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a stone tied to the end of a string 80 cm long is...

Question:
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

Answer:

Length of the string, l = 80 cm = 0.8 m

Number of revolutions = 14

Time taken = 25 s

Frequency, v = Number of revolutions / time taken = 14 / 25 Hz

Angular frequency, ω = 2πν

= 2 x 22/7 x 14/25 = 88 / 25 rad s^-1

Centripetal acceleration, a_c = ω² r

= (88 / 25)² x 0.8

= 9.91 m/s²

The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

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Question:
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

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Question: A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

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Question:
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?