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A rollercoaster travels through a loop at [tex]$16.3 \, m/s$[/tex] and has a centripetal acceleration of [tex]$12.4 \, m/s^2$[/tex].

What is the radius of the loop?

[tex] r = [?] \, m [/tex]


Sagot :

To determine the radius of the loop in which the rollercoaster is traveling, we need to relate the given quantities using the formula for centripetal acceleration. The formula for centripetal acceleration [tex]\( a \)[/tex] is given by:

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

where:
- [tex]\( a \)[/tex] is the centripetal acceleration,
- [tex]\( v \)[/tex] is the velocity, and
- [tex]\( r \)[/tex] is the radius of the loop.

We are given:
- Velocity, [tex]\( v = 16.3 \, \text{m/s} \)[/tex]
- Centripetal acceleration, [tex]\( a = 12.4 \, \text{m/s}^2 \)[/tex]

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

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

Now substitute the given values into the equation:

[tex]\[ r = \frac{(16.3)^2}{12.4} \][/tex]

Carrying out the calculations, first compute [tex]\( 16.3^2 \)[/tex]:

[tex]\[ 16.3^2 = 265.69 \][/tex]

Next, divide this result by 12.4:

[tex]\[ r = \frac{265.69}{12.4} \approx 21.426612903225806 \][/tex]

Thus, the radius of the loop is:

[tex]\[ r \approx 21.43 \, \text{m} \][/tex]