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Archy [21]
3 years ago
6

A particle moves at a constant speed in a circular path with a radius of r=2.06 cm. If the particle makes four revolutions each

t = 1 second, what is the magnitude of its acceleration?
Physics
1 answer:
nataly862011 [7]3 years ago
8 0

The centripetal acceleration is 13.0 m/s^2

Explanation:

For an object in uniform circular motion, the centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the speed of the object

r is the radius of the circle

The speed of the object is equal to the ratio between the length of the circumference (2\pi r) and the period of revolution (T), so it can be rewritten as

v=\frac{2\pi r}{T}

Therefore we can rewrite the acceleration as

a=\frac{4\pi^2 r}{T^2}

For the particle in this problem,

r = 2.06 cm = 0.0206 m

While it makes 4 revolutions each second, so the period is

T=\frac{1}{4}s = 0.25 s

Substituting into the equation, we find the acceleration:

a=\frac{4\pi^2 (0.0206)}{0.25^2}=13.0 m/s^2

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

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