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A tire 0.650 m in radius rotates at a constant rate of 210 rev/min. Find the speed of a small stone lodged in the tread of the tire (on its outer edge). (Hint: In one revolution, the stone travels a distance equal to the circumference of its path, 2πr.)

Respuesta :

Answer:

v = 14.3 m/s

Explanation:

As we know that frequency of rotation is given by

[tex]f = 210 rev/min[/tex]

[tex]f = 210\times \frac{1}{60} hz[/tex]

[tex]f = 3.5 hz[/tex]

now the angular speed is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(3.5 hz)[/tex]

[tex]\omega = 7\pi rad/s[/tex]

now the tangential speed is given as

[tex]v = r\omega[/tex]

[tex]v = (0.650 m)(7\pi) m/s[/tex]

[tex]v = 14.3 m/s[/tex]

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