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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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Question 2.
boyakko [2]

The tensile stress of the wire supporting 2 kg mass is determined as 6.1 x 10⁷ N/m².

<h3>Tensile stress of the wire</h3>

The tensile stress of the wire is calculated as follows;

σ = F/A

where;

  • A is area of the wire

A = πr² = πD²/4

where;

  • D is diameter = 0.64 mm

A = π x (0.64 x 10⁻³)²/4

A = 3.22 x 10⁻⁷ m²

σ = F/A = (mg)/A = (2 x 9.8)/( 3.22 x 10⁻⁷)

σ = 6.1 x 10⁷ N/m²

Learn  more about tensile stress here: brainly.com/question/25748369

#SPJ1

3 0
2 years ago
A. In one short sentence, explain why we call the force of gravity an attractive force.
kondor19780726 [428]

Answer:

Explanation:

(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.

(b) Yes, it does and the formula for force of gravity between any two object is

F = G\frac{m1m2}{r}

where m1 and m2 are masses of the first and second object respectively

r is the distance between the center of the two masses

G is the gravitational constant

3 0
3 years ago
Explore the role of friction when creating safety ramps and escalators. "Why is it important for people and object to move slowl
faust18 [17]

The role of friction is of great importance when creating safety ramps and escalators because with the help of friction things move.

<h3>Why is it important to move objects slowly on ramps and escalator?</h3>

It is important to move objects slowly on ramps and escalator because the ramps and escalator moves object in the opposite direction of gravity. If we did not move objects slowly, then the objects or a person get damaged.

So we can conclude that the role of friction is of great importance when creating safety ramps and escalators because with the help of friction things move.

Learn more about friction here: brainly.com/question/24338873#SPJ1

3 0
2 years ago
Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr
lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time  of computer A is CT_{A} =200 ps

Clock time  of computer B is CT_{B} =250 ps

Effective CPI of computer A is CPI_{A} =1.5

Effective CPI of computer B isCPI_{B} =1.7

CPU time of A is

CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec

CPU time of B is

CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec

Hence Computer A is Faster by \frac{425}{300} =1.41

Computer A is 1.41 times faster than the Computer B

4 0
3 years ago
Which describes increasing the efficiency of energy resource
vova2212 [387]

Making cars that get better gas mileage

5 0
3 years ago
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