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GalinKa [24]
3 years ago
12

A point on the rim of a 0.75 m diameter grinding wheel changes speed uniformly from 11 m/s to 20 m/s in 8.1 s. what is the avera

ge angular acceleration of the wheel during this interval?
Physics
1 answer:
Lynna [10]3 years ago
4 0
<span>3.0 sâ»Â˛ We can calculate the tangential acceleration by dividing the change in speed by the elapsed time, e.g. (vâ‚ - vâ‚€) / t = (20 m/s - 11 m/s) / 8.1 s = 1.111 m/s² Now we can calculate the angular acceleration by dividing by the radius. The radius of the wheel is half its diameter, so r = 0.75 m / 2 = 0.375 m Finally, the angular acceleration is 1.111 m/s² / 0.375m = 3.0 sâ»Â˛ The unit sâ»Â˛ is just an alternate way of writing 1/s².</span>
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B.

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Given that a student runs up a flight of stairs which info is not needed to calculate the rate of the student is doing work against gravity A the height of the flight of stairs B the length of flight of stairs C the time taken to run up the stairs D the weight of the student

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