Test #2-solutions

# Test #2-solutions - Version 081/ABBAB Test #2 Antoniewicz...

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Version 081/ABBAB – Test #2 – Antoniewicz – (56445) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 6.0 points Planet X has a mass 6 . 67 times that oF the Earth and a radius 2 . 56 times the radius oF the Earth. What is the ratio oF the acceleration due to gravity on the surFace oF Planet X to the acceleration due to gravity on the surFace oF the Earth? 1. 0.896017 2. 0.38741 3. 0.240725 4. 0.598082 5. 0.173322 6. 0.255921 7. 1.01776 8. 0.144287 9. 0.730147 10. 0.515553 Correct answer: 1 . 01776. Explanation: Let : M X =6 . 67 M E and R X =2 . 56 R E . The acceleration due to gravity is a = GM R 2 M R 2 , so g X g E = R 2 E R X 2 M X M E g X g E = ± R E 2 . 56 R E ² 2 ± 6 . 67 M E M E ² = 1 . 01776 . 002 6.0 points The Following fgure shows a ±erris wheel that rotates 3 times each minute and has a diame- ter oF 21 m. The acceleration oF gravity is 9 . 8m / s 2 . R ω What Force does the seat exert on a 37 kg rider at the lowest point oF the ride? 1. 366.94 2. 400.943 3. 586.103 4. 645.964 5. 593.759 6. 716.886 7. 391.884 8. 820.206 9. 761.85 10. 422.616 Correct answer: 400 . 943 N. Explanation: Let : r = D 2 =10 . 5m , and T = one minute n = (60 s) (3) =20s , The speed oF the wheel is v = 2 π r T = 2 π (10 . 5m) 20 s =3 . 29867 m / s , so the centripetal acceleration is a = v 2 r

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Version 081/ABBAB – Test #2 – Antoniewicz – (56445) 2 = (3 . 29867 m / s) 2 (10 . 5m) =1 . 03631 m / s 2 . The force exerted by the seat balances the gravity and provides the centripetal force, so F l = m [ g + a ] =(37kg)(9 . 8m / s 2 +1 . 03631 m / s 2 ) = 400 . 943 N . 003 6.0 points Aballofmass380ghangsfromaspringwhose sti f ness is 120 N / m. A string is attached to the ball and you pull the string to the right, so that the ball hangs motionless, as shown in the Fgure. In this situation, the spring is stretched, and its length is 16 cm. What would be the relaxed length of the spring, if it were detached from the ball and laid on a table? Use g =9 . 8m / s 2 . 16 cm 8cm 1. 0.118341 2. 0.0999493 3. 0.124166 4. 0.115433 5. 0.0888241 6. 0.0898903 7. 0.102742 8. 0.101 9. 0.110527 10. 0.112141 Correct answer: 0 . 124166 m. Explanation: Let θ be the angle between the spring and the vertical, and let s represent the amount by which the spring has stretched under the given circumstances. Then sin θ = 8cm 16 cm and cos θ = 13 . 8564 cm 16 cm , where 13 . 8564 cm is obtained from the Pythagorean theorem. The vertical compo- nent of the spring tension must null out the ball’s weight, so, using Hooke’s Law, we can write mg = k s s cos θ s = mg k s 16 cm 13 . 8564 cm = (380 g)(9 . 8m / s 2 )(16 cm) (120 N / m)(13 . 8564 cm) =0 . 0358342 m . So the original length of the spring is
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## This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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Test #2-solutions - Version 081/ABBAB Test #2 Antoniewicz...

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