quiz 1 - Version 034 – Quiz 1 – coker –(58245 1 This...

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Unformatted text preview: Version 034 – Quiz 1 – coker – (58245) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This quiz covers ONLY topics covered in class lecture in Chs. 2 through 5. 001 10.0 points A dragster maintains a speedometer reading of 100 km/h and passes through a curve with a constant radius. Which statement is true? 1. The dragster moved along a straight line at a constant velocity of 100 km/h. 2. The dragster rounded the curve at a changing velocity of 100 km/h. correct 3. The dragster rounded the curve at a con- stant velocity of 100 km/h. 4. The dragster rounded the curve at a changing speed of 100 km/h. 5. All are wrong. Explanation: Constant velocity means not only constant speed, but constant direction, so the velocity is changing. A car rounding a curve changes its direction of motion, regardless of its speed. 002 10.0 points You and your friend throw balloons filled with water from the roof of a several story apart- ment house. You simply drop a balloon from rest. A second balloon is thrown downward by your friend 3 . 6 s later with an initial speed of 70 . 56 m / s. They hit the ground simultane- ously. The acceleration of gravity is 9 . 8 m / s 2 . You can neglect air resistance. How high is the apartment house? 1. 159.201 2. 53.361 3. 127.449 4. 58.3222 5. 112.896 6. 150.932 7. 142.884 8. 48.6202 9. 99.225 10. 80.3723 Correct answer: 142 . 884 m. Explanation: For balloon 1: y 1 = 1 2 g t 2 1 For balloon 2: y 2 = v t 2 + g t 2 2 2 Balloon 2 is thrown Δ t seconds after balloon 1 and they hit the ground at the same time, so: t 1 = t 2 + Δ t Since the height of the house is the same for both of the balloons, y 1 = y 2 g t 2 1 2 = v ( t 1 − Δ t ) + g 2 ( t 1 − Δ t ) 2 ( v − g Δ t ) t 1 = v Δ t − g 2 Δ t 2 t 1 = v Δ t − g 2 Δ t 2 v − g Δ t = Δ t 2 + v v − g Δ t Δ t 2 For Δ t = 2 s t 1 = 1 + v v − 2 g Version 034 – Quiz 1 – coker – (58245) 2 For v = 30 m/s, Δ t = 2 s t 1 = 5 . 4 s From the above, once t 1 is known, the height of the house is given simply by y = g t 2 1 2 For t 1 = 5 . 4 s, y = 142 . 884 m 003 10.0 points A mass slides with negligible friction on a circular track of 1 m radius oriented vertically. Its speed at the position shown in the figure is 3.13 m/s. The acceleration of gravity is 9.8 m/s 2 . v At the position shown in the figure, which of the labeled arrows best represents the di- rection of the acceleration of the mass? 1. The mass is traveling at a constant veloc- ity, therefore it has no acceleration. 2. correct 3. 4. 5. 6. 7. 8. 9. Explanation: The magnitude of the centripetal is a r = v 2 r = (3 . 13 m / s) 2 1 m ≈ 9 . 8 m / s 2 ....
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This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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quiz 1 - Version 034 – Quiz 1 – coker –(58245 1 This...

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