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Unformatted text preview: madrid (tmm2353) – HW 61 – Antoniewicz – (56445) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 2.0 points An electron has mass 9 . 1 × 10 31 kg. The speed of light is 3 × 10 8 m / s. If the electron’s speed is 0 . 986 c (that is, v/c = 0 . 986), what is its particle energy? Correct answer: 4 . 91168 × 10 13 J. Explanation: The electron’s particle energy, or total en ergy, is given by E = γ mc 2 = 1 radicalBigg 1 − parenleftbigg v c parenrightbigg 2 mc 2 = 1 radicalbig 1 − . 986 2 (9 . 1 × 10 31 kg) c 2 = 4 . 91168 × 10 13 J . 002 (part 2 of 3) 2.0 points What is its rest energy? Correct answer: 8 . 19 × 10 14 J. Explanation: The rest energy is given by E r = mc 2 = (9 . 1 × 10 31 kg)(3 × 10 8 m / s) 2 = 8 . 19 × 10 14 J . 003 (part 3 of 3) 1.0 points What is its kinetic energy? Correct answer: 4 . 09268 × 10 13 J. Explanation: The kinetic energy is just the difference be tween the particle energy and the rest energy: K = E − E r = 4 . 91168 × 10 13 J − 8 . 19 × 10 14 J = 4 . 09268 × 10 13 J . 004 (part 1 of 4) 2.0 points The point of this question is to compare rest energy E r and kinetic energy KE at low speeds. A baseball is moving at a speed of 17 m / s. Its mass is 142 g (0 . 142 kg). The speed of light is 3 × 10 8 m / s. What is its rest energy? Correct answer: 1 . 278 × 10 16 J. Explanation: The rest energy is given by E rest = mc 2 = (142 g)(3 × 10 8 m / s) 2 = 1 . 278 × 10 16 J . 005 (part 2 of 4) 1.0 points Is it okay to calculate its kinetic energy using the formula 1 2 mv 2 ? 1. Yes, because  vectorv  ≪ c . correct 2. Yes, because accuracy isn’t too important here. 3. No, because baseballs are not subatomic particles 4. No, because  vectorv  ≪ c . Explanation: It’s okay to use the approximate formula here, because the baseball’s speed is so small compared to the speed of light. 006 (part 3 of 4) 1.0 points What is its kinetic energy? Correct answer: 20 . 519 J. Explanation: madrid (tmm2353) – HW 61 – Antoniewicz – (56445) 2 Since we can use the approximate formula from Newtonian mechanics, the kinetic energy is KE = 1 2 m  vectorv  2 = 1 2 (142 g)(17 m / s) 2 = 20 . 519 J ....
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This note was uploaded on 11/15/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner

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