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ioda
3 years ago
11

The earth has a magnetic field much like a bar magnet. Explain this statement. what does a bar magnet look like? what does its f

ield look like? where are the poles? how does it relate to the earth?
Physics
2 answers:
OlgaM077 [116]3 years ago
8 0

Explanation:

The earth has a magnetic field. It is much like a bar magnet. Imagine a gigantic bar magnet inside the Earth. But there is no giant magnet inside it.

To have a pretty good idea what earth's magnetic field is shaped like we imagine a bar magnet inside the earth.

The magnetic field is made by the motion of molten iron in earth's outer core.  The swirling motion of molten iron changes all the time. Therefore, the magnetic fields will also get change. Then, the magnet poles also move.

The North pole and the south pole are two geographic poles of earth.  These poles are the places on the earth's surface that earth's imaginary spin axis passes through.

There are two magnetic poles of the earth: North magnetic pole and South magnetic pole.

Earth's magnetic field is tilted a little bit. If we imagine that earth's magnetic field is made by a giant bar magnet. Then, the bar magnet would make an with earth's spin axis.

The geographic poles and the magnetic poles are not in the same place.

If we are standing at one magnetic poles then the magnetic field lines would be straight up and down.

amm18123 years ago
7 0
The Earth magnetic fields helps balance out the planet  by the core from the north pole to the south pole in order to keep earth gravity at stand even during revolving around the solar system
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What is (a) the wavenumber and (b) the wavelength of the radiation used by an fm radio transmitter broadcasting at 92. 0 mhz?
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The wavenumber and (b) the wavelength of the radiation used by an fm radio transmitter broadcasting at 92. 0 mhz will be  31.25 * 10^{2} m^{-1} and 0.032 * 10^{2} m respectively

Forms of electromagnetic radiation like radio waves, light waves or infrared (heat) waves make characteristic patterns as they travel through space. Each wave has a certain shape and length. The distance between peaks (high points) is called wavelength.

Wavenumber, also called wave number, a unit of frequency, often used in atomic, molecular, and nuclear spectroscopy, equal to the true frequency divided by the speed of the wave and thus equal to the number of waves in a unit distance.

wavelength = ?

frequency = 92 m Hz = 92 * 10^{6} Hz

speed of light = 3 * 10^{8} m/s

speed of light = frequency * wavelength

wavelength = speed of light  / frequency

                     = 3 * 10^{8}  / 92 * 10^{6}

                     = 0.032 * 10^{2} m

wavenumber = 1 / wavelength

                      = 1 / 0.032 * 10^{2} m

                      = 31.25 * 10^{2} m^{-1}

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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
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6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

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