6 . 41782 × 10 11 m / s 2 . The vertical dispacement is Δ y = v 0 t + 1 2 a t 2 = 1 2 a t 2 since v o = 0, so Δ y Δ y = 1 2 ( 6 . 41782 × 10 11 m / s 2 ) × (9 . 31034 × 10 - 10 s) 2 × 1000 mm 1 m = 0 . 000278156 mm . 003 (part 3 of 3) 10.0 points What is the proton’s speed after being in the electric field for that time?
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