Ch6-H3-solutions - chauhan(mmc2762 – Ch6-H3 – turner –(56910 1 This print-out should have 15 questions Multiple-choice questions may continue

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Unformatted text preview: chauhan (mmc2762) – Ch6-H3 – turner – (56910) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A two-dimensional force vector F = F x ˆ ı + F y ˆ moves a particle to a new location that is vectorr = r x ˆ ı + r y ˆ from where it was originally located. If F x = 5 N, F y = − 3 N, r x = 4 m, and r y = 2 m, what is the work done on the particle by the force? Correct answer: 14 J. Explanation: The work is given by W = vector F · Δ vectorr = F x r x + F y r y = (5 N) (4 m) + ( − 3 N) (2 m) = 14 J . 002 (part 2 of 2) 10.0 points Find the angle between vector F and vectorr . Correct answer: 57 . 5288 ◦ . Explanation: Since W = vector F · Δ vectorr = vextendsingle vextendsingle vextendsingle vector F vextendsingle vextendsingle vextendsingle | Δ vectorr | cos θ θ = cos − 1 W vextendsingle vextendsingle vextendsingle vector F vextendsingle vextendsingle vextendsingle | Δ vectorr | . vextendsingle vextendsingle vextendsingle vector F vextendsingle vextendsingle vextendsingle = radicalBig F 2 x + F 2 y = radicalBig (5 N) 2 + ( − 3 N) 2 = 5 . 83095 N and | Δ vectorr | = radicalBig r 2 x + r 2 y = radicalbig 4 m 2 + 2 m 2 = 4 . 47214 m , so θ = cos − 1 bracketleftbigg 14 J (5 . 83095 N) (4 . 47214 m) bracketrightbigg = 57 . 5288 ◦ . 003 (part 1 of 6) 10.0 points A lithium nucleus has mass 5 . 1 × 10 − 27 kg. If its speed is 0 . 983 c (that is, v/c = 0 . 983), what is the particle energy? Use c = 3 × 10 8 m / s. Correct answer: 2 . 49992 × 10 − 9 J. Explanation: The particle energy is just the total energy, given by E = γ m c 2 . So we have E = 1 radicalBigg 1 − parenleftbigg v c parenrightbigg 2 m c 2 = 1 radicalbig 1 − . 983 2 (5 . 1 × 10 − 27 kg) c 2 = 2 . 49992 × 10 − 9 J . 004 (part 2 of 6) 10.0 points What is the rest energy? Correct answer: 4 . 59 × 10 − 10 J. Explanation: The rest energy is just E rest = m c 2 = (5 . 1 × 10 − 27 kg) c 2 = 4 . 59 × 10 − 10 J . 005 (part 3 of 6) 10.0 points What is the kinetic energy? Correct answer: 2 . 04092 × 10 − 9 J. Explanation: The kinetic energy is given by KE = E − E rest = 2 . 49992 × 10 − 9 J − 4 . 59 × 10 − 10 J = 2 . 04092 × 10 − 9 J ....
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This note was uploaded on 04/03/2012 for the course PHYSICS 303K taught by Professor Antoniewicz during the Spring '11 term at University of Texas at Austin.

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Ch6-H3-solutions - chauhan(mmc2762 – Ch6-H3 – turner –(56910 1 This print-out should have 15 questions Multiple-choice questions may continue

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