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Unformatted text preview: Pickering, Tracey – Quiz 1 – Due: Feb 15 2006, 10:00 pm – Inst: Drummond 1 This printout should have 27 questions. Multiplechoice 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 ] = Nm 3. [ G ] = Js / kg 4. [ G ] = Nm / 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 : µ kgm s 2 ¶ m 2 (kg) 2 = m 3 (kgs 2 ) . Thecommonexpressionis[ 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 1wk 7days · 1day 24hr · 1hr 60min ≡ wk min . A conversion factor for m 2 acre can also be de rived µ 36in 1yd · 2 . 54cm 1in · 1m 100cm ¶ 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 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? Pickering, Tracey – Quiz 1 – Due: Feb 15 2006, 10:00 pm – Inst: Drummond 2 1. t x t Q t R t S t P 2. t x t Q t R t S t P 3. t x t Q t R t S t P 4. t x t Q t R t S t P 5. t x t Q t R t S t P correct 6. None of these graphs are correct. 7. t x t Q t R t S t P 8. t x t Q t R t S t P 9. t x 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....
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This note was uploaded on 06/23/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner

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