the arc from point a to the north pole of a planet subtends a central angle of 45°, as shown in the figure…

the arc from point a to the north pole of a planet subtends a central angle of 45°, as shown in the figure to the right. the radius of the planet is 4250 mi. any point on the surface of the planet (except at the poles) makes one revolution (2π radians) about the axis of the planet in 20 hours. what are the angular and linear velocities for point a with respect to its rotation around the axis of the planet? (simplify your answer. type an exact answer, using π as needed. use integers or fractions for any numbers in the expression.) the linear velocity is miles per hour. (round to one decimal place as needed.)

the arc from point a to the north pole of a planet subtends a central angle of 45°, as shown in the figure to the right. the radius of the planet is 4250 mi. any point on the surface of the planet (except at the poles) makes one revolution (2π radians) about the axis of the planet in 20 hours. what are the angular and linear velocities for point a with respect to its rotation around the axis of the planet? (simplify your answer. type an exact answer, using π as needed. use integers or fractions for any numbers in the expression.) the linear velocity is miles per hour. (round to one decimal place as needed.)

Answer

Explanation:

Step1: Find the angular velocity

The formula for angular velocity (\omega) is (\omega=\frac{\theta}{t}), where (\theta) is the central angle and (t) is the time. Given (\theta = 2\pi) radians and (t = 20) hours. (\omega=\frac{2\pi}{20}=\frac{\pi}{10}) radians per hour.

Step2: Find the radius of the circular path of Point A

The radius (r) of the circular path of Point A is (r = 4250\cos45^{\circ}) (using the property of right - triangle in the cross - section of the planet). Since (\cos45^{\circ}=\frac{\sqrt{2}}{2}), then (r = 4250\times\frac{\sqrt{2}}{2}=2125\sqrt{2}) miles.

Step3: Find the linear velocity

The formula for linear velocity (v) is (v = r\omega). Substitute (r = 2125\sqrt{2}) and (\omega=\frac{\pi}{10}) into the formula. (v=(2125\sqrt{2})\times\frac{\pi}{10}=\frac{425\sqrt{2}\pi}{2}) (exact value). For the approximate value: (v=\frac{425\times1.414\times3.14}{2}) First, (425\times1.414 = 425\times(1 + 0.4+0.014)=425+170 + 5.95=595 + 5.95 = 600.95) Then (600.95\times3.14=(600 + 0.95)\times3.14=1884+2.983 = 1886.983) (v=\frac{1886.983}{2}=943.5) (approximate value)

Answer:

The exact linear velocity is (\frac{425\sqrt{2}\pi}{2}) miles per hour and the approximate linear velocity is (943.5) miles per hour.