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A cubical surface surrounds a point charge q . Describe what happens to the total flux through the surface if (c) the surface is

valentina_108 [34]

Answer:

Gauss law states that the electric flux is defined as the electric field multiplied by the area of the surface in a plane perpendicular to the field.

Explanation:

Mathematically,

Φ=Q ϵo

Where;

Q is enclosed charge

ϵo is the permittivity of the free space

According to Gauss law, which states that the electric flux is defined as the electric field multiplied by the area of the surface in a plane perpendicular to the field.

Φ=Q ϵo

Where;

Q is enclosed charge

ϵo is the permittivity of the free space

If the cube is transformed into a sphere the total flux in the electric field remains unchanged or remains the same. This is because the gaussian law does not postulate that electric flux is dependent on the object in a plane. Hence, the transformation of the cube to a sphere does not affect the electric flux generated in the field.

To learn more about how the total flux through a sphere relates to the surface change, click brainly.com/question/4362789

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3 0

1 year ago

In moving out of a dormitory at the end of the semester, a student does 1.82 x 104 J of work. In the process, his internal energ

emmainna [20.7K]

Answer:

Explanation:

(a) Work done, W = 1.82 x 10^4 J

(b) internal energy, U = - 4.07 x 10^4 J ( as it decreases)

(c) According to the first law of thermodynamics

Q = W + U

Q = 1.82 x 10^4 - 4.07 x 10^4

Q = - 2.25 x 10^4 J

4 0

3 years ago

A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above

Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed $v_{1}= 2\ m/s$

Speed $v_{2}=7\ m/s$

Pressure in main pipe $P_{1}=2\times10^{5}\ Pa$

(I). We need to calculate the diameter

Using equation of continuity

$Av_{1}=Av_{2}$

$\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}$

$(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7$

$d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}$

$d_{2}=5.345\ cm$

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

$P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})$

$P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)$

$P_{2}=1.28500\times10^{5}\ Pa$

(III). If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}$

$P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})$

Here, $h_{3}=0$

Put the value in the equation

$P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17$

$P_{3}=3.5400\times10^{5}\ Pa$

Hence, This is required solution.

7 0

3 years ago

An auto, moving too fast on a horizontal stretch of mountain road, slides off the road, falling into deep snow 43.9 m below the

allsm [11]

Ok so it is 42.8 because if you - 43.9 by 1.1 by the way don’t listen to any of the I just want point

7 0

3 years ago

How do I find deceleration

Mars2501 [29]

Deceleration isn't a term in physics; you can have positive acceleration which is speeding up and negative acceleration which is slowing down
Use newton's law F = ma
a = F/m

4 0

3 years ago