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Kaylis [27]
3 years ago
9

Which is the best procedure to make a permanent magnet?

Physics
2 answers:
notka56 [123]3 years ago
7 0

placing a magnetically hard material in a strong magnetic field

fgiga [73]3 years ago
3 0
<h3><u>Answer;</u></h3>

<em>Placing a magnetically hard material in a strong magnetic field</em>

<h3><u>Explanation</u>;</h3>
  • <em><u>Magnets</u></em> are those materials that generate magnetic field. Materials may either be <em><u>magnetic materials or non-magnetic materials</u></em>. <em><u>Non magnetic materials</u></em> are those materials that can not be attracted by a magnet while <em><u>magnetic materials</u></em> are those that can be attracted by a magnet.
  • <em><u>Magnetic materials</u></em> may be classified as either soft magnetic materials or hard magnetic materials based on the ease of magnetization. <em><u>Soft magnetic materials </u></em>are those materials that are easily magnetized and easily demagnetized while <em><u>hard magnetic materials </u></em>are hard to magnetize and demagnetized.
  • <em><u>Hard magnetic materials</u></em> are used to make permanent magnets. This is done by placing them in a strong magnetic field.
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If two lines in a system of linear equations have the same slope and same y-intercept, how many solutions will the system have?
arsen [322]

Answer:

We will have <u>infinite solutions </u>to the system of linear equations.

Explanation:

Well, when we have two lines with the <u>same slopes and the same y-interception</u>, both of them <u>are overlapped, </u>so we will have <u>infinite solutions </u>to the system of linear equations.

This kind of system is called <u>dependent system.</u>

I hope it helps you!

8 0
3 years ago
Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy
erma4kov [3.2K]

Answer:

Explanation:

⁵⁷Co₂₇  + e⁻¹  =  ²⁷Fe₂₆

mass defect = 56.936296 + .00055 - 56.935399

= .001447 u

equivalent energy

= 931.5 x .001447 MeV

= 1.3479 MeV .

= 1.35 MeV

energy of gamma ray photons = .14  + .017

= .157 MeV .

Rest of the energy goes to neutrino .

energy going to neutrino .

= 1.35 - .157

= 1.193 MeV.

5 0
3 years ago
An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

7 0
4 years ago
What happens if two plates move away from each other?
Sloan [31]

Answer:

They will sometimes crash into other plates in the process and will rub while they are moving creating earthquakes

Explanation:

7 0
3 years ago
Read 2 more answers
What is the energy of an electromagnetic wave that has a frequency of
sukhopar [10]

Answer:

Energy, \; E = 2.6504 * 10^{-34} \; Joules

Explanation:

Given the following data;

Frequency = 4.0 x 10⁹ Hz

Planck's constant, h = 6.626 x 10-34 J·s.

To find the energy of the electromagnetic wave;

Mathematically, the energy of an electromagnetic wave is given by the formula;

E = hf

Where;

E is the energy possessed by a wave.

h represents Planck's constant.

f is the frequency of a wave.

Substituting the values into the formula, we have;

Energy, \; E = 4.0 x 10^{9} * 6.626 x 10^{-34}

Energy, \; E = 2.6504 * 10^{-34} \; Joules

8 0
3 years ago
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