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A 35 kg child moves with uniform circular motion while riding a horse on a carousel. The horse is 3.2 m from the carousel's axis of rotation and has a tangential speed of 2.6 m/s. what is the child's centripetal acceleration?

Answer 1

William Audy

The child's centripetal acceleration is #2.1125m \/s^2#.

Newton's second law relates force #F#, mass #m# and acceleration #a#.

#F=ma#

The centripetal force required to keep an object of mass #m# with tangential velocity #v# at a radius #r# is:

#F=(mv^2)/r#

Combining these two equations gives:

#ma=(mv^2)/r#

Mass #m# is a factor of both sides and cancels out.

#a=v^2/r#

In the case of the child #v=2.6# and #r=3.2#. So the centripetal acceleration #a# is:

#a=2.6^2/3.2=2.1125#

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Answer 2

Carson Alexander

To find the centripetal acceleration of the child, you can use the formula:

$a_c = \frac{v^2}{r}$

Where:

$a_c$ is the centripetal acceleration,
$v$ is the tangential speed,
$r$ is the radius of the circular path.

In this scenario, the tangential speed $v$ of the horse is given as 2.6 m/s, and the radius $r$ of the circular path is 3.2 m.

Substitute the given values into the formula:

$a_c = \frac{(2.6 \, \text{m/s})^2}{3.2 \, \text{m}}$

$a_c = \frac{6.76 \, \text{m}^2/\text{s}^2}{3.2 \, \text{m}}$

$a_c = 2.1125 \, \text{m/s}^2$

So, the child's centripetal acceleration is $2.1125 \, \text{m/s}^2$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.