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Suppose that Uranus rotates on its axis once every 17.2 hours. The equator lies on a circle with a diameter of 31,762 miles.
(a) Find the angular speed of a point on its equator in radians per day (24 hours).
(b) Find the linear speed of a point on the equator in miles per day.
Do not round any intermediate computations, and round your answer to the nearest whole number.

Respuesta :

a) Angular speed of a point on Uranus equator in radians per day (24 hours) is 9 rad/day

b) The linear speed of a point on the equator in miles per day is 278457 miles/day.

The angular speed is able to quantify how much a body turns around an axis. It is equal to the change in angular displacement over time or the product of the frequency and 2π radians:

                                     w = 2Πf   or,   w = 2Π/t       (1)

The linear speed (v) is the product of the angular speed (ω) and the radius (r):

                                     v = r* ω             (2)

We have been given,   t = 17.2hrs.

                                        r = 31762miles

a) For the angular speed we will use the formula given in equation (1).

                            w = 2Π/17.2  rad/hr.

                            w = (2Π/17.2)*(24hrs./1day)

                             w = (2*3.1415927*24)/17.2 rad/day

                             w = 8.767 rad/day ≈ 9 rad/day

b) For the linear speed we will use the formula given in equation (2).

                            v = 31762 * 8.767 miles/day

                            v = 278457.454 miles /day ≈ 278457 miles per day

Hence angular speed is 9 rad/day and linear speed is 278457miles/day.

Learn more about angular speed here : https://brainly.com/question/20432894

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