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jUs8T8dvOVPo_1207550327_hma293

jUs8T8dvOVPo_1207550327_hma293 - test 01 ALLEN HUNTER Due...

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test 01 – ALLEN, HUNTER – Due: Feb 7 2008, 6:00 pm 1 Question 1, chap 2, sect 5. part 1 of 1 10 points Suppose the position equation for a moving object is given by s ( t ) = 3 t 2 + 2 t + 5 , where s is measured in meters and t is mea- sured in seconds. Find the velocity of the object when t = 2. 1. None of these 2. 14 m/sec correct 3. 10 m/sec 4. 13 m/sec 5. 6 m/sec Explanation: s ( t ) = 3 t 2 + 2 t + 5 . The velocity is v ( t ) = s ( t ) = 6 t + 2 and v (2) = 14. Question 2, chap 4, sect 5. part 1 of 1 10 points Given: The battleship and enemy ships A and B lie along a straight line. Neglect air friction. A battleship simultaneously fires two shells (with the same muzzle velocity) at these two enemy ships. battleship A B If the shells follow the parabolic trajectories shown in the figure, which ship gets hit first? 1. A 2. need more information 3. B correct 4. both at the same time Explanation: The time interval for the entire projectile motion is given by t trip = t rise + t fall = 2 t rise , where t rise is the rising time from 0 to the maximum height, and t fall the falling time from h to 0. In the absence of air resistance t rise = t fall , h = 1 2 g t 2 fall , or t trip = 2 radicalBigg 2 h g . So the smaller is h , the smaller is t trip . In other words, enemy ship B will get hit first. Question 3, chap 2, sect 7. part 1 of 1 10 points A car, moving along a straight stretch of highway, begins to accelerate at 0 . 0103 m / s 2 . It takes the car 24 . 7 s to cover 1 km. How fast was the car going when it first began to accelerate? 1. 32 . 3248 m / s 2. 33 . 3653 m / s 3. 34 . 4299 m / s 4. 35 . 5375 m / s 5. 36 . 6408 m / s 6. 37 . 9031 m / s 7. 39 . 0844 m / s 8. 40 . 3586 m / s correct 9. 41 . 6629 m / s 10. 43 . 1448 m / s Explanation: We can describe this situation with the equation d = v 0 t + 1 2 at 2 Given the time, distance and acceleration we simply solve for the initial velocity v 0 = d t - 1 2 at

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test 01 – ALLEN, HUNTER – Due: Feb 7 2008, 6:00 pm 2 converting the distance from km to m to ob- tain the proper units. Question 4, chap 5, sect 9. part 1 of 1 10 points An empty 220 kg elevator accelerates up- ward at 1 . 2 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . What is the tension in the cable that lifts the elevator cab? 1. 1904 N 2. 1972 N 3. 2034 N 4. 2109 N 5. 2185 N 6. 2260 N 7. 2331 N 8. 2420 N correct 9. 2508 N 10. 2599 N Explanation: Let : g = 9 . 8 m / s 2 , m = 220 kg , and a = 1 . 2 m / s 2 , so W = m g W = (220 kg) (9 . 8 m / s 2 ) = 2156 N . T W a m According to Newton’s second law, F net = T - m g = m a T = m ( g + a ) = (220 kg) bracketleftBig (9 . 8 m / s 2 ) + (1 . 2 m / s 2 ) bracketrightBig = 2420 N . Question 5, chap 5, sect 9. part 1 of 1 10 points A block of mass 2 . 57 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 7 . 08 kg which hangs in the air, as shown on the following picture. Assume the cord to be light (massless and weightless) and unstretchable and the the pulley to have no friction and no rotational inertia.
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