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Quiz1 - Pickering Tracey Quiz 1 Due 10:00 pm Inst Drummond...

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Pickering, Tracey – Quiz 1 – Due: Feb 15 2006, 10:00 pm – Inst: Drummond 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Newton’s law of universal gravitation is F = G M m r 2 . Here, M and m are masses and r is the sep- aration distance. The dimension of force is specified by the equation F = ma . What are the SI units of the constant G ? 1. [ G ] = m / kg / s 2 2. [ G ] = N m 3. [ G ] = J s / kg 4. [ G ] = N m / s 2 5. [ G ] = m 2 / kg 2 / s 2 6. [ G ] = kg / m 2 / s 2 7. [ G ] = m 3 / kg / s 2 correct 8. [ G ] = W / m 3 9. [ G ] = m 2 / kg 10. [ G ] = m 3 / kg 2 / s 2 Explanation: F = G M m r 2 so G = F r 2 M m . Dimensional analysis of G : kg m s 2 m 2 (kg) 2 = m 3 (kg s 2 ) . The common expression is [ G ] = m 3 / kg / s 2 . 002 (part 1 of 1) 10 points Given: An acre is an area equivalent to that of a rectangle 60 . 5 yd wide and 80 yd long. There are 36 inches in one yard. There are 2.54 cm in one inch. In May 1998, forest fires in southern Mexico and Guatemala spread smoke all the way to Austin. Those fires consumed forest land at a rate of 23500 acres / week. On the average, how many square meters of forest are burned down every minute? Correct answer: 9434 . 64 m 2 / min. Explanation: A conversion factor for week min can be easily derived 1 wk 7 days · 1 day 24 hr · 1 hr 60 min wk min . A conversion factor for m 2 acre can also be de- rived 36 in 1 yd · 2 . 54 cm 1 in · 1 m 100 cm 2 yd 2 acre m 2 acre , where yd 2 acre is given in the problem. Therefore the rate R in square meters per minutes is R = m 2 / acre week / min (23500 acres / week) = (0 . 836127 m 2 / yd 2 )(4840 yd 2 / acre) (10080 min / week) × (23500 acres / week) = 9434 . 64 m 2 / min . 003 (part 1 of 1) 10 points The graph below shows the velocity v as a function of time t for an object moving in a straight line. t v 0 t Q t R t S t P Which of the following graphs shows the corresponding displacement x as a function of time t for the same time interval?

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Pickering, Tracey – Quiz 1 – Due: Feb 15 2006, 10:00 pm – Inst: Drummond 2 1. t x 0 t Q t R t S t P 2. t x 0 t Q t R t S t P 3. t x 0 t Q t R t S t P 4. t x 0 t Q t R t S t P 5. t x 0 t Q t R t S t P correct 6. None of these graphs are correct. 7. t x 0 t Q t R t S t P 8. t x 0 t Q t R t S t P 9. t x 0 t Q t R t S t P Explanation: The displacement is the integral of the ve- locity with respect to time: ~x = Z ~v dt . Because the velocity increases linearly from zero at first, then remains constant, then de- creases linearly to zero, the displacement will increase at first proportional to time squared, then increase linearly, and then increase pro- portional to negative time squared. From these facts, we can obtain the correct answer. t x 0 t Q t R t S t P 004 (part 1 of 3) 10 points Consider the plot below describing motion along a straight line with an initial position of x 0 = 10 m.
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