HW 17 - johnson (mjj622) HW 17 Coker (56625) 1 This...

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Unformatted text preview: johnson (mjj622) HW 17 Coker (56625) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. This is the last chapter of material to be covered on Quiz 4. Remember that no Quiz 4 material will appear on the final to be given on December 11. 001 (part 1 of 2) 10.0 points A 0 . 8 kg physics book with dimensions of 28 . 4 cm by 19 . 7 cm is on a table. What force does the book apply to the table? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 7 . 84 N. Explanation: Let : m = 0 . 8 kg and g = 9 . 8 m / s 2 . The force is F = mg = (0 . 8 kg) (9 . 8 m / s 2 ) = 7 . 84 N . 002 (part 2 of 2) 10.0 points What pressure does the book apply? Correct answer: 140 . 13 Pa. Explanation: Let : = 28 . 4 cm and w = 19 . 7 cm . The pressure is P = F A = F w = 7 . 84 N (28 . 4 cm) (19 . 7 cm) parenleftbigg 100 cm 1 m parenrightbigg 2 = 140 . 13 Pa . 003 10.0 points The air pressure above the liquid in figure is 1 . 21 atm . The depth of the air bubble in the liquid is 41 . 6 cm and the liquids density is 832 kg / m 3 . 41 . 6 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 . 25965 10 5 Pa. Explanation: Let : = 832 kg / m 3 , P = 1 . 21 atm , g = 9 . 8 m / s 2 , and h = 41 . 6 cm = 0 . 416 m . P = P + g h = (1 . 21 atm) 1 . 013 10 5 Pa atm + (832 kg / m 3 ) (9 . 8 m / s 2 ) (0 . 416 m) = 1 . 25965 10 5 Pa . 004 10.0 points A heavy liquid with a density 13 g / cm 3 is poured into a U-tube as shown in the left- hand figure below. The left-hand arm of the tube has a cross-sectional area of 13 cm 2 , and the right-hand arm has a cross-sectional area of 3 . 95 cm 2 . A quantity of 107 g of a light liquid with a density 1 . 7 g / cm 3 is then poured into the right-hand arm as shown in the right- hand figure below. johnson (mjj622) HW 17 Coker (56625) 2 h 1 h 2 13 cm 2 3 . 95 cm 2 heavy liquid 13 g / cm 3 L 13 cm 2 3 . 95 cm 2 light liquid 1 . 7 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 . 485591 cm. Explanation: Let : m = 107 g , A 1 = 13 cm 2 , A 2 = 3 . 95 cm 2 , = 1 . 7 g / cm 3 , and h = 13 g / cm 3 . Using the definition of density = m V = m A 2 L L = m A 2 = 107 g (3 . 95 cm 2 ) (1 . 7 g / cm 3 ) = 15 . 9345 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 ....
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HW 17 - johnson (mjj622) HW 17 Coker (56625) 1 This...

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