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mina [271]
3 years ago
14

A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol

lowing distances from the axis of the rod, where distances are measured perpendicular to the rod's axis.
Physics
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula E = \dfrac{kq }{d}

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}

E = 1461.95 N/C

c) The electric field E is calculated as:

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}

E = 239.76 N/C

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4 0
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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3 years ago
A car starts from rest and drives at 45 m/s in 20 s. how far did it travel
Schach [20]

Answer:

distance=speed×time.

45×20=900m.

6 0
3 years ago
Which statement is true about crystal lattice energy?
irinina [24]

The answer is:

It is a measure of the strength of the bonds between ions.

Lattice energy is an estimate of the strength of the bonds formed by ionic compounds.

The first two choices are wrong because it is actually the opposite.

  • As the ion size increases, lattice energy <u>decreases</u>.
  • As charge of ions increases, lattice energy <u>increases</u>

As for the other third option, it is wrong because lattice energy is the energy RELEASED not absorbed.

3 0
3 years ago
Read 2 more answers
the solubility product of lead fluoride is 3.6 x 10–8. what is its solubility in 0.10M NaF solution, in grams per liter
BartSMP [9]

Answer:

8.8 × 10⁻³ g/L

Explanation:

NaF is a strong electrolyte that ionizes according to the following reaction.

NaF(aq) → Na⁺(aq) + F⁻(aq)

Then, the concentration of F⁻ will also be 0.10 M.

In order to find the solubility of PbF₂ (S), we will use an ICE Chart.

        PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)

I                           0                0.10

C                         +S               +2S

E                          S              0.10 + 2S

The solubility product (Kps) is:

Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²

In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.

Kps = 3.6 × 10⁻⁸ = S . (0.10)²

S = 3.6 × 10⁻⁵ M

The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:

3.6 × 10⁻⁵ mol/L × 245.20 g/mol = 8.8 × 10⁻³ g/L

4 0
3 years ago
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