HW11 - boland(kbb544 – HW11 – mackie –(10611 1 This print-out should have 12 questions Multiple-choice questions may continue on the next

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Unformatted text preview: boland (kbb544) – HW11 – mackie – (10611) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. NOTE THE SAT DEADLINE 001 10.0 points Three masses are arranged in the ( x,y ) plane as shown. 8 kg 7 kg 6 kg y (m) 1 3 5 7 9 x (m) 1 3 5 7 9 What is the magnitude of the resulting force on the 8 kg mass at the origin? The value of the universal gravitational constant is 6 . 6726 × 10- 11 N · m 2 / kg 2 . Correct answer: 8 . 04335 × 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 ) = (7 m , 0 m) , and m b = 6 kg , ( x b ,y b ) = (0 m , 2 m) . Applying Newton’s universal gravitational law for m o and m a , 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) (7 m) 2 + (0 m) 2 = 7 . 62583 × 10- 11 N acts along the x axis. Applying the gravitational 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) (6 kg) (0 m) 2 + (2 m) 2 = 8 . 00712 × 10- 10 N acts along the y axis. The magnitude of the resultant force is F = radicalBig F 2 ao + F 2 bo = bracketleftbig (7 . 62583 × 10- 11 N) 2 + (8 . 00712 × 10- 10 N) 2 bracketrightbig 1 / 2 = 8 . 04335 × 10- 10 N . 002 10.0 points The planet Krypton has a mass of 5 . 6 × 10 23 kg and radius of 3 . 7 × 10 6 m. What is the acceleration of an object in free fall near the surface of Krypton? The gravita- tional constant is 6 . 6726 × 10- 11 N · m 2 / kg 2 . Correct answer: 2 . 72948 m / s 2 . Explanation: Let : M = 5 . 6 × 10 23 kg , R = 3 . 7 × 10 6 m , and G = 6 . 6726 × 10- 11 N · m 2 / kg 2 . Near the surface of Krypton, the gravita- tion force on an object of mass m is F = G M m R 2 , so the acceleration a of a free-fall object is a = g Krypton = F m = G M R 2 = (6 . 6726 × 10- 11 N · m 2 / kg 2 ) × 5 . 6 × 10 23 kg (3 . 7 × 10 6 m) 2 = 2 . 72948 m / s 2 . boland (kbb544) – HW11 – mackie – (10611) 2...
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This note was uploaded on 02/20/2011 for the course PHYS 1062 taught by Professor Mackie during the Spring '11 term at Temple.

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HW11 - boland(kbb544 – HW11 – mackie –(10611 1 This print-out should have 12 questions Multiple-choice questions may continue on the next

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