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3 years ago

Five properties of magnet

Physics

1 answer:

Olenka [21]3 years ago

3 0

Answer:

1. The magnet is magnetic and can attract iron articles.

2. The magnet has magnetic poles. Each magnet has two kinds of poles: N pole and S pole. They are in pairs.

3. Temporary magnet and permanent magnet: when the ferromagnetic material is magnetized, it is easy to lose the magnetic property, which is called temporary magnet (for example: iron); when the ferromagnetic material is magnetized, it is not easy to lose the magnetic property, which is called permanent magnet (for example: steel).

4. When two magnets are close to each other, the same poles will repel and push away from each other, and the different poles will attract and stick to each other. Therefore: the same pole repels each other, the different pole attracts each other.

5. The attraction of a magnetic object is called magnetism. An object is surrounded by a magnetic material. The area affected by the magnetic force is called the magnetic field.

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Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, $\rho = 10.49 g/cm^{3}$

Density of Pd, $\rho = 12.02 g/cm^{3}$

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, $\rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}$

where

$V_{c} = a^{3}$ = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

$\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}$

Therefore, the length of the unit cell is given as:

$a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}$ (1)

Average atomic weight is given as:

$A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}$

where

$C_{Ag}$ = 79 %

$A_{Ag}$ = 107

$C_{Pd}$ = 21%

$A_{Pd}$ = 106

Therefore,

$A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78$

In the similar way, average density is given as:

$\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}$

$\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}$

Therefore, edge length is given by eqn (1) as:

$a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm$

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3 years ago

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

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Answer: $37^{\circ}$ North of west

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Given

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3 years ago

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3 years ago

Five properties of magnet​

Five properties of magnet