Answer:
f = 3.97 Hz
Explanation:
Given that,
Centripetal acceleration, [tex]a=13\ m/s^2[/tex]
The radius of motion is 0.02 m
The formula for the centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{13\times 0.02} \\\\v=0.5\ m/s[/tex]
The speed of an object in a circular path is given by :
[tex]v=\dfrac{2\pi r}{t}[/tex]
t is time period
Also, f=1/t (f is frequency)
[tex]f=\dfrac{v}{2\pi r}\\\\f=\dfrac{0.5}{2\pi \times 0.02}\\\\f=3.97\ Hz[/tex]
Hence, the frequency of motion s 3.97 Hz.
The frequency of the motion is 4.1 Hz.
The linear velocity of the of the object is calculated as follows;
[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{ar} \\\\v = \sqrt{13 \times 0.02} \\\\v = 0.51 \ m/s[/tex]
The angular speed of the object is calculated as follows;
[tex]\omega =\frac{v}{r} \\\\\omega = \frac{0.51}{0.02} \\\\\omega = 25.5 \ rad/s[/tex]
The frequency of the motion is calculated as follows;
[tex]\omega = 2\pi f\\\\f = \frac{\omega }{2\pi} \\\\f = \frac{25.5}{2\pi } \\\\f = 4.1 \ Hz[/tex]
Thus, the frequency of the motion is 4.1 Hz.
Learn more about frequency of the circular motion here: https://brainly.com/question/20481765
