Quiz 5 Solutions - schroeder(cs39963 Newtons Laws of Motion...

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schroeder (cs39963) – Newton’s Laws of Motion – balasubramanya – (1401310) 1 This print-out should have 54 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Stokes law says F = 6 πrηv . F = m a is a force, r the radius and v the velocity. The parameter η has the dimension of 1. [ η ] = M T 2. [ η ] = T 2 L 2 M 3. [ η ] = T M 4. [ η ] = M T L 2 5. [ η ] = M T 2 L 2 6. [ η ] = T L M 7. [ η ] = T L 2 M 8. [ η ] = T 2 L M 9. [ η ] = M T L correct 10. [ η ] = M T 2 L Explanation: Recall that the dimension of force is [ F ] = [ m a ] = M L T 2 . The number 6 π is dimensionless. The dimen- sions of the radius and velocity are [ r ] = L ; [ v ] = L T so [ F ] = [6 π r η v ] M L T 2 = L L T [ η ] Solving for [ η ] yields [ η ] = M LT . 002 10.0 points Newton’s law of universal gravitation is rep- resented by F = G M m r 2 Here F is the gravitational force, M and m are masses, and r is a length. Force has the SI units kg · m / s 2 . What are the SI units of the proportionality constant G ? 1. W / m 3 2. N · m / s 2 3. kg / (m 2 · s 2 ) 4. N · m 5. m 3 / (kg 2 · s 2 ) 6. m 2 / kg 7. m 3 / (kg · s 2 ) correct 8. J · s / kg 9. m 2 / (kg 2 · s 2 ) 10. m / (kg · s 2 ) Explanation: F = G M m r 2 G = F r 2 M m Dimensional analysis of G : parenleftbigg kg · m s 2 parenrightbigg · m 2 (kg) 2 = m 3 kg · s 2
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schroeder (cs39963) – Newton’s Laws of Motion – balasubramanya – (1401310) 2 003 10.0 points Is it correct to say that no force acts on a body at rest? 1. No force acts on a body at rest; all forces cancel each other. 2. No net force acts on a body at rest; when the net force is zero, the body is in static equilibrium. correct 3. No net force acts on a body at rest; no force acts on the body at all. 4. No force acts on a body at rest; if at least one force acted on it the body would move. 5. All are wrong. Explanation: There may be any number of forces that act to produce a zero net force. When the net force is zero, the body is in static equilibrium. 004 (part 1 of 2) 10.0 points Consider the mass and a frictionless ramp shown in the figure. The mass is pushed upward with some initial velocity. Because of the influence of gravity it moves upward and reaches the highest point. It then turns around and moves downward. Starting from point A , along the way it passes consecutively the points B , C , D and E at equal time intervals. Denote by Δ vectorv if the change in ve- locities between any pair of points (referred to as the initial and the final points); i.e. , vectorv f = vectorv i + Δ vectorv if , where vectorv i and vectorv f are the cor- responding velocity vectors at the two points. A B C D E The direction of Δ vectorv AB ; i.e. , the change of velocity vectors vectorv B vectorv A is 1. Downhill along the ramp. correct 2. Uphill along the ramp if v A > 0 . 098 m/s, and downhill if v A < 0 . 098 m/s. 3. Uphill along the ramp. 4. 0. 5. Uphill along the ramp if v A < 0 . 098 m/s, and downhill if v A > 0 . 098 m/s.
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