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Practice Exam 1

# Practice Exam 1 - practicework 01 – BAUTISTA ALDO – Due...

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Unformatted text preview: practicework 01 – BAUTISTA, ALDO – Due: Feb 13 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 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 ] = J s / kg 2. [ G ] = m 2 / kg 3. [ G ] = m / kg / s 2 4. [ G ] = m 3 / kg / s 2 correct 5. [ G ] = N m / s 2 6. [ G ] = m 2 / kg 2 / s 2 7. [ G ] = N m 8. [ G ] = kg / m 2 / s 2 9. [ G ] = m 3 / kg 2 / s 2 10. [ G ] = W / m 3 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 . Question 2 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 21600 acres / week. On the average, how many square meters of forest are burned down every minute? Correct answer: 8671 . 84 m 2 / min (tolerance ± 1 %). 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 (21600 acres / week) = (0 . 836127 m 2 / yd 2 )(4840 yd 2 / acre) (10080 min / week) × (21600 acres / week) = 8671 . 84 m 2 / min . Question 3 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 practicework 01 – BAUTISTA, ALDO – Due: Feb 13 2006, 4:00 am 2 Which of the following graphs shows the corresponding displacement x as a function of time t for the same time interval? 1. t x t Q t R t S t P 2. t x t Q t R t S t P 3. None of these graphs are correct. 4. t x t Q t R t S t P 5. t x t Q t R t S t P 6. t x t Q t R t S t P 7. t x t Q t R t S t P correct 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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Practice Exam 1 - practicework 01 – BAUTISTA ALDO – Due...

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