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3 years ago

11

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.

2 answers:

olga_2 [115]3 years ago

8 0

Answer:

the answer is 84

Explanation:

Marat540 [252]3 years ago

5 0

Answer:

Centripetal acceleration = 83.77m/s²

Explanation:

<u>Given the following data;</u>

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;

$Centripetal \; acceleration, a = \frac {3.3^{2}}{0.13}$

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

<em>Centripetal acceleration = 83.77m/s²</em>

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

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