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Unformatted text preview: yoon (jy4326) – HW10 – markert – (58840) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The crosssectional area of the output piston in a hydraulic device is 17 times the input piston’s area. By how much will the device multiply the input force? Correct answer: 17. Explanation: Let : n = 17 . P in = P out F in A in = F out A out = F out n A in F out = n F in = 17 F in . 002 (part 2 of 2) 10.0 points By what factor will the output piston move compared to the distance the input piston is moved? Correct answer: 0 . 0588235. Explanation: F in d in = F out d out = n F in d out d out = 1 n d in = 1 17 d in = 0 . 0588235 d in . 003 10.0 points The air pressure above the liquid in figure is 1 . 02 atm . The depth of the air bubble in the liquid is 40 . 8 cm and the liquid’s density is 753 kg / m 3 . 40 . 8 cm air air Determine the air pressure in the bubble suspended in the liquid. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 06337 × 10 5 Pa. Explanation: Let : ρ = 753 kg / m 3 , P = 1 . 02 atm , g = 9 . 8 m / s 2 , and h = 40 . 8 cm = 0 . 408 m . P = P + ρ g h = (1 . 02 atm) · 1 . 013 × 10 5 Pa atm + (753 kg / m 3 ) (9 . 8 m / s 2 ) (0 . 408 m) = 1 . 06337 × 10 5 Pa . 004 (part 1 of 2) 10.0 points A heavy liquid with a density 11 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 15 . 6 cm 2 , and the righthand arm has a crosssectional area of 6 . 01 cm 2 . A quantity of 93 . 7 g of a light liquid with a density 1 . 2 g / cm 3 is then poured into the righthand arm as shown in the righthand figure below. h 1 h 2 15 . 6 cm 2 6 . 01 cm 2 heavy liquid 11 g / cm 3 L 15 . 6 cm 2 6 . 01 cm 2 light liquid 1 . 2 g / cm 3 yoon (jy4326) – HW10 – markert – (58840) 2 Determine the height L of the light liquid in the column in the right arm of the Utube, as shown in the second figure above. Correct answer: 12 . 9922 cm. Explanation: Let : m ℓ = 93 . 7 g , A 1 = 15 . 6 cm 2 , A 2 = 6 . 01 cm 2 , ρ ℓ = 1 . 2 g / cm 3 , and ρ h = 11 g / cm 3 . Using the definition of density ρ ℓ = m ℓ V ℓ = m ℓ A 2 L L = m ℓ A 2 ρ ℓ = 93 . 7 g (6 . 01 cm 2 ) (1 . 2 g / cm 3 ) = 12 . 9922 cm . 005 (part 2 of 2) 10.0 points If the density of the heavy liquid is 11 g / cm 3 , by what height h 1 does the heavy liquid rise in the left arm? Correct answer: 0 . 394178 cm. Explanation: 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 ....
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This note was uploaded on 01/08/2010 for the course CH 53675 taught by Professor Bocknack during the Fall '09 term at University of Texas.
 Fall '09
 BOCKNACK
 Organic chemistry

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