All categories

3 years ago

Which is the best procedure to make a permanent magnet?

Physics

2 answers:

notka56 [123]3 years ago

7 0

placing a magnetically hard material in a strong magnetic field

fgiga [73]3 years ago

3 0

<h3>Answer;</h3>

Placing a magnetically hard material in a strong magnetic field

<h3>Explanation;</h3>

Magnets are those materials that generate magnetic field. Materials may either be magnetic materials or non-magnetic materials. Non magnetic materials are those materials that can not be attracted by a magnet while magnetic materials are those that can be attracted by a magnet.
Magnetic materials may be classified as either soft magnetic materials or hard magnetic materials based on the ease of magnetization. Soft magnetic materials are those materials that are easily magnetized and easily demagnetized while hard magnetic materials are hard to magnetize and demagnetized.
Hard magnetic materials are used to make permanent magnets. This is done by placing them in a strong magnetic field.

You might be interested in

At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?

Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$$m / s^2$

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$$m/s^2$ is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}$

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

brainly.com/question/14669575

#SPJ4

4 0

2 years ago

How to figure out the question?

Vladimir [108]

Does this help at all to answer the question

7 0

3 years ago

When the pendulum bob reaches the mean position, the net force acting on it is zero. Why then does it swing past the mean positi

ryzh [129]

Answer:

The pendulum bob swing past the mean position because:

When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to the restoring toque the bob starts to accelerates towards its mean postion. The maximum acceleration of the pendulum's bob is $-w^{2} A$ and the the acceleration decreases as $-w^{2} x$ towards the mean position.

The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.

6 0

3 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

3 years ago

A bullet is fired into the air at an angle of 45°. How far does it travel before it is 1,000 feet above the ground? (Assume that

Readme [11.4K]

Answer:

It travels 1414 feets.

Explanation:

Let's take the length the bullet travels l as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height h it reached, we can calculate l using the sin(α) function:

$sin(\alpha )=\frac{opposite side}{hypotenuse}\\sin(\alpha)=\frac{h}{l}\\l=\frac{h}{sin(\alpha)}$

Replacing:

$l=\frac{1000ft}{sin(\frac{\pi}{4})}$

Solving and roundin to the nearest foot:

$l=1414 ft$

3 0

3 years ago