This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 031 – 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 5 . 3 × 10 23 kg and radius of 4 . 3 × 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. 1.91264 2. 46.8736 3. 4.52245 4. 19.8562 5. 14.7767 6. 1.62645 7. 13.4931 8. 2.8262 9. 5.87125 10. 1.26867 Correct answer: 1 . 91264 m / s 2 . Explanation: Let : M = 5 . 3 × 10 23 kg , R = 4 . 3 × 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 ) × 5 . 3 × 10 23 kg (4 . 3 × 10 6 m) 2 = 1 . 91264 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 4 kg is located at the origin, the mass 7 kg is located on the x coordinate, and the mass 9 kg is located on the y coordinate. x y 9 kg 4 kg 7 kg 13 m 5 m Find the distance of the center of gravity from the origin. 1. 5.07592 2. 3.52942 3. 3.22696 4. 3.03718 5. 3.92562 6. 5.23927 7. 3.80173 8. 3.47519 9. 1.73277 10. 6.10574 Correct answer: 5 . 07592 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) (13 m) (4 kg) + (7 kg) + (9 kg) = 4 . 55 m and y c = m 0 + m 2 y 2 m + m 1 + m 2 = (9 kg) (5 m) (4 kg) + (7 kg) + (9 kg) = 2 . 25 m . Then the distance of the center of gravity from the origin is d = radicalBig x 2 c + y 2 c Version 031 – Quiz3 – gentle – (56245) 2 = radicalBig (4 . 55 m) 2 + (2 . 25 m) 2 = 5 . 07592 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 5 6 m moves with a speed v 5 / 6 = v 5 to the left. m v before 5 6 m v 5 1 6 m v 1 / 6 after The speed v f = bardbl vectorv 1 / 6 bardbl of the other piece the object is 1. v f = 5 v . 2. v f = 2 v . 3. v f = 6 v . 4. v f = 10 v . 5. v f = 9 v . 6. v f = 7 v . correct 7. v f = 4 v . 8. v f = 8 v . 9. v f = 3 v . 10. None of these is correct....
View
Full
Document
This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas.
 Fall '10
 MARKERT
 mechanics, Mass

Click to edit the document details