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A girl whirls a rock is a horizontal circle of diameter 5.2 m and at height 8.56 m above level ground. The string breaks and the rock flies off tangentially and strikes the ground after travelling a horizontal distance of 23.52 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?

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The magnitude of the centripetal acceleration of the stone while in circular motion is 31.7 m/s².

First, we need to find the initial velocity of the rock before it breaks free. We can use the formula for the velocity of a point on the circumference of a circle:

V = ωr

Where V is the velocity, ω is the angular velocity, and r is the radius. Since the rock strikes the ground after traveling a horizontal distance of 23.52 m, we can set up the following equation:

23.52 = ω(5.2/2)

Simplifying, we get:

ω = (23.52 × 2)/5.2 = 9 m/s

Now, we can find the time it takes for the rock to fall to the ground using the formula:

t = √(2h/g)

Where t is the time, h is the height, and g is the acceleration due to gravity. Plugging in the values:

t = √(2 × 8.56/9.8) = 1.5 s

The centripetal acceleration is given by the formula:

a = v² / r

Where a is the acceleration, v is the velocity, and r is the radius. Plugging in the values:

a = 9² / (5.2/2) = 31.7 m/s²

So, the magnitude of the centripetal acceleration of the stone while in circular motion is 31.7 m/s².

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