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Unformatted text preview: quiz 01 VENNES, ROSS Due: Sep 19 2007, 10:00 pm 1 Question 1, chap 1, sect 6. part 1 of 1 10 points A plastic tube allows a flow of 13 . 6 cm 3 / s of water through it. How long will it take to fill a 204 cm 3 bottle with water? 1. 13 . 6538 s 2. 14 . 0881 s 3. 14 . 539 s 4. 15 s correct 5. 15 . 4861 s 6. 15 . 969 s 7. 16 . 4748 s 8. 16 . 9853 s 9. 17 . 5188 s 10. 18 . 0645 s Explanation: Let : r = 13 . 6 cm 3 / s and V = 204 cm 3 . cm 3 cm 3 s = cm 3 s cm 3 = s , so the time is defined by t = V r = 204 cm 3 13 . 6 cm 3 / s = 15 s . Question 2, chap 4, sect 4. part 1 of 1 10 points A brick is thrown upward from the top of a building at an angle of 26 . 2 above the hori zontal and with an initial speed of 7 . 17 m / s. The acceleration of gravity is 9 . 8 m / s 2 . If the brick is in flight for 2 . 7 s, how tall is the building? 1. 20 . 5568 m 2. 21 . 1956 m 3. 21 . 8537 m 4. 22 . 5344 m 5. 23 . 236 m 6. 23 . 9744 m 7. 24 . 7275 m 8. 25 . 5136 m 9. 26 . 3134 m 10. 27 . 1739 m correct Explanation: Basic Concept The height of the building is determined by the vertical motion with gravity acting down and an initial velocity acting upward: y = y + v y t 1 2 g t 2 Solution Choose the origin at the base of the build ing. The initial position of the brick is y = h , the vertical component of the initial velocity is v y = v sin directed upward, and y = 0 when the brick reaches the ground, so 0 = h + v y t 1 2 g t 2 h = v y t + 1 2 g t 2 = (3 . 1656 m / s) (2 . 7 s) + 1 2 ( 9 . 8 m / s 2 ) (2 . 7 s) 2 = 27 . 1739 m . Question 3, chap 4, sect 4. part 1 of 1 10 points Denote the initial speed of a cannon ball fired from a battleship as v . When the initial projectile angle is 45 with respect to the horizontal, it gives a maximum range R . y x 4 5 R v The time of flight t max of the cannonball for this maximum range R is given by 1. t max = 3 v g 2. t max = 1 3 v g 3. t max = v g quiz 01 VENNES, ROSS Due: Sep 19 2007, 10:00 pm 2 4. t max = 2 v g 5. t max = 2 3 v g 6. t max = 4 v g 7. t max = 2 v g correct 8. t max = 1 2 v g 9. t max = 1 4 v g 10. t max = 1 2 v g Explanation: The cannonballs time of flight is t = 2 v y g = 2 v sin 45 g = 2 v g . Question 4, chap 3, sect 3. part 1 of 2 10 points A skier squats low and races down a(n) 11 ski slope. During a 6 s interval, the skier accelerates at 2 . 2 m / s 2 . a) What is the horizontal component of the skiers acceleration (perpendicular to the direction of free fall)? 1. 2 . 15958 m / s 2 correct 2. 2 . 23168 m / s 2 3. 2 . 30703 m / s 2 4. 2 . 39076 m / s 2 5. 2 . 47275 m / s 2 6. 2 . 55223 m / s 2 7. 2 . 641 m / s 2 8. 2 . 72824 m / s 2 9. 2 . 81386 m / s 2 10. 2 . 91089 m / s 2 Explanation: a y a x 2 . 2 m / s 2 11 Note: Figure is not drawn to scale....
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This homework help was uploaded on 04/18/2008 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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