{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 10.0points 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 G M 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 ⇒ − G M m r i + 1 2 m v 2 i = 0 + 1 2 m v 2 f v f = radicalbigg v 2 i 2 G M R = 24 . 2487 m / s . 002(part1of2)8.0points A fan cart of mass 0 . 81 kg initially has a velocity of vectorv i = ( 0 . 91 , 0 , 0 ) m / s . Then the fan is turned on, and the air exerts a constant force of vector F = (− 0 . 42 , 0 , 0 ) 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: 0 . 63 kg · m / s. Explanation: The change in momentum is given by Δ vectorp = vector F net Δ t = ( (− 0 . 42 , 0 , 0 ) N)(1 . 5 s) = (− 0 . 63 , 0 , 0 ) kg · m / s . So the x component is 0 . 63 kg · m / s . 003(part2of2)7.0points 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: 0 . 3283 J.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 = (− 0 . 63 , 0 , 0 ) kg · m / s 0 . 81 kg = (− 0 . 777778 , 0 , 0 ) m / s . So the final velocity is vectorv f = vectorv i + Δ vectorv = ( 0 . 132222 , 0 , 0 ) m / s . Now, the change in kinetic energy is given by Δ KE = KE f KE i = 1 2 m v 2 f 1 2 m v 2 i = 1 2 (0 . 81 kg) (0 . 132222 m / s) 2 1 2 (0 . 81 kg) (0 . 91 m / s) 2 = 0 . 3283 J . 004 10.0points 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 . 1. P = ρ ˙ V gh 2 k 2. P = ˙ V gh correct 3. P = ρ ˙ V gh k 4. P = ρgh 2 k ˙ V 5. P = kρgh 2 ˙ V 6. P = kρgh ˙ V 7. P = ˙ V gh 2 8. P = ρgh k ˙ V Explanation: The electrical power of a hydroelectric plant is the potential energy converted to electrical energy per second. The mass of water flowing through the plant per second is ˙ M = ρ ˙ V . The potential energy change of this mass is: Δ U = ˙ Mgh = ρ ˙ V gh.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern