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Unformatted text preview: 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 t + 1 2 at 2 Given the time, distance and acceleration we simply solve for the initial velocity v = d t 1 2 at 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 = mg W = (220 kg) (9 . 8 m / s 2 ) = 2156 N . T W a m According to Newtons second law, F net = T mg = ma T = m ( g + a ) = (220 kg) bracketleftBig (9 . 8 m / s 2 ) + (1 . 2 m / s 2 ) bracketrightBig = 2420 N ....
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This note was uploaded on 04/06/2008 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas at Austin.
 Spring '07
 KOPP
 Physics

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