A test tube in a centrifuge is pivoted so that it swings out horizontally as the machine builds up speed. If the bottom of the tube is 180.0 mm from the central spin axis, and if the machine hits 54000 rev/min, what would be the centripetal force exerted on a giant amoeba of mass 8.00E-9 kg at the bottom of the tube?

To find the centripetal force exerted on the giant amoeba, we can use the following formula:

Centripetal Force (Fc) = mass (m) x radial distance (r) x angular velocity (ω)^2

First, let's convert the angular velocity from rev/min to rad/s. We know that 1 rev/min is equal to 2π rad/min. Therefore:

54000 rev/min * (2π rad/1 rev) * (1 min/60 s) = 5669.9 rad/s

Now we can calculate the centripetal force:

Fc = (8.00E-9 kg) * (180.0 mm) * (0.001 m/1 mm) * (5669.9 rad/s)^2

Fc = 8.00E-9 kg * 0.1800 m * 5669.9^2 rad^2/s^2

Fc ≈ 8.00E-9 kg * 0.1800 m * 32,115,757 rad^2/s^2

Fc ≈ 4.61 N

Therefore, the centripetal force exerted on the giant amoeba at the bottom of the tube would be approximately 4.61 Newtons.