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3 years ago

What is the difference between a permanent magnet and a temporary magnet

2 answers:

7 0

A permanent magnet does not loose its magnetism easily. It can only be lost by permanent destruction say by fire. Example of this is magnetism from a bar magnet. This magnetism is not induced by say electric current.

A temporary magnet looses its magnetism easily. The magnetism is gained or loosed for the action of carrying out a particular task. This magnetism can be achieved by the gain or loss by passing of electric current over say an iron core.

lara31 [8.8K]3 years ago

4 0

Pernament magnet is one that retains its magnetic properties for a long period of time. Temporary megnet only maintains its magnetism while in a magnetic field produced by permanent magnet or electric current.

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Elabora una tabla, como la del ejemplo, con los resultados obtenidos en los test que desarrollaste en actividades anteriores. Lu

rodikova [14]

Explanation:

Clear rendering reads;

"Make a table, like the one in the example, with the results obtained in the tests you carried out in previous activities. Then answer: What do these results indicate regarding your physical condition and how do they relate to your health? From your answers, consider the challenge of improving your physical condition by maintaining or improving your exercise routine or permanent practice of physical activity".

So the incomplete text above it seems you've been instructed to perform an experiment and then state your result/analysis.

8 0

3 years ago

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

r < a:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

a < r < b:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

r = b:

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

b < r < c:

E = 0

r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

r < c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

r<a:

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

At r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

At a < r < b:

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

$E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = b:

$E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

Again, the minus sign indicates the direction of the field towards the center.

At b < r < c:

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

At r < c:

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

Can someone help with me 1,2,3 please I will mark brainless .

Yuki888 [10]

Answer:

1) A. .33 hr

2) B. 6ft

3) A. 58mi/hr

6 0

2 years ago

A boy kicks a soccer ball, giving it an initial speed of 7.5m/s at an angle of 27° above the horizontal(=ground). How high will

Gelneren [198K]

When the initial speed given is 7.5m/s at an angle of 27° , ball will go

4.637 meters.

Assume no air opposition to the ball ;

Vertical component of ball is sin 27° = 0.453

0.453* 7.5 = 3.404 meters /sec

Time taken to reach ground is :

3.404 = -3.404+9.8*t

t= 6.808/9.8= 0.694 sec

Horizontal component is 7.5*cos27°= 6.682m/s

Distance = speed * time

=6.682 * 0.694

=4.637 meters

Horizontal distance it can cover in 0.694 sec is 4.637 meters

So range of ball is 4.637 meters.

Form of motion experienced by an object or particle that is projected near surface of the earth and moves along a curve is called Projectile motion. Three types of projectile motion are Horizontal projectile motion. Oblique projectile motion and Projectile motion on an inclined plane.

To know more about projectile motion, refer

brainly.com/question/24216590

#SPJ13

3 0

1 year ago

Help!!Help!!Help!!Help!!

DerKrebs [107]

Answer:

y=8

Explanation:

every time you multiply x by 3 you divide y by 3.

x=2, multiply it by 3: x=6

y=24, divide it by 3: y=8

8 0

3 years ago