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

## Answers

Answered by

15

**Answer:**

Length of string = 80cm= 0.8m

Stone makes 14 revolution in 25 sec

therefore, Stone makes 14/25 revolution in 1 sec

frequency= 14/25 sec^-1

We know

Angular speed (ω) = 2π×frequency

=2π×14/25

=28π/25 rad/sec

Now acceleration(a) =ω²r

where r is the radius { length of the string }

Acceleration =0.8 × {28π/25}²

= 9.91 m/s²

**Explanation:**

*Hope** **this** **helps** *

Answered by

0

**Answer:**

Acceleration of the stone is 9.91 m/s² and direction along the radius towards the Centre.

**Explanation:**

r = 80 cm = 0.8 m

f = 14/25 = 0.56

ω = 2πf = 2π ×0.56 = 3.52

= rω² = 0.8×(3.52)² = 9.91 m/s²

= centripetal acceleration ⇒ along the radius towards the Centre.

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