the arc from point a to the north pole of a planet subtends a central angle of ( 45^{circ} ), as shown in…

the arc from point a to the north pole of a planet subtends a central angle of ( 45^{circ} ), 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 ( ( 2 pi ) 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? the angular velocity is ( \frac{pi}{10} ) radians per hour. (simplify your answer. type an exact answer, using ( pi ) as needed. use integers or fractions for any numbers in the expression.) the linear velocity is ( square ) miles per hour. (round to one decimal place as needed.)
Answer
Explanation:
Step1: Find the radius of the circular path of Point A
The radius (r) of the circular path of Point A is given by (r = 4250\cos45^{\circ}). Since (\cos45^{\circ}=\frac{\sqrt{2}}{2}), then (r = 4250\times\frac{\sqrt{2}}{2}= 2125\sqrt{2}) miles.
Step2: Use the formula for linear velocity (v=\omega r)
We know that the angular velocity (\omega=\frac{\pi}{10}) radians per hour (given). Substitute (\omega=\frac{\pi}{10}) and (r = 2125\sqrt{2}) into the formula (v=\omega r). (v=\frac{\pi}{10}\times2125\sqrt{2}) (v=\frac{2125\sqrt{2}\pi}{10}) (v=\frac{425\sqrt{2}\pi}{2}) Now, calculate the numerical value: (v=\frac{425\times1.4142\times3.1416}{2}) First, (425\times1.4142 = 600.035) Then (600.035\times3.1416\approx1884.57) Finally, (\frac{1884.57}{2}\approx942.3)
Answer:
The linear velocity is (942.3) miles per hour.