hw2_solns

# Hw2_solns - Answer Key – Homework 2 – David McIntyre 1 This print-out should have 14 questions check that it is complete Multiple-choice

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Unformatted text preview: Answer, Key – Homework 2 – David McIntyre 1 This print-out should have 14 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. Chapter 15 problems. 001 (part 1 of 2) 5 points A simple U-tube that is open at both ends is partially filled with a liquid of density 553 kg / m 3 . Water is then poured into one arm of the tube, forming a column 11 cm in height, as shown in the following diagram. The density of water is 1000 kg / m 3 . water liquid h l 1 What is the difference, h 1 , in the heights of the two liquid surfaces? Correct answer: 8 . 8915 cm. Explanation: Basic Concepts: gauge pressure, varia- tion of pressure with depth Because the liquid in the U-tube is static, the pressure exerted by the water column of height l in the right branch of the tube must balance the pressure exerted by the liquid of height h 1 + l in the left branch. Therefore, P + ( l + h 1 ) ρ ‘ g = P + l ρ w g Solving for h 1 , h 1 = l µ ρ w ρ ‘- 1 ¶ = 11 cm µ 1000 kg / m 3 553 kg / m 3- 1 ¶ = 8 . 8915 cm . 002 (part 2 of 2) 5 points A simple U-tube that is open at both ends is partially filled with water. A liquid of density 553 kg / m 3 is then poured into one arm of the tube, forming a column 11 cm in height, as shown in the following diagram. water liquid h l 2 What is the difference, h 2 , in the heights of the two liquid surfaces? Correct answer: 4 . 917 cm. Explanation: Because the liquid in the U-tube is static, the pressure exerted by the water column of height l- h 2 in the left branch of the tube must balance the pressure exerted by the liq- uid of height l poured into the right branch. Therefore, P + ( l- h 2 ) ρ w g = P + l ρ ‘ g Solving for h 2 , h 2 = l µ 1- ρ ‘ ρ w ¶ = 11 cm µ 1- 553 kg / m 3 1000 kg / m 3 ¶ = 4 . 917 cm 003 (part 1 of 1) 0 points Consider an object that floats in water but sinks in oil. When the object floats in a glass of water, half of the object is submerged. We now slowly pour oil into the glass so it com- pletely covers the object. oil Answer, Key – Homework 2 – David McIntyre 2 Note: The second figure does not necessarily show the correct changes, only the experimen- tal setup. Compared to the water level, the object 1. moves up correct 2. moves down 3. stays in the same place Explanation: Consider the point where we have poured in an amount of oil exactly enough to cover the object: We can infer that it does not matter how much more oil we pour into the glass after this point, since it will exert an evenly distributed pressure on both water and object. Before this point, however, the oil added will be on top of the water only. Therefore, the oil will push the water down, while not pushing the object down. The extra pressure from the oil will be transferred to the bottom of the object, so it must push the object up....
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## This note was uploaded on 04/07/2010 for the course PHYS 212 taught by Professor Mu during the Spring '09 term at Oregon State.

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Hw2_solns - Answer Key – Homework 2 – David McIntyre 1 This print-out should have 14 questions check that it is complete Multiple-choice

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