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Unformatted text preview: wilson (kw22588) HW1 Mackie (20224) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A particle of mass 21 g and charge 14 C is released from rest when it is 55 cm from a second particle of charge 15 C. Determine the magnitude of the initial ac celeration of the 21 g particle. Correct answer: 297 . 107 m / s 2 . Explanation: Let : m = 21 g , q = 14 C = 1 . 4 10 5 C , d = 55 cm = 0 . 55 m , Q = 15 C = 1 . 5 10 5 C , and k e = 8 . 9875 10 9 . The force exerted on the particle is F = k e  q 1  q 2  r 2 = m a bardbl vectora bardbl = k e bardbl vectorq bardblbardbl vector Q bardbl m d 2 = k e vextendsingle vextendsingle 1 . 4 10 5 C vextendsingle vextendsingle vextendsingle vextendsingle 1 . 5 10 5 C vextendsingle vextendsingle (0 . 021 kg) (0 . 55 m 2 ) = 297 . 107 m / s 2 . 002 10.0 points Four point charges, each of magnitude 3 . 11 C, are placed at the corners of a square 81 . 8 cm on a side. The value of Coulombs constant is 8 . 98755 10 9 N m 2 / C 2 . If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. Correct answer: 0 . 248683 N. Explanation: Let : k e = 8 . 98755 10 9 N m 2 / C 2 , d = 81 . 8 cm = 0 . 818 m , and Q = 3 . 11 C = 3 . 11 10 6 C . F 24 F 34 F 14 + 3 + 1 + 2 4 The forces are F 34 = F 24 = k e Q 2 d 2 = ( 8 . 98755 10 9 N m 2 / C 2 ) ( 3 . 11 10 6 C ) 2 (0 . 818 m) 2 = 0 . 129914 N F 14 = k e Q 2 ( 2 d ) 2 = 1 2 k e Q 2 d 2 = 1 2 F 34 The vector sum of F 24 and F 34 is in the same direction as F 14 , so F = F 14 + radicalBig F 34 2 + F 24 2 = 1 2 F 34 + radicalBig 2 F 34 2 = parenleftbigg 1 2 + 2 parenrightbigg F 34 = parenleftbigg 1 2 + 2 parenrightbigg (0 . 129914 N) = . 248683 N . 003 10.0 points A molecule of DNA (deoxyribonucleic acid) lies along a straight line. It is 1 . 17 m long. The ends of the molecule become singly ion ized; negative on one end, positive on the other. The helical molecule acts as a spring and compresses 0 . 8% upon becoming charged. The Coulomb constant is 8 . 99 10 9 N m 2 / C 2 . Determine the effective spring constant of the molecule. Take into account the com pressed length when calculating the distance between the ends of the molecule. wilson (kw22588) HW1 Mackie (20224) 2 Correct answer: 1 . 83025 10 8 N / m. Explanation: Let : = 1 . 17 m = 1 . 17 10 6 m , = 0 . 008 = 9 . 36 10 9 m , k C = 8 . 99 10 9 N m 2 / C 2 , and q e = 1 . 60218 10 19 C ....
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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Charge, Mass

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