1) Allison's angular velocity: 0.262 rad/h
2) Allison's linear velocity: 837.6 mi/h
Step-by-step explanation:
1)
The angular velocity of a body is the rate of change of the angular position. Mathematically:
![\omega=\frac{\Delta \theta}{\Delta t}](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B%5CDelta%20%5Ctheta%7D%7B%5CDelta%20t%7D)
where
is the angular displacement
is the time taken
Allison makes one complete revolution around the Earth's axis in 1 day (24 hours), so
![\Delta t = 24 h](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%2024%20h)
The angular displacement instead is
![\Delta \theta=2\pi rad](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%3D2%5Cpi%20rad)
which corresponds to the angle of one complete revolution. Therefore, Allison's angular velocity is
![\omega=\frac{2\pi}{24}=0.262 rad/h](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%5Cpi%7D%7B24%7D%3D0.262%20rad%2Fh)
2)
The linear velocity of a body in circular motion is given by
![v=\omega r](https://tex.z-dn.net/?f=v%3D%5Comega%20r)
where
is the angular velocity
r is the distance from the axis of rotation
For Allison, we have
![\omega=0.262 rad/h](https://tex.z-dn.net/?f=%5Comega%3D0.262%20rad%2Fh)
Now we have to find r, the distance of Allison from the Earth's axis.
We know that her latitude is
36.1627° N
Which is the angle between the Earth's radius and the equator, and the Earth radius is
R = 3960 miles
Therefore, Allison's distance from the Earth's axis can be found using trigonometry:
![r=R cos \theta = (3960)cos(36.1627^{\circ})=3197 mi](https://tex.z-dn.net/?f=r%3DR%20cos%20%5Ctheta%20%3D%20%283960%29cos%2836.1627%5E%7B%5Ccirc%7D%29%3D3197%20mi)
And so, Allison's linear velocity is:
![v=(0.262)(3197)=837.6 mi/h](https://tex.z-dn.net/?f=v%3D%280.262%29%283197%29%3D837.6%20mi%2Fh)
Learn more about rotational motion:
brainly.com/question/9575487
brainly.com/question/9329700
brainly.com/question/2506028
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