A stone is whirled at the end of a string. It moves at constant speed in a circle of radius 106 cm, making a complete circle once every 1.38 seconds. What is the centripetal acceleration of the stone?
A silver cube has sides that measure 2.5 meters. The cube is heated from 15°C to 30°C. What is the change in its volume?
To find the centripetal acceleration of the stone, we need to use the formula for centripetal acceleration, which is given by:
a = (v^2) / r
where:
a = centripetal acceleration
v = tangential velocity
r = radius of the circle
First, let's find the tangential velocity (v) of the stone. The formula for tangential velocity is given by:
v = (2πr) / T
where:
π = pi (approximately 3.14)
T = time period (time taken to complete one revolution)
From the question, we are given that the time period (T) is 1.38 seconds and the radius (r) is 106 cm.
Substituting these values into the formula, we get:
v = (2 * 3.14 * 106) / 1.38
Now, let's calculate the tangential velocity.
v = 1623.188 cm/s (rounded to three decimal places)
Next, we can substitute the tangential velocity (v) and the radius (r) into the formula for centripetal acceleration:
a = (v^2) / r
Plugging in the values, we have:
a = (1623.188^2) / 106
Now, let's calculate the centripetal acceleration.
a = 24,877.3556 cm/s^2 (rounded to four decimal places)
Therefore, the centripetal acceleration of the stone is approximately 24,877.3556 cm/s^2.