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Svetllana [295]
2 years ago
9

Which actions most likely cause the domains in a ferromagnetic material to align?

Physics
2 answers:
Alexxx [7]2 years ago
8 0

Answer:

A ferromagnetic material is a temporary magnet. The domains in a ferromagnetic material are randomly arranged. Under certain actions, the domains align in a particular direction and the material acts as a magnet. The actions that can cause alignment of domains in a ferromagnetic material are:

  • rubbing the material against a magnet would cause the alignment of domains in the same direction as of the magnet.
  • passing electricity around the material would generate magnetic field which would cause domains to align along the direction of the field.
  • placing the material near a strong magnet would cause the alignment of domains in the direction of the field generated by the strong magnet.

Other actions like heating the material,  placing the material in a magnetic field of opposite polarity and hitting the material would lead to demagnetization of the magnetic material.

patriot [66]2 years ago
8 0

Answer:

-rubbing the material against a magnet

-passing electricity around the material would generate magnetic field

-placing the material near a strong magnet

Explanation:

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If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal
iVinArrow [24]

Answer:

xcritical = d− m1 /m2 ( L /2−d)

Explanation: the precursor to this question will had been this

the precursor to the question can be found online.

ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)

. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces

smallest possible value of x such that the bar remains stable (call it xcritical)

∑τA = 0 = m2g(d− xcritical)− m1g( −d)

xcritical = d− m1 /m2 ( L /2−d)

6 0
2 years ago
Does the rankine degree represent a larger or smaller temperature unit than the kelvin degree
zvonat [6]
The best and most correct answer among the choices provided by the question is the first choice, larger.

Rankine is Fahrenheit + 460 , while Kelvin is Celsius + 273. We all know that Fahrenheit has larger number compared to kelvin , thus rankine is much larger.
Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
8 0
2 years ago
In a power plant, pipes transporting superheated vapor are very common. Superheated vapor flows at a rate of 0.3 kg/s inside a p
grigory [225]

Answer:h=160.84 W/m^2-K

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied Q=mc\Delta T=0.3\times 2190\times (30)

Heat due to convection =hA(100-30)

A=\pi d\cdot L

A=\pi 0.05\times 10=1.571 m^2

Q_{convection}=h\times 1.571\times (100-22)=122.538 h

Q=Q_{convection}

19,710=122.538 h

h=160.84 W/m^2-K

5 0
2 years ago
The wind blows a jay bird south with a force of 300 Newtons. The
adoni [48]

Answer:

F = 316.22 N

Explanation:

Given that,

The wind blows a jay bird south with a force of 300 Newtons.

The  jay bird flies north, against the wind, with a force of 100  newtons.

Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

F=\sqrt{300^2+100^2} \\\\F=316.22\ N

Hence, the net force on the jay bird is 316.22 N.

6 0
2 years ago
The hammer throw was one of the earliest Olympic events. In this event, a heavy ball attached to a chain is swung several times
Aleonysh [2.5K]

Answer:

Given that

T= 0.43 s

Radius of the ball path's , r=2.1 m

a)

We know that

f= 1/T

Here f= frequency

      T= Time period

Now by putting the values

f= 1/T

T= 0.43 s

f= 1/0.43

f=2.32 Hz

b)

We know that

V= ω r

ω = 2 π f

ω=Angular speed

V= Linear speed

ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s

V= ω r= 14.60 x 2.1 = 30.66 m/s

c)

Acceleration ,a

a =ω ² r

a= 14.6 ² x 2.1 = 447.63 m/s²

We know that g = 10 m/s²

So

a= a/g= 447.63/10 = 44.7 g m/s²

a= 44.7 g m/s²

7 0
2 years ago
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