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Old_Midterm-2C

Old_Midterm-2C - quiz 02 BROWNING ADAM Due 10:00 pm...

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quiz 02 – BROWNING, ADAM – Due: Oct 17 2007, 10:00 pm 1 Question 1, chap 6, sect 1. part 1 of 1 10 points A box weighing 710 N is pushed along a horizontal floor at constant velocity with a force of 260 N parallel to the floor. What is the coefficient of kinetic friction between the box and the floor? 1. 0 . 34375 2. 0 . 354839 3. 0 . 366197 correct 4. 0 . 378378 5. 0 . 390625 6. 0 . 402778 7. 0 . 415385 8. 0 . 428571 9. 0 . 442623 10. 0 . 457627 Explanation: Let : W = 710 N and F app = 260 N . vector F app vector f k vector N m vector W Because the box is moving with constant velocity, its acceleration is zero and the net foce acting on it is zero. Applying summationdisplay F y = 0 to the box, N - W = 0 N = W = 710 N . Applying summationdisplay F x = 0 to the box, F app - f k = 0 f k = F app μ k N = F app μ k = F app N = 260 N 710 N = 0 . 366197 . Question 2, chap 7, sect 1. part 1 of 1 10 points Starting from rest at a height equal to the radius of the circular track, a block of mass 23 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient μ ). The radius of the track is 25 m. The acceleration of gravity is 9 . 8 m / s 2 . 25 m 23 kg θ If the kinetic energy of the block at the bottom of the track is 2100 J, what is the work done against friction? 1. 2641 . 6 J 2. 2738 J 3. 2827 . 6 J 4. 2915 . 2 J 5. 3006 . 4 J 6. 3105 J 7. 3214 J 8. 3320 . 8 J 9. 3423 . 8 J 10. 3535 J correct Explanation: W = W f + K W f = m g R - K = (23 kg) (9 . 8 m / s 2 ) (25 m) - (2100 J) = 3535 J . Question 3, chap 8, sect 5. part 1 of 1 10 points A rain cloud contains 7.74 × 10 7 kg of water vapor. The acceleration of gravity is 9 . 81 m / s 2 . How long would it take for a 3.59 kW pump to raise the same amount of water to the cloud’s altitude of 2.45 km?

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quiz 02 – BROWNING, ADAM – Due: Oct 17 2007, 10:00 pm 2 1. 4 . 85746 × 10 8 s 2. 5 . 01948 × 10 8 s 3. 5 . 18181 × 10 8 s correct 4. 5 . 35718 × 10 8 s 5. 5 . 53175 × 10 8 s 6. 5 . 76078 × 10 8 s 7. 5 . 95765 × 10 8 s 8. 6 . 17526 × 10 8 s 9. 6 . 37904 × 10 8 s 10. 6 . 60288 × 10 8 s Explanation: Basic Concepts: W = P Δ t W = Fd cos θ = Fd = mgd since θ = 0 cos θ = 1. Given: m = 7 . 74 × 10 7 kg P = 3 . 59 kW d = 2 . 45 km g = 9 . 81 m / s 2 Solution: Δ t = W P = mgd P = (7 . 74 × 10 7 kg)(9 . 81 m / s 2 )(2450 m) 3590 W = 5 . 18181 × 10 8 s Question 4, chap 5, sect 6. part 1 of 1 10 points Consider the following system of two masses and two pulleysl The pulleys that are massless and frictionless, and the mass on the left accelerates upward while the mass on the right accelerates downward. The acceleration of gravity is 9 . 8 m / s 2 . 6 kg 34 kg a Find the acceleration of the mass on the left. 1. 6 . 88649 m / s 2 2. 7 . 12727 m / s 2 3. 7 . 43448 m / s 2 correct 4. 7 . 66957 m / s 2 5. 8 . 00563 m / s 2 6. 8 . 29231 m / s 2 7. 8 . 575 m / s 2 8. 8 . 90909 m / s 2 9. 9 . 22353 m / s 2 10. 9 . 52 m / s 2 Explanation: Let : m 1 = 6 kg m 2 = 34 kg and g = 9 . 8 m / s 2 . m 1 m 2 1 2 T T T a Let a denote the upward acceleration of the m 1 mass on the left and let T be the tension of the long string attached to the m 1 . To understand the motion of the right mass m 2 , consider the free-body diagram for the right pulley: The pulley is massless, so regardless of its acceleration all forces acting on the pulley must balance; the tension of the short string connecting the right pulley to the m 2 mass must be 2 T rather than T . At the same time, the unstretchability of the long string requires the pulley — and hence the m 2 mass
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Old_Midterm-2C - quiz 02 BROWNING ADAM Due 10:00 pm...

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