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Unformatted text preview: brown (twb493) – oldmidterm 01 – turner – (57340) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Suppose the volume V of some object happens to depend on time t according to the equation V ( t ) = At 3 + B t 2 , where A and B are some constants. Let L and T denote dimensions of length and time, respectively. What is the dimension of the constant A ? 1. L / T 3 2. L 3 / T 3 correct 3. L 3 · T 3 4. L 2 / T 5. L / T Explanation: The term At 3 has dimensions of volume, so [ V ] = [ A ] [ t ] 3 [ A ] = [ V ] [ t ] 3 = L 3 T 3 . 002 10.0 points Consider a car which is traveling along a straight road with constant acceleration a . There are two checkpoints A and B which are a distance 115 m apart. The time it takes for the car to travel from A to B is 5 . 31 s. A B 6 . 49 m / s 2 115 m Find the velocity v B for the case where the acceleration is 6 . 49 m / s 2 . Correct answer: 38 . 8882 m / s. Explanation: Let : Δ x = 115 m , Δ t = 5 . 31 s , and a = 6 . 49 m / s 2 . ¯ v = v A + v B = Δ x Δ t v A = 2 Δ x Δ t − v B and v B = v A + a Δ t = parenleftbigg 2 Δ x Δ t − v B parenrightbigg + a Δ t 2 v B = 2 Δ x Δ t + a Δ t v B = Δ x Δ t + a Δ t 2 = 115 m 5 . 31 s + (6 . 49 m / s 2 ) (5 . 31 s) 2 = 38 . 8882 m / s . 003 10.0 points A tennis ball with a speed of 22 . 1 m / s is moving perpendicular to a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 15 . 8015 m / s. If the ball is in contact with the wall for . 0123 s, what is the average acceleration of the ball while it is in contact with the wall? Take “toward the wall” to be the positive direction. Correct answer: − 3081 . 42 m / s 2 . Explanation: Let : v i = 22 . 1 m / s , v f = − 15 . 8015 m / s , and Δ t = 0 . 0123 s . a av = Δ v Δ t = v f − v i Δ t = − 15 . 8015 m / s − 22 . 1 m / s . 0123 s = − 3081 . 42 m / s 2 . 004 10.0 points brown (twb493) – oldmidterm 01 – turner – (57340) 2 Runner A is initially 4 mi west of a flagpole and is running with a constant velocity of 7 mi / h due east. Runner B is initially 3 . 9 mi east of the flagpole and is running with a constant velocity of 5 . 1 mi / h due west. What is the displacement of runner B from the flagpole when their paths cross? Consider East to be the positive direction. Correct answer: 0 . 570248 mi. Explanation: Let : v A = 7 mi / h , x i,A = − 4 mi , v B = − 5 . 1 mi / h , x i,B = 3 . 9 mi , and a A = a B = 0 . x = x i + v i t + 1 2 a t 2 = x i + v i t. They meet when x A = x B x i,A + v A t = x i,B + v B t ( v A − v B ) t = x i,B − x i,A t = x i,B − x i,A v A − v B = 3 . 9 mi − ( − 4 mi) 7 mi / h − ( − 5 . 1 mi / h) = 0 . 652893 h , so x A = x i + v Δ t = − 4 mi + (7 mi / h)(0 . 652893 h) = . 570248 mi ....
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This note was uploaded on 04/13/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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