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lidiya [134]
3 years ago
13

There are many uses for permanent magnets and temporary magnets like an electromagnet. Electric appliances with electric motors

use magnets to turn electricity into motion. Other examples include electric toothbrushes, fans, lawnmowers, and anything else containing a motor.
Magnets are used to hold doors closed, such as in refrigerators, kitchen cabinets and others. Magnets are also used to read and write data on a computer's hard drive or on old-fashioned cassette tapes. There are more magnets in headphones and stereo speakers which help to turn stored music back into the sounds you can hear.
In summary, magnetism is another property of some kinds of matter. There are two poles to a magnet, the south pole and the north pole. Like poles repel and unlike poles repel. There are two different kinds of magnets, temporary and permanent. A temporary magnet can be made by using electricity. All magnets contain an invisible electromagnetic field which surrounds the magnet. There are many uses for magnets, from holding doors closed to helping music sound better with the magnets inside of speakers.

Based on the information provided above how would you describe a magnet?
Physics
1 answer:
Naily [24]3 years ago
3 0

Answer:

A magnet is a piece of metal with a strong attraction to another metal object. The attraction a magnet produces is called a "magnetic field."

This magnetic field is invisible but is responsible for the most notable property of a magnet: a force that pulls on other ferromagnetic materials, such as iron, steel, nickel, cobalt, and so on.  and attracts or repels other magnets.

Most magnets are made of iron, and magnets are used in many common items like cassette tapes, credit cards, toys, and compasses. Magnets are used to make a tight seal on the doors to refrigerators and freezers.

mag·net a peice of iron etc

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Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m
faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>
  • It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
  • We have the expression for slit width w as,

                           w=\frac{2*wavelength*d}{a}

where, d is the distance between slit and the screen, and a is the slit width.

  • Thus, distance between slit and the screen is,

                           d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

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What are the effects of global warming in simple words?
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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

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