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Marina86 [1]
2 years ago
9

Why does hitting a magnet with a hammer cause the magnetism to be reduced?

Physics
1 answer:
Katarina [22]2 years ago
8 0

Hitting a magnet with a hammer cause the magnetism to be Physical disruption and vibration damage the material's order, demagnetizing it.

<h3>Explanation:</h3>

There are numerous methods for demagnetizing a magnet.

A magnet, as we know, has magnetic moments, which are the arrangements of molecules in a specific direction.

The hammer causes the magnetic poles of the magnet to point in opposite directions, causing the magnet to bend.

When we repeatedly hammer on a magnet, the magnetic dipoles inside the magnet are released from their ordered configuration.

Magnetism is known to be caused by the presence of magnetic moments.

As a result, when we hammer it, the dipoles are perturbed, lose their orientation, and magnetic moments cease to exist. As a result, the magnet will become demagnetized.

<h3>What is Magnetism?</h3>

The force that magnets use to either attract or repel one another is known as magnetism.

Electric charges in motion are what generate magnetism. Every substance is made of small particles known as atoms.

Electrons, which are charged particles, are found in every atom. Electrons spin like tops around an atom's nucleus, or center. Their mobility causes an electric current to flow, causing each electron to act as a miniature magnet.

Another strongly magnetic substance must enter the magnetic field of an existing magnet in order to be magnetized. A magnet's magnetic field is the area of magnetic force that surrounds it.

An electric current can magnetize some compounds.

To learn more about magnetism, visit:

brainly.com/question/13026686

#SPJ4

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EleoNora [17]
I think that the answer is A
4 0
4 years ago
A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 50 i +
nikklg [1K]

Answer:

The ball lands 154.3 ft from the origin at an angle of 13.6° from the eastern direction toward the south.

Explanation:

Hi there!

The position vector of the ball is described by the following equation:

r = (x0 + v0x · t + 1/2 · ax · t², y0 + v0y · t + 1/2 · ay · t², z0 + v0z · t + 1/2 · g · t²)

Where:

r =  poisition vector of the ball at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity (eastward).

t = time.

ax = horizontal acceleration (eastward).

y0 = initial horizontal position.

v0y = initial horizontal velocity (southward).

ay = horizontal acceleration (southward)

z0 = initial vertical position.

v0z = initial vertical velocity.

g = acceleration due to gravity.

We have to find at which time the vertical component of the position vector is zero (the ball is on the ground) and then we can calculate the horizontal distance traveled by the ball at that time, using the equations of the horizontal components of the position vector.

Let´s place the origin of the system of reference at the throwing point so that x0 and y0 and z0 = 0.

y =  z0 + v0z · t + 1/2 · g · t²            (z0 = 0)

0 = 48 ft/s · t - 1/2 · 32 ft/s² · t²

0 = t (48 ft/s - 16 ft / s² · t)                 (t= 0, the origin point)

0 = 48 ft/s - 16 ft / s² · t

- 48 ft/s / -16 f/s² = t

t = 3.0 s

Now, we can calculate how much distance the ball traveled in that time.

First, let´s calculate the distance traveled in the eastward direction:

x = x0 + v0x · t + 1/2 · ax · t²              (x0 = 0, ax = 0 there is no eastward acceleration)

x = 50 ft/s · 3 s

x = 150 ft

And now let´s calculate the distance traveled in southward direction:

y = y0 + v0y · t + 1/2 · ay · t²   (y0 = 0 and v0y = 0, initially, the ball does not have a southward velocity).

y =  1/2 · ay · t²

y = 1/2 · (-8 ft/s²) · (3 s)²

y = -36 ft

Then, the final position vector will be:

r = (150 ft, -36 ft, 0)

The traveled distance is the magnitude of the position vector:

|r| = \sqrt{(150ft)^{2} + (-36ft)^{2}} = 154.3 ft

To calculate the angle, we have to use trigonometry (see attached figure):

cos angle  = adjacent side / hypotenuse

cos α = x/r

cos α = 150 ft / 154.3 ft

α = 13.6°

The ball lands 154.3 ft from the origin at an angle of 13.5° from the eastern direction toward the south.

8 0
3 years ago
If a person is walking along at 1.4m/s. How long will it take that person to walk one time around a high school track?​
ratelena [41]

Answer:

285.7s

Explanation:

Given parameters:

Speed of the person  = 1.4m/s

Unknown:

Time it takes to walk one time round the track = ?

Solution:

The circumference of the track is 400m for a standard pitch.

Now;

 Speed  = \frac{distance}{time}

 Time  = \frac{distance }{speed }  

 Now insert the parameters;

Time  = \frac{400}{1.4}  = 285.7s

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3 years ago
a 70 kg desk sits on the floor motionless If gravity is pulling it with an acceleration of 9.8 m/s/s how much force is gravity e
adell [148]

Since force is mass*acceleration,

F = 70kg * 9.8 m/s

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Standing waves are set up on two strings fixed at each end, as shown in the drawing. The two strings have the same tension and m
uysha [10]

The beat frequency produced by the two standing waves is 13 Hz.

<h3>The wavelength of the shorter string</h3>

The wavelength of the shorter string is calculated as follows;

L = \frac{\lambda}{2} \\\\\lambda = 2L\\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{41.9}{225} \\\\\lambda = 0.186 \ m\\\\\lambda = 18.6 \ cm\\\\L= \frac{\lambda }{2} \\\\L = \frac{18.6 \ cm}{2} = 9.3\ cm

<h3>The length of the longer string</h3>

L_2 = 0.58 \ cm \ + 9.3 \ cm\\\\L_2 = 9.88 \ cm \\\\\lambda _2 = 2L_2\\\\\lambda _2 = 2(9.88 \ cm)\\\\\lambda_2 = 19.76 \ cm = 0.1976 \ m

The frequency of the longer string is calculated as follows;

v_1 = v_2\\\\f_2 = \frac{v_2}{\lambda_2} \\\\f_2 = \frac{41.9}{0.1976} \\\\f_2 = 212 \ Hz

<h3>Beat frequency</h3>

The beat frequency produced by the two standing waves is calculated as follows;

F_b = 225 \ Hz \ - \ 212 \ Hz\\\\F_b = 13 \ Hz

Learn more about beat frequency here: brainly.com/question/3086912

8 0
3 years ago
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