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LUCKY_DIMON [66]
3 years ago
10

You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe

re, there is a strong electric field surrounding your device. Other researchers in your laboratory are complaining that your electric field is affecting their equipment. You think about how you can obtain the large electric field that you need close to the sphere but prohibit the field from reaching your colleagues. You decide to surround your device with a spherical transparent plastic shell. The nonconducting shell is given a uniform charge distribution. Required:a. The shell is placed so that the small sphere is at the exact center of the shell. Determine the charge that must be placed on the shell to completely eliminate the electric field outside of the shell. b. What if the shell moves? Does the small sphere have to be at the center of the shell for this scheme to work?
Physics
1 answer:
madam [21]3 years ago
5 0

Answer:

the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

Explanation:

We can answer this exercise using Gauss's law

      Ф = ∫ e . dA = q_{int} / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

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A boat takes off from the dock at 2.5m/a and speeds up at 4.2m/s/s for 6.0 s. How far has the boat traveled? Round answer to nea
natima [27]

Answer:

91 m

Explanation:

v₀ = 2.5 m/s

a = 4.2 m/s²

t = 6.0 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.5 m/s) (6.0 s) + ½ (4.2 m/s²) (6.0 s)²

Δx = 90.6 m

Rounded to the nearest whole number, the distance traveled is 91 m.

6 0
4 years ago
A boy pulls a wagon with a force of 6 N east as another boy pushes it with a force of 4 N east. What is the net force?
Angelina_Jolie [31]
The net force with be 10 N east. As both boys are pushing a wagon to the same direction, you just add the numbers together. If the directions were opposite like if it was the boy pushing 6 N east and the other boy pushing 4 N west...then it would come out as 2 N east (you subtract and go to the direction with the biggest number) if you have any confusions, feel free to ask
4 0
3 years ago
Friction between solids can be most accurately defined as the force that
ki77a [65]

Answer:

True

Explanation:

Friction between solids can be most accurately defined as the force that

opposes the sliding motion of two surfaces that are touching each other.

6 0
3 years ago
A. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.
never [62]

Answer:

a) F₁ = F₂,  F₃ = F₄,  b)  the correct answer is 3

Explanation:

a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies

Forces 1 and 2 are action and reaction forces F₁ = F₂

Forces 3 and 4 are action and reaction forces F₃ = F₄

as it indicates that the

b) how the car increases if speed implies that force 1> force3

      F₁ > F₃

therefore the correct answer is 3

3 0
3 years ago
At the very end of Wagner's series of operas The Ring of Nibelung, Brunnhilde takes the golden ring form the finger of the dead
Blababa [14]

Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

n_{air}=1\\ n_{water}=1.33=4/3

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3

At critical angle B = 90°

\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)

Therefore

A = 48.6^\circ

With this, we can find the radius of the circle (refer to my diagram)

h* tan (A) = R\\R =11.3 m

And with that we can find the area

A = \pi R^2=404\ m^2

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

v_{water} = \frac{c}{n_{water}} = 0.75c

to appear as if it has traveled with the same duration as light with speed c

In order for this to happen our brain perceive shortened length  which is the apparent depth.

To put it in mathematical term

t_{travel}=\frac{h_{apparent}}{v_{water}} =\frac{h}{c}

So we get apparent depth

h_{apparent}=0.75h = 7.5\ m

4 0
3 years ago
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