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Kamila [148]
3 years ago
10

An aluminum pie plate of radius 0.12 m is spins and accelerates at a constant rate from 13 rad/s to 29 rad/s in 5.4 s. What was

its angular displacement during that time

Physics
1 answer:
kondor19780726 [428]3 years ago
7 0

Explanation:

Below is an attachment containing the solution.

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A bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s. What is the angular acceleration of the tires?
Nimfa-mama [501]

The angular acceleration of a bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s will be 2. 57 rad/s²

<h3>What is angular acceleration?</h3>

Angular acceleration can be defined as the time it takes for a change in angular velocity. It is denoted as 'α' with a unit of rads/s²

It is expressed thus;

α= Δω ÷ Δt

Where α = angular acceleration

Δw = change in angular velocity = velocity ÷ radius

Δ t = change in time

<h3>How to calculate the angular acceleration</h3>

Using the formula:

α= Δω ÷ Δt

v = 5. 33m/s, r = 0. 33m and t = 6.27s

Substitute the values to get the angular velocity

Δw = v÷ r = 5. 33 ÷ 0.330 = 16. 15 m/s

Substitute the value of Δw into the equation

α= Δω ÷ Δt = 16. 15 ÷ 6. 27 = 2. 57 rad/s²

Therefore, the angular acceleration of a bike with tires radius of 0. 330m, speed of 5. 33mls in 6. 27s is 2. 57 rad/s²

Learn more about angular acceleration here:

brainly.com/question/21278452

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Answer:

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Explanation:

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