Respuesta :
The centripetal acceleration is [tex]13.0 m/s^2[/tex]
Explanation:
For an object in uniform circular motion, the centripetal acceleration is given by
[tex]a=\frac{v^2}{r}[/tex]
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 ([tex]2\pi r[/tex]) and the period of revolution (T), so it can be rewritten as
[tex]v=\frac{2\pi r}{T}[/tex]
Therefore we can rewrite the acceleration as
[tex]a=\frac{4\pi^2 r}{T^2}[/tex]
For the particle in this problem,
r = 2.06 cm = 0.0206 m
While it makes 4 revolutions each second, so the period is
[tex]T=\frac{1}{4}s = 0.25 s[/tex]
Substituting into the equation, we find the acceleration:
[tex]a=\frac{4\pi^2 (0.0206)}{0.25^2}=13.0 m/s^2[/tex]
Learn more about centripetal acceleration:
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The direction of the particle changes constantly when it is moving in a circular path, which results in the change of velocity of the particle.
The centripetal acceleration of the particle is 13.12 m/s2.
What is the centripetal acceleration?
The centripetal acceleration is defined as the change in the velocity of the object with the change in the direction of the object when it is moving a circular path.
Given that the radius of the circular path is r=2.06 cm.
For the uniform circular motion v, the centripetal acceleration a is given by
[tex]a = \dfrac {v^2}{r}[/tex]
The particle makes four revolutions each t = 1 second, hence for 1 revolution, the time period will be,
[tex]T = \dfrac {1}{4} \;\rm s[/tex]
[tex]T = 0.25 \;\rm s[/tex]
The speed of the particle is given as below.
[tex]v = \dfrac {2\pi r}{T}[/tex]
[tex]v= \dfrac {2\times 3.14 \times 0.0206}{0.25}[/tex]
[tex]v = 0.52 \;\rm m/s[/tex]
The acceleration of the particle is given below.
[tex]a = \dfrac {0.52^2}{0.0206}[/tex]
[tex]a = 13.12 \;\rm m/s^2[/tex]
Hence we can conclude that the centripetal acceleration of the particle is 13.12 m/s2.
To know more about the centripetal acceleration, follow the link given below.
https://brainly.com/question/17689540.
