Answer :
The point on the centrifuge makes Δθ = 5.0×10^3 rev (five thousand revolutions) during the given time period of 147 s.
To solve this problem, we can use the kinematic equation for angular motion: θ = ω_i t + (1/2) α t^2, where θ is the angular displacement, ω_i is the initial angular velocity, α is the angular acceleration, and t is the time.
Given that the centrifuge starts from rest (ω_i = 0) and the point is located 8.5 cm (0.085 m) from the axis of rotation, we can calculate the angular acceleration α using the equation v = ω r, where v is the linear velocity and r is the distance from the axis of rotation.
Substituting the given values (v = 147 m/s, r = 0.085 m) and solving for α, we find α ≈ 2065 rad/s^2.
Now, we can use the equation Δθ = ω_i t + (1/2) α t^2 with ω_i = 0 to calculate the angular displacement. Plugging in the values (t = 147 s, α ≈ 2065 rad/s^2), we find Δθ ≈ 5.0×10^3 rev (five thousand revolutions).
Therefore, during the given time period, the point on the centrifuge makes approximately five thousand revolutions.
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