Question

1. the ferris wheel at the texas fair has a radius of 15 m and is turning with a velocity of 4 m/s. what is its centripetal acceleration? 5) a car tire spins with a centripetal acceleration of 22 m/s² and has a radius of 0.5 m. what is the tangential velocity of the tire?

Explanation:

Question 4

Step1: Recall centripetal acceleration formula

The formula for centripetal acceleration \( a_c \) is \( a_c = \frac{v^2}{r} \), where \( v \) is the tangential velocity and \( r \) is the radius of the circular path.

Step2: Substitute given values

Given \( v = 4 \, \text{m/s} \) and \( r = 15 \, \text{m} \). Substitute these into the formula: \( a_c = \frac{(4)^2}{15} = \frac{16}{15} \approx 1.07 \, \text{m/s}^2 \).

Step1: Recall centripetal acceleration formula

We use the same formula \( a_c = \frac{v^2}{r} \), and we need to solve for \( v \). Rearranging the formula gives \( v = \sqrt{a_c \cdot r} \).

Step2: Substitute given values

Given \( a_c = 22 \, \text{m/s}^2 \) and \( r = 0.5 \, \text{m} \). Substitute these into the rearranged formula: \( v = \sqrt{22 \times 0.5} = \sqrt{11} \approx 3.32 \, \text{m/s} \).

Answer:

The centripetal acceleration is approximately \( \boldsymbol{1.07 \, \text{m/s}^2} \) (or \( \frac{16}{15} \, \text{m/s}^2 \)).

Question 5