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ozzi
3 years ago
10

How should you draw the field lines for Earth's magnetic field?

Physics
2 answers:
Juliette [100K]3 years ago
7 0

Answer:

Coming out near the South Pole and going in near the North Pole

Explanation:

yuradex [85]3 years ago
3 0
A. Coming out near the South Pole and going in near the North Pole
You might be interested in
Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r
cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

                                         E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

                                 E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

                                  E_net = E_2 + E_4

                                  E_2 = E_4

                                  E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

                                  r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

                                  r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

                                  E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

                                 E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

                                  E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

                                  Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

                                  E_net = 2*(99.518)*cos(27.12)

                                  E_net = 177.151 N / C

               

5 0
3 years ago
Question 1<br> 2 pts<br> Explain what causes a solution to be a strong acid.
lubasha [3.4K]

Answer:

Cuanto más fuerte es el ácido, más rápido se disocia para generar H +start superscript, plus, end superscript. Por ejemplo, el ácido clorhídrico (HCl) se disocia completamente en iones hidrógeno y cloruro cuando se mezcla con agua, por lo que se considera un ácido fuerte.

5 0
3 years ago
11. A sensitive thermometer is one which
Marta_Voda [28]

Answer:

A. ls sensitive to heat

this is correct answer

I hope it's helpful for you.......

4 0
3 years ago
a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800
Olegator [25]

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

3 0
3 years ago
If a ball rolls at constant velocity, it is rolling at constant speed? explain
Shtirlitz [24]
Velocity is speed with direction. So, if velocity varies directly with speed, that statement would be true. A constant velocity would resort in a constant speed. They are connected and are dependant on each other.

I hope this helps!
~kaikers
7 0
3 years ago
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