# exam 4 - Version 176 – FOURTH EXAM – Kleinman –(58225...

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Unformatted text preview: Version 176 – FOURTH EXAM – Kleinman – (58225) 1 This print-out should have 15 questions. Multiple-choice 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 string’s 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 , ℓ = 6 . 25 m , v s = 343 m / s , and μ = 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 ℓ radicalBigg T μ By the assumption of the problem, we have f string 1 = f well 1 , so 1 2 ℓ radicalBigg T μ = v s 4 d , from which we immediately find T = μ parenleftbigg ℓ 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 Boltzmann’s 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. 1.43125e-20 2. 9.64648e-21 3. 1.59254e-20 4. 8.57717e-21 5. 5.21467e-21 6. 2.34843e-20 7. 1.13251e-20 8. 8.9024e-21 9. 6.2535e-21 10. 1.19448e-20 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 ....
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## This note was uploaded on 05/07/2009 for the course PHY 58225 taught by Professor Klienman during the Spring '09 term at University of Texas.

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exam 4 - Version 176 – FOURTH EXAM – Kleinman –(58225...

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