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# Test1 - midterm 01 RAMSEY TAYLOR Due 4:00 am Version number...

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midterm 01 – RAMSEY, TAYLOR – Due: Sep 21 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:1, V3:2, V4:3, V5:1. Question 1 part 1 of 1 10 points Convert the volume 7 . 02 in . 3 to m 3 , recall- ing that 1 in. = 2.54 cm and 100 cm = 1 m. 1. 8 . 79985 × 10 - 5 m 3 2. 9 . 07843 × 10 - 5 m 3 3. 9 . 40617 × 10 - 5 m 3 4. 9 . 78308 × 10 - 5 m 3 5. 0 . 000100944 m 3 6. 0 . 000104222 m 3 7. 0 . 000107499 m 3 8. 0 . 000111432 m 3 9. 0 . 000115037 m 3 correct 10. 0 . 00011897 m 3 Explanation: Basic Concept: Unit Conversion Solution: Convert the units from English units to SI units: V = 7 . 02 in . 3 · 2 . 54 cm 1 in 3 · 1 m 100 cm 3 = 0 . 000115037 m 3 . Question 2 part 1 of 1 10 points A train car moves along a long straight track. The graph shows the position as a function of time for this train. position time The graph shows that the train 1. moves at a constant velocity all of the time. 2. slows down during the first part of the time and speeds up during the last part of the time. 3. speeds up during the first part of the time and slows down during the last part of the time. 4. speeds up all the time. 5. slows down all the time. correct Explanation: The slope of the curve diminishes as time increases. Among the choices give, The train “slows down all of the time” best describes the situ- ation. Question 3 part 1 of 1 10 points Assume: The ball is thrown so that the man’s muscles give the ball the same speed in each case. Neglect the height of the man. A man can throw a ball a maximum hori- zontal distance of 60 . 7 m. The acceleration of gravity is 9 . 8 m / s 2 . How far can he throw the same ball verti- cally upward? 1. 28 . 45 m 2. 29 . 4 m 3. 30 . 35 m correct 4. 31 . 3 m 5. 32 . 3 m 6. 33 . 3 m 7. 34 . 35 m 8. 35 . 45 m 9. 36 . 55 m 10. 37 . 7 m Explanation: The range of a particle is given by the ex- pression R = v 2 0 sin 2 θ g . The maximum horizontal distance is obtained when the ball is thrown at an angle θ = 45 and sin 2 θ = 1 . Solving for v 0 , v 0 = p g R .

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midterm 01 – RAMSEY, TAYLOR – Due: Sep 21 2006, 4:00 am 2 When the ball is thrown upward with this speed, the maximum height is obtained from the equation v 2 f = v 2 0 - 2 g h . Let v f = 0, and solve for h h = R 2 = v 2 0 2 g . Question 4 part 1 of 1 10 points A cylinder, 18 cm long and 5 cm in radius, is made of two different metals bonded end- to-end to make a single bar. The densities are 4 . 9 g / cm 3 and 6 . 5 g / cm 3 . 18 cm 5 cm What length of the lighter metal is needed if the total mass is 7651 g? 1. 9 . 1522 cm 2. 9 . 45556 cm 3. 9 . 76995 cm 4. 10 . 0838 cm 5. 10 . 4003 cm 6. 10 . 7222 cm 7. 11 . 0658 cm 8. 11 . 437 cm 9. 11 . 8415 cm 10. 12 . 2403 cm correct Explanation: Let : = 18 cm , r = 5 cm , ρ 1 = 4 . 9 g / cm 3 , ρ 2 = 6 . 5 g / cm 3 , and m = 7651 g . Volume of a bar of radius r and length is V = π r 2 and its density is ρ = m V = m π r 2 so that m = ρ π r 2 x - x r Let x be the length of the lighter metal; then - x is the length of the heavier metal. Thus, m = m 1 + m 2 = ρ 1 π r 2 x + ρ 2 π r 2 ( - x ) = ρ 1 π r 2 x + ρ 2 π r 2 - ρ 2 π r 2 x .
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Test1 - midterm 01 RAMSEY TAYLOR Due 4:00 am Version number...

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