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# hw 4 - wilson(kw22588 HW4 Mackie(20224 1 This print-out...

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Unformatted text preview: wilson (kw22588) HW4 Mackie (20224) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Two alpha particles (helium nuclei), each consisting of two protons and two neu- trons, have an electrical potential energy of 6 . 39 10 19 J . Given: k e = 8 . 98755 10 9 N m 2 / C 2 , q p = 1 . 6021 10 19 C , and g = 9 . 8 m / s 2 . What is the distance between these parti- cles at this time? Correct answer: 1 . 44404 10 9 m. Explanation: Let: U electric = 6 . 39 10 19 J , k e = 8 . 98755 10 9 N m 2 / C 2 , q p = 1 . 6021 10 19 C , q = 2 q p = 3 . 2042 10 19 C , and q n = 0 C . q 1 = q 2 = 2 q p + 2 q n = 2 (1 . 6021 10 19 C) + 2 (0 C) = 3 . 2042 10 19 C U electric = k e q 1 q 2 r r = k e q 1 q 2 U electric = k e q 2 1 U electric = (8 . 99 10 9 N m 2 / C 2 ) (3 . 2042 10 19 C) 2 6 . 39 10 19 J = 1 . 44404 10 9 m . 002 (part 1 of 2) 10.0 points An electric field does 13 J of work on a . 0002 C charge. What is the voltage change? Correct answer: 65000 V. Explanation: Let : W = 13 J and q = 0 . 0002 C . Work is W = qV V = W q = 13 J . 0002 C = 65000 V . 003 (part 2 of 2) 10.0 points The same electric field does 26 J of work on a . 0004 C charge. What is the voltage change? Correct answer: 65000 V. Explanation: Let : W = 26 J and q = 0 . 0004 C . The voltage change is V = W q = 26 J . 0004 C = 65000 V . 004 10.0 points Two insulating spheres having radii 0 . 22 cm and 0 . 42 cm, masses 0 . 14 kg and 0 . 68 kg, and charges 3 C and 2 C are released from rest when their centers are separated by 1 . 2 m . How fast is the smaller sphere moving when they collide? Correct answer: 9 . 96558 m / s. Explanation: Let : k e = 8 . 99 10 9 N m 2 / C 2 , r 1 = 0 . 22 cm = 0 . 0022 m , wilson (kw22588) HW4 Mackie (20224) 2 r 2 = 0 . 42 cm = 0 . 0042 m , m 1 = 0 . 14 kg , m 2 = 0 . 68 kg , q 1 = 3 C = 3 10 6 C , q 2 = 2 C = 2 10 6 C , and d = 1 . 2 m . By conservation of momentum 0 = m 1 v 1 m 2 v 2 v 2 = m 1 v 1 m 2 By conservation of energy ( K + U ) i = ( K + U ) f 0 + k e q 1 q 2 d = 1 2 m 1 v 2 1 + 1 2 m 2 v 2 2 + k e q 1 q 2 r 1 + r 2 k e q 1 q 2 d = 1 2 m 1 v 2 1 + 1 2 m 2 parenleftbigg m 1 v 1 m 2 parenrightbigg 2 + k e q 1 q 2 r 1 + r 2 k e q 1 q 2 parenleftbigg 1 r 1 + r 2 1 d parenrightbigg = m 1 v 2 1 2 m 2 ( m 1 + m 2 ) . So v 2 1 = 2 m 2 k e q 1 q 2 m 1 ( m 1 + m 2 ) parenleftbigg 1 r 1 + r 2 1 d parenrightbigg = 2 (0 . 68 kg) ( 8 . 99 10 9 N m 2 / C 2 ) (0 . 14 kg) (0 . 14 kg + 0 . 68 kg) ( 3 10 6 C) (2 10 6 C) parenleftbigg 1 . 0022 m + 0 . 0042 m 1 1 . 2 m parenrightbigg = 99 . 3129 m 2 / s 2 ....
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hw 4 - wilson(kw22588 HW4 Mackie(20224 1 This print-out...

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