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Kazeer [188]
2 years ago
8

a_______ is a region where the magnetic fields of a large number of atoms are lined up parallel to a magnets field.

Physics
2 answers:
patriot [66]2 years ago
5 0

Answer:

Magnetic Domain

Explanation:

guapka [62]2 years ago
5 0
Magnetic domain is the answer
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The path the moon's umbra traces across earth's surface is the path of totality.What would you see if you were standing in the p
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You would see the darkest part of the moon
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3 years ago
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari
kolbaska11 [484]

Answer:

Final volumen first process V_{2} = 98,44 cm^{3}

Final Pressure second process P_{3} = 1,317 * 10^{10} Pa

Explanation:

Using the Ideal Gases Law yoy have for pressure:

P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

n_{1} = n_{2} = n

In an isobaric process the pressure is constant so:

P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

Replacing : T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}

V_{2} = 98,44 cm^{3}

Replacing on the ideal gases formula the pressure at this piont is:

P_{2} = 3,92 * 10^{9} Pa

For Temperature the ideal gases formula is:

T = \frac{P V }{n R }

For the second process you have that T_{2} = T_{3}  So:

\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }

P_{2} V_{2}  = P_{3} V_{3}

P_{3} = \frac{P_{2} V_{2}}{V_{3}}

P_{3} = 1,317 * 10^{10} Pa

7 0
3 years ago
2. Use physics terms to explain the benefits of crumple zones in modern cars.
Elden [556K]

Answer:

They decrease trauma by allowing for a more gradual change in velocity

Explanation:

6 0
3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him
juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal  direction

m_{b}v_{b}\cos\theta= m_{c}v_{c}

v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}

Put the value into the formula

v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}

v_{c}=0.0345\ m/s

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0
3 years ago
Evidence that the cosmic background radiation really is the remnant of a Big Bang comes from predicting characteristics of remna
mario62 [17]

Answer:

B) The cosmic background radiation is expected to contain spectral lines of hydrogen and helium, and it does.

Explanation:

3 0
3 years ago
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