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 176 FOURTH EXAM Kleinman (58225) 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 string is tied across the opening of a deep well, at a height 5 m above the water level. The string has length 6 . 25 m and linear den sity 1 . 26 g / m; any time there is a wind, the string vibrates like mad. The speed of sound in air is 343 m / s. Assuming that the resonance causing this vibration involves the lowest harmonics of both the string and the well, calculate the strings tension. 1. 57.9054 2. 108.889 3. 205.285 4. 72.8016 5. 100.944 6. 114.764 7. 125.855 8. 82.8612 9. 393.088 10. 153.146 Correct answer: 57 . 9054 N. Explanation: Let : d = 5 m , = 6 . 25 m , v s = 343 m / s , and = 1 . 26 g / m . When the wind blows, it makes noises and the well amplifies the sound frequencies that resonate with its air column. If the string resonates with one of those frequencies, this sound sets up strong vibrations of the string. The air column in the well is open at one end and closed at the other, so its resonant frequencies are odd multiples of the lowest harmonic f well 1 = v s 4 d where v s is the speed of sound in the air. The resonant frequencies of the string are integral multiples of its lowest harmonic f string 1 = 1 2 radicalBigg T By the assumption of the problem, we have f string 1 = f well 1 , so 1 2 radicalBigg T = v s 4 d , from which we immediately find T = parenleftbigg v s 2 d parenrightbigg 2 = (1 . 26 g / m) bracketleftbigg (6 . 25 m) (343 m / s) 2 (5 m) bracketrightbigg 2 = 57 . 9054 N . 002 10.0 points A vessel with a capacity of 5 . 9 L contains . 159 mol of an ideal gas at 1 . 7 atm. The Boltzmanns constant is 1 . 38066 10 23 J / K, and the universal gas constant is 8 . 31 J / K mol. What is the average translational kinetic energy of a single molecule? 1. 1.43125e20 2. 9.64648e21 3. 1.59254e20 4. 8.57717e21 5. 5.21467e21 6. 2.34843e20 7. 1.13251e20 8. 8.9024e21 9. 6.2535e21 10. 1.19448e20 Correct answer: 1 . 59254 10 20 J. Explanation: Let : P = 1 . 7 atm = 1 . 7221 10 5 Pa , V = 5 . 9 L = 0 . 0059 m 3 , n = 0 . 159 mol , R = 8 . 31 J / K mol and k B = 1 . 38066 10 23 J / K . Version 176 FOURTH EXAM Kleinman (58225) 2 From the equation of state for an ideal gas we obtain T = P V n R = (1 . 7221 10 5 Pa) (0 . 0059 m 3 ) (0 . 159 mol) (8 . 31 J / K mol) = 768 . 975 K According to the equipartition theorem, the average translational energy of a single molecule equals E t = 1 2 mv 2 = 3 2 k B T = 3 2 (1 . 38066 10 23 J / K) (768 . 975 K) = 1 . 59254 10 20 J 003 10.0 points A horizontal pipe of diameter 0 . 77 m has a smooth constriction to a section of diameter . 462 m . The density of oil flowing in the pipe is 821 kg / m 3 ....
View
Full
Document
 Spring '09
 KLIENMAN

Click to edit the document details