Question

What angular speed (in revolutions/second) is needed for a centrifuge to produce an acceleration of 1 000 g at a radius arm of 15.0 cm?

a. 40.7 rev's

b. 75.4 rev/s

c. 81.4 rev/s

d. 151 rev/s

Answer 1

Answer:

a. 40.7 rev's

Explanation:

"Centrifugal force" is the apparent force that a non-inertial observer seems to perceive as a result of the non-inertiality of his reference system. Therefore, in this case we must work with an equivalent force, that is, the centripetal force, which is given by:

$F_c=ma_c$

The centripetal acceleration is defined as:

$a_c=\frac{v^2}{r}$

The linear speed is given by:

$v=\omega r$

So, we have:

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2 r\\\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{1000*9.8\frac{m}{s^2}}{15*10^{-2}m}}\\\omega=255.60\frac{rad}{s}\\\omega=\frac{255.60\frac{rad}{s}}{2\pi}\\\omega=40.7\frac{rev}{s}$

Answer 2

Option A is correct. The angular speed needed will be  40.7 rev/s.The rate of change of the angular distance is known as the angular speed.

What is angular speed?

The angular velocity is the rate at which the angular distance changes. It is measured in rad/sec.

The linear speed is given as;

$\rm v= \omega r$

The value of centripetal acceleration is given as;

$\rm a_c = \frac{v^2}{r}$

$\rm a_c= \frac{(\omega r)^2}{r} \\\\ \rm a_c= \omega^2r \\\\ \omega=\sqrt{\frac{a_c}{r} } \\\\ \omega=\sqrt{\frac{1000\times9.81}{15\times10^{-2}} } \\\\ \omega=40.7 rev/sec$

Hence the angular speed needed will be 40.7 revs.

To learn more about the angular speed refer to the link;

https://brainly.com/question/9684874