Get the information you need quickly and easily with IDNLearn.com. Ask anything and receive immediate, well-informed answers from our dedicated community of experts.

A roller coaster travels through a loop at [tex]$14.2 \, \text{m/s}$[/tex] and has a centripetal acceleration of [tex]$13.6 \, \text{m/s}^2$[/tex].

What is the radius of the loop?
[tex]\[ r = [?] \, \text{m} \][/tex]


Sagot :

To solve for the radius of the loop in which the rollercoaster travels, we need to use the formula for centripetal acceleration. The formula for centripetal acceleration [tex]\(a_c\)[/tex] is given by:

[tex]\[ a_c = \frac{v^2}{r} \][/tex]

Where:
- [tex]\(a_c\)[/tex] is the centripetal acceleration,
- [tex]\(v\)[/tex] is the velocity of the object,
- [tex]\(r\)[/tex] is the radius of the circular path.

In the given problem:
- The velocity [tex]\(v\)[/tex] of the rollercoaster is [tex]\(14.2 \, \text{m/s}\)[/tex],
- The centripetal acceleration [tex]\(a_c\)[/tex] is [tex]\(13.6 \, \text{m/s}^2\)[/tex].

To find the radius [tex]\(r\)[/tex], we rearrange the centripetal acceleration formula to solve for [tex]\(r\)[/tex]:

[tex]\[ r = \frac{v^2}{a_c} \][/tex]

Now, we can plug in the given values:

[tex]\[ r = \frac{(14.2 \, \text{m/s})^2}{13.6 \, \text{m/s}^2} \][/tex]

Performing the calculation step-by-step:
1. Square the velocity:
[tex]\[ (14.2 \, \text{m/s})^2 = 201.64 \, \text{m}^2/\text{s}^2 \][/tex]

2. Divide the result by the centripetal acceleration:
[tex]\[ r = \frac{201.64 \, \text{m}^2/\text{s}^2}{13.6 \, \text{m/s}^2} = 14.826470588235294 \, \text{m} \][/tex]

So, the radius of the loop is approximately:

[tex]\[ r \approx 14.83 \, \text{m} \][/tex]

Hence, the radius [tex]\(r\)[/tex] of the loop is [tex]\(14.83\)[/tex] meters.
Your engagement is important to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.