Find the linear speed of the bottom of a test tube in a centrifuge if the centripetal acceleration there is 5.5×104 times the acceleration of gravity. The distance from the axis of rotation to the bottom of the test tube is 8.0 .
my butt
To find the linear speed of the bottom of the test tube in the centrifuge, we can use the formula for centripetal acceleration:
a = ω²r
where:
a = centripetal acceleration
ω = angular velocity
r = distance from the axis of rotation
In this case, we are given that the centripetal acceleration is 5.5×104 times the acceleration of gravity. So we can rewrite the equation as:
5.5×104g = ω²r
where g is the acceleration due to gravity.
We can rearrange the equation to solve for ω:
ω = √(5.5×104g/r)
Next, we can use the formula for linear speed:
v = ωr
Substituting the value of ω, we get:
v = √(5.5×104g/r) * r
Simplifying further:
v = √(5.5×104g * r)
Finally, substituting the given distance from the axis of rotation:
v = √(5.5×104 * 9.8 * 8.0)
Calculating the value gives:
v ≈ 374.038 m/s
Therefore, the linear speed of the bottom of the test tube in the centrifuge is approximately 374.038 m/s.
V^2/R = 5.5*10^4*9.8 m/s^2.
R must be in meters. You do not say what the units are.
Solve the equation above for V.