Answer:
![W=\frac{p_0V_0-p_1V_1}{\gamma-1}](https://tex.z-dn.net/?f=W%3D%5Cfrac%7Bp_0V_0-p_1V_1%7D%7B%5Cgamma-1%7D)
Explanation:
An adiabatic process refers to one where there is no exchange of heat.
The equation of state of an adiabatic process is given by,
![pV^{\gamma}=k](https://tex.z-dn.net/?f=pV%5E%7B%5Cgamma%7D%3Dk)
where,
= pressure
= volume
![\gamma=\frac{C_p}{C_V}](https://tex.z-dn.net/?f=%5Cgamma%3D%5Cfrac%7BC_p%7D%7BC_V%7D)
= constant
Therefore, work done by the gas during expansion is,
![W=\int\limits^{V_1}_{V_0} {p} \, dV](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5E%7BV_1%7D_%7BV_0%7D%20%7Bp%7D%20%5C%2C%20dV)
![=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV](https://tex.z-dn.net/?f=%3Dk%5Cint%5Climits%5E%7BV_1%7D_%7BV_0%7D%20%7BV%5E%7B-%5Cgamma%7D%7D%20%5C%2C%20dV)
![=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bk%7D%7B%5Cgamma%20-1%7D%20%28V_0%5E%7B1-%5Cgamma%7D-V_1%5E%7B1-%5Cgamma%7D%29%5C%5C)
(using
)
![=\frac{p_0V_0-p_1V_1}{\gamma-1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bp_0V_0-p_1V_1%7D%7B%5Cgamma-1%7D)
Answer:
1 / f = 1 / p + 1 / q
Explanation:
For this exercise we will use two methods, an analytical method which is to use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, positive for converging lenses and negative for diverging lenses, p and q are the distance to the object and the image respectively
m = h’/ h = - q / p
where m is the magnification, h ’and h are the height of the image and the object.
We will also use a graphical method, where three rays will be traded
1) A ray that passes through the center of the lens and should not
2) a ray that passes through the focal length and comes out parallel to the lens
3) A ray that is horizontal and comes out through the focal from the other side
for this second method see the attachment
Answer:
Waves
Explanation:
Subatomic particles are particles that are of the subdivision of an atom and are smaller that atom, they are also described as waves in quantum mechanics.
Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance moved against the field of intensity, r = 30 cm
= 0.3 m
The electric field, E = 50 N/C
The formula for energy stored in the charge is,
V = k q/r
Where,
![k=\frac{1}{4\pi\epsilon_{0} }](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_%7B0%7D%20%20%7D)
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0