Question

The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. The centripetal acceleration of the edge of the disc is ?m/s2.

Answer 1

the answer is 84

Step-by-step explanation:

Answer 2

Centripetal acceleration = 83.77m/s²

Step-by-step explanation:

Given the following data;

Radius, r = 0.13m

Velocity, v = 3.3m/s

To find centripetal acceleration;

Centripetal acceleration is given by the formula;

`Acceleration, a = \frac {v^(2)}{r}`

Substituting into the equation, we have;

`Centripetal \; acceleration, a = \frac {3.3^(2)}{0.13}`

`Centripetal \; acceleration, a = \frac {10.89}{0.13}`

Centripetal acceleration = 83.77m/s²

Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s².