Answer:
(a) 31.44 m/s (b) 164.74 m/s²
Explanation:
Given that,
The diameter of a disk, d = 12 cm
Radius, r = 6 cm
Angular speed = 5.24 rad/s
(a) Linear speed,
![v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s](https://tex.z-dn.net/?f=v%3Dr%5Comega%5C%5C%5C%5Cv%3D6%5Ctimes%205.24%5C%5C%5C%5Cv%3D31.44%5C%20m%2Fs)
(b) Centripetal acceleration,
![a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv%5E2%7D%7Br%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7B31.44%5E2%7D%7B6%7D%5C%5C%5C%5Ca%3D164.74%5C%20m%2Fs%5E2)
MAKE ME...................................................................
The electric flow is defined as the change of the charge in relation to the constant of the vacuum permittivity constant.
That is to say that mathematically this expression could be given as
![\Phi = \frac{q}{\epsilon_0}](https://tex.z-dn.net/?f=%5CPhi%20%3D%20%5Cfrac%7Bq%7D%7B%5Cepsilon_0%7D)
Here,
= Electric Flux
q = Charge
= Vacuum Permittivity
Replacing with our values and solving to find q, we have,
![-1850N\cdot m^2/C = \frac{q}{8.854*10^{-12} C\cdot N^{-1} \cdot m^{-2}}](https://tex.z-dn.net/?f=-1850N%5Ccdot%20m%5E2%2FC%20%3D%20%5Cfrac%7Bq%7D%7B8.854%2A10%5E%7B-12%7D%20C%5Ccdot%20N%5E%7B-1%7D%20%5Ccdot%20m%5E%7B-2%7D%7D)
![q = - 1.63799*10^{-8}C](https://tex.z-dn.net/?f=q%20%3D%20-%201.63799%2A10%5E%7B-8%7DC)
Therefore the charge is enclose within the octahedron with a charge of ![- 1.63799*10^{-8}C](https://tex.z-dn.net/?f=-%201.63799%2A10%5E%7B-8%7DC)
Answer:
Explanation:
1.{1,3}2. adn as if bruh ma dude
To solve this problem it is necessary to apply the concepts related to pressure as a unit that measures the force applied in a specific area as well as pressure as a measurement of the density of the liquid to which it is subjected, its depth and the respective gravity.
The two definitions of pressure can be enclosed under the following equations
![P = \frac{F}{A}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BF%7D%7BA%7D)
Where
F= Force
A = Area
![P = \rho gh](https://tex.z-dn.net/?f=P%20%3D%20%5Crho%20gh)
Where,
Density
g = Gravity
h = Height
Our values are given as,
![d = 0.8m \rightarrow r = 0.4m](https://tex.z-dn.net/?f=d%20%3D%200.8m%20%5Crightarrow%20r%20%3D%200.4m)
![A = \pi r^2 = \pi * 0.4^2 = 0.503m^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20r%5E2%20%3D%20%5Cpi%20%2A%200.4%5E2%20%3D%200.503m%5E2)
If we make a comparison between the lid and the tube, the diameter of the tube becomes negligible.
Matching the two previous expressions we have to
![\frac{F}{A} = \rho g h](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D%20%3D%20%5Crho%20g%20h)
Re-arrange to find h
![h = \frac{F}{A\rho g}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7BF%7D%7BA%5Crho%20g%7D)
![h = \frac{390}{(0.503)(1000)(9.8)}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B390%7D%7B%280.503%29%281000%29%289.8%29%7D)
![h = 0.079m](https://tex.z-dn.net/?f=h%20%3D%200.079m)
![h = 7.9cm](https://tex.z-dn.net/?f=h%20%3D%207.9cm)
Therefore the height of water in the tube is 7.9cm