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Natali [406]
4 years ago
9

What are artificial magnets and natural magnets ? examples​

Physics
2 answers:
Stells [14]4 years ago
7 0
Example of artificial magnets: columbite,ferrite and pyrrhotite
examples of natural magnets:
magnetite or lodestone
nevsk [136]4 years ago
5 0

Answer:

The name is basically self explanatory.

Natural magnets are magnets that occurs naturally in nature. All natural magnets are permanent magnets, meaning they will never lose their magnetic power. lodestones is one example of a natural magnet.

Artificial magnets are basically man made magnets. They have extra-strong magnetic power. One example of artificial magnets are the magnets you stick at your refrigerator.

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Unpolarized light passes through two ideal Polaroid sheets. The axis of the first is vertical and the axis of the second is at 3
andrew-mc [135]

Answer:

63.78%

Explanation:

The equation that relates the angle of the polarization and the intensity of ligth transmitted is I=Io*cos²(x), where I is the intensity of the incident light, Io is the intensity of the transmitted light, and x is the angle of polarization. The fraction of incident light will be I/Io, so the answer in given by cos²(37) = 0.6378

3 0
3 years ago
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aleksklad [387]

Answer:

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There is a decrease in power. It can be concluded that there has also been a change in: A)current or voltage.
liraira [26]

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Which list shows the development of atomic models in chronological order?
ikadub [295]
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8 0
3 years ago
Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 3.44 kg and rotate with
Free_Kalibri [48]

Answer:

(a) 20,154.1 J

(b) 95,223.5 J

Explanation:

The expression for the moment of inertia for the uniform solid cylinder is as follows;

I= \frac{1}{2}mr^{2}

Here, I is the moment of inertia, r is the radius and m is the mass of the object.

The expression for the rotational kinetic energy is as follows;

K= \frac{1}{2}I\omega ^{2}

Here, K is the rotational kinetic energy and \omega is the angular velocity.

(a)

Calculate the moment of inertia of the smaller solid cylinder.

I= \frac{1}{2}mr^{2}

Put m= 3.44 kg and r= 0.356 m.

I= \frac{1}{2}(3.44)(0.356)^{2}

I= 0.218 kg m^{2}

Calculate the rotational kinetic energy of the smaller cylinder.

K= \frac{1}{2}I\omega ^{2}

Put I= 0.218 kg m^{2} and \omega = 430 rads^{-1}.

K= \frac{1}{2}(0.218)(430) ^{2}

K= 20,154.1 J

Therefore, the rotational kinetic energy for the smaller cylinder is 20,154.1 J.

(b)

Calculate the moment of inertia of the larger solid cylinder.

I= \frac{1}{2}mr^{2}

Put m= 3.44 kg and r= 0.775 m.

I= \frac{1}{2}(3.44)(0.775)^{2}

I= 1.03 kg m^{2}

Calculate the rotational kinetic energy of the smaller cylinder.

K= \frac{1}{2}I\omega ^{2}

Put I= 1.03 kg m^{2} and \omega = 430 rads^{-1}.

K= \frac{1}{2}(1.03)(430) ^{2}

K= 95,223.5 J

Therefore, the rotational kinetic energy for the larger cylinder is 95,223.5 J.

7 0
4 years ago
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