Homework 4-solutions - im(gi768 Homework 4 markert(55325...

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im (gi768) – Homework 4 – markert – (55325) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points The potential difference between a storm cloud and the ground is 2 . 49 × 10 8 V. If a bolt carrying 3 C falls from a cloud to Earth, what is the magnitude of the change of potential energy of the charge? Correct answer: 7 . 47 × 10 8 J. Explanation: Let : V = 2 . 49 × 10 8 V and q = 3 C . The potential energy is U = V q = (2 . 49 × 10 8 V )(3 C ) = 7 . 47 × 10 8 J . 002 10.0points In a charging process, 5 × 10 13 electrons are removed from one small metal sphere and placed on a second identical sphere. Initially both metal spheres were neutral. After the charging process the electrical potential en- ergy associated with the two spheres is found to be 0 . 06 J . What is the distance between the two spheres? The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 and the elemental charge is 1 . 6 × 10 19 C . Correct answer: 9 . 58672 m. Explanation: Let : n = 5 × 10 13 , q e = 1 . 6 × 10 19 C , U e = 0 . 06 J , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . q 1 = n q e and q 2 = n q e , so U e = k e q 1 q 2 r r = k e q 1 q 2 U e = k e ( n q e ) 2 U e = (8 . 98755 × 10 9 N · m 2 / C 2 ) × (5 × 10 13 ) 2 (1 . 6 × 10 19 C) 2 0 . 06 J = 9 . 58672 m . 003(part1of2)10.0points Find the speed of an electron that has a kinetic energy of 2 . 02 eV. 1 eV= 1 . 602 × 10 19 J . Correct answer: 8 . 4292 × 10 5 m / s. Explanation: Let : E k = 2 . 02 eV and m e = 9 . 109 × 10 31 kg . The kinetic energy is E k = 1 2 m e v 2 e v e = radicalbigg 2 K e m e = radicalBigg 2(2 . 02 eV) 9 . 109 × 10 31 kg 1 . 602 × 10 19 J 1 eV = 8 . 4292 × 10 5 m / s . 004(part2of2)10.0points Calculate the speed of a proton with a kinetic energy of 2 . 02 eV. Correct answer: 19671 m / s. Explanation: Let : E k = 2 . 02 eV and m p = 1 . 6726 × 10 27 kg .
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im (gi768) – Homework 4 – markert – (55325) 2 The kinetic energy is E k = 1 2 m p v 2 p v p = radicalBigg 2 E k m p = radicalBigg 2(2 . 02 eV) 1 . 6726 × 10 27 kg 1 . 602 × 10 19 J 1 eV = 19671 m / s . 005 10.0points Four charges are fixed at the corners of a square centered at the origin as follows: q at ( a, + a ); 2 q at (+ a, + a ); 3 q at (+ a, a ); and 6 q at ( a, a ). A fifth charge + q with mass m is placed at the origin and released from rest. Find the speed when it is a great distance from the origin, where the potential energy of the fifth charge due to the four point charges is negligible. 1. bardbl vectorv bardbl = q radicalBigg 6 6 k m a 2. bardbl vectorv bardbl = q radicalBigg 6 5 k m a 3. bardbl vectorv bardbl = q radicalBigg 6 3 k m a 4. bardbl vectorv bardbl = q radicalBigg 6 2 k m a correct 5. bardbl vectorv bardbl = q radicalBigg 3 5 k m a 6. bardbl vectorv bardbl = q radicalBigg 2 5 k m a 7. bardbl vectorv bardbl = q radicalBigg 3 2 k m a 8. bardbl vectorv bardbl = q radicalBigg 3 6 k m a 9. bardbl vectorv bardbl = q radicalBigg 3 3 k m a 10. bardbl vectorv bardbl = q radicalBigg 2 2 k m a Explanation: x a +6 q a 3 q +2 q + q + q, m 2 a The initial energy of the charge is E i = K i + U i = U i = q bracketleftbigg k q 2 a + 2 k q 2 a + ( 3 q ) k 2 a + 6 k q 2 a
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