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Vilka [71]
3 years ago
5

Describe the sun. Include its structure (layers), relative size, location, and effects of its magnetic field. Where does the sun

get its power?
Physics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer: Nuclear fusion.

Explanation: The sun is a medium-sized star, its radius is  695.510 km and its mass is equivalent to that obtained by bringing together about 110 planets equal to Earth (6371 km is its radius).

It has six layers: The core, the radioactive zone, the convective zone, the photosphere, the chromosphere and the corona.

Magnetic field disruptions near active regions can generate strong explosions in the sun such as sun flashes and coronal mass ejections. The degree of complexity of the sun´s magnetic field increases and decreases with the course of each sunspot cycle.  

Sir Arthur Eddington was the first to evaluate all the data and dared to conjecture that nuclear fusion, the process that creates heavy elements from the fusion of lighter ones, could be responsible for the great production of the sun´s energy; this process make the sun´s energy was taken for the earth and the planet get back to the sun recycled energy. The sun has a very large and complex magnetic field; the average magnetic field of the sun is approximately 1 Gauss, almost twice as strong as the average magnetic field of the Earth´s surface (approximately 0.5 Gauss). Because the surface of the sun is more than 12.000 times larger than the Earth, the overall influence of the sun´s magnetic field is immensely large.

You might be interested in
Two small spheres assumed to be identical conductors are placed at 30 cm from each other on a horizontal axis. the first S1 is l
charle [14.2K]

a) The electric force exerted by S1 on S2 is 21.58μN.

In this case we are talking about two different types of charges, a positive charge and a negative charge, therefore, they are sensing a force of attraction.  

The magnitude of the force is determined by using the following formula:

F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}

where:

= Electric force [N]

= Electric constant ()

= First charge [C]

= Second charge [C]

r =  distance between the two charges

So, in this case, the force can be calculated like this:

F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}

So the force will be equal to:

F=21.58x10^{-6}N

which is the same as:

F=21.58 \micro N

b) The electric field created by S1 at the level of S2 is 1.20 \frac{kN}{C}

The electric field tells us how many Newtons of force can be applied on a given point in space per unit of charge caused by an existing electric charge. From the concept, we can take the following formula for the electric field.

E_{S1}=\frac{F_{e}}{q_{2}}

where:

= electric field generated by the first sphere.

 

E_{S1}=\frac{1.20 x10^{-6}N}{18x10^{-9}C}

which yields:

E_{S1}=1.20x10^{3} \frac{N}{C}

E_{S1}=1.20 \frac{kN}{C}

When talking about electric fields, we know what their direction is if we suppose the electric field is always affecting a positive charge in the given point in space. In this case, since S1 is positive, we can asume the electric field is in a direction away from S1.

c)

The electric potential created by S1 at the level or S2 is 360V

Electric potential is defined to be the amount of energy you will have at a given point per electric charge. This electric potential can be found by using the following formula:

V=Er

Where V is the electric potential and it is given in volts.

  • Volts are defined to be 1 Joule per Coulomb. Energy by electric charge.

So we can use the data found in the previous sections to find the electric potential:

V=(1.20x10^{3} \frac{N}{C})(30x10^{-2}m)

V=360V

d)  The force exerted by S2 on S1 will be the same in magnitude as the force exerted by S1 on S2 but oposite in direction. This is because the force will depend on the two charges, and the distance between them, so:

The electric force exerted by S1 on S2 is 21.58μN.

 

The magnitude of the force is determined by using the following formula:

F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}

F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}

So the force will be equal to:

F=21.58x10^{-6}N

which is the same as:

F=21.58 \micro N

e) The electric field generated by S1 in the middle of S1 and S2 is 4.79 \frac{kN}{C}

In order to find the electric field generated by S1, we can make use of the following formula

E=k_{e} \frac{q_{1}}{r_{1}^{2}}

E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}})

which yields:

E=4.79 \frac{kN}{C}

f)  The electric field in the middle of S1 and S2 is 11.99 \frac{kN}{C}

In order to find the electric field generated by two different charges at a given point is found by using the following formula:

E=k_{e} \sum \frac{q_{i}}{r_{i}^{2}}

where:

q_{i}= each of the charges in the system

r_{i}= the distance between each of the charges and the point we are analyzing.

Since the electric field is a vector, we need to take into account the individual electric fields' directions. In this case we suppose we have a positive test charge between the two charges. We can see that the positive test charge will sense a force in the same direction independently on if the force is excerted by the positive charge or the negative charge. Therefore both electric fields will have the same direction. We'll suppose the electric fields will be positive then, so:

E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})[\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}}+\frac{18x10^{-9}C}{(15x10^{-2}m)^{2}}]

which yields:

E=11.99 \frac{kN}{C}

g) The electric potential in the middle of S1 and S2 is 1.80 kV

Since we know what the electric field is from the previous question, we can make use of the same formula we used before to find the electric potential in the middle of S1 and S2

So let's take the formula:

V=Er

So we can use the data found in the previous sections to find the electric potential:

V=(11.99x10^{3} \frac{N}{C})(15x10^{-2}m)

V=1.80kV

h)

The electric potential generated by S2 on the position of S1 is 539.4V and can be found by using the following formula:

V=k_{e}\frac{q_{2}}{r}

So we can use the data provided by the problem to find the electric potential.

V=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{18x10^{-9}C}{30x10^{-2}m})

V=539.4V

8 0
3 years ago
1.5 x 10^3 standard notation
poizon [28]

Answer:

1500

Explanation:

3 0
3 years ago
Suppose that you hear a clap of thunder 16.2 s (seconds) after seeing the associated lightning stroke. The speed of sound waves
spayn [35]
Since the speed of sound is very high, we consider the time period of thunder and lightening negligible.

Now, d = vt

thus, d = 343 x 16.2

thus, d = 5556.6 metres

Thus, we are 5556.6 metres away from lightening ( approximately)

5 0
3 years ago
Convert 162 km/h into m/s.
patriot [66]

Answer:

45 m/s

Explanation:

162 / 3.6 = 45 m/s

Divide by 3.6 to convert km/h to m/s.

4 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
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