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marta [7]
3 years ago
10

Why is Newton’s cradle (Newton’s balls) described as an “almost-ideal” closed system? Explain your reasoning in one or two sente

nces.
Physics
1 answer:
zysi [14]3 years ago
3 0

Answer:

ikr

Explanation:

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A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the
Aleksandr-060686 [28]

Answer:

(a):  The magnitude of the electric force on the small sphere = \dfrac{q\sigma}{2\epsilon_o}.

(b): Shown below.

Explanation:

<u>Given:</u>

  • m = mass of the small sphere.
  • q = charge on the small sphere.
  • L = length of the silk fiber.
  • \sigma = surface charge density of the large vertical insulating sheet.

<h2>(a):</h2>

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

\rm E = \dfrac{\sigma}{2\epsilon_o}.

<em>where,</em>

\epsilon_o is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

\rm F_e=qE.

Thus, the magnitude of the electric force on the small sphere is given by

\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

<h2>(b):</h2>

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

\rm \tan \theta = \dfrac{F_e}{W}.

<em>where,</em>

  • \rm F_e = electric force on the sphere, acting along left.
  • \rm W = weight of the sphere, acting vertically downwards.

<em />

\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .

g is the acceleration due to gravity.

6 0
3 years ago
why do you believe in social learning theory and how does learning many of our morals and practice from our family uniil we grow
Hitman42 [59]
Ion know. What story is this?
4 0
3 years ago
Just think about this.
diamong [38]
This ain’t the place, bud. If you have a QUESTION, then you can post it here.
8 0
3 years ago
Read 2 more answers
In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm, then what amount of elastic potential energy
myrzilka [38]
<h2>Answer:</h2>

4E

<h2>Explanation:</h2>

The elastic potential energy of an elastic material (e.g a spring, a wire), is the energy stored when the material is stretched or compressed. It is given by

U = \frac{1}{2}kx^2               --------------------(i)

Where;

U = potential energy stored

k = spring constant of the material

x = elongation (extension or compression of the material).

<em>From the first statement;</em>

<em>when elongation (x) is 4cm, energy stored (U) is E</em>

<em>Substitute these values into equation (i) as follows;</em>

E = \frac{1}{2}k(4)^2

E = 8k

<em>Make k subject of the formula</em>    

k = \frac{E}{8}   [measured in J/cm]

<em>From the second statement;</em>

<em>It is stretched by 4cm.</em>

This means that total elongation will be 4cm + 4cm = 8cm.

The potential energy stored will be found by substituting the value of x = 8cm and k = \frac{E}{8} into equation (i) as follows;

U = \frac{1}{2}\frac{E}{8} (8)^2  

U = \frac{1}{2}{8E}

U = {4E}

Therefore, the potential energy stored will now be 4 times the original one.

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3 years ago
What are the two main factors that affect how quickly a coastline erodes?
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Factors Affecting the Rate of Erosion<span>. The biggest factor affecting coastal </span>erosion<span> is the </span>strength<span> of the waves breaking along the coastline. A wave's </span>strength<span> is controlled by its fetch and the </span>wind speed<span>.</span>
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3 years ago
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