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riadik2000 [5.3K]
2 years ago
9

State some important uses of magnet​

Physics
2 answers:
Anastaziya [24]2 years ago
4 0
Magnets are used in magnetic compass, doorbells, refrigerators. Magnets are used in dynamos, motors, loudspeakers, microphones etc. Ceramic magnets are used in computers. Magnets are used in toys to give a magic effect.
GarryVolchara [31]2 years ago
3 0

Use of magnet:

  • Magnets are used in magnetic compass, doorbells, refrigerators.
  • Magnets are used in dynamos, motors, loudspeakers, microphones etc.
  • Ceramic magnets are used in computers.
  • Magnets are used in toys to give a magic effect.
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What are the main characteristics of S.H.M​
agasfer [191]

Answer:

•→ The motion of a particle or body in S.H.M acts towards a fixed point.

•→ Acceleration of the body under S.H.M is proportional to its displacement.

•→ This motion is periodic.

•→ Mechanical energy is conserved in S.H.M

Explanation:

S.H.M is Simple Harmonic Motion

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2 years ago
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An electric field around two charged objects is shown.
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<u></u>

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according to newton's first law of motion what is the reason for a ball throwing up in the airfall back to earth
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3 years ago
The de broglie wavelength of a ________ will have the shortest wavelength when traveling at 30 cm/s.
Airida [17]

Answer: H/MV

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5 0
3 years ago
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(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
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