Answer:
![1.475\times 10^{-13}\ C/m^3](https://tex.z-dn.net/?f=1.475%5Ctimes%2010%5E%7B-13%7D%5C%20C%2Fm%5E3)
Explanation:
= Permittivity of free space = ![8.85\times 10^{-12}\ F/m](https://tex.z-dn.net/?f=8.85%5Ctimes%2010%5E%7B-12%7D%5C%20F%2Fm)
A = Area
h = Altitude = 600 m
Electric flux through the top would be
(negative as the electric field is going into the volume)
At the bottom
![120A](https://tex.z-dn.net/?f=120A)
Total flux through the volume
![\phi=120-110\\\Rightarrow \phi=10A](https://tex.z-dn.net/?f=%5Cphi%3D120-110%5C%5C%5CRightarrow%20%5Cphi%3D10A)
Electric flux is given by
![\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0](https://tex.z-dn.net/?f=%5Cphi%3D%5Cdfrac%7Bq%7D%7B%5Cepsilon_0%7D%5C%5C%5CRightarrow%20q%3D%5Cphi%5Cepsilon_0%5C%5C%5CRightarrow%20q%3D10A%5Cepsilon_0)
Charge per volume is given by
![\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfrac%7Bq%7D%7Bv%7D%5C%5C%5CRightarrow%20%5Crho%3D%5Cdfrac%7B10A%5Cepsilon_0%7D%7BAh%7D%5C%5C%5CRightarrow%20%5Crho%3Ddfrac%7B10%5Cepsilon_0%7D%7Bh%7D%5C%5C%5CRightarrow%20%5Crho%3D%5Cdfrac%7B10%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%7D%7B600%7D%5C%5C%5CRightarrow%20%5Crho%3D1.475%5Ctimes%2010%5E%7B-13%7D%5C%20C%2Fm%5E3)
The volume charge density is ![1.475\times 10^{-13}\ C/m^3](https://tex.z-dn.net/?f=1.475%5Ctimes%2010%5E%7B-13%7D%5C%20C%2Fm%5E3)
Hypothesis testing is basically testing the results of a experiment to see weather your results are valid or not.
Answer: 9.9%
Explanation: efficiency = (work output /work input) × 100
Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.
Work output = 9200 j and work input = 92048 j
Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%
Answer:
The intensity of the electric field is
![|E|=10654.37 \:N/C](https://tex.z-dn.net/?f=%7CE%7C%3D10654.37%20%5C%3AN%2FC)
Explanation:
The electric field equation is given by:
![|E|=k\frac{q}{d^{2}}](https://tex.z-dn.net/?f=%7CE%7C%3Dk%5Cfrac%7Bq%7D%7Bd%5E%7B2%7D%7D)
Where:
- k is the Coulomb constant
- q is the charge at 0.4100 m from the balloon
- d is the distance from the charge to the balloon
As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.
Then, let's use the equation written above.
![|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}](https://tex.z-dn.net/?f=%7CE%7C%3D%289%2A10%5E%7B9%7D%29%5Cfrac%7B1.99%2A10%5E%7B-7%7D%7D%7B0.41%5E%7B2%7D%7D)
![|E|=10654.37 \:N/C](https://tex.z-dn.net/?f=%7CE%7C%3D10654.37%20%5C%3AN%2FC)
I hope it helps you!
The answer is C. an electron in an orbit has a fixed energy.