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Physics Test1 Solutions

# Physics Test1 Solutions - quiz 01 VENNES ROSS Due 10:00 pm...

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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 0 + v 0 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 0 = h , the vertical component of the initial velocity is v 0 y = v 0 sin θ directed upward, and y = 0 when the brick reaches the ground, so 0 = h + v 0 y t - 1 2 g t 2 h = - v 0 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 0 . When the initial projectile angle is 45 with respect to the horizontal, it gives a maximum range R . y x 45 R v 0 The time of flight t max of the cannonball for this maximum range R is given by 1. t max = 3 v 0 g 2. t max = 1 3 v 0 g 3. t max = v 0 g

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quiz 01 – VENNES, ROSS – Due: Sep 19 2007, 10:00 pm 2 4. t max = 2 v 0 g 5. t max = 2 3 v 0 g 6. t max = 4 v 0 g 7. t max = 2 v 0 g correct 8. t max = 1 2 v 0 g 9. t max = 1 4 v 0 g 10. t max = 1 2 v 0 g Explanation: The cannonball’s time of flight is t = 2 v 0 y g = 2 v 0 sin 45 g = 2 v 0 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 skier’s 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. Basic Concept: a x = a (cos θ ) Given: θ = - 11 a = 2 . 2 m / s 2 Solution: a x = (2 . 2 m / s 2 )[cos( - 11 )] = 2 . 15958 m / s 2 Question 5, chap 3, sect 3. part 2 of 2 10 points b) What is the vertical component of the skier’s acceleration? 1. - 0 . 41978 m / s 2 correct 2. - 0 . 404912 m / s 2 3. - 0 . 388229 m / s 2 4. - 0 . 374241 m / s 2 5. - 0 . 362883 m / s 2 6. - 0 . 350846 m / s 2 7. - 0 . 338691 m / s 2 8. - 0 . 324375 m / s 2 9. - 0 . 312567 m / s 2 10. - 0 . 295202 m / s 2 Explanation: Basic Concept: a y = a (sin θ ) Solution: a y = (2 . 2 m / s 2 )[sin( - 11 )] = - 0 . 41978 m / s 2 Question 6, chap 2, sect 2.
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Physics Test1 Solutions - quiz 01 VENNES ROSS Due 10:00 pm...

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