# Test 1 - quiz 01 – GAUTHIER JOSEPH – Due 10:00 pm 1...

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Unformatted text preview: quiz 01 – GAUTHIER, JOSEPH – 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 12 . 6 cm 3 / s of water through it. How long will it take to fill a 246 cm 3 bottle with water? 1. 14 . 7183 s 2. 15 . 1923 s 3. 15 . 6688 s 4. 16 . 1654 s 5. 16 . 6667 s 6. 17 . 1942 s 7. 17 . 7273 s 8. 18 . 2787 s 9. 18 . 8976 s 10. 19 . 5238 s correct Explanation: Let : r = 12 . 6 cm 3 / s and V = 246 cm 3 . cm 3 ÷ cm 3 s = cm 3 · s cm 3 = s , so the time is defined by t = V r = 246 cm 3 12 . 6 cm 3 / s = 19 . 5238 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 20 . 9 ◦ above the hori- zontal and with an initial speed of 8 . 07 m / s. The acceleration of gravity is 9 . 8 m / s 2 . If the brick is in flight for 3 . 1 s, how tall is the building? 1. 28 . 9078 m 2. 29 . 8397 m 3. 30 . 7634 m 4. 31 . 7161 m 5. 32 . 7087 m 6. 33 . 7285 m 7. 34 . 7859 m 8. 35 . 8737 m 9. 37 . 0004 m 10. 38 . 1645 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 θ 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 =- (2 . 87888 m / s) (3 . 1 s) + 1 2 ( 9 . 8 m / s 2 ) (3 . 1 s) 2 = 38 . 1645 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 = 1 √ 2 v g 2. t max = v g 3. t max = √ 3 v g quiz 01 – GAUTHIER, JOSEPH – Due: Sep 19 2007, 10:00 pm 2 4. t max = 2 v g 5. t max = 2 3 v g 6. t max = 1 2 v g 7. t max = 1 4 v g 8. t max = √ 2 v g correct 9. t max = 1 √ 3 v g 10. t max = 4 v g Explanation: The cannonball’s 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 5 s interval, the skier accelerates at 4 . 3 m / s 2 . a) What is the horizontal component of the skier’s acceleration (perpendicular to the direction of free fall)? 1. 3 . 15138 m / s 2 2. 3 . 25144 m / s 2 3. 3 . 36442 m / s 2 4. 3 . 47733 m / s 2 5. 3 . 59009 m / s 2 6. 3 . 70261 m / s 2 7. 3 . 82835 m / s 2 8. 3 . 9603 m / s 2 9. 4 . 09235 m / s 2 10. 4 . 221 m / s 2 correct Explanation: a y a x 4 . 3 m / s 2- 11 ◦ Note: Figure is not drawn to scale....
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Test 1 - quiz 01 – GAUTHIER JOSEPH – Due 10:00 pm 1...

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