Answer:
V = 9.33 V
Explanation:
By definition, the capacitance of a capacitor, is the proportion between the charge on one of charged surfaces and the potential difference between these surfaces, as follows:
![C = \frac{Q}{V}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7BQ%7D%7BV%7D)
for a parallel-plate type capacitor, it can be showed, applying Gauss'Law to the a gaussian surface with the shape of a pillbox parallel to the surface of one of the plates, half outside the surface, half inside it, that the capacitance can be expressed as follows:
![C =\frac{\epsilon*A}{d}](https://tex.z-dn.net/?f=C%20%3D%5Cfrac%7B%5Cepsilon%2AA%7D%7Bd%7D)
As we can see, if a slab of paraffin, with dielectric constant of 2.25, is inserted, the capacitance will be increased in the same factor:
![Cf = 2.25 * Co](https://tex.z-dn.net/?f=Cf%20%3D%202.25%20%2A%20Co)
Now, if the capacitor, once charged, is disconnected from the battery, the charge Q will remain constant.
So, if the capacitance will be increased 2.25 times, the only way to do this is that the voltage between plates be reduced in the same factor:
![Cf = 2.25 * Co = 2.25* \frac{Q}{Vo} = \frac{Q}{Vf} \\ Vf= \frac{Vo}{2.25} =\frac{21.0V}{2.25} = 9.33 V](https://tex.z-dn.net/?f=Cf%20%3D%202.25%20%2A%20Co%20%3D%202.25%2A%20%5Cfrac%7BQ%7D%7BVo%7D%20%3D%20%5Cfrac%7BQ%7D%7BVf%7D%20%5C%5C%20Vf%3D%20%5Cfrac%7BVo%7D%7B2.25%7D%20%3D%5Cfrac%7B21.0V%7D%7B2.25%7D%20%3D%209.33%20V)
Answer:
Gold metal is a noble metal,i.e.,it is a non-reactive metal. As such,it does not form compounds with any other elements naturally. So it is found in its native or pure state in nature.
Simple, A. None of the other answers make sense so, A.
Answer:
I do agree with the student. Because imagene you have a tissue and a stick. And you throw them at the same speed , the stick will go farther because it' s heavier but the tissue would just blow around and not go forward because it is not heavy enought.
Explanation:
Hope this is correct have a terrific day.
Answer:
Velocity of a proton,
Explanation:
It is given that,
Potential difference, ![V=15\ kV=15\times 10^3\ V](https://tex.z-dn.net/?f=V%3D15%5C%20kV%3D15%5Ctimes%2010%5E3%5C%20V)
Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.
Using the conservation of energy as :
![\dfrac{1}{2}mv^2=qV](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%3DqV)
q is the charge of proton
m is the mass of proton
![v=\sqrt{\dfrac{2qV}{m}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cdfrac%7B2qV%7D%7Bm%7D%7D)
![v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cdfrac%7B2%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5C%20C%5Ctimes%2015%5Ctimes%2010%5E3%5C%20V%7D%7B1.67%5Ctimes%2010%5E%7B-27%7D%5C%20kg%7D)
![v=1695361.75\ m/s](https://tex.z-dn.net/?f=v%3D1695361.75%5C%20m%2Fs)
![v=1.69\times 10^6\ m/s](https://tex.z-dn.net/?f=v%3D1.69%5Ctimes%2010%5E6%5C%20m%2Fs)
or
![v=1.7\times 10^6\ m/s](https://tex.z-dn.net/?f=v%3D1.7%5Ctimes%2010%5E6%5C%20m%2Fs)
So, the velocity of a proton is
. Hence, this is the required solution.