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3 years ago

A magnetically soft material is placed in a strong magnetic field. What is the most likely outcome?

Physics

2 answers:

Svet_ta [14]3 years ago

9 0

D-It will become a temporary magnet because the domains will easily realign.

natka813 [3]3 years ago

5 0

Answer: The correct answer is D.

Explanation:

When the iron is placed in the magnetic field then it will magnetize and behave like a temporary magnet. The domains of the iron will align.

Magnetically soft material: It is an temporary magnet which is made from a material that is magnetized. It creates its own magnetic field. For example, magnetically soft material is an electromagnet. The direction of this material can be changed easily.

In the given problem, a magnetically soft material is placed in a strong magnetic field. The domains of the magnetically soft material are already aligned but when this material is placed in the magnetic field then the domains will easily realign.

Therefore, the magnetically soft material will become a temporary magnet because the domains will easily realign.

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A CAR ACCElerates FROM REST To

Ilia_Sergeevich [38]

Answer:

666.6 kW

Explanation:

Power Formula

Power = Force × Velocity

Calculating Force

F = ma
F = m(v - u / t) [From the equation v = u + at]
F = 2,000 (50/7.5)
F = 2,000 (20/3)
F = 40,000/3
F = 13333.3 N

Solving for Power

P = 13333.3 × 50
P = 666666.6 W
P = 666.6 kW

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2 years ago

a fly wheel of mass 12kg and radius 0.5m with wrapped around it.on the end of the rope is 2kg load.intially it is at rest.calcul

leva [86]

Answer:

19.6 N of torque. The 2kg load is being affected by acceleration due to gravity which is 9.8 m/s^s

Explanation:

2×9.8=19.6

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2 years ago

A ray diagram is shown.

alexandr402 [8]

Answer:

Explanation:

The purple and yellow rays from the same point on the object are shown converging at point Z, where the image will be.

7 0

3 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

3 years ago

All the following statements about DC current are true except:

trapecia [35]

Answer:

Is considered as not as dangerous as AC

6 0

4 years ago