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HW 17 - johnson(mjj622 HW 17 Coker(56625 This print-out...

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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 = m g = (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 ℓ 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 liquid’s 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 : ρ = 832 kg / m 3 , P 0 = 1 . 21 atm , g = 9 . 8 m / s 2 , and h = 41 . 6 cm = 0 . 416 m . P = P 0 + ρ 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.
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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 , ρ = 1 . 7 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 = 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 . At the level of the heavy-light 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 parenleftbigg h 1 + A 1 A 2 h 1 parenrightbigg = ρ h g h 1 parenleftbigg 1 + A 1 A 2 parenrightbigg .
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