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What would be the resulting period of rotation of the space station? You may assume that the space station has a circumference of approximately 3000 $\mathrm{m} .$

A. 0.6 $\mathrm{s}$B. 3 $\mathrm{s}$C. 42 $\mathrm{s}$D. 60 $\mathrm{s}$E. 380 $\mathrm{s}$

See solution.

Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

University of Michigan - Ann Arbor

University of Washington

Simon Fraser University

Hope College

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So if you think that the space station is rotating at a constant speed, then we know that if we have a certain like this and let's say the space station is around here, it's rotating around the center. Then the circumference if we call the circumference, is C. So that's the total distance that the space station is covering for one time period. And if we divide that total circumference by by the time period e, we see that for a constant velocity that equals to the velocity. From there we can solve for the time period which is see over fee, Where is the speed of the space station and c the circumference? Since we're already given this our conference and speed, we can use the numbers directly. So for the circumference, we know it's 3,000 meters and the velocity is 50. Meet us for a second. From there we see that the time period is 60 seconds. So the correct answer will be number D. Thank you

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