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lukranit [14]
3 years ago
12

A 3.00 × 10^-6 kg ant rides the second hand of an enormous vertical wall clock. The ant smoothly sweeps a circle of radius 50.0

m. What is the magnitude of the upward force felt by the ant due to the second hand when it is exactly at 6 AM?
Physics
1 answer:
kotykmax [81]3 years ago
4 0

Answer:

1.65×10⁻⁶ N

Explanation:

m = Mass of ant = 3\times 10^{-6}\ kg

r = Radius = 50 m

t = Time taken to complete on rotation = 60 seconds

Angular velocity

\omega=\frac{2\pi}{t}\\\Rightarrow \omega=\frac{2\pi}{60}

Centripetal acceleration is force that is acting outward

F_c=m\omega^2r\\\Rightarrow F_c=3\times 10^{-6}\times \left(\frac{2\pi}{60}\right)^2\times 50\\\Rightarrow F_c=1.65\times 10^{-6}\ N

The magnitude of the upward force felt by the ant due to the second hand would be 1.65×10⁻⁶ N

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