Final answer:
The period and frequency of the disk can be determined using the given rotational speed. The speed of a speck of dust on the outside edge of the disk can also be calculated. The acceleration experienced by the speck of dust depends on the angular acceleration, which is not provided in the question.
Step-by-step explanation:
(a) The period of a rotating disk is the time taken for one complete revolution. In this case, the disk is rotating at 10,000 rpm (revolutions per minute), so we need to convert it to revolutions per second.
To convert from rpm to rev/s, we can use the formula:
1 rev/s = (1 min / 60 s) x (1 rev / 1 min)
So the frequency of the disk is:
Frequency = 10,000 rpm x (1 min / 60 s) x (1 rev / 1 min) = 166.67 rev/s
(b) The speed of a speck of dust on the outside edge of the disk can be calculated using the formula:
Speed = circumference x angular velocity
Since the disk has a diameter of 12 cm, its circumference is:
Circumference = π x diameter = 3.14 x 12 cm = 37.68 cm
And the angular velocity of the disk can be converted from rpm to rev/s using the same formula as above:
Angular velocity = 10,000 rpm x (1 min / 60 s) x (1 rev / 1 min) = 166.67 rev/s
So the speed of the speck of dust is:
Speed = 37.68 cm x 166.67 rev/s = 6270.33 cm/s
(c) The acceleration experienced by the speck of dust can be calculated using the formula:
Acceleration = angular acceleration x radius
Since the angular acceleration is not given in the question, we cannot calculate the exact acceleration in units of g without additional information.