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3 years ago

5

A 0.30-m-radius automobile tire accelerates from rest at a constant 2.0 rad/s2. What is the centripetal acceleration of a point

on the outer edge of the tire after 5.0 s

1 answer:

lianna [129]3 years ago

8 0

Answer:

The centripetal acceleration is 30 m/s²

Explanation:

Given that,

Radius = 0.30 m

Acceleration = 2.0 rad/s²

Time = 5.0

We need to calculate the centripetal acceleration

Using formula of angular velocity

$\omega=a\times t$

Where,

a = acceleration

t = time

Put the value into the formula

$\omega=2.0\times5.0$

$\omega=10\ rad/s$

Using formula of centripetal acceleration

$a_{c}=r\omega^2$

$a_{c}=0.30\times10^2$

$a_{c}=30\ m/s^2$

Hence, The centripetal acceleration is 30 m/s²

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