Acomputer hard disk 8.0 cm in diameter is initially at rest. a small dot is painted on the edge of the disk. the disk accelerates at 610 rad/s2 for 12s, then coasts at a steady angular velocity for another 12s.
a. what is the speed of the dot at t = 1.0 s? 12m/s
b. through how many revolutions has it turned?

a) 282.8 m/s

b) 13812 revolutions

Explanation:

a) The aceleration(a) is the angular velocity(w) divide by the time. The disk had cosated for 12 s in a constant angular velocity, so it will be:

a = w/t

610 = w/12

w = 7230 rad/s

The angular velocity is related by the linear velocity:

V = w*r, where r is the radius of the circumference ( r = 0.08/2 = 0.04 m)

V = 7230*0.04 = 292.8 m/s

b) The number of revolutions is the frequency (f) of the moviment, which is related to the angular velocity as:

w = 2π*f

f = w/2π

f = 7230/6.28

f = 1151 revolutions/s

So, the total revolutions (n) is 12 s was:

n = 12*1151 = 13812 revolutions

layer a is the exosphere and layer b is the thermosphere.   is the answer i think but i hope this .

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