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cricket20 [7]
3 years ago
13

Do you think it is possible to control the magnetic properties of a magnet? Can a magnet be turned on and off? Explain your answ

er.
Physics
1 answer:
Brrunno [24]3 years ago
8 0

Answer:

Yes it is possible to control to some extent

Explanation:

Yes it is possible to control to some extent. Explanation: In general there are two types of magnets : permanent and temporary (electromagnets). ... On the other hand permanent magnet cannot be switched on and off but the magnetic properties can be altered event to an extent when it loses all its magnetic properties.

You might be interested in
Then it is called
DiKsa [7]

Answer:

It is called force of friction

Explanation:

The force of friction is a force that acts between two objects whose surfaces are in contact with each other.

Consider the typical case of an object sliding along a certain surface. There are two types of frictions:

- Static friction: this is the force of friction that acts when the object is not in motion yet. If you push the object forward with a force F, the object will not move immediately, but it will "oppose" to this motion with a force of static friction exactly equal to the push applied:

F_f = F

However, this force of static friction has a maximum value, which is given by

F_{max} = \mu_s N

where

\mu_s is the coefficient of static friction

N is the normal reaction exerted by the surface on the object

So, when F becomes greater than F_{max}, the static friction is no longer able to balance the push applied, and the object will start sliding forward.

- Kinetic friction: this is the force of friction that acts when the object is already in motion. Its magnitude is given by

F_f = \mu_k N

where

\mu_k is the coefficient of kinetic friction, and its value is generally smaller than \mu_s. The direction of this force is also opposite to the direction of motion of the object.

8 0
3 years ago
Set up differential equation of angular S.H.M​
3241004551 [841]

Answer:

{ \sf{ \omega \: is \: the \: angular \: velocity}} \\ { \sf{ \theta \: is \: the \: angular \: displacement}}

3 0
3 years ago
A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away
alekssr [168]

Answer:

31.2 m/s

Explanation:

f_{app} = Frequency of approach = 480 Hz

f_{aw} = Frequency of going away = 400 Hz

V = Speed of sound in air = 343 m/s

v = Speed of truck

Frequency of approach is given as

f_{app} = \frac{Vf}{V - v}                           eq-1

Frequency of moving awayy is given as

f_{aw} = \frac{Vf}{V + v}                          eq-2

Dividing eq-1 by eq-2

\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}

\frac{480}{400} = \frac{343 + v}{343 - v}

v = 31.2 m/s

7 0
3 years ago
Match each fossil fuel with its common use.
nika2105 [10]

Answer:

gasoline and natural gas

electricity and coal

heating homes and gas

Explanation:

7 0
3 years ago
21. A toy car starts from rest and begins to
Bond [772]

Answer:

v = 66 m/s

Explanation:

Given that,

The initial velocity of a car, u = 0

Acceleration of the car, a = 11 m/s²

We need to find the final velocity of the toy after 6 seconds.

Let v is the final velocity. It can be calculated using first equation of motion. It is given by :

v = u +at

v = 0 + 11 m/s² × 6 s

v = 66 m/s

So, the final velocity of the car is 66 m/s.

8 0
3 years ago
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