a ferris wheel rotates in a circular motion w...
a ferris wheel rotates in a circular motion with a speed of 4.62 m/s. at the top of the ferris wheel, passengers experience a centripetal acceleration of 8.11 m/s². what is the radius of its motion? ? m
Answer
# Explanation:
## Step1: Recall centripetal - acceleration formula
The formula for centripetal acceleration is $a_c=\frac{v^{2}}{r}$, where $a_c$ is centripetal acceleration, $v$ is speed, and $r$ is the radius of the circular path.
## Step2: Rearrange the formula to solve for $r$
We can rewrite the formula $a_c = \frac{v^{2}}{r}$ as $r=\frac{v^{2}}{a_c}$.
## Step3: Substitute given values
Given $v = 4.62\ m/s$ and $a_c=8.11\ m/s^{2}$, we substitute these values into the formula: $r=\frac{(4.62)^{2}}{8.11}$.
First, calculate $(4.62)^{2}=4.62\times4.62 = 21.3444$.
Then, $r=\frac{21.3444}{8.11}\approx2.63\ m$.
# Answer:
$2.63$