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Snowcat [4.5K]
3 years ago
12

Ferromagnetic materials-

Physics
1 answer:
QveST [7]3 years ago
5 0

Answer:

materials which exhibit a spontaneous net magnetization at the atomic level, even in the absence of an external magnetic field.

Explanation:

When a material is placed within a magnetic field, the magnetic forces of the material's electrons will be affected. This effect is known as Faraday's Law of Magnetic Induction. However, materials can react quite differently to the presence of an external magnetic field. This reaction is dependent on a number of factors, such as the atomic and molecular structure of the material, and the net magnetic field associated with the atoms. The magnetic moments associated with atoms have three origins. These are the electron motion, the change in motion caused by an external magnetic field, and the spin of the electrons.

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Okay i'm totally stuck and nobody I know really gets it either, so i've turned to Yahoo for help :)
OlgaM077 [116]

Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:

                     (weight) x (distance from the pivot) <u>on one side</u>
is equal to
                     (weight) x (distance from the pivot) <u>on the other side</u>.

That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.

       (Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).


So now we come to the strange beings on the alien planet.
There are three choices right away that both work:

<u>#1).</u>
(400 N) in the middle-seat, facing (200 N) in the end-seat.

       (400) x (1)  =    (200) x (2)

<u>#2).</u>
(200 N) in the middle-seat, facing (100 N) in the end-seat.

       (200) x (1)  =    (100) x (2)

<u>#3).</u>

On one side:  (300 N) in the end-seat       (300) x (2) = <u>600</u>

On the other side:
                      (400 N) in the middle-seat  (400) x (1) = 400
           and     (100 N) in the end-seat      (100) x (2) = 200
                                                    Total . . . . . . . . . . . . <u>600</u> 


These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.


5 0
3 years ago
Read 2 more answers
A hockey puck with mass 0.3 kg is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After t
JulsSmile [24]

Answer:

The net friction force is 8.01 N

Explanation:

Net friction force = mass of hockey puck × acceleration

From the equations of motion

v^2 = u^2 + 2as

v = 40 m/s

u = 0 m/s (puck was initially at rest)

s = 30 m

40^2 = 0^2 + 2×a×30

60a = 1600

a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

Net friction force = 0.3 × 26.7 = 8.01 N

3 0
3 years ago
Read 2 more answers
1) Which of the following is not a dwarf planet?
vodomira [7]
1 io is a moon not a planet.
2)C,The Milky Way is a spiral galaxy
3)D,this is called the particle wave duality,stuff can act like particles and waves at the same time.
4)the white dwarfs
5)using distance and compass direction(I’m not sure about this one though)
5 0
3 years ago
Read 2 more answers
a body is dropped from a height of 240m. (i)How long does it take to reach the ground. (ii) What is the velocity with which it w
kramer

Answer:

7.0 s, 69 m/s

Explanation:

If we take down to be positive, then the time to reach the ground is:

x = x₀ + v₀ t + ½ at²

240 m = (0 m) + (0 m/s) t + ½ (9.8 m/s²) t²

t = 7.0 seconds

The final velocity is:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²) (240 m - 0 m)

v = 69 m/s

5 0
3 years ago
A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va
ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

U=\dfrac{1}{2}CV^2

C=\dfrac{\varepsilon _oA}{d}

U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2

If energy become double U' = 2 U then d'

U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

7 0
3 years ago
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