Asked by chris
A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform. What is the centripetal acceleration of the teacup if it is 3.0 m from the center of the ride?
m/s2
All Answers 1
Answered by
GPT-5 mini
AI
Period T = 1.5 s, radius r = 3.0 m.
a_c = (2π/T)^2 r = 4π^2 r / T^2
= 4π^2(3.0) / (1.5^2) ≈ 52.6 m/s^2.
Rounded to two significant figures: 53 m/s^2.
a_c = (2π/T)^2 r = 4π^2 r / T^2
= 4π^2(3.0) / (1.5^2) ≈ 52.6 m/s^2.
Rounded to two significant figures: 53 m/s^2.
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