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nirvana33 [79]
2 years ago
13

List down different layers of the sun. Rank these layers based on their distance from the sun’s center

Physics
1 answer:
vfiekz [6]2 years ago
6 0

Answer:

Core

Radiative zone

Convective zone

Photosphere

Chromosphere

Transient region

Corona

Ranks of layers based on their distance from the sun’s center

1st-corona

2nd-Transient region

3rd-chromosphere

4th-Photosphere

5th-convective zone

6th-radiative zone

7th-core

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A bromine atom has an atomic number of 35 and an atomic mass of 80. What is the structure of this atom?
Brums [2.3K]

Answer:

D. It has a central nucleus composed of 35 protons and 45 neutrons,

surrounded by an electron cloud containing 35 electrons.

hope this was helpful ! <3

3 0
2 years ago
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10. During 4th period we put Klaudia in a box because she was talking too much. We still heard her voice through the box so we d
Alex787 [66]

Answer: Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

Explanation:

Given that;

friction force of ground box = 68 N

student of 7th grade = n

Whitmore can apply a force of 25 N

every other 7th grade student can apply a force of 6 N.

now

friction force = forced applied by whitmore + total force ny 7th grade student

we substitute

68 = 25 + 6n

6n = 68 - 25

6n = 43

n = 43/6

n = 7.17

Therefore Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

6 0
2 years ago
What two factors affect the gravitational force between two objects
Marina86 [1]

Answer:

When dealing with the force of gravity between two objects, there are only two things that are important – mass, and distance. The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them.Explanation:

7 0
2 years ago
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A basketball player can jump 1.6 m off the hardwood floor. With what upward velocity did he leave the floor?
irinina [24]
First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2

= 31.36

Sq root = 5.6
3 0
3 years ago
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A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci
MA_775_DIABLO [31]

Answer:

v_B=3.78\times 10^5\ m/s

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is 6.63\times 10^{-27}\ kg

Speed of nucleus at A is v_A=6.2\times 10^5\ m/s

Potential at point A, V_A=1.5\times 10^3\ V

Potential at point B, V_B=4\times 10^3\ V

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s

So, the speed at point B is 3.78\times 10^5\ m/s.

7 0
3 years ago
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