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Unformatted text preview: Version 024 – midterm 01 – turner – (57340) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points When Maggie applies the brakes of her car, the car slows uniformly from 15.3 m/s to 0 m/s in 2.36 s. How far ahead of a stop sign must she apply her brakes in order to stop at the sign? 1. 18.557 2. 15.336 3. 17.784 4. 16.555 5. 17.352 6. 18.96 7. 18.054 8. 16.095 9. 17.017 10. 16.756 Correct answer: 18 . 054 m. Explanation: Let : v i = 15 . 3 m / s , v f = 0 m / s , and Δ t = 2 . 36 s . The equation simplifies to Δ x = v i + v f 2 Δ t = 1 2 v i Δ t = 1 2 (15 . 3 m / s) (2 . 36 s) = 18 . 054 m . 002 10.0 points The mass of a certain neutron star is 5 × 10 30 kg (2 . 5 solar masses) and its radius is 2000 m. What is the acceleration of gravity at the surface of this condensed, burnedout star? The value of the universal gravitational con stant is 6 . 67 × 10 − 11 N · m 2 / kg 2 . 1. 13340000000000.0 2. 32669400000000.0 3. 1.334e+14 4. 18527800000000.0 5. 11116700000000.0 6. 12969400000000.0 7. 24012000000000.0 8. 29644400000000.0 9. 83375000000000.0 10. 1.0672e+14 Correct answer: 8 . 3375 × 10 13 m / s 2 . Explanation: Let : M = 5 × 10 30 kg , r = 2000 m , and G = 6 . 67 × 10 − 11 N · m 2 / kg 2 . The gravitational acceleration is g = GM r 2 = ( 6 . 67 × 10 − 11 N · m 2 / kg 2 ) (5 × 10 30 kg) (2000 m) 2 = 8 . 3375 × 10 13 m / s 2 , several hundreds of billion times g on earth. 003 10.0 points Consider four vectors vector F 1 , vector F 2 , vector F 3 , and vector F 4 , with magnitudes are F 1 = 31 N, F 2 = 26 N, F 3 = 25 N, and F 4 = 51 N, and θ 1 = 140 ◦ , θ 2 = − 160 ◦ , θ 3 = 36 ◦ , and θ 4 = − 70 ◦ , measured from the positive x axis with the counter clockwise angular direction as positive. What is the magnitude of the resultant vec tor bardbl vector F bardbl , where vector F = vector F 1 + vector F 2 + vector F 3 + vector F 4 ? 1. 42.9132 2. 22.8643 3. 30.0104 4. 13.7651 5. 12.293 6. 11.3251 7. 44.9152 8. 31.3876 9. 16.697 10. 24.5588 Correct answer: 24 . 5588 N. Version 024 – midterm 01 – turner – (57340) 2 Explanation: The x components are F x = F cos θ , so F 1 x = (31 N) cos(140 ◦ ) = − 23 . 7474 N F 2 x = (26 N) cos( − 160 ◦ ) = − 24 . 432 N F 3 x = (25 N) cos(36 ◦ ) = 20 . 2254 N F 4 x = (51 N) cos( − 70 ◦ ) = 17 . 443 N . and the y components are F y = F sin θ , so F 1 y = (31 N) sin(140 ◦ ) = 19 . 9264 N F 2 y = (26 N) sin( − 160 ◦ ) = − 8 . 8925 N F 3 y = (25 N) sin(36 ◦ ) = 14 . 6946 N F 4 y = (51 N) sin( − 70 ◦ ) = − 47 . 9243 N . The x and y components of the resultant vec tor vector F are F x = F 1 x + F 2 x + F 3 x + F 4 x = ( − 23 . 7474 N) + ( − 24 . 432 N) + (20 . 2254 N) + (17 . 443 N) = − 10 . 511 N and F y = F 1 y + F 2 y + F 3 y + F 4 y = (19 . 9264 N) + ( − 8 . 8925 N) + (14 . 6946 N) + ( − 47 . 9243 N) = − 22 . 1958 N...
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This note was uploaded on 04/13/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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