Answer:
The charge stored is ![Q = 4.25 *10^{-7 } \ C](https://tex.z-dn.net/?f=Q%20%3D%20%204.25%20%2A10%5E%7B-7%20%7D%20%5C%20C)
The energy stored is
Explanation:
From the question we are told that
The area of the plates is
The separation between the plate is ![d = 0.10 \ mm =0.0001 \ m](https://tex.z-dn.net/?f=d%20%3D%20%200.10%20%5C%20mm%20%3D0.0001%20%5C%20m)
The potential difference is ![V = 12 \ V](https://tex.z-dn.net/?f=V%20%3D%20%2012%20%5C%20V)
The permitivity of free space is ![\epsilon_o = 8.85 *10^{-12} C^2 \cdot N^{-1} \cdot m^2](https://tex.z-dn.net/?f=%5Cepsilon_o%20%20%3D%208.85%20%2A10%5E%7B-12%7D%20C%5E2%20%5Ccdot%20N%5E%7B-1%7D%20%5Ccdot%20m%5E2)
The dielectric constant of glass is K = 5.0
Generally the capacitance of this capacitor is
![C = \frac{\epsilon_o * A}{d}](https://tex.z-dn.net/?f=C%20%3D%20%20%5Cfrac%7B%5Cepsilon_o%20%20%2A%20A%7D%7Bd%7D)
substituting values
![C = \frac{8.85*10^{-12} * 0.40}{0.0001}](https://tex.z-dn.net/?f=C%20%3D%20%20%5Cfrac%7B8.85%2A10%5E%7B-12%7D%20%20%2A%200.40%7D%7B0.0001%7D)
![C = 3.34 *10^{-8} \ C](https://tex.z-dn.net/?f=C%20%3D%20%20%203.34%20%2A10%5E%7B-8%7D%20%5C%20C)
The charge stored is mathematically evaluated as
![Q = CV](https://tex.z-dn.net/?f=Q%20%3D%20CV)
substituting values
![Q = (3.54*10^{-8} * 12)](https://tex.z-dn.net/?f=Q%20%3D%20%20%283.54%2A10%5E%7B-8%7D%20%2A%2012%29)
![Q = 4.25 *10^{-7 } \ C](https://tex.z-dn.net/?f=Q%20%3D%20%204.25%20%2A10%5E%7B-7%20%7D%20%5C%20C)
The energy stored is
![E = 0.5 * CV^2](https://tex.z-dn.net/?f=E%20%3D%20%200.5%20%2A%20%20CV%5E2)
substituting values
![E = 0.5 * (4.25 *10^{-7} * 12^2)](https://tex.z-dn.net/?f=E%20%3D%20%200.5%20%2A%20%20%284.25%20%2A10%5E%7B-7%7D%20%2A%2012%5E2%29)
![E = 2.55*10^{-6} \ J](https://tex.z-dn.net/?f=E%20%3D%202.55%2A10%5E%7B-6%7D%20%5C%20J)
There are NO true statements on the list you provided.
Answer:
1st – Place the film canister on the <u>scale</u>.
2nd – Slide the large <u>weight </u>to the right until the arm drops below the line and then move it back one notch.
3rd – Repeat this process with the <u>top</u> weight. When the arm moves below the line, back it up one groove.
4th – Slide the <u>small </u>weight on the front beam until the <u>lines</u> match up.
5th – Add the amounts on each beam to find the total <u>mass </u>to the nearest tenth of a gram.
Explanation:
The triple beam balance is an instrument that is used in measuring the mass of substances to a very high degree of precision. The reading error is given by ±0.05 grams. The triple beam balance as the name implies has three beams that measure substances of different mass levels.
The beams are categorized as small, medium, and large. There is a balance on which the substance to be weighed is placed directly upon. To use this measuring device, the procedures mentioned above are followed.
Answer:37 J
Explanation:
Given
Step :1
Heat added Q=44 J
Work done=-20 J
![\Delta E_1=Q+W=44-20=24 J](https://tex.z-dn.net/?f=%5CDelta%20E_1%3DQ%2BW%3D44-20%3D24%20J)
Step :2
Heat added Q=-61 J
work done ![W_2](https://tex.z-dn.net/?f=W_2)
![\Delta E_2=Q+W_2](https://tex.z-dn.net/?f=%5CDelta%20E_2%3DQ%2BW_2)
![\Delta E_2=61+W_2](https://tex.z-dn.net/?f=%5CDelta%20E_2%3D61%2BW_2)
![\Delta E_1+\Delta E_2=0](https://tex.z-dn.net/?f=%5CDelta%20E_1%2B%5CDelta%20E_2%3D0)
as the process is cyclic
![44-20-61+W_2=0](https://tex.z-dn.net/?f=44-20-61%2BW_2%3D0)
![W_2=37 J](https://tex.z-dn.net/?f=W_2%3D37%20J)
work done in compression is 37 J