Answer:
ω = 15275.25 rad/s
Explanation:
Given that,
Radial acceleration of an ultracentrifuge is, [tex]a=5\times 10^5g[/tex]
Distance from the axis, r = 2.1 cm = 0.021 m
g is the free-fall acceleration such that g = 9.8 m/s²
We need to find the angular speed of an ultracentrifuge. The formula that is used to find the angular speed is given by formula as follows :
[tex]a=r\omega^2[/tex]
Putting all the values,
[tex]\omega=\sqrt{\dfrac{a}{r}} \\\\\omega=\sqrt{\dfrac{5\times 10^5\times 9.8}{0.021}} \\\\\omega=15275.25\ rad/s[/tex]
So, the required angular speed, ω, of an ultracentrifuge is 15275.25 rad/s.
