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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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Answer:

Explanation:

For third normal mode of vibration

l = \frac{3\lambda}{2}  , λ is wavelength , l is length of string .

.4 = \frac{3\lambda}{2}

λ = .267 m

velocity = \sqrt{\frac{T}{m} }

T is tension and m is mass unit length

m = .5 x 10⁻³ / 40 x 10⁻²

= .00125 kg / m

Putting the values

velocity = \sqrt{\frac{80}{.00125} }

= 253 m /s

frequency

= velocity / λ

= 253 / .267

= 947.5 Hz .

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An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a
Blababa [14]

Answer:

\frac{\Delta L}{L} =5.37\times 10^{-4}

Explanation:

Given:

  • cross sectional area of the wire, A=5.75\times 10^{-6}\ m^2
  • density of aluminium wire, \rho=2.7\times 10^3\ kg.m^{-3}
  • young's modulus of the material, E=7\times 10^{10}\ N.m^{-2}
  • wave speed, v=118\ m.s^{-1}

<u>We have mathematical expression for strain as:</u>

\frac{\Delta L}{L} =\frac{\sigma}{E} ...............................(1)

and since, \sigma =\frac{T}{A}

where, T = tension force in the wire

equation (1) becomes:

\frac{\Delta L}{L} =\frac{T}{A.E} ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

v=\sqrt{\frac{T}{\mu} } ...................................(3)

where: \mu= linear mass density of the wire

\therefore \mu=\rho \times A

Now, equation (3) becomes

v=\sqrt{\frac{T}{\rho \times A} }

T=v^2.\rho \times A ............................(4)

Using eq. (2) & (4) for tension T

v^2.\rho \times A=A.E\times \frac{\Delta L}{L}

\frac{\Delta L}{L} =\frac{v^2.\rho}{E}

putting the respective values

\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}

\frac{\Delta L}{L} =5.37\times 10^{-4}

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Answer:

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Explanation:

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