HW6-solutions

# HW6-solutions - juhl(jj23837 HW6 ditmire(58216 This...

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juhl (jj23837) – HW6 – ditmire – (58216) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Three masses are arranged in the ( x, y ) plane as shown. y [m] - 5 - 3 - 1 1 3 5 x [m] - 5 - 3 - 1 1 3 5 8 kg 7 kg 1 kg What is the magnitude of the result- ing force on the 8 kg mass at the ori- gin? The universal gravitational constant is 6 . 6726 × 10 - 11 N · m 2 / kg 2 . Correct answer: 1 . 40866 × 10 - 10 N. Explanation: Let: m o = 8 kg , ( x o , y o ) = (0 m , 0 m) , m a = 7 kg , ( x a , y a ) = (3 m , - 3 m) , and m b = 1 kg , ( x b , y b ) = ( - 2 m , 2 m) . Applying Newton’s universal gravitational law for m o and, F ao = G m o m a ( x a - x o ) 2 + ( y a - y o ) 2 = (6 . 6726 × 10 - 11 N · m 2 / kg 2 ) × (8 kg) (7 kg) (3 m) 2 + ( - 3 m) 2 = 2 . 07592 × 10 - 10 N , where tan θ a = y a x a θ a = arctan y a x a = arctan - 3 m 3 m = 315 , so the components of this force are F a x = F a cos θ a = ( 2 . 07592 × 10 - 10 N ) cos 315 = 1 . 4679 × 10 - 10 N and F a y = F a sin θ a = ( 2 . 07592 × 10 - 10 N ) sin 315 = - 1 . 4679 × 10 - 10 N . Applying Newton’s law for m o and m b , F bo = G m o m b ( x b - x o ) 2 + ( y b - y o ) 2 = (6 . 6726 × 10 - 11 N · m 2 / kg 2 ) × (8 kg) (1 kg) ( - 2 m) 2 + (2 m) 2 = 6 . 6726 × 10 - 11 N , where θ b = arctan y b x b = arctan 2 m - 2 m = 135 , so the components of this force are F b x = F b cos θ b = ( 6 . 6726 × 10 - 11 N ) cos 135 = - 4 . 71824 × 10 - 11 N and F b y = F b sin θ b = ( 6 . 6726 × 10 - 11 N ) sin 135 = 4 . 71824 × 10 - 11 N .

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juhl (jj23837) – HW6 – ditmire – (58216) 2 The magnitude of the resultant force is F = F 2 x + F 2 y = ( F a x + F b x ) 2 + ( F a y + F b y ) 2 = ( 1 . 4679 × 10 - 10 N + - 4 . 71824 × 10 - 11 N ) 2 + ( - 1 . 4679 × 10 - 10 N 4 . 71824 × 10 - 11 N ) 2 1 / 2 = 1 . 40866 × 10 - 10 N . 002 (part 2 of 2) 10.0 points Select the figure showing the direction of the resultant force on the 8 kg mass at the origin. 1. 195 2. 15 3. 255 4. 135 5. 315 correct 6. 75 7. None of these Explanation: The direction θ as measured in a counter- clockwise direction from the positive x axis is θ = arctan F y F x = arctan - 9 . 96073 × 10 - 11 N 9 . 96073 × 10 - 11 N = 315 . 003 10.0 points If the gravitational force between the elec- tron (of mass 9 . 11 × 10 - 31 kg) and the proton (of mass 1 . 67 × 10 - 27 kg) in a hydrogen atom is 5 × 10 - 48 N, how far apart are the two par- ticles? The universal gravitational constant is 6 . 673 × 10 - 11 N · m 2 / kg 2 .
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