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Unformatted text preview: Johnson, Matthew – Quiz 4 – Due: May 4 2005, 10:00 pm – Inst: Kleinman 1 This printout should have 19 questions. Multiplechoice 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 heavy liquid with a density 13 g / cm 3 is poured into a Utube as shown in the left hand figure below. The lefthand arm of the tube has a crosssectional area of 9 . 54 cm 2 , and the righthand arm has a crosssectional area of 6 . 34 cm 2 . A quantity of 109 g of a light liquid with a density 1 . 4 g / cm 3 is then poured into the righthand arm as shown in the righthand figure below. h 1 h 2 9 . 54 cm 2 6 . 34 cm 2 heavy liquid 13 g / cm 3 L 9 . 54 cm 2 6 . 34 cm 2 light liquid 1 . 4 g / cm 3 If the density of the heavy liquid is 13 g / cm 3 , by what height h 1 does the heavy liquid rise in the left arm? Correct answer: 0 . 527998 cm. Explanation: Let : m ‘ = 109 g , A 1 = 9 . 54 cm 2 , A 2 = 6 . 34 cm 2 , ρ ‘ = 1 . 4 g / cm 3 , and ρ h = 13 g / cm 3 . Using the definition of density ρ ‘ = m ‘ V ‘ = m ‘ A 2 L L = m ‘ A 2 ρ ‘ = 12 . 2803 cm . After the light liquid has been added to the right side of the tube, a volume A 2 h 2 of heavy liquid is displaced to the left side, raising the heavy liquid on the left side by a height of h 1 with a displaced volume of A 1 h 1 . Since the volume of heavy liquid is not changed, we have A 1 h 1 = A 2 h 2 . At the level of the heavylight liquid interface in the right side, the absolute pressure is P = P atm + ρ ‘ g L, and at the same level in the left tube, P = P atm + ρ h g [ h 1 + h 2 ] . Equating these two values, we obtain ρ ‘ g L = ρ h g [ h 1 + h 2 ] = ρ h g • h 1 + A 1 A 2 h 1 ‚ = ρ h g h 1 • 1 + A 1 A 2 ‚ . Solving for h 1 , we have h 1 = ρ ‘ L ρ h • 1 + A 1 A 2 ‚ = (1 . 4 g / cm 3 )(12 . 2803 cm) (13 g / cm 3 ) • 1 + (9 . 54 cm 2 ) (6 . 34 cm 2 ) ‚ = . 527998 cm . 002 (part 1 of 2) 0 points Given: The speed of sound in air is 343 m / s. The length of a string is 408 cm. The string is held fixed at each end. The string vibrates in six sections; i.e. , the string has six antinodes. Johnson, Matthew – Quiz 4 – Due: May 4 2005, 10:00 pm – Inst: Kleinman 2 Determine the frequency of vibrations in the string. Correct answer: 252 . 206 Hz. Explanation: Let : N = 6 = number of antinodes v = 343 m / s , and ‘ = 408 cm . When a string is held fixed at each end, the wavelength is λ = 2 ‘ N , where N = 1 , 2 , 3 , ··· = 2(408 cm) (6) = (136 cm)(0 . 01 m / cm) = 1 . 36 m . f = v λ = (343 m / s) (1 . 36 m) = 252 . 206 Hz . 003 (part 2 of 2) 0 points What is the fundamental frequency; i.e. , the lowest frequency the string can sustain?...
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 Spring '08
 Turner
 Wavelength, Correct Answer, Orders of magnitude

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