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1 year ago

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

1 answer:

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

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The speed of a wave on a violin A string is 288 m/s and on the G string is 128 m/s. The force exerted on the ends of the string

Katyanochek1 [597]

Answer:

$\dfrac{\mu_A}{\mu_G}=0.197$

Explanation:

given,

Speed of a wave on violin A = 288 m/s

Speed on the G string = 128 m/s

Force at the end of string G = 110 N

Force at the end of string A = 350 N

the ratio of mass per unit length of the strings (A/G). = ?

speed for string A

$v_A = \sqrt{\dfrac{F_A}{\mu_A}}$ .......(1)

speed for string G

$v_G = \sqrt{\dfrac{F_G}{\mu_G}}$ ........(2)

Assuming force is same in both the string

now,

dividing equation (2)/(1)

$\dfrac{v_G}{v_A}=\dfrac{\sqrt{\dfrac{F_G}{\mu_G}}}{\sqrt{\dfrac{F_A}{\mu_A}}}$

$\dfrac{v_G}{v_A}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}$

$\dfrac{128}{288}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}$

$\dfrac{\mu_A}{\mu_G}=0.197$

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3 years ago

A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an

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Answer:

$v_{2f} = \frac{2vm_1}{m_2 + m_1}$

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

$P_i = m_1v$

After the collision

$P_f = m_1v_{1f} + m_2v_{2f}$

So using the law of momentum conservation

$P_i = P_f$

$m_1v = m_1v_{1f} + m_2v_{2f}$

We can solve for the speed of ball 1 post collision in terms of others:

$v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

Their kinetic energy is also conserved before and after collision

$m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2$

$m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2$

From here we can plug in $v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

$m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2$

$m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2$

$m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2$

$v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0$

$v_{2f}(1 + \frac{m_2}{m_1}) = 2v$

$v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}$

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