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Ierofanga [76]
3 years ago
7

The magnetic fields of Earth and several other planets are produced by _____. magnetic substances conductive materials in motion

extreme temperatures extreme pressures
Physics
2 answers:
MakcuM [25]3 years ago
7 0
The answer is conductive materials in motion
Alina [70]3 years ago
4 0

Answer: magnetic substances

Explanation:

The magnetic field of Earth and several other planets are produced by magnetic substances such as Iron and nickel present in the molten form in their core. With their motion, the domains of the magnetic material align in a single direction generating huge magnetic field.

The magnetic field helps protect from high speed solar particles and helps in navigation. Auroras are caused due to magnetic fields. Auroras can be seen of Earth, Saturn, Jupiter indicating that these planets have magnetic fields.

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MIDDLE SCHOOL SCIENCE- <br> Please help, I will give brainliest to best answer.
tatyana61 [14]

Answer:

1.) Waves carry energy through empty space or through a medium without transporting matter. While all waves can transmit energy through a medium, certain waves can also transmit energy through empty space. ... When waves travel through a medium, the particles of the medium are not carried along with the wave.

2.) Mechanical Waves are waves which propagate through a material medium (solid, liquid, or gas) at a wave speed which depends on the elastic and inertial properties of that medium. There are two basic types of wave motion for mechanical waves: longitudinal waves and transverse waves. Longitudinal waves vibrating in the direction of propagation while Transverse waves vibrate at right angles to the direction of its propagation.

3.) They can carry a little energy or a lot of energy. They can be transverse or longitudinal. However, all waves have common properties—amplitude, wavelength, frequency, and speed. Amplitude describes how far the medium in a wave moves.

I hope this helps!

5 0
2 years ago
You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe
madam [21]

Answer:

the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

Explanation:

We can answer this exercise using Gauss's law

      Ф = ∫ e . dA = q_{int} / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

5 0
3 years ago
Pls help me guys simple question​
Airida [17]

Answer:

please write neater

Explanation:

can you write neater so I can answer th question but also is a equal to b

7 0
4 years ago
Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
Elena L [17]

Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C

Volume charge density is given by

\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

8 0
3 years ago
How are lasers used to determine the distance from earth to the moon?
mezya [45]
A beam of laser is directed at a reflecting surface put on the moon when the beam of laser is reflected a receiver on the each measure the time since the beam was sent till it was received. Laser is simply light so it has constant velocity in vacuum ~ air  (c = 3 x 10^8 m/s)

to find the distance:

t : time measured between launching the beam and receiving it 

d : distance

d = ct 

5 0
3 years ago
Read 2 more answers
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