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Unformatted text preview: Version 028/AABDA midterm 01 Turner (58220) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A cat chases a mouse across a 0.94 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 1.0 m from the edge of the table. The acceleration of gravity is 9 . 81 m / s 2 . What was the cats speed when it slid off the table? 1. 2.42611 2. 2.28431 3. 4.24654 4. 2.24614 5. 3.58982 6. 4.09007 7. 5.63425 8. 3.49639 9. 2.75587 10. 7.44296 Correct answer: 2 . 28431 m / s. Explanation: Basic Concept: x = v x t y = 1 2 g ( t ) 2 Given: y = . 94 m x = 1 . 0 m g = 9 . 81 m / s 2 . Solution: t = radicalBigg 2 y g = x v x v x = radicalbigg g 2 y x = radicalBigg (9 . 81 m / s 2 ) 2 ( . 94 m) (1 m) = 2 . 28431 m / s . 002 10.0 points When an object falls through air, there is a drag force ( with dimension M L / T 2 ) that depends on the product of the surface area of the object and the square of its velocity; i.e. , F air = C A v 2 , where C is a constant. What is the dimension for constant C ? 1. [ C ] = M T 2. [ C ] = M L 3 correct 3. [ C ] = M L 2 4. [ C ] = T L M 5. [ C ] = T 2 L 2 M 6. [ C ] = T L 2 M 7. [ C ] = T 2 L M 8. [ C ] = T M 9. [ C ] = M T L 2 10. [ C ] = M T 2 L 2 Explanation: [ F ] = M L / T 2 , [ A ] = L 2 , and [ v ] = L / T , so [ C ] = [ F ] [ A ] [ v ] 2 = M L / T 2 L 2 (L / T) 2 = M L T 2 T 2 L 4 = M L 3 . 003 10.0 points Version 028/AABDA midterm 01 Turner (58220) 2 A cylinder, 16 cm long and 3 cm in radius, is made of two different metals bonded endto end to make a single bar. The densities are 4 . 4 g / cm 3 and 6 . 3 g / cm 3 . 1 6 c m 3 cm , radius What length of the lighterdensity part of the bar is needed if the total mass is 2377 g? 1. 9.92098 2. 13.1373 3. 7.52775 4. 12.5493 5. 11.7805 6. 10.4417 7. 9.5286 8. 9.69549 9. 8.8057 10. 10.0839 Correct answer: 8 . 8057 cm. Explanation: Let : = 16 cm , r = 3 cm , 1 = 4 . 4 g / cm 3 , 2 = 6 . 3 g / cm 3 , and m = 2377 g . Volume of a bar of radius r and length is V = r 2 and its density is = m V = m r 2 so that m = r 2 x x r Let x be the length of the lighter metal; then x is the length of the heavier metal and m = m 1 + m 2 = 1 r 2 x + 2 r 2 ( x ) = ( 1 2 ) r 2 x + 2 r 2 x r 2 ( 1 2 ) = m 2 r 2 x = m 2 r 2 r 2 ( 1 2 ) = 2377 g (6 . 3 g / cm 3 ) (3 cm) 2 (16 cm) (3 cm) 2 (4 . 4 g / cm 3 6 . 3 g / cm 3 ) = 8 . 8057 cm ....
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 Spring '08
 Turner
 Physics

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