Frequency=1/Period
Frequency=1/0.03s
Frequency=100/3 Hz
Frequency=33.33 Hz
Answer:
the final pressure = 1120 torr
Explanation:
In the starting, when a 10-L drum consist 6 L of ether at 18°C
Volume of gas ![= ( 10 L - 6 L ) = 4 L](https://tex.z-dn.net/?f=%3D%20%28%2010%20L%20-%206%20L%20%29%20%3D%204%20L)
Total pressure of gas which is equal to Atmospheric Pressure = 760 Torr
vapor pressure of liquid (ether) = 400 Torr
vapor pressure of air
![= ( 760 Torr - 400 Torr ) = 360 Torr](https://tex.z-dn.net/?f=%20%3D%20%28%20760%20Torr%20-%20400%20Torr%20%29%20%3D%20360%20Torr)
when the drum top is sealed & dented to 8 L capacity
Volume of gas ![= ( 8 L - 6 L ) = 2 L](https://tex.z-dn.net/?f=%20%3D%20%28%208%20L%20-%206%20L%20%29%20%3D%202%20L)
Total pressure of gasses
![= 760 Torr x ( 4 L / 2 L ) = 1560 Torr](https://tex.z-dn.net/?f=%20%3D%20760%20Torr%20x%20%28%204%20L%20%2F%202%20L%20%29%20%3D%201560%20Torr)
vapor pressure of ether ![= 400 Torr x 2 = 800 Torr](https://tex.z-dn.net/?f=%3D%20400%20Torr%20x%202%20%3D%20800%20Torr)
vapor pressure of air ![= 360 Torr x 2 = 720 Torr](https://tex.z-dn.net/?f=%3D%20360%20Torr%20x%202%20%3D%20720%20Torr)
But some of the liquid (ether) vapor will condensed into liquid just to maintain the vapor pressure of ether at 400 Torr )
therefore, the final pressure
![= [ 400\ Torr (ether ) + 720\ Torr (air) ] = 1120 Torr](https://tex.z-dn.net/?f=%3D%20%20%5B%20400%5C%20Torr%20%28ether%20%29%20%2B%20720%5C%20Torr%20%28air%29%20%5D%20%3D%201120%20Torr)
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
So,
a) 0 < r < r1 :
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
Hence, E = 0 for r < r1
b) r1 < r < r2:
Electric field =?
Let, us consider the Gaussian Surface,
E x 4
= ![\frac{Q1}{E_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7BQ1%7D%7BE_%7B0%7D%20%7D)
So,
Rearranging the above equation to get Electric field, we will get:
E = ![\frac{Q1}{E_{0} . 4 \pi. r^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BQ1%7D%7BE_%7B0%7D%20.%204%20%5Cpi.%20r%5E%7B2%7D%20%20%20%7D)
Multiply and divide by
E =
x ![\frac{r1^{2} }{r1^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7Br1%5E%7B2%7D%20%7D%7Br1%5E%7B2%7D%20%7D)
Rearranging the above equation, we will get Electric Field for r1 < r < r2:
E= (σ1 x
) /(
x
)
c) r > r2 :
Electric Field = ?
E x 4
= ![\frac{Q1 + Q2}{E_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7BQ1%20%2B%20Q2%7D%7BE_%7B0%7D%20%7D)
Rearranging the above equation for E:
E = ![\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BQ1%2BQ2%7D%7BE_%7B0%7D%20.%204%20%5Cpi.%20r%5E%7B2%7D%20%20%20%7D)
E =
+ ![\frac{Q2}{E_{0} . 4 \pi. r^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BQ2%7D%7BE_%7B0%7D%20.%204%20%5Cpi.%20r%5E%7B2%7D%20%20%20%7D)
As we know from above, that:
= (σ1 x
) /(
x
)
Then, Similarly,
= (σ2 x
) /(
x
)
So,
E =
+ ![\frac{Q2}{E_{0} . 4 \pi. r^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BQ2%7D%7BE_%7B0%7D%20.%204%20%5Cpi.%20r%5E%7B2%7D%20%20%20%7D)
Replacing the above equations to get E:
E = (σ1 x
) /(
x
) + (σ2 x
) /(
x
)
Now, for
d) Under what conditions, E = 0, for r > r2?
For r > r2, E =0 if
σ1 x
= - σ2 x ![r2^{2}](https://tex.z-dn.net/?f=r2%5E%7B2%7D)
A believe that’s called a reference point.