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miss Akunina [59]
3 years ago
7

A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax

es of a rectangular coordinate system and the corner formed by these two sides is at the origin. An electric field of magnitude E=bx2 directed along the +x axis exists in this region.
1.How much charge is contained in the volume of the shell?
2.Express your answer in terms of the variables a, b, and the permittivity constant ϵ0.
Physics
1 answer:
fomenos3 years ago
3 0

Answer: Q = ba⁴ * ε₀

Explanation:

Remember, Gauss's Law states that:

"The net electric flux through any hypothetical closed surface is equal to 1/ε₀ times the net electric charge within that closed surface" i.e

flux Φ = Q / ε₀ where

ε₀ = 8.85e-12 C²/N·m²

Also, the flux, Φ = EAcosθ, where

E = magnitude of the electric field in V/m

A = area of the surface in m²

θ = angle between the electric field lines and the normal (perpendicular) to S.

The field is directed along the x-axis, so all of the flux passes through the side of the cube at x = a. This means that θ = 0º. Thus,

Φ = EA

Then, E = bx² and we're interested in the point where x = a, so if we substitute x for a, we have

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = ba² * a²

so

Q = ba⁴ * ε₀

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Discuss Joule-Thompson effect with relevant examples and formulae.
Delicious77 [7]

Answer:

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

Explanation:

Joule -Thompson effect

 Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process  

Let

 P_1,T_1 Pressure and temperature at inlet and

 P_2,T_2 Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

\mu _j=\left(\frac{\partial T}{\partial p}\right)_h

Now from T-ds equation

dh=Tds=vdp

So

Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp

⇒dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

So Joule -Thompson coefficient

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

\dfrac{\partial v}{\partial T}=\dfrac{v}{T}

So by putting the values in

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

\mu _j=0 For ideal gas.

6 0
3 years ago
Can you please help me ASAPPP !!
Natali5045456 [20]

Answer:

D

Explanation:

Look at the two of them, if you put them together the image is identical to that of the unknown sample.

5 0
3 years ago
Help me please, help
Alisiya [41]

Answer:

455,000 Pa

Explanation:

PV = nRT

If n is constant:

PV / T = PV / T

(101,325 Pa) (718 mL) / (273 K) = P (175 mL) / (26 + 273) K

P = 455,000 Pa

6 0
3 years ago
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The speed of both cars is the same ... 80 km per hour.

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3 years ago
The force needed to move an object 14 meters with 83 joules of work is what?
AleksandrR [38]
W=F*D
83J=F*14
83/14=F
5.92N
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