This question involves the concepts of tension force, centripetal force, and time period.
The tension force in the string is "0.88 N".
We will find the angular speed of the ball:
[tex]\omega = \frac{2\pi}{T}[/tex]
where,
ω = angualar speed = ?
T = time period = 0.9 s
[tex]\omega = \frac{2\pi}{0.9\ s}[/tex]
ω = 6.98 rad/s
We will find the linear speed of the ball:
v = rω
whre,
v = linear speed = ?
r = radius of circular path = length of string = 0.3 m
Therefore,
v = (0.3 m)(6.98 rad/s)
v = 2.09 m/s
We will find the tension force in the string:
In order for the ball to stay on the circular path the tension force in the string must be equal to the centripetal force:
[tex]Tension = Centripetal\ Force\\\\ T = \frac{mv^2}{r}=\frac{(0.06\ kg)(2.09\ m/s)^2}{0.3\ m}[/tex]
T = 0.88 N
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