Exam 4 - Morby Grant – Quiz 4 – Due May 3 2006 11:00 pm...

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Unformatted text preview: Morby, Grant – Quiz 4 – Due: May 3 2006, 11:00 pm – Inst: Drummond 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 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 5 . 3 h, has a mass of 1 . 9 × 10 27 kg and a radius of 6 . 99 × 10 7 m. Given that G = 6 . 67 × 10- 11 Nm 2 / kg 2 , calculate how far above Jupiter’s surface the satellite must be. Correct answer: 3 . 54316 × 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 = ‰ G M T 2 4 π 2 ¾ 1 3 = { 6 . 67 × 10- 11 Nm 2 / kg 2 } 1 3 × ‰ (1 . 9 × 10 27 kg)(19080 s) 2 4(3 . 1415926536 ) 2 ¾ 1 3 = 1 . 05332 × 10 8 m . Now, the altitude h of the satellite (measured from the surface of Jupiter) is h = r- R = (1 . 05332 × 10 8 m)- (6 . 99 × 10 7 m) = 3 . 54316 × 10 7 m . 002 (part 1 of 1) 10 points It is found that a 5 m segment of a long string contains 3 complete wavelengths and has a mass of 190 g. At one point it is vibrating sinusoidally with a frequency of 53 Hz and a peak to valley displacement of 0 . 74 m. What power is being transmitted along the string? Correct answer: 25479 . 6 W. Explanation: since k ≡ 2 π λ , ω ≡ 2 π f , v ≡ ω k , P = 1 2 μω 2 A 2 v = 1 2 µ M L ¶ ω 2 A 2 ‡ ω k · = 25479 . 6 W . 003 (part 1 of 1) 10 points At t = 0, a transverse wave pulse in a wire is described by the function y = 5 e- x 2 / 2 , where x and t are given in SI units. Which of the following functions describes this wave if it is traveling in the negative x direction with a speed of 4 . 5 m / s . 1. y = 5 e- x 2 / 2 2. y = 5 e- ( x +4 . 5 t ) 2 / 2 correct 3. y = 5 e- ( x 2 +4 . 5 t ) / 2 4. y = 5 e- ( x 2- 4 . 5 t ) / 2 5. y = 5 e- x 2 / 2 + 4 . 5 t 6. y = 5 e- ( x- 4 . 5 t ) 2 / 2 7. y = 5 e- x 2 / 2- 4 . 5 t Explanation: For the wave traveling in the negative di- rection with the speed of 4 . 5 m / s , we need to replace x with x + 4 . 5 t which gives y = 5 e- ( x +4 . 5 t ) 2 / 2 . 004 (part 1 of 1) 10 points The length of a hollow pipe is 43 . 2 cm. There is a standing wave in the pipe with a wavelength of 19 . 2 cm. Morby, Grant – Quiz 4 – Due: May 3 2006, 11:00 pm – Inst: Drummond 2 Which figure schematically represents this standing wave? 1. 2. 3. 4. 5. 6. correct 7. 8. 9. 10. Explanation: Let : N = number of nodes λ = 19 . 2 cm , and ‘ = 43 . 2 cm ....
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Exam 4 - Morby Grant – Quiz 4 – Due May 3 2006 11:00 pm...

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