A carousel at the local carnival rotates once every 45 seconds. (a) What is the linear speed of an outer horse on the carousel, which is 2.75 m from the axis of rotation?
(b) What is the linear speed of an inner horse that is 1.75 m from the axis of rotation?

(a) The linear speed of an outer horse on the carousel is 0.384 m/s.

(b) The linear speed of an inner horse on the carousel is 0.244 m/s.

Given data:

The time interval for the rotation of carousel is, t = 45 s.

The distance of the outer horse from the axis of rotation is, r = 2.75 m.

The distance of an inner horse from the axis of rotation is, r' = 1.75 m.

(a)

The linear speed in this problem can be obtained from the concept of rotational mechanic, in which the ratio of the circumference and the time gives required linear speed. So,

v = 2 π r/t

Solving as,

v = 2 π (2.75) / 45

v = 0.384 m/s

Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.384 m/s.

(b)

Now similarly the linear speed of an inner horse is calculated as,

v' = 2 π r' / t

Solving as,

v' = 2 π (1.75) / 45

v' = 0.244 m/s

Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.244 m/s.

Learn more about the rotational mechanics here:

link

We know that the carousel does a complete rotation in 45 seconds.

Then the frequency of this carousel will be f =  1/45 seconds.

And the angular frequency will be 2*pi times the frequency, then we have:

angular frequency = w = 2*3.14*(1/45s) = 0.1396 s^-1

Now, the linear speed of an object that rotates with a radius R, and an angular frequency W is:

S = R*W

then:

a) in this case the radius is 2.75m, then the linear speed is:

S = 2.75m*0.1396 s^-1 = 0.3839 m/s

b) in this case the radius is 1.75m, then the linear speed here is:

S = 1.75m*0.1396 s^-1 = 0.2443 m/s

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg