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wariber [46]
3 years ago
7

A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform. What is the centripet

al acceleration of the teacup if it is 3.0 m from the center of the ride?
Physics
2 answers:
deff fn [24]3 years ago
6 0

Answer:

Centripetal acceleration, 52.6 m/s²

Explanation:

It is given that,

A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform.

Distance covered by the particle in circular path is, d=2\pi r

We have to find centripetal acceleration of the teacup, a_c=\dfrac{v^2}{r}

Where, v is the velocity v=\dfrac{2\pi r}{t}

a_c=\dfrac{4\pi^2 r}{t^2}

a_c=\dfrac{4\pi^2\times 3}{(1.5)^2}

a_c=52.6\ m/s^2

Hence, centripetal acceleration of the teacup if it is 3.0 m from the center of the ride is 52.6 m/s²

arlik [135]3 years ago
5 0
The correct answer is 53 meters per second squared 
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