Answer:
Qualitative, Quantitative, Qualitative, Quantitative, and Qualitative.
Explanation:
To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.
![I = MR^2](https://tex.z-dn.net/?f=I%20%3D%20MR%5E2)
Here,
M = Mass
R = Radius of the hoop
The precession frequency is given as
![\Omega = \frac{Mgd}{I\omega}](https://tex.z-dn.net/?f=%5COmega%20%3D%20%5Cfrac%7BMgd%7D%7BI%5Comega%7D)
Here,
M = Mass
g= Acceleration due to gravity
d = Distance of center of mass from pivot
I = Moment of inertia
= Angular velocity
Replacing the value for moment of inertia
![\Omega= \frac{MgR}{MR^2 \omega}](https://tex.z-dn.net/?f=%5COmega%3D%20%5Cfrac%7BMgR%7D%7BMR%5E2%20%5Comega%7D)
![\Omega = \frac{g}{R\omega}](https://tex.z-dn.net/?f=%5COmega%20%3D%20%5Cfrac%7Bg%7D%7BR%5Comega%7D)
The value for our angular velocity is not in SI, then
![\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})](https://tex.z-dn.net/?f=%5Comega%20%3D%201000rpm%20%28%5Cfrac%7B2%5Cpi%20rad%7D%7B1%20rev%7D%29%28%5Cfrac%7B1min%7D%7B60s%7D%29)
![\omega = 104.7rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20104.7rad%2Fs)
Replacing our values we have that
![\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}](https://tex.z-dn.net/?f=%5COmega%20%3D%20%5Cfrac%7B9.8m%2Fs%5E2%7D%7B%288%2A10%5E%7B-2%7Dm%29%28104.7rad%29%7D)
![\Omega = 1.17rad/s](https://tex.z-dn.net/?f=%5COmega%20%3D%201.17rad%2Fs)
The precession frequency is
![\Omega = \frac{2\pi rad}{T}](https://tex.z-dn.net/?f=%5COmega%20%3D%20%5Cfrac%7B2%5Cpi%20rad%7D%7BT%7D)
![T = \frac{2\pi rad}{\Omega}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20rad%7D%7B%5COmega%7D)
![T = \frac{2\pi}{1.17}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B1.17%7D)
![T = 5.4 s](https://tex.z-dn.net/?f=T%20%3D%205.4%20s)
Therefore the precession period is 5.4s
In a fluid, all the forces exerted by the individual particles combine to make up the pressure exerted by the fluid
Due to fundamental nature of fluids, a fluid cannot remain at rest under the presence of shear stress. However, fluids can exert pressure normal to any contacting surface. If a point in the fluid is thought of as a small cube, then it follows from the principles of equilibrium that the pressure on every side of this unit of fluid must be equal. but if this were not a case, the fluid would move in the directions of the resulting force, So the pressure on a fluid at rest is isotropic.
Hope This Helps :D <span />
ಠ_ಠ Hey, hang on.. you might've made a discovery. Nobody has tested it so how do we know? ಠ_ಠ