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9966 [12]
3 years ago
13

What is the charge of an atom with 21 protons and 6 electrons

Physics
2 answers:
enot [183]3 years ago
6 0

Answer:

i think it's 15

Explanation:

because the charge of a proton is +1 and for an electron it's -1

Svetach [21]3 years ago
4 0

Answer:

Scandium with an ion charge of +3

Explanation:

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What conclusions can you draw about the connection between the Sun’s altitude and the latitude of the observer on seasonal chang
Cerrena [4.2K]

Answer:

Explanation:

Altitude of the Sun and the latitude position on the earth play an important role in the season change on the earth.

When the altitude of the sun is high then the average temperature of the earth is higher because the luminous intensity of the sun rays is higher due to the focusing of high energy sun rays over a small area.

But when the sun is at higher altitudes we receive less denser rays of the sun and hence we have less heat on the earth on an average.

  • But despite of the altitude some places on the earth have distinct temperature than the other place at the same time of the year. This is due to their latitudinal location. The places near the equator are warmer most of the times throughout the year because they receive the most direct rays while the poles receive slanting rays and hence are colder even in summer when the earth is at lower altitudes.
6 0
3 years ago
A player hits a ball with a force of 150 N by the bat. How much force does the ball exerts on the bat?
Reptile [31]

Answer:

The force on the ball is the same

3 0
3 years ago
Why can’t you hit a feather in mid air with a force of 200 N<br> (This is all the info I was given)
Aloiza [94]
You cannot for real ehhhh
5 0
2 years ago
Mess up my notes part 2 00:01:40And the white snapdragon will have the genotype small r and small r. Now, when the alleles combi
Semmy [17]

Answer:

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Explanation:

8 0
2 years ago
A spherical shell has inner radius Rin and outer radius Rout. The shell contains total charge Q, uniformly distributed. The inte
Tju [1.3M]

Answer:

  E = Q / 4 π ε₀  (r-R_int) / (R_put³ -R_int³)

Explanation:

For this exercise we can use Gauss's law

          Ф = ∫ E. dA = q_{int} / ε₀

Where q is the charge inside the surface.

In this case the surface must be a sphere, the electric field lines and the radius of the sphere are parallel, so the scalar product is reduced to the algebraic product

           ∫ E dA = q_{int}/ ε₀

The area of ​​a sphere is

           A = 4π r²

- The electric field for a distance r < R_int

The charge inside is zero, so the electric field

          E = 0        r <R_in t

- The field for a radio inside the shell

   Let's use the concept of density

         ρ = Q / V

         q = ρ (4/3 π r³)

         dq = ρ 4π r² dr

We substitute in the Gaussian equation

     E ∫ dA = ρ 4π r² dr / ε₀

    E 4π r² = ρ 4π/ε₀   r³ / 3

     E = ρ / 3ε₀ r

We evaluate between the lower limit r = R_int, E = 0 and the upper limit r = r, E = E

      E- 0 = ρ / 3ε₀ (r –R_int)

Density  is

      ρ = q / 4/3 π (R_out³ - R_int³)

Where R <r

     E = Q / 4 π ε₀  (r-R_int) / (R_put³ -R_int³)

5 0
3 years ago
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