Question

The top of a building sways around in a circle. OA = 50 cm. T = 9.3 seconds. Then v = _____ m/s.

Answer 1

The speed of the top of the building swaying in a circle with a radius of 0.50 meters and a period of 9.3 seconds is approximately 0.107 meters per second.

Step-by-step explanation:

In order to calculate the speed (v) of the top of a building swaying around in a circle with a radius of OA = 0.50 meters (since 50 cm equals 0.50 meters) and a period (T) of 9.3 seconds, we have to use the formula for the speed of an object moving in a circle: v = 2πr/T.

Now, using the given values:

1. Convert the radius to meters: r = 0.50 meters.
2. Use the period T = 9.3 seconds.
3. Calculate the speed using the formula v = 2π(0.50 meters) / 9.3 seconds.

Therefore, the speed of the top of the building swaying is:

v = (2π × 0.50 m) / 9.3 s

≈ (3.1416 × 1) / 9.3 s

≈ 0.107 m/s.

The top of the building sways at approximately 0.107 meters per second.