To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.
Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.
The average angular acceleration
![\alpha = \frac{\omega_f - \omega_0}{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Comega_f%20-%20%5Comega_0%7D%7Bt%7D)
Here
= Angular acceleration
Initial and final angular velocity
There is not initial angular velocity,then
![\alpha = \frac{\omega_f}{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Comega_f%7D%7Bt%7D)
We know that the relation between the tangential velocity with the angular velocity is given by,
![v = r\omega](https://tex.z-dn.net/?f=v%20%3D%20r%5Comega)
Here,
r = Radius
= Angular velocity,
Rearranging to find the angular velocity
![\omega = \frac{v}{r}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7Bv%7D%7Br%7D%7D)
Remember that the radius is half te diameter.
Now replacing this expression at the first equation we have,
![\alpha = \frac{30}{0.20*6}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B30%7D%7B0.20%2A6%7D)
![\alpha = 25 rad /s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%2025%20rad%20%2Fs%5E2)
Therefore teh average angular acceleration of each wheel is ![25rad/s^2](https://tex.z-dn.net/?f=25rad%2Fs%5E2)
C.) <span>The total mass of an object can be assumed to be focused at one point, which is called its center of "Mass"
Hope this helps!</span>
Hi! I think the answer is C. as we have never visited the planet we can’t be 100% sure but we can have an idea of what the planet might me like. With experimenting with the simulation it can help you get an understanding or an idea of what the planet might be like exactly like option C. I hope this helped Goodluck :)
As the headwind comes at the head of the plane, therefore 50 mph headwind slows the plane down.
Thus actual speed of plane with headwind, ![v=550- 50=500 mph](https://tex.z-dn.net/?f=v%3D550-%2050%3D500%20mph)
Now using the formula , ![velocity= \frac{distance}{time}](https://tex.z-dn.net/?f=velocity%3D%20%5Cfrac%7Bdistance%7D%7Btime%7D)
![500 mph = \frac{1250 mile}{time}](https://tex.z-dn.net/?f=500%20mph%20%3D%20%5Cfrac%7B1250%20mile%7D%7Btime%7D)
![time = \frac{1250 mil}{500 mph}](https://tex.z-dn.net/?f=time%20%3D%20%5Cfrac%7B1250%20mil%7D%7B500%20mph%7D)
![time = 2.5 hour](https://tex.z-dn.net/?f=time%20%3D%202.5%20hour)
Hence plane will take time to complete the distance 1250 mile in 2.5 hours