8TWuuJMD88AF_1207550346_hma293

8TWuuJMD88AF_1207550346_hma293 - test 02 ALLEN, HUNTER Due:...

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Unformatted text preview: test 02 ALLEN, HUNTER Due: Mar 20 2008, 6:00 pm 1 Question 1, chap 9, sect 11. part 1 of 1 10 points An open cart on a level surface rolls without frictional loss through a downpour of rain. The rain falls vertically downward as shown below. As the cart rolls, an appreciable amount of rain water accumulates in the cart. v rain rain water cart The speed of the cart will 1. decrease because of conservation of mo- mentum. correct 2. remain the same because the raindrops are falling perpendicular to the direction of carts motion. 3. increase because of conservation of mo- mentum. 4. increase because of conservation of me- chanic energy. 5. decrease because of conservation of me- chanic energy. Explanation: This is an inelastic collision in the direction along which the cart is rolling. Only mo- mentum vectorp along that direction is conserved. Because the raindrops fall vertically, they do not carry momentum horizontally. Assume m of rain water accumulates on the cart: p i = p f mv = ( m + m ) v . Therefore v = m m + m v v < v . The speed of the cart will decrease because of conservation of momentum. Question 2, chap 8, sect 4. part 1 of 1 10 points Tim, with mass 74 kg, climbs a gymnasium rope a distance of 5 . 1 m. The acceleration of gravity is 9 . 8 m / s 2 . How much potential energy does Tim gain? 1. 2782 . 42 J 2. 2870 . 22 J 3. 2960 . 78 J 4. 3057 . 6 J 5. 3156 . 38 J 6. 3256 . 34 J 7. 3368 . 46 J 8. 3476 . 65 J 9. 3584 . 84 J 10. 3698 . 52 J correct Explanation: Let : m = 74 kg , h = 5 . 1 m , and g = 9 . 8 m / s 2 . Potential energy is U = mg h = (74 kg) (9 . 8 m / s 2 ) (5 . 1 m) = 3698 . 52 J . Question 3, chap 10, sect 4. part 1 of 1 10 points A potters wheel moves from rest to an angular speed of 0.42 rev/s in 35.9 s. Assuming constant angular acceleration, what is its angular acceleration in rad/s 2 ? 1. . 0626669 rad / s 2 2. . 0647751 rad / s 2 3. . 066798 rad / s 2 4. . 0688854 rad / s 2 5. . 0712526 rad / s 2 6. . 073508 rad / s 2 correct test 02 ALLEN, HUNTER Due: Mar 20 2008, 6:00 pm 2 7. . 0757972 rad / s 2 8. . 0781779 rad / s 2 9. . 0807015 rad / s 2 10. . 083202 rad / s 2 Explanation: Basic Concept: f = i + t = t since i = 0 rad / s. Given: f = 0 . 42 rev / s t = 35 . 9 s Solution: = f t = . 42 rev / s 35 . 9 s 2 rad 1 rev = 0 . 073508 rad / s 2 Question 4, chap 10, sect 10. part 1 of 1 10 points A 26 kg mass and a 15 kg mass are sus- pended by a pulley that has a radius of 8 . 3 cm and a mass of 7 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 4 . 1 m apart. Treat the pulley as a uniform disk....
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8TWuuJMD88AF_1207550346_hma293 - test 02 ALLEN, HUNTER Due:...

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