the turntable of a record player rotates at a rate of 11 1/3. A record with a radius of 15cm is being played. a punk kid throws a piece of gum that lands right on the edge of the record. What linear distance in meters does the gum travel if it stays on the record for 10 minutes? what will its acceleration be during this time?
What are the units on the rate you gave us? Meters per second, radians per second, or revolutions per second, perhaps?
To find the linear distance traveled by the gum, we need to determine the circumference of the record and multiply it by the number of revolutions the turntable makes in 10 minutes.
The rate of the turntable rotation you provided, 11 1/3, needs to be converted to a specific unit in order to proceed with the calculations. Assuming this rate represents the number of revolutions per minute, we can convert it to revolutions per second by dividing it by 60.
11 1/3 revolutions per minute = (11 + 1/3) revolutions per minute = (11 + 1/3) / 60 revolutions per second
Now, we can find the circumference of the record using the formula C = 2πr, where r is the radius of the record.
C = 2π(15cm) = 2π(0.15m) = 0.3π meters
Next, we calculate the number of revolutions the turntable makes in 10 minutes, which is the product of the rate and the time.
Number of revolutions = (rate) * (time) = [(11 + 1/3) / 60] * 10 minutes
Since we want to find the linear distance traveled by the gum, we multiply the circumference by the number of revolutions.
Linear distance = (circumference) * (number of revolutions)
Linear distance = (0.3π meters) * [(11 + 1/3) / 60] * 10 minutes
To calculate the acceleration of the gum, we need to know the initial and final velocities. Without this information, we cannot directly determine the acceleration. However, if we assume the gum starts from rest and ends at rest, then the acceleration will be zero.
Therefore, the linear distance traveled by the gum is (0.3π meters) * [(11 + 1/3) / 60] * 10 minutes, and its acceleration, assuming it starts and stops at rest, is zero.