HW6 Solution - salazar(ggs332 – HW06 – TSOI –(58160 1...

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Unformatted text preview: salazar (ggs332) – HW06 – TSOI – (58160) 1 This print-out should have 24 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 4 kg masses are located at points in the xy plane as shown. 50 cm 58 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? The universal gravita- tional constant is 6 . 6726 × 10 − 11 N · m 2 / kg 2 . Correct answer: 5 . 32061 × 10 − 9 N. Explanation: Let : m = 4 kg , x = 50 cm = 0 . 5 m , y = 58 cm = 0 . 58 m , and G = 6 . 6726 × 10 − 11 N · m 2 / kg 2 . The force from the mass on the right points in the x direction and has magnitude F 1 = G mm x 2 = Gm 2 x 2 = (6 . 6726 × 10 − 11 N · m 2 / kg 2 ) (4 kg) 2 (0 . 5 m) 2 = 4 . 27046 × 10 − 9 N . The other force points in the y direction and has magnitude F 2 = (6 . 6726 × 10 − 11 N · m 2 / kg 2 ) (4 kg) 2 (0 . 58 m) 2 = 3 . 17365 × 10 − 9 N . F 2 F 1 F θ The magnitude of the resultant force is F = radicalBig F 2 1 + F 2 2 = bracketleftBig (4 . 27046 × 10 − 9 N) 2 + (3 . 17365 × 10 − 9 N) 2 bracketrightBig 1 / 2 = 5 . 32061 × 10 − 9 N . 002 (part 2 of 2) 10.0 points At what angle from the positive x-axis will the resultant force point? Let counterclockwise be positive, within the limits − 180 ◦ to 180 ◦ . Correct answer: 36 . 6184 ◦ . Explanation: The angle θ shown is θ = arctan parenleftbigg f 2 f 1 parenrightbigg = arctan parenleftbigg 3 . 17365 × 10 − 9 N 4 . 27046 × 10 − 9 N parenrightbigg = 36 . 6184 ◦ . 003 (part 1 of 3) 10.0 points Given: G = 6 . 67259 × 10 − 11 N m 2 / kg 2 A uniform solid sphere of mass m 2 = 108 kg and radius R 2 = 1 . 26 m is inside and concen- tric with a spherical shell of mass m 1 = 233 kg and radius R 1 = 2 . 39 m (see the figure). salazar (ggs332) – HW06 – TSOI – (58160) 2 R R a b c m m 1 1 2 2 Find the magnitude of the gravitational force exerted by the sphere and spherical shell on a particle of mass 1 kg located at a dis- tance 0 . 58 m from the center of the sphere and spherical shell. Correct answer: 2 . 08946 × 10 − 9 N. Explanation: In this case the distance a = 0 . 58 m from the particle of mass m = 1 kg to the center of the sphere (whose mass is m 2 = 108 kg) is smaller than the radius R 2 = 1 . 26 m of the sphere. Thus, the particle is inside the sphere and the gravitational force F a exerted on it increases linearly with the distance from the particle to the center of the sphere F a = G m 2 m R 3 2 a = (108 kg) (1 kg) (1 . 26 m) 3 (0 . 58 m) × 6 . 67259 × 10 − 11 N m 2 / kg 2 = 2 . 08946 × 10 − 9 N . 004 (part 2 of 3) 10.0 points Find the magnitude of the gravitational force exerted by the sphere and spherical shell on a particle of mass 1 kg located at a distance 1 . 86 m from the center of the sphere and spherical shell....
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This note was uploaded on 11/03/2010 for the course PHYSICS 58160 taught by Professor Tsoi during the Spring '10 term at University of Texas.

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HW6 Solution - salazar(ggs332 – HW06 – TSOI –(58160 1...

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