This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 007 – TEST02 – TSOI – (58160) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A “synchronous” satellite, which always re mains above the same point on a planet’s equator, is put in orbit about a planet similar to Jupiter. This planet rotates once every 8 . 2 h, has a mass of 2 × 10 27 kg and a radius of 6 . 99 × 10 7 m. Given that G = 6 . 67 × 10 − 11 N m 2 / kg 2 , calculate how far above Jupiter’s surface the satellite must be. 1. 88696000.0 2. 71005000.0 3. 73431900.0 4. 48904400.0 5. 37248100.0 6. 55685900.0 7. 62742400.0 8. 81477200.0 9. 65214500.0 10. 43238300.0 Correct answer: 7 . 34319 × 10 7 m. Explanation: Basic Concepts: Solution: According to Kepler’s third law: T 2 = 4 π 2 GM r 3 where r is the radius of the satellite’s orbit. Thus, solving for r r = braceleftbigg G M T 2 4 π 2 bracerightbigg 1 3 = { 6 . 67 × 10 − 11 N m 2 / kg 2 } 1 3 × braceleftbigg (2 × 10 27 kg) (29520 s) 2 4 (3 . 14159) 2 bracerightbigg 1 3 = 1 . 43332 × 10 8 m . Now, the altitude h of the satellite (measured from the surface of Jupiter) is h = r R = (1 . 43332 × 10 8 m) (6 . 99 × 10 7 m) = 7 . 34319 × 10 7 m . 002 10.0 points An applied force varies with position ac cording to F = k 1 x n k 2 , where n = 3, k 1 = 8 . 3 N / m 3 , and k 2 = 55 N. How much work is done by this force on an object that moves from x i = 5 . 12 m to x f = 7 . 52 m? 1. 972.821 2. 537.473 3. 5.07781 4. 7.53789 5. 45.615 6. 641.969 7. 8.86427 8. 35.2888 9. 1211.84 10. 19.0098 Correct answer: 5 . 07781 kJ. Explanation: Basic Concepts: W = integraldisplay vector F · dvectors Solution: The work done by a varying force is W = integraldisplay x 2 x 1 vector F · dvectors. which adds up all the little vector F · vectors parts along the path, taking into account the changing force. Here all the motion is in the ˆ ı direction so ds = dx . W = integraldisplay x f x i F · dx = integraldisplay 7 . 52 m 5 . 12 m bracketleftbig (8 . 3 N / m 3 ) x 3 (55 N) bracketrightbig · dx = bracketleftbigg (8 . 3 N / m 3 ) x 4 4 (55 N) x bracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle 7 . 52 m 5 . 12 m = parenleftbigg 8 . 3 N / m 3 4 parenrightbigg bracketleftbig (7 . 52 m) 4 (5 . 12 m) 4 bracketrightbig (55 N) (7 . 52 m 5 . 12 m) = 5077 . 81 J = 5 . 07781 kJ . Version 007 – TEST02 – TSOI – (58160) 2 003 (part 1 of 2) 10.0 points The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The block of mass m 1 lies on a rough horizontal surface with a constant coefficient of kinetic friction μ . This block is connected to a spring with spring constant k . The second block has a mass m 2 ....
View
Full
Document
This note was uploaded on 05/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

Click to edit the document details