Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
Answer:
The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.
Explanation:
Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon (
,
) are
and
, respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:
![\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7Bx_%7BE%7D%5Ccdot%20m_%7BE%7D%2Bx_%7BM%7D%5Ccdot%20m_%7BM%7D%7D%7Bm_%7BE%7D%2Bm_%7BM%7D%7D)
If
and
, then:
![\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B%280%5C%2Ckm%29%5Ccdot%20%285.972%5Ctimes%2010%5E%7B24%7D%5C%2Ckg%29%2B%28384%2C403%5C%2Ckm%29%5Ccdot%20%287.349%5Ctimes%2010%5E%7B22%7D%5C%2Ckg%29%7D%7B5.972%5Ctimes%2010%5E%7B24%7D%5C%2Ckg%2B7.349%5Ctimes%2010%5E%7B22%7D%5C%2Ckg%7D)
![\bar x = 4.673\,km](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%204.673%5C%2Ckm)
The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.
The speed of a proton after it accelerates from rest through a potential difference of 350 V is
.
Initial velocity of the proton ![u = 0](https://tex.z-dn.net/?f=u%20%3D%200)
Given potential difference ![\Delta V = 350V](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20350V)
let's assume that the speed of the proton is
,
Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge
when accelerated with a potential difference
is,
![U = q \Delta V](https://tex.z-dn.net/?f=U%20%3D%20q%20%5CDelta%20V)
Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e
![\Delta K = \Delta V](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20%5CDelta%20V)
If the initial and final velocity of the proton is
and
respectively then,
change in Kinetic Energy ![\implies \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0](https://tex.z-dn.net/?f=%5Cimplies%20%20%5CDelta%20K%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20-%5Cfrac%7B1%7D%7B2%7Dmu%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20-%200)
change in Potential Energy ![\implies \Delta U = q\Delta V](https://tex.z-dn.net/?f=%5Cimplies%20%5CDelta%20U%20%3D%20q%5CDelta%20V)
from conservation of energy,
![v= \sqrt{\frac{2q\Delta V}{m}}](https://tex.z-dn.net/?f=v%3D%20%5Csqrt%7B%5Cfrac%7B2q%5CDelta%20V%7D%7Bm%7D%7D)
so, ![v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B2%5Ctimes%20350%20%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B1.67%20%5Ctimes%2010%5E%7B-27%7D%7D)
![= 25.86 \times 10^4 ~m/s](https://tex.z-dn.net/?f=%3D%2025.86%20%5Ctimes%2010%5E4%20~m%2Fs)
To read more about the conservation of energy, please go to brainly.com/question/14668053