A student analyzes data of the motion of a planet as it orbits a star that is in deep space. The orbit of the planet is consider
2 answers:
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Answer:
0.453 m/s
Explanation:
Assuming the handle has diameter of 0.4 m while inner part diameter is 0.1 m then the circumference of outer part is where d is diameter and subscript h denote handle. By substituting 0.4 for the handle's diameter then cirxumference of outer part is
The rate of rotation will then be 1.81/1.256=1.441 rev/s
Similarly, circumference of inner part will be where subscript i represent inner. Substituting 0.1 for inner diameter then
The rate of rotation found for outer handle applies for inner hence speed will be 0.3142*1.441=0.453 m/s