test2_09Soln - LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS...

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LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS TEST 2 Time: 50 minutes; Calculators & formulae provided at the end = only aid; Total = 20 points. 1)[5] A crate ( m = 1 . 0 × 10 2 kg) needs to be pulled across a smooth floor by John and Bob as shown in the figure. The friction coefficients are known as μ s = 0 . 25 k = 0 . 10. The crate location at time t = 0 is shown, John pulls with F J = 1 . 0 × 10 2 N, and Bob with F B = 2 . 0 × 10 2 N at the angles indicated. Provide answers for x ( t ) and y ( t ), i.e., for the position vector of the motion. Start with a free-body diagram (include friction!). Will the crate move? 1
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2) [5] Derive the formula for the centripetal acceleration ( a cp = v 2 r ) from the position vector describing uniform circular motion (formula sheet!), and show the direction for the acceleration vector. 2
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of gravity and Newton’s 2 nd law. Assume d S - E = 1 . 5 × 10 11 m, and M S = 2 . 0 × 10 30 kg. Then calculate the length of a year from one orbit. 3
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test2_09Soln - LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS...

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