Alarge centrifuge is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. at what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation

ω = 2.55 rad/sec

Explanation:

Assuming no other external forces acting in the horizontal plane, the only force keeping the  rider in a circular path of a radius equal to his distance to the center of rotation, is the centripetal force.

According to Newton's 2nd law, in the horizontal direction, we have:

F = Fc = m*a = m*ω²*r

We know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m.

Replacing these values in (1), and solving for ω, we get:

ω = √98.0m/s²/ 15.0 m = 2.55 rad/sec

as we probably am aware that our atmosphere or air have various levels, each level has different gasses, and each gas has another porousness. at the point when the light originates from lower porousness to higher penetrability, it alters its course. with the goal that implies that when the light originates from the moon to the earth, it alters it's course a lot more occasions. it essentially implies that the state of the light piece of the moon is changed on the earth.

not copied too! hope this !

joke:

h o h o hope you have a merry christmas!

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm