Answer:
1 N
Explanation:
Buoyant Force: This is also called upthrust, It can be defined as the force which act upward exerted by a fluid when an object is placed in it.
The S.I unit is Newton.
From the question,
Buoyant force = Weight of the object in air- weight of the object when submerged in water.
U = W-W'.......................... Equation 1
Where U = upthrust, W = weight in air, W' = weight when submerged in water.
Given: W = 3 N, W' = 2 N
Substitute into equation 1
U = 3-2
W = 1 N
Goggles, a lab coat, chemical gloves, and close- toed shoes
Answer:
what is the hunting age? i've never heard of it
Answer:
B
Explanation:
OOf we are doing this stuff atm
So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B
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