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Lemuel physics 2 6

Lemuel physics 2 6 - Wong Lemuel Homework 6 Due Mar 2 2007...

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Wong, Lemuel – Homework 6 – Due: Mar 2 2007, 11:00 pm – Inst: Matt Mackie 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. NOTE: HW is due 11pm central time, which is midnight our time. 001 (part 1 of 1) 10 points Two alpha particles (helium nuclei), each con- sisting of two protons and two neutrons, have an electrical potential energy of 6 × 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 . 5379 × 10 - 9 m. Explanation: Let: U electric = 6 × 10 - 19 J , k e = 8 . 98755 × 10 9 N m 2 / C 2 , q p = 1 . 6021 × 10 - 19 C , q α = 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 × 10 - 19 J = 1 . 5379 × 10 - 9 m . keywords: 002 (part 1 of 2) 10 points An electric field does 11 J of work on a 0 . 0009 C charge. What is the voltage change? Correct answer: 12222 . 2 V. Explanation: Let : W = 11 J and q = 0 . 0009 C . Work is W = qV V = W q = 11 J 0 . 0009 C = 12222 . 2 V . 003 (part 2 of 2) 10 points The same electric field does 22 J of work on a 0 . 0018 C charge. What is the voltage change? Correct answer: 12222 . 2 V. Explanation: Let : W = 22 J and q = 0 . 0018 C . The voltage change is V = W q = 22 J 0 . 0018 C = 12222 . 2 V . keywords: 004 (part 1 of 1) 10 points Two insulating spheres having radii 0 . 12 cm and 0 . 32 cm, masses 0 . 11 kg and 0 . 41 kg, and charges - 4 μ C and 2 μ 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: 15 . 2795 m / s. Explanation:

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Wong, Lemuel – Homework 6 – Due: Mar 2 2007, 11:00 pm – Inst: Matt Mackie 2 Let : k e = 8 . 99 × 10 9 N · m 2 / C 2 , r 1 = 0 . 12 cm = 0 . 0012 m , r 2 = 0 . 32 cm = 0 . 0032 m , m 1 = 0 . 11 kg , m 2 = 0 . 41 kg , q 1 = - 4 μ C = - 4 × 10 - 6 C , q 2 = 2 μ 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 m 1 v 1 m 2 2 + k e q 1 q 2 r 1 + r 2 - k e q 1 q 2 1 r 1 + r 2 - 1 d = 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 ) 1 r 1 + r 2 - 1 d = - 2 (0 . 41 kg) ( 8 . 99 × 10 9 N · m 2 / C 2 ) (0 . 11 kg) (0 . 11 kg + 0 . 41 kg) × ( - 4 × 10 - 6 C) (2 × 10 - 6 C) × 1 0 . 0012 m + 0 . 0032 m - 1 1 . 2 m = 233 . 464 m 2 / s 2 .
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Lemuel physics 2 6 - Wong Lemuel Homework 6 Due Mar 2 2007...

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