QTest #4-solutions

# QTest #4-solutions - 1 1067.0 2 781.0 3 946.0 4 715.0 5...

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Version 012/AAADA – QTest #4 – Antoniewicz – (57420) 1 This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Consider a conical pendulum, where a string with length is attached to a mass m . The angle between the string and the ver- tical is θ . The orbit is in the horizontal plane with radius r and tangential velocity vV . V r g m θ What is the Free body diagram For mass m ? The acceleration oF gravity is 9 . 8 m / s 2 . 1. θ correct 2. θ 3. θ 4. θ 5. θ Explanation: There are only two Forces exerted on the ball, the gravity and the tension on the string. T m g θ 002 (part 1 of 2) 10.0 points An elevator accelerates upward at 1 . 2 m / s 2 . The acceleration oF gravity is 9 . 8 m / s 2 . What is the upward Force exerted by the ±oor oF the elevator on a(n) 99 kg passenger?

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Unformatted text preview: 1. 1067.0 2. 781.0 3. 946.0 4. 715.0 5. 1056.0 6. 704.0 7. 957.0 8. 561.0 9. 748.0 10. 1089.0 Correct answer: 1089 N. Explanation: Version 012/AAADA – QTest #4 – Antoniewicz – (57420) 2 N a mg When the elevator is accelerating upward, F net = m a = N -m g N = m a + m g. 003 (part 2 of 2) 10.0 points If the same elevator accelerates downwards with an acceleration of 1 . 2 m / s 2 , what is the upward force exerted by the elevator Foor on the passenger? 1. 438.6 2. 688.0 3. 731.0 4. 627.8 5. 739.6 6. 825.6 7. 713.8 8. 851.4 9. 808.4 10. 447.2 Correct answer: 851 . 4 N. Explanation: N a mg When the elevator is accelerating down-ward, F net = m a = m g-N 2 N 2 = m g-m a....
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QTest #4-solutions - 1 1067.0 2 781.0 3 946.0 4 715.0 5...

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