HW20.sol - homework 20 RAHMAN TARIQUE Due 3:00 am Question...

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homework 20 – RAHMAN, TARIQUE – Due: Apr 18 2008, 3:00 am 1 Question 1, chap 14, sect 4. part 1 of 1 10 points A steel cable with a cross-sectional area of 1 . 5 cm 2 has a mass of 2 . 95 kg / m . Its Young’s modulus is 1 . 27 × 10 11 N / m 2 . The acceleration of gravity is 9 . 8 m / s 2 . If 762 m of the cable is hung over a vertical cliF, how much does the cable stretch under its own weight? Correct answer: 44 . 0588 cm (tolerance ± 1 %). Explanation: Basic Concepts: Δ L L = F Y A Let : A = 1 . 5 cm 2 = 0 . 00015 m 2 , Y = 1 . 27 × 10 11 N / m 2 , L = 762 m , λ = 2 . 95 kg / m , The tension in this cable is not uniform, so this becomes a fairly di±cult problem. ²rom the de³nition of Young’s modulus, we have Δ L L = F Y A . F = mg = λy g weight of the cable at posi- tion y . Δ y = i L 0 0 Δ L L dy = λg Y A i L 0 0 y dy = 1 2 λg L 2 0 Y A = 1 2 (2 . 95 kg / m) (9 . 8 m / s 2 ) (762 m) 2 (1 . 27 × 10 11 N / m 2 ) (0 . 00015 m 2 ) = 0 . 440588 m = 44 . 0588 cm . Question 2, chap 14, sect 4. part 1 of 1 10 points Use bulk modulus B = 2 . 1 × 10 9 Pa . What increase of pressure is required to change the volume of a sample of water by 0 . 71 %? Correct answer: 1 . 491 × 10 7 Pa (tolerance ± 1 %). Explanation: ²rom the de³nition of bulk modulus, the change in pressure is Δ P = - Δ V V B. ²or our situation, Δ V V = 0 . 0071 . Then, Δ P = - (0 . 0071) B = - (0 . 0071) 2 . 1 × 10 9 Pa = 1 . 491 × 10 7 Pa . Question 3, chap 15, sect 1. part 1 of 1 10 points When an object oscillating in simple har- monic motion is at its maximum displacement from the equilibrium position, which of the following is true of the values of its speed and the magnitude of the restoring force? Magnitude of Speed Restoring Force 1. Zero Maximum correct 2. Maximum Zero 3. 1 2 maximum 1 2 maximum 4. Zero Zero 5. Maximum 1 2 maximum Explanation: The maximum displacement occurs at the turning points, which are the points where the velocity or speed is zero.
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homework 20 – RAHMAN, TARIQUE – Due: Apr 18 2008, 3:00 am 2 The magnitude of the restoring force is given by Hooke’s law F = - k x, where k is the spring constant and x is the displacement. Since x is a maximum, F is a maximum. From a di±erent perspective, the displace- ment from the equilibrium position can be written as y = A sin θ , where θ is the phase of the oscillation. When the object is at its maximum dis- placement, θ = π 2 . The speed is then v = ω A cos θ = ω A cos π 2 = 0 and the restoring force is F = mA ω 2 sin θ = mA ω 2 π 2 = mA ω 2 , at its maximum value. Question 4, chap 15, sect 1. part 1 of 1 10 points A mass attached to a spring oscillates back and forth as indicated in the position vs. time plot below.
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HW20.sol - homework 20 RAHMAN TARIQUE Due 3:00 am Question...

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