It looks like the diagram is telling you that there is a net force of 8 N + 4 N = 12 N pointed to the right. By Newton's second law, the magnitude of the net force <em>F</em> is equal to the mass <em>m</em> times the acceleration <em>a</em> :
<em>F</em> = <em>m</em> <em>a</em>
12 N = (4 kg) <em>a</em>
<em>a</em> = (12 N) / (4 kg)
<em>a</em> = 3 m/s²
-Positively charged nucleus
-Empty spaced
-Dense core
Answer:
The surface charge density on planes A and B respectively is
![\sigma__{A}} } = 2\sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%202%5Csigma)
and
![\sigma__{B}} = \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BB%7D%7D%20%3D%20%20%20%5Csigma)
Explanation:
From the question we are told that
The electric field in region to the left of A is ![E_i = \frac{3 \sigma}{2 \epsilon_o}](https://tex.z-dn.net/?f=E_i%20%3D%20%20%5Cfrac%7B3%20%5Csigma%7D%7B2%20%5Cepsilon_o%7D)
The direction of the electric field is left
The electric field in the region to the right of B is ![E_f = \frac{3 \sigma}{2 \epsilon_o}](https://tex.z-dn.net/?f=E_f%20%3D%20%20%5Cfrac%7B3%20%5Csigma%7D%7B2%20%5Cepsilon_o%7D)
The direction of the electric field is right
The electric field in the region between the two planes is ![E_m = \frac{\sigma }{2 \epsilon_o }](https://tex.z-dn.net/?f=E_m%20%20%3D%20%20%5Cfrac%7B%5Csigma%20%7D%7B2%20%5Cepsilon_o%20%7D)
The direction of the electric field is right
Let the surface charge density on planes A and B be represented as ![\sigma__{A}} \ \ and \ \ \sigma__{B}} \ \ \ respectively](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%20%5C%20%5C%20and%20%20%5C%20%5C%20%5Csigma__%7BB%7D%7D%20%5C%20%5C%20%5C%20%20respectively)
From the question we see that
![E_i = E_f](https://tex.z-dn.net/?f=E_i%20%3D%20E_f)
Generally the electric to the right and to the left is due to the combined electric field generated by plane A and B so
![E_i = E_f = \frac{3\sigma }{2\epsilon} = \frac{\sigma_A }{ 2 \epsilon_o } + \frac{\sigma_B }{ 2 \epsilon_o }](https://tex.z-dn.net/?f=E_i%20%3D%20E_f%20%3D%20%20%5Cfrac%7B3%5Csigma%20%7D%7B2%5Cepsilon%7D%20%20%3D%20%20%5Cfrac%7B%5Csigma_A%20%7D%7B%202%20%5Cepsilon_o%20%7D%20%2B%20%20%5Cfrac%7B%5Csigma_B%20%7D%7B%202%20%5Cepsilon_o%20%7D)
=> ![\sigma__{A}} + \sigma__{B}} = 3 \sigma -- -(1)](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%2B%20%5Csigma__%7BB%7D%7D%20%3D%20%203%20%5Csigma%20%20--%20-%281%29)
Generally the electric field at the middle of the plane A and B is due to the diffencence in electric field generated by plane A and B
i.e
![\frac{\sigma }{2 \epsilon_o } = \frac{\sigma_A }{ 2 \epsilon_o } - \frac{\sigma_B }{ 2 \epsilon_o }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma%20%7D%7B2%20%5Cepsilon_o%20%7D%20%20%3D%20%20%5Cfrac%7B%5Csigma_A%20%7D%7B%202%20%5Cepsilon_o%20%7D%20-%20%20%20%5Cfrac%7B%5Csigma_B%20%7D%7B%202%20%5Cepsilon_o%20%7D)
=> ![\sigma__{A}} - \sigma__{B}} = \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20-%20%20%5Csigma__%7BB%7D%7D%20%3D%20%20%5Csigma)
=> ![\sigma__{A}} } = \sigma + \sigma__{B}](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%20%5Csigma%20%2B%20%5Csigma__%7BB%7D)
From equation 1
![\sigma + \sigma__{B}}+ \sigma__{B}} = 3 \sigma](https://tex.z-dn.net/?f=%5Csigma%20%2B%20%5Csigma__%7BB%7D%7D%2B%20%5Csigma__%7BB%7D%7D%20%3D%20%203%20%5Csigma)
=> ![\sigma__{B}} = \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BB%7D%7D%20%3D%20%20%20%5Csigma)
So
![\sigma__{A}} } = \sigma + \sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%20%5Csigma%20%2B%20%5Csigma)
=> ![\sigma__{A}} } = 2\sigma](https://tex.z-dn.net/?f=%5Csigma__%7BA%7D%7D%20%7D%20%3D%20%202%5Csigma)
Answer:
given,
mass of the block = m₁
mass of the another block = 3 m₁
initial Amplitude, A = 6 cm
final amplitude = 6 cm
total mechanical energy = 12 J
total energy of the block spring
![E = \dfrac{1}{2}kA^2](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7B1%7D%7B2%7DkA%5E2)
A is the amplitude and k is spring constant
initial energy is equal to 12 J
from the above expression we can say that
Energy of the given system depends up on the magnitude of spring constant and the amplitude.
so, energy of both the system will be same.
we know,
![E = \dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![12= \dfrac{1}{2}\times 3 m_1 v^2](https://tex.z-dn.net/?f=12%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%203%20m_1%20v%5E2)
![v^2 = \dfrac{8}{m_1}](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cdfrac%7B8%7D%7Bm_1%7D)
![v = \sqrt{ \dfrac{8}{m_1}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%20%5Cdfrac%7B8%7D%7Bm_1%7D%7D)
Answer:
A. 98,000 J
Explanation:
The gravitational potential energy of an object is given by
U = mgh
where
m is the mass of the object
g is the gravitational acceleration
h is the heigth above the ground
In this problem,
m = 2000 kg
g = 9.8 m/s^2
h = 5.0 m
Substituting into the equation, we find
![U=(2000 kg)(9.8 m/s^2)(5.0 m)=98,000 J](https://tex.z-dn.net/?f=U%3D%282000%20kg%29%289.8%20m%2Fs%5E2%29%285.0%20m%29%3D98%2C000%20J)