Post a New Question

Physics

posted by on .

A typical laboratory centrifuge rotates at 4100 rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations.
(a) What is the acceleration at the end of a test tube that is 11 cm from the axis of rotation?

dropped from a height of 1.2 m and stopped in a 1.0 ms long encounter with a hard floor?

  • Physics - ,

    acceleration in the centifruge
    w^2r=(4100rpm*1min/60 sec * 2PIrad/sec)^2*.11m = appx 275 g's. Work it out. check my math.

    Now dropping it,
    1/2 mv^2=mgh
    v=sqrt(2gh)=sqrt(2*9.8*1.2)=4.95m/s

    a=changevelocity/time=4.95/.001=4950m/s^2=about 500 g's

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question