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Stels [109]
2 years ago
5

Calculate the total electric flux leaving the cubical surface formed by the six planes x, y, z = ±5 if the charge distribution i

s: (a) twopoint charges, 0.1 μC at (1,− 2, 3) and 17μC at (−1, 2,−2)
Physics
1 answer:
pashok25 [27]2 years ago
7 0

If I correctly recall my E&M studies from 1958,
I suspect that this is a chump question.

The E flux through ANY closed surface is equal to the net
charge inside, or maybe it's the net charge divided by  ε₀ .
Yep, I think that's it.

Your charges are both positive, so if they're inside your cube,
the net charge inside will be  17.1μC .

Your cube has all faces 5 units from the axes, and neither charge
has an x, y, or z component with absolute value greater or equal
to 5.  So both charges are well inside the cube.

Seems to me, then, like the total E flux through the cubical
surface is

                       17.1μC  / ε₀  .

I don't really care whether this is helpful or not. 
I just hope it's correct.

You might be interested in
A jet engine gets its thrust by taking in air, heating and compressing it, and
nadya68 [22]

Answer:

F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw

Explanation:

Thrust is known as a reaction force which appears when a system expels or accelerates mass in one specific direction. If we know the acceleration and the mass of the air expelled by the jet engine, we can compute the thrust .

The acceleration is calculated by using the dynamics formula

\displaystyle a=\frac{v_f-v_o}{t}

The values are  

v_f=500\ m/s,\ v_o=100\ m/s,\ t=1\ sec

\displaystyle a=\frac{500-100}{1}=400\ m/s^2

The thrust is

F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw

4 0
2 years ago
7. A local sign company needs to install a new billboard. The signpost is 30 m tall, and the ladder truck is parked 24 m away fr
wolverine [178]
<h2>Solution :</h2>

Here ,

• Height of sign post = 30 m

• Distance between signpost and truck = 24 m

Let the

• Top of signpost = A

• Bottom of signpost = B

• The end of truck facing sign post be = C

Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .

Where

• AB = Height of signpost = 30 m

• BC = distance between both = 24 m

• AC = Minimum length of ladder

→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )

→ AC² = (30)² + (24)²

→ AC² = 900 + 576

→ AC² = 1476

→ AC = 38.41875

or AC apx = 38.42

So minimum height of ladder = 38.42

6 0
2 years ago
This is really urgent
hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is n=\frac{c}{v}

26) Critical angle: [tex]sin \theta_c = \frac{n_2}{n_1}[/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

with

n_1, n_2 are the refractive index of 1st and 2nd medium

\theta_1, \theta_2 are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

  • A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
  • A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

  • A material with refractive index of 1.1
  • A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1 (1)

For light moving from air to water:

n_1 \sim 1.00 is the index of refraction of air

n_2 = 1.33 is the index of refraction ofwater

In this case, \frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75

For light moving from air to glass,

n_2 = 1.51 is the index of refraction of glass

And so

\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66

From eq.(1), we see that the angle of refraction \theta_2 is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

n=\frac{c}{v}

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

  • The higher the index of refraction, the slower the light
  • The lower the index of refraction, the faster the light

26)

From Snell's law,

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, n_1 > n_2, so \frac{n_1}{n_2}>1, and since sin \theta_2 cannot be larger than 1, there exists a maximum value of the angle of incidence \theta_c (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

sin \theta_c = \frac{n_2}{n_1}

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

sin \theta_c = \frac{n_2}{n_1}

For the interface glass-air,

n_1 \sim 1.51\\n_2 = 1.00

The critical angle is

\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}

For the interface glass-water,

n_1 \sim 1.51\\n_2 = 1.33

The critical angle is

\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

7 0
2 years ago
Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f
stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2

When s = Db, t = 2t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2

Dividing the two equations

\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db

Hence, Da=(1/4)Db

3 0
2 years ago
A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force
boyakko [2]

Answer:

Explanation:

D

7 0
1 year ago
Read 2 more answers
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