The magnitude of the string's tension as it is swung in a horizontal circle is 0.744 Newton
Given the data in the question;
- Mass of the ball;[tex]m = 16g = 0.016kg[/tex]
- String is swung in a horizontal circle, Radius; [tex]r = 1.4m[/tex]
- Period of revolution; [tex]T = 1.09s[/tex]
- Magnitude of the string's tension; [tex]F_T = ?[/tex]
First, we need to get magnitude of the centripetal acceleration:
[tex]a_c = \frac{4\pi ^2r}{T^2}[/tex]
We substitute in our given values
[tex]a_c = \frac{4\pi ^2\ *\ 1.4m}{(1.09s)^2} \\\\a_c = 46.519 m/s^2[/tex]
Now, from Newton's Second Law of Motion:
[tex]F = m\ * \ a[/tex]
Where F is the force, m is the mass of the object and a is the acceleration
We substitute into the equation to get the magnitude of the string's tension.
[tex]F = 0.016kg \ *\ 46.519m/s^2[/tex]
[tex]F = 0.744 kg.m/s^2\\\\F = 0.744 N[/tex]
Therefore, the magnitude of the string's tension as it is swung in a horizontal circle is 0.744 Newton
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