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vodka [1.7K]
3 years ago
10

What are the methods of avoiding demagnetization​

Physics
1 answer:
lbvjy [14]3 years ago
4 0

\\

The best method to use in prevention of self-demagnetization of a permanent magnet is by use of a keeper. Keepers were also commonly known as amateurs. They help in storing magnets safely, especially those that have low coercivity ( magnets that are extremely susceptible to stray fields).

\\

Hope It Helps!

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Hey, guys I kinda need help with this question, 1. The number of atoms of each element are not equal on both sides of the equati
Xelga [282]

Answer:

The same number of atoms of each element must appear on both sides of a chemical equation. However, simply writing down the chemical formulas of reactants and products does not always result in equal numbers of atoms. You have to balance the equation to make the number of atoms equal on each side of an equation.

Explanation:

I hope thats what u needed.

5 0
3 years ago
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The mass of an electron is ____. A higher than the mass of the proton or the neutron
Harrizon [31]
Mass of an electron = 9.110 x 10⁻³¹ kg.
Mass of a proton = 1.6727 x 10⁻²⁷ kg

∴ mass of a proton/mass of an electron = 1.6727 x 10⁻²⁷ kg/9.110 x 10⁻³¹ kg.
                                                               ~1836

∴ mass of a proton = 1836 x mass of an electron.
∴ mass of an electron is insignificant to the mass of an atom.

∴mass of an atom = mass of protons + mass of neutrons
5 0
3 years ago
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A steel rope is used to lift a 26 kg slab of concrete from the ground to a height of 20 m. Assume that the rope moves the concre
Ne4ueva [31]

Answer:

5200 Joules

Explanation:

Work Formula:

W = F . D

W = (26.10) . 20

W = 260 . 20

W = 5200 Joules

5 0
3 years ago
How does the configuration of the electric field occur between a "parallel plate" setup in a lab?What is the effect of conductor
sleet_krkn [62]

Explanation:

The magnitude of the electric field between the plates is given by

E = -ΔV/d

minus sign indicates Potential decreases in the direction of electric field

where

ΔV is the potential difference between the plates

D is the distance between the plates.

The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface.  That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.

8 0
3 years ago
Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu
lora16 [44]

The initial force between the two charges is given by:

F=k \frac{q_1 q_2}{d^2}

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

q_1' = q_1\\q_2' = q_2\\d' = 2d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F

So the force has increased by a factor 6.

8 0
3 years ago
Read 2 more answers
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