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3 years ago

How is an electromagnet different from a magnet?

Physics

1 answer:

RSB [31]3 years ago

5 0

Answer: b . An electromagnet can be changed by changing the amount of current.

Explanation:

In an electromagnet, the magnetic field is generated by an electric current in a coil of wire which reinforces an iron core. But when the electric current applied on the mentioned core is disconnected, it immediately loses its magnetization. This also means that the intensity of the magnetic field can be controlled (increased or decreased) by changing the intensity of the electric current flowing through the nucleus.

On the other hand, the permanent magnet does not depend on electric current to magnetize, because due to its composition (a ferromagnetic material) the magnetization capacity continues even after not being in contact with an external magnetic field.

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How can I get brainly ads back? It only gives me the option to get the subscription.

allochka39001 [22]

One way i got mine back was answering questions of others.

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3 years ago

What is the difference between the experimental group and the control group? its like confusing

Doss [256]

An experimental group is the group in the experiment that receives the variable that is being tested. Only one variable is tested at a time. The control group does not receive the test variable. the experimental groups are used to find the answers in a experiment

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3 years ago

A 2.0-kg ball is at rest when a horizontal force of 5.0 N is applied. In the absence of friction, what is the speed of the ball

Rzqust [24]

Answer:

v = 7.1 m/s

Explanation:

Work applied will change kinetic energy

Fd = ½mv²

v = √(2Fd/m) = √(2(5.0)(10) / 2.0) = √50 = 7.07... m/s

7 0

3 years ago

In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0

3 years ago

A projectile is fired upward with an initial speed vo on an airless world. A short time later, it comes back down and has a fina

Zinaida [17]

Answer:

$W_{grav} < 0$

Explanation:

When a projectile is fired upwards with some initial speed then the it reaches the top of the projectile and then falls back to the ground.

According to the question we need to find the work done by the gravity which is acting downwards for the projectile when it is at a position just about to hit the ground in course of falling down.

As we know that work is given as:

$W=F.s\cos\theta$

here:

$F=$ force of gravity on the object (which is acting downwards)

$s=$ displacement of the object

Here the work done by the gravity at an instant just before the projectile hits the earth will be negative as the displacement is in the direction opposite to the force of gravity.

7 0

4 years ago