Posted by **Sasha** on Tuesday, November 3, 2009 at 8:39pm.

a child sitting 1.10 m from the center of a merry go round moves with a speed of 1.25 m/s. calculate the centripetal acceleration of the child and the net horizontal force exerted on the child (mass = 25 kg).

- PHYSICS -
**drwls**, Tuesday, November 3, 2009 at 9:05pm
The centripetal acceleration is

a = V^2/R and the force is

F = m a = m V^2/R.

m = 25 kg

V = 1.25 m/s

R = 1.10 m

Do the numbers.

- PHYSICS -
**Sasha**, Tuesday, November 3, 2009 at 9:18pm
so i did a = 1.25^/1.1 = 1.42 m/s^2

then F =ma = 25(1.42) = 35.5N is that right?

## Answer this Question

## Related Questions

- Physics - A child sitting 1.39 m from the center of a merry-go-round moves with ...
- physics - A merry go round makes 2 revolutions every 2 s. A child is sitting 1....
- physics - A child sitting 1.50m from the center of a merry-go-around moves with...
- Physics - A child on a merry-go-round is moving with a speed of 1.40 m/s when 1....
- Physics-circular force - A child on a merry-go-round is moving with a speed of 1...
- 11th grade - A child sitting 1.08 m from the center of a merry-go-round moves ...
- 11th grade - A child sitting 1.08 m from the center of a merry-go-round moves ...
- physics - A child moves with a speed of 5.40 m/s when positioned 14.0 m from the...
- physics - 1.A merry-go-round makes one complete revo- lution in 12.2 s. A 49.4 ...
- physics - is the acceleration of a child sitting near the center of a merry go ...

More Related Questions