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Ch3-HW2-solutions

# Ch3-HW2-solutions - louey(cal2859 – Ch3-HW2 – florin...

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Unformatted text preview: louey (cal2859) – Ch3-HW2 – florin – (56930) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points At what height above the surface of the Earth is there a 4% difference between the approximate magnitude of the gravitational field (9 . 8 N / kg) and the actual magnitude of the gravitational field at that location ( r e = 6 . 4 × 10 6 m)? Correct answer: 1 . 31973 × 10 5 m. Explanation: The actual magnitude of the gravitational field at a height y above the surface of the Earth is given by g actual = GM E ( R + y ) 2 . The approximate magnitude is given at the surface, i.e. , when y = 0. So a glance at each expression will reveal the fact that the actual magnitude will be smaller than the approximate surface value at heights above the surface of the Earth. Thus for a 4% difference, we take the actual value to be 96% of the approximate value and solve for y : g actual = (0 . 96) · g approx ⇒ GM E ( R + y ) 2 = (0 . 96) GM E R 2 ⇒ 1 . 96 R 2 = ( R + y ) 2 ⇒ y 2 + 2 Ry − . 04 . 96 R 2 = 0 . We can use the quadratic formula and take the positive solution to obtain y : y = − 2 R + radicalBigg (2 R ) 2 − 4 parenleftbigg − . 04 . 96 R 2 parenrightbigg 2 = parenleftBigg − 1 + radicalbigg 1 . 96 parenrightBigg R = parenleftBigg − 1 + radicalbigg 1 . 96 parenrightBigg · (6 . 4 × 10 6 m) = 1 . 31973 × 10 5 m . 002 10.0 points A steel ball of mass m falls from a height h onto a scale calibrated in Newtons. The ball rebounds repeatedly to nearly the same height h . The scale is sluggish in its response to the intermittent hits and displays an average force F avg , such that F avg T = F Δ t, where F Δ t is the brief impulse that the ball imparts to the scale on every hit, and T is the time between hits. Calculate this average force in terms of m , h , and other physical constants. 1. F avg = mgh 2. F avg = mg T 3. F avg = mg h 4. F avg = mgT 5. F avg = mg correct Explanation: The time T between hits is the same as the duration of the ball’s fall from rest back to the same height again at rest. The net change in momentum over this period is 0. Then, since we know that the force of gravity can be taken to be constant, we can now apply the momentum principle: louey (cal2859) – Ch3-HW2 – florin – (56930) 2 Δ P = F net Δ t 0 = ( F avg + F g ) · T 0 = ( F avg − mg ) · T ⇒ F avg = mg . 003 (part 1 of 7) 4.0 points In outer space, far from other objects, block 1 of mass 40 kg is at position vectorr 1 = ( 6 , 8 , ) m , and block 2 of mass 1000 kg is at position vectorr 2 = ( 17 , 13 , ) m ....
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Ch3-HW2-solutions - louey(cal2859 – Ch3-HW2 – florin...

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