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Serjik [45]
3 years ago
12

Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 8.7 nC. Using insulatin

g handles, the blocks are moved so they touch each other. After touching for a few seconds, the blocks are separated (again using insulating handles).(a) What is the final charge of block A?
(b) What happened while the blocks were in contact with each other?
Physics
2 answers:
vodka [1.7K]3 years ago
8 0

Answer:

a) final charge on block A, Q_A=4.35\times 10^{-9}\ C

b) There is a flow of current when the two blocks were brought in contact since there is imbalance of the charge distribution and both are identical metal masses.

Explanation:

Given:

  • initial charge on metal block A, Q_a=0\ C
  • initial charge on metal block B, Q_b=8.7\times 10^{-9}\ C

After the two are brought in contact with each other final charge on each of the block will be equal because the two bodies are identical.

So,

a)

Final charge on block A and block B, Q_A=Q_B=\frac{8.7\times 10^{-9}}{2} =4.35\times 10^{-9}\ C

b)

When the two blocks are brought into physical contact of together then the charges from the charged block move to the neutral block until there are equal charges on both the identical blocks. This flow of charge creates a flow of current for a short time till the charges get balanced on both the blocks to attain equilibrium. Charges flow due to the potential difference.

JulsSmile [24]3 years ago
6 0

Explanation:

(a)   Since, it is given that the blocks are identical so distribution of charge will be uniform on both the blocks.

Hence, final charge on block A will be calculated as follows.

         Charge on block A = \frac{(8.7 + 0 nC}{2}

                                           = 4.35 nC

Therefore, final charge on the block A is 4.35 nC.

(b)  As it is given that the positive charge is coming on block A . This means that movement of electrons will be from A to B.

Thus, we can conclude that while the blocks were in contact with each other then electrons will flow from A to B.

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A stone is thrown horizontally at a speed of 10.0 m/s from the top of a cliff 139.4 m high.
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3 years ago
The normal eye, myopic eye and old age
yanalaym [24]

Answer:

1)    f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

        \frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}

        f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

           1 / f = 1/15 + 1 / 1.5

           1 / f = 0.733

            f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

         f’₀ / f = 1.5 / 1.36

         f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

          a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

          tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

          sin θ = y / f

where f is the distance of the lens (eye)

           y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

           y / f = 1.22 λ / D

           y = 1.22 λ f / D

where D is the diameter of the eye

          D = 2R₀

          D = 2 0.1

          D = 0.2 cm

           

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

         

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

      \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

       \frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}

       \frac{1}{f}= 0.7066

        f = 1.415 cm

therefore the diffraction is

        y = 1.22  550 10⁻⁹  1.415  / 0.2

        y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

       d = 2y

       d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

       \frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}

        \frac{1}{f}= 0.733

         f = 1.36 cm

diffraction is

        y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

        y = 4.56 10-6 m

the diffraction diameter is

        d_myope = 2y

         d_myope = 9.16 10-6 m

         \frac{d_{normal}}{d_{myope}} = 9.49 /9.16

        \frac{d_{normal}}{d_{myope}} =  1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

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