Test #3-solutions - Version 080/ABBAA – Test #3 –...

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Unformatted text preview: Version 080/ABBAA – Test #3 – Antoniewicz – (56445) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The escape speed from an asteroid whose ra- dius is 12 km is only 14 m / s. If you throw a rock away from the asteroid at a speed of 28 m / s, what will be its final speed? Use G = 6 . 7 × 10 − 11 N · m 2 / kg 2 . 1. 20.7846 2. 15.5885 3. 24.2487 4. 13.8564 5. 8.66025 6. 22.5167 7. 10.3923 8. 17.3205 9. 12.1244 10. 19.0526 Correct answer: 24 . 2487 m / s. Explanation: First use the escape speed to get the mass of the asteroid: v esc = radicalbigg 2 GM R ⇒ M = 1 2 v 2 R G = 1 . 75522 × 10 16 kg . Now, if v i = 28 m / s, then v f is found from E i = E f : U i + K i = U f + K f ⇒ − GM m r i + 1 2 mv 2 i = 0 + 1 2 mv 2 f ⇒ v f = radicalbigg v 2 i − 2 GM R = 24 . 2487 m / s . 002 (part 1 of 2) 8.0 points A fan cart of mass 0 . 81 kg initially has a velocity of vectorv i = ( . 91 , , ) m / s . Then the fan is turned on, and the air exerts a constant force of vector F = (− . 42 , , ) N on the cart for 1 . 5 s. What is the change in the x component of momentum of the fan cart over this 1 . 5 s time interval? (Since the force is only applied along the x direction, the other components of momentum will not change.) 1. -0.84 2. -0.525 3. -0.36 4. -0.54 5. -0.465 6. -0.705 7. -0.57 8. -0.45 9. -0.78 10. -0.63 Correct answer: − . 63 kg · m / s. Explanation: The change in momentum is given by Δ vectorp = vector F net Δ t = ( (− . 42 , , ) N)(1 . 5 s) = (− . 63 , , ) kg · m / s . So the x component is − . 63 kg · m / s . 003 (part 2 of 2) 7.0 points What is the change in kinetic energy of the fan cart over this 1 . 5 s time interval? 1. -0.278918 2. -0.342384 3. -0.315222 4. -0.3283 5. -0.269644 6. -0.420704 7. -0.409629 8. -0.337629 9. -0.373283 10. -0.366849 Correct answer: − . 3283 J. Version 080/ABBAA – Test #3 – Antoniewicz – (56445) 2 Explanation: To find the change in kinetic energy, we need to know the initial and final velocities. We know the initial velocity; to get the final, we first find Δ vectorv : Δ vectorv = Δ vectorp m = (− . 63 , , ) kg · m / s . 81 kg = (− . 777778 , , ) m / s . So the final velocity is vectorv f = vectorv i + Δ vectorv = ( . 132222 , , ) m / s . Now, the change in kinetic energy is given by Δ KE = KE f − KE i = 1 2 mv 2 f − 1 2 mv 2 i = 1 2 (0 . 81 kg) (0 . 132222 m / s) 2 − 1 2 (0 . 81 kg) (0 . 91 m / s) 2 = − . 3283 J . 004 10.0 points In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. If the volume flow rate is ˙ V and the height of the falls is h , what is the amount of electrical power produced, assum- ing the plant’s efficiency is k ? Let the density of water be notated by ρ ....
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This note was uploaded on 11/15/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Test #3-solutions - Version 080/ABBAA – Test #3 –...

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