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Quiz 4 - Version 002 Quiz 4 Coker(56625 This print-out...

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Version 002 – Quiz 4 – Coker – (56625) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This quiz covers the simple harmonic oscil- lator, waves, sound, and static and dynamic systems of liquids and gases. None of this ma- terial will be re-covered on the Final Exam. 001 10.0 points A wave on a string is reflected from a fixed end. The reflected wave 1. has a larger speed than the original wave. 2. has a larger amplitude than the original wave. 3. is 180 out of phase with the original wave at the end. correct 4. cannot be transverse. 5. is in phase with the original wave at the end. Explanation: At a fixed end, the incident and reflected wave must always add to zero (since the end is fixed and cannot move), so they must always be 180 out of phase with each other at that end. 002 10.0 points Which Second Law expression describes a simple harmonic oscillator? Take B to be a positive nonzero constant. 1. a x = + B x 2. a x = B x 3. a x = B x 2 4. α = + B θ 5. α = B θ correct 6. a x = + B x 2 7. α = + B parenleftbigg θ 2 2 parenrightbigg 8. α = B sin ( θ ) Explanation: A simple harmonic oscillator must have a Second Law such that the second time deriva- tive of a displacement parameter is equal to a negative constant times the same displace- ment parameter. The only expression with this form is α = B θ . 003 10.0 points Consider a steel ball floating on the surface of mercury in a half-filled container. What happens when the rest of the container is filled with water? Mercury is denser than steel, and both are denser than water. Mer- cury and water do not mix. 1. The ball sinks to the bottom of the con- tainer 2. Mercury floats on top of water, and the ball floats on mercury’s surface 3. The ball floats to the surface of the water.
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Version 002 – Quiz 4 – Coker – (56625) 2 4. The ball remains partially submerged in mercury but not as deeply as before. correct 5. The ball remains partially submerged in mercury to exactly the same depth as before. 6. The ball remains partially submerged in mercury to a greater depth then before. Explanation: Mercury is much denser than the water, so it stays in the bottom half of the container while water fills the top half. The steel ball floats in mercury but sinks in water, so it has to float at the mercury-water interface, partially submerged in mercury and partially in water. The only non-trivial question is whether adding water makes the ball go deeper into mercury or less deep. To answer this question, note that the ball floating at the interface is subject to two buoyant forces: B M = ρ M g V in mercury and B W = ρ W g V in water , so the equilibrium condition is M ball g = B M + B W = ρ M g V in mercury + ρ W g V in water . (1) Although the water is much less dense then mercury, it does provide a bit of buoyant force, so in equilibrium V in mercury [after] < M ball ρ M (2) after adding the water.
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