Question

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.)

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.