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KiRa [710]
3 years ago
11

How many neutrons does element x have if it’s atomic number is 23 and it’s mass is 90

Physics
1 answer:
Oxana [17]3 years ago
3 0

The amount of neutrons would be 23.

The atomic number represents the amount of neutrons, so all you have to do is look at it to know. The atomic mass was put here simply to mislead you. B.T.W, the element is Valadium.

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1. How would you describe the area in which you live? Use terms such as desert, prairie, or
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Answer:

forest rual area

Explanation:

because thats where i live

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With the aim of a diagram describe how a rain bow is formed​
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Answer:

Water droplets acts as tiny prism in the sky. The sunlight when enters these tiny droplets undergo internal reflection and also refract these rays which are dispersed causing a band of seven colors called rainbow.

Explanation:

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A solenoid coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long an
horsena [70]

Answer:

The average magnetic flux through each turn of the inner solenoid is 11.486\times10^{-8}\ Wb

Explanation:

Given that,

Number of turns = 22 turns

Number of turns another coil = 330 turns

Length of solenoid = 21.0 cm

Diameter = 2.30 cm

Current in inner solenoid = 0.140 A

Rate = 1800 A/s

Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid

We need to calculate the magnetic flux

Using formula of magnetic flux

\phi=BA

\phi=\dfrac{\mu_{0}N_{2}I}{l}\times\pi r^2

Put the value into the formula

\phi=\dfrac{4\pi\times10^{-7}\times330\times0.140}{21.0\times10^{-2}}\times\pi\times(\dfrac{2.30\times10^{-2}}{2})^2

\phi=11.486\times10^{-8}\ Wb

Hence, The average magnetic flux through each turn of the inner solenoid is 11.486\times10^{-8}\ Wb

7 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

8 0
3 years ago
Sonia was experimenting with electric charges. She tied two inflated balloons together, held them next to each other, and rubbed
lions [1.4K]

Answer: They will repel each other.

Explanation:

Two inflated balloons when rubbed with woolen cloth will lead to repeal each other because of the similar charges on both the balloons.

Rubbing both the balloons together by the woolen cloth will introduce negative charge in the balloons.

As, we know that the same charges repeal each other both of the balloons with be apart from each other.

This is due to the static electricity, the negatively charged particles jump to the positive one. When balloons are rubbed they become negatively charged.

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3 years ago
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