a) Their angular speeds are the same
b) Andrea's tangential speed is twice the value of Chuck's tangential speed
Explanation:
a)
The angular speed of Andrea and Chuck is the same.
Let's call
the angular speed at which the merry-go-round is rotating. We know that the angular speed is defined as:

where
is the angular displacement covered in one revolution
T is the period of revolution
The merry go round is a rigid body, so all its point cover the same angular displacement in the same time: this means that it doesn't matter where Andrea and Chuck are located along the merry-go-round, their angular speed will still be the same.
b)
For an object in circular motion, the tangential speed is given by

where
is the angular speed
r is the distance from the centre of rotation
Here let's call
the distance at which Chuck is rotating, so his tangential speed is

Now we know that Andrea is rotating twice as far from the centre, so at a distance of

So his tangential speed is

So, Andrea's tangential speed is twice the value of Chuck's tangential speed.
Learn more about circular motion:
brainly.com/question/2562955
brainly.com/question/6372960
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