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3 years ago

Help please!! will give brainly for full answers

1 answer:

7 0

1)it starts moving /performing random action
2)once the paper clips is pulled away from the nail it will not perform the same action ,it not hold them together it changes to force field instead of contact force

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1)A sequência de uma entrevista depende da intenção pretendida do entrevistador e do meio de comunicação que será publicado. Nor

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Answer:

saya tak tau bahasa iggris harus Indonesia janga dia diabaikan begitu saja itu

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3 years ago

Give a real-world example of how energy is transformed from electrical energy to thermal energy. Describe how the heat can be tr

timofeeve [1]

Answer:

A microwave

Explanation:

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Lightbulb is a really good one

Same with the sun. That one has Chemical as well

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2 years ago

The density of a sample can be obtained by dividing its __________ by its _______________.

mixer [17]

<span>Density can be calculated and found by dividing the sample's mass by its volume. D=m/v</span>

3 0

3 years ago

a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance

aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b) Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

= $\frac{KQ}{r^{2} }$ for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }$

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }$

E = - 0.77 x 10³ V/m

8 0

3 years ago

How can you make the potential energy as high as possible in a magnetic field between one electromagnet and one piece of iron?

harkovskaia [24]

In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

$U = \mu B$

where;

$\mu$ is the dipole moment
B is the magnetic field

$B = \frac{\mu_0 I}{2\pi r}$

$U = \mu\times (\frac{\mu_0 I}{2\pi r} )$

Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

In step 1, to increase the potential energy, the iron will move towards the electromagnet.
In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.

Learn more about potential energy in magnetic field here: brainly.com/question/14383738

7 0

2 years ago