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3 years ago

What atoms are not grouped in magnetic domains

Physics

1 answer:

Crazy boy [7]3 years ago

8 0

Answer:

Electrons are teeny tiny magnets. They have a north and a south pole, too, and spin around an axis. This spinning results in a very tiny but extremely significant magnetic field. Every electron has one of two possible orientations for its axis.In most materials, atoms are arranged in such a way that the magnetic orientation of one electron cancels out the orientation of another. Iron and other ferromagnetic substances, though, are different (ferrummeans iron in Latin). Their atomic makeup is such that smaller groups of atoms band together into areas called domains, in which all the electrons have the same magnetic orientation. Below is an applet that shows you how these domains respond to an outside magnetic field.

Explanation:

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Water flows over the edge of a waterfall at a rate of 1.2 x 10^6 kg/s. There are 50.0 m between the top and bottom of the waterf

Anit [1.1K]

Answer:

5.88×10⁸ W

Explanation:

Power = change in energy / time

P = mgh / t

P = (m/t) gh

P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)

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3 years ago

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A 1451 kg car is traveling at 48.0 km/h. How much kinetic energy does it possess? K.E. =

jarptica [38.1K]

   Kinetic energy = (1/2) (mass) (speed)²

BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second. So we'll have to
make that conversion.

KE = (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²

   = (725.5) · (48 · 1000 · 1 / 3,600)² (kg) · (km·m·hr / hr·km·sec)²

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2 years ago

How did Koch choose candidates?

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2 years ago

TRUE/FALSE: Water balance is the balance between water intake and water excretion?

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3 years ago

Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32.1 kj/mol.

Novosadov [1.4K]

Answer:

389.78681 K

Explanation:

$P_1$ = Initial pressure = 55.1 mmHg

$P_2$ = Final pressure = 1 atm = 760 mmHg

$T_2$ = Boiling point

$T_1$ = Initial temperature = 35°C

$\Delta H_{vap}$ = Heat of vaporization = 32.1 kJ/mol

From the Clausius-Claperyon equation

$ln\dfrac{P_2}{P_1}=(-\dfrac{\Delta H_{vap}}{R})(\dfrac{1}{T_2}-\dfrac{1}{T_1})\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{P_2}{P_1}\dfrac{R}{\Delta H_{vap}}+\dfrac{1}{T_1}\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{760}{55.1}\dfrac{8.314}{32.1\times 10^{3}}+\dfrac{1}{273.15+35}\\\Rightarrow T_2=\left(-ln\left(\frac{760}{55.1}\right)\frac{8.314}{32.1\times \:10^3}+\frac{1}{273.15+35}\right)^{-1}\\\Rightarrow T_2=389.78681\ K$

The normal boiling point of the substance is 389.78681 K

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3 years ago