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# hw1 - wilson(kw22588 HW1 Mackie(20224 This print-out should...

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wilson (kw22588) – HW1 – Mackie – (20224) 1 This print-out should have 19 questions. Multiple-choice 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 μ C is released from rest when it is 55 cm from a second particle of charge 15 μ 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 μ C = 1 . 4 × 10 5 C , d = 55 cm = 0 . 55 m , Q = 15 μ 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 bardbl bardbl 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 μ C, are placed at the corners of a square 81 . 8 cm on a side. The value of Coulomb’s 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 μ 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) = 0 . 248683 N . 003 10.0 points A molecule of DNA (deoxyribonucleic acid) lies along a straight line. It is 1 . 17 μ 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.

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wilson (kw22588) – HW1 – Mackie – (20224) 2 Correct answer: 1 . 83025 × 10 8 N / m. Explanation: Let : = 1 . 17 μ m = 1 . 17 × 10 6 m , Δ = 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 . The compressed length is r = Δ = 1 . 16064 × 10 6 m and the compression force is F compression = k C q e q e r 2 . The effective force constant is k = F Δ = k C q 2 e r 2 Δ = 8 . 99 × 10 9 N · m 2 / C 2 × ( 1 . 60218 × 10 19 C ) 2 (1 . 16064 × 10 6 m) 2 (9 . 36 × 10 9 m) = 1 . 83025 × 10 8 N / m . 004 (part 1 of 2) 10.0 points Three charges are arranged in the ( x, y ) plane (as shown in the figure below, where the scale is in meters).
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hw1 - wilson(kw22588 HW1 Mackie(20224 This print-out should...

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