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Midterm 1

# Midterm 1 - Version 028/AABDA midterm 01 Turner(58220 This...

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Version 028/AABDA – midterm 01 – Turner – (58220) 1 This print-out should have 18 questions. Multiple-choice 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 cat’s 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: Δ x = v x Δ t Δ y = 1 2 g t ) 2 Given: Δ y = 0 . 94 m Δ x = 1 . 0 m g = 9 . 81 m / s 2 . Solution: Δ t = radicalBigg 2 Δ y g = Δ x v x v x = radicalbigg g y Δ x = radicalBigg (9 . 81 m / s 2 ) 2 ( 0 . 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

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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 end-to- end to make a single bar. The densities are 4 . 4 g / cm 3 and 6 . 3 g / cm 3 . 16 cm 3 cm , radius What length of the lighter-density 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 , ρ 1 = 4 . 4 g / cm 3 , ρ 2 = 6 . 3 g / cm 3 , and m = 2377 g . Volume of a bar of radius r and length is V = π r 2 and its density is ρ = m V = m π r 2 so that m = ρ π 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 = ρ 1 π r 2 x + ρ 2 π r 2 ( x ) = ( ρ 1 ρ 2 ) π r 2 x + ρ 2 π r 2 x π r 2 ( ρ 1 ρ 2 ) = m ρ 2 π r 2 x = m ρ 2 π r 2 π r 2 ( ρ 1 ρ 2 ) = 2377 g (6 . 3 g / cm 3 ) π (3 cm) 2 (16 cm) π (3 cm) 2 (4 . 4 g / cm 3 6 . 3 g / cm 3 ) = 8 . 8057 cm . 004 10.0 points A particle moving at a velocity of 5 . 1 m / s in the positive x direction is given an accelera- tion of 0 . 9 m / s 2 in the positive y direction for 2 . 7 s. What is the final speed of the particle? 1. 24.0302 2. 55.3482 3. 32.6049 4. 5.64933 5. 21.0535 6. 9.91889 7. 46.269 8. 1.34614 9. 10.7713 10. 21.6668 Correct answer: 5 . 64933 m / s. Explanation:
Version 028/AABDA – midterm 01 – Turner – (58220) 3 Let : v xf = v xi = 5 . 1 m / s . a y = 0 . 9 m / s 2 , and t = 2 . 7 s . The vertical velocity undergoes constant ac- celeration: v yf = v yi + a t = 0 + (0 . 9 m / s 2 )(2 . 7 s) = 2 . 43 m / s . Thus v f = radicalBig v 2 xf + v 2 yf = radicalBig (5 . 1 m / s) 2 + (2 . 43 m / s) 2 = 5 . 64933 m / s .

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Midterm 1 - Version 028/AABDA midterm 01 Turner(58220 This...

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