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In the depths of space, asteroid miners attach a cable between their 3,694.-kg spaceship and a relatively small 1476-kg asteroid. The ship and asteroid are initially at rest 751 m apart. They begin to "reel in" the asteroid under a constant tension of 512.0 N. We are going to calculate how long it takes take for them to meet. This is a somewhat complicated question involving several different concepts, so let's just take it step by step. 11. Draw free-body diagrams for the asteroid, and for the spaceship. 12. Calculate the magnitudes of the accelerations of the asteroid, and of the spaceship, respectively. (a) aa = 0.3469 m/s²; a, = 0.1386 m/s² (b) aq = 0.1386 m/s²; a, = 0.3469 m/s² (c) aa = 0.1388 m/s²; a, = 0.2775 m/s² (d) aa = 0.2775 m/s²; a, = 0.1388 m/s² 13. Calculate the magnitude of the acceleration of the asteroid, relative to the spaceship. (a) 0.2083 m/s² (b) 0.4163 m/s² (c) 0.1387 m/s² (d) 0.4855 m/s² 14. Calculate the time it takes for the asteroid and the spaceship to arrive at the same location. (a) 104 s (b) 84.9 s (c) 60.1 s (d) 55.6 s Find the gravitational force of attraction between the space-tugboat and the asteroid in the previous question, when they are at their initial separation. Should we have included this force in our considerations?

In the depths of space, asteroid miners attach a cable between their 3,694.-kg spaceship and a relatively small 1476-kg asteroid. The ship and asteroid are initially at rest 751 m apart. They begin to "reel in" the asteroid under a constant tension of 512.0 N. We are going to calculate how long it takes take for them to meet. This is a somewhat complicated question involving several different concepts, so let's just take it step by step. 11. Draw free-body diagrams for the asteroid, and for the spaceship. 12. Calculate the magnitudes of the accelerations of the asteroid, and of the spaceship, respectively. (a) aa = 0.3469 m/s²; a, = 0.1386 m/s² (b) aq = 0.1386 m/s²; a, = 0.3469 m/s² (c) aa = 0.1388 m/s²; a, = 0.2775 m/s² (d) aa = 0.2775 m/s²; a, = 0.1388 m/s² 13. Calculate the magnitude of the acceleration of the asteroid, relative to the spaceship. (a) 0.2083 m/s² (b) 0.4163 m/s² (c) 0.1387 m/s² (d) 0.4855 m/s² 14. Calculate the time it takes for the asteroid and the spaceship to arrive at the same location. (a) 104 s (b) 84.9 s (c) 60.1 s (d) 55.6 s Find the gravitational force of attraction between the space-tugboat and the asteroid in the previous question, when they are at their initial separation. Should we have included this force in our considerations?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Hello I need help with the statement below question 14. Please and thank you
### Space Physics Problem Set In the depths of space, asteroid miners attach a cable between their 3,694-kg spaceship and a relatively small 1,476-kg asteroid. The ship and asteroid are initially at rest 751 m apart. They begin to “reel in” the asteroid under a constant tension of 512.0 N. We are going to calculate how long it takes for them to meet. This is a somewhat complicated question involving several different concepts, so let’s just take it step by step. --- #### 11. Draw Free-body Diagrams **Question:** Draw free-body diagrams for the asteroid, and for the spaceship. --- #### 12. Calculate Accelerations **Question:** Calculate the magnitudes of the accelerations of the asteroid, and of the spaceship, respectively. **Options:** - (a) \( a_a = 0.3469 \, m/s^2 \); \( a_s = 0.1386 \, m/s^2 \) - (b) \( a_a = 0.1386 \, m/s^2 \); \( a_s = 0.3469 \, m/s^2 \) - (c) \( a_a = 0.1388 \, m/s^2 \); \( a_s = 0.2775 \, m/s^2 \) - (d) \( a_a = 0.2775 \, m/s^2 \); \( a_s = 0.1388 \, m/s^2 \) --- #### 13. Relative Acceleration **Question:** Calculate the magnitude of the acceleration of the asteroid, relative to the spaceship. **Options:** - (a) \( 0.2083 \, m/s^2 \) - (b) \( 0.4163 \, m/s^2 \) - (c) \( 0.1387 \, m/s^2 \) - (d) \( 0.4855 \, m/s^2 \) --- #### 14. Time for Meeting **Question:** Calculate the time it takes for the asteroid and the spaceship to arrive at the same location. **Options:** - (a) 104 s - (b) 84.9 s - (c) 60.1 s - (d) 55.6 s --- #### Gravit
Transcribed Image Text:### Space Physics Problem Set In the depths of space, asteroid miners attach a cable between their 3,694-kg spaceship and a relatively small 1,476-kg asteroid. The ship and asteroid are initially at rest 751 m apart. They begin to “reel in” the asteroid under a constant tension of 512.0 N. We are going to calculate how long it takes for them to meet. This is a somewhat complicated question involving several different concepts, so let’s just take it step by step. --- #### 11. Draw Free-body Diagrams **Question:** Draw free-body diagrams for the asteroid, and for the spaceship. --- #### 12. Calculate Accelerations **Question:** Calculate the magnitudes of the accelerations of the asteroid, and of the spaceship, respectively. **Options:** - (a) \( a_a = 0.3469 \, m/s^2 \); \( a_s = 0.1386 \, m/s^2 \) - (b) \( a_a = 0.1386 \, m/s^2 \); \( a_s = 0.3469 \, m/s^2 \) - (c) \( a_a = 0.1388 \, m/s^2 \); \( a_s = 0.2775 \, m/s^2 \) - (d) \( a_a = 0.2775 \, m/s^2 \); \( a_s = 0.1388 \, m/s^2 \) --- #### 13. Relative Acceleration **Question:** Calculate the magnitude of the acceleration of the asteroid, relative to the spaceship. **Options:** - (a) \( 0.2083 \, m/s^2 \) - (b) \( 0.4163 \, m/s^2 \) - (c) \( 0.1387 \, m/s^2 \) - (d) \( 0.4855 \, m/s^2 \) --- #### 14. Time for Meeting **Question:** Calculate the time it takes for the asteroid and the spaceship to arrive at the same location. **Options:** - (a) 104 s - (b) 84.9 s - (c) 60.1 s - (d) 55.6 s --- #### Gravit
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