Between 1911 and 1990, the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of 1.2 mm/y. The tower is 55 m tall. In radians per second, what is the average angular speed of the tower’s top about its base?

Average angular speed = 6.91 x 10^(-13) rad/s

Explanation:

Linear velocity(v) = 1.2 mm/y or 0.0012 m/y

Height(h) = 55 m

Now, for angular speed, it has to be a circular motion. Thus, since the top of the tower is rotating around it's base, we can say that the height of this tower is the radius of the circular motion at the top.

Thus; from angular velocity equation ;angular velocity is given as;

ω = v/r

So, ω = 0.0012/55 = 2.182 x 10^(-5) rad/y

Now, the SI unit for angular velocity is rad/s

Mow let's convert the answer from rad/year to rad/seconds;

= [2.182 x 10^(-5)] / (365 x 24 x 60 x 60) = [2.182 x 10^(-5)] / 31536000 =

6.91 x 10^(-13) rad/s

i think the answer would be reserve.