This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: nguyen (jmn727) oldhomework 06 Turner (59070) 1 This printout should have 11 questions. Multiplechoice 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 8 . 7 10 6 m / s in the horizontal direction. It enters a uniform electric field of 28400 N / C directed vertically. Ignoring gravitational effects, find the time it takes the proton to travel 0 . 105 m horizon tally. Correct answer: 12 . 069 ns. Explanation: Let : v x = 8 . 7 10 6 m / s , E = 28400 N / C , and x = 0 . 105 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 ydirection, with no component in the x direction. Hence, the field can exert no force on the proton in the xdirection. This implies a constant speed in the xdirection. Conse quently, x = v x t t = x v x = . 105 m 8 . 7 10 6 m / s 10 9 ns s = 12 . 069 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 . 198126 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 ) (28400 N / C) 1 . 67262 10 27 kg = 2 . 72039 10 12 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 ( 2 . 72039 10 12 m / s 2 ) (1 . 2069 10 8 s) 2 1000 mm 1 m = . 198126 mm ....
View
Full
Document
 Spring '08
 Turner
 Work

Click to edit the document details