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Unformatted text preview: Version 030 – Quiz3 – gentle – (56245) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The planet Krypton has a mass of 7 . 8 × 10 23 kg and radius of 2 . 1 × 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 . 1. 5.78746 2. 0.955711 3. 8.80412 4. 11.8019 5. 9.78832 6. 3.26563 7. 4.8486 8. 1.26867 9. 4.28863 10. 2.38165 Correct answer: 11 . 8019 m / s 2 . Explanation: Let : M = 7 . 8 × 10 23 kg , R = 2 . 1 × 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 freefall object is a = g Krypton = F m = G M R 2 = (6 . 6726 × 10 − 11 N · m 2 / kg 2 ) × 7 . 8 × 10 23 kg (2 . 1 × 10 6 m) 2 = 11 . 8019 m / s 2 . 002 10.0 points Three particles are located in a coordinate system as shown in the figure below. The acceleration of gravity is 9 . 8 m / s 2 . The mass 7 kg is located at the origin, the mass 7 kg is located on the x coordinate, and the mass 6 kg is located on the y coordinate. x y 6 kg 7 kg 7 kg 9 m 4 m Find the distance of the center of gravity from the origin. 1. 3.37083 2. 4.45982 3. 4.16854 4. 2.88111 5. 3.29983 6. 3.92214 7. 4.96433 8. 2.58466 9. 4.38292 10. 3.51141 Correct answer: 3 . 37083 m. Explanation: The center of gravity of the x and y compo nents from the origin respectively are x c = m 0 + m 1 x 1 m + m 1 + m 2 = (7 kg) (9 m) (7 kg) + (7 kg) + (6 kg) = 3 . 15 m and y c = m 0 + m 2 y 2 m + m 1 + m 2 = (6 kg) (4 m) (7 kg) + (7 kg) + (6 kg) = 1 . 2 m . Then the distance of the center of gravity from the origin is d = radicalBig x 2 c + y 2 c = radicalBig (3 . 15 m) 2 + (1 . 2 m) 2 Version 030 – Quiz3 – gentle – (56245) 2 = 3 . 37083 m . 003 10.0 points An object of mass m is moving with speed v to the right on a horizontal frictionless surface, as shown, when it explodes into two pieces. Subsequently, one piece of mass 32 37 m moves with a speed v 32 / 37 = v 4 to the left. m v before 32 37 m v 4 5 37 m v 5 / 37 after The speed v f = bardbl vectorv 5 / 37 bardbl of the other piece the object is 1. v f = 3 v . 2. v f = 6 v . 3. v f = 4 v . 4. v f = 8 v . 5. v f = 5 v . 6. v f = 10 v . 7. v f = 7 v . 8. v f = 9 v . correct 9. v f = 2 v . 10. None of these is correct....
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This note was uploaded on 12/05/2011 for the course PHY 301 taught by Professor Swinney during the Fall '07 term at University of Texas.
 Fall '07
 Swinney
 mechanics, Mass

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