All categories

2 years ago

Which statement describes the magnetic field inside a bar magnet?

1 answer:

5 0

The statement that describes the magnetic field inside a bar magnet is as follows: it points from north to south.

<h3>What is a bar magnet?</h3>

A bar magnet is a permanent magnet of rectangular shape.

A magnet generally possess a magnetic field, which is a condition in the space around a magnet which there is a detectable magnetic force and the presence of two magnetic poles.

A bar magnet like every other magnet possesses a magnetic field that points from the north pole to the south pole.

Learn more about magnets at: brainly.com/question/13026686

#SPJ1

You might be interested in

La luz roja visible tiene una longitud de onda de 680 nanómetros (6,8 x 10-7 m). La velocidad de la luz es de 3.0 x108 m / s. ¿C

Lina20 [59]

Answer:

Frequency, $f=4.41\times 10^{14}\ Hz$

Explanation:

Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?

It is given that,

Wavelength of a visible red light is, $\lambda=6.8\times 10^{-7}\ m$

Speed of light is, $c=3\times 10^8\ m/s$

We need to find the frequency of visible red light. It can be calculated using below relation.

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.8\times 10^{-7}}\\\\f=4.41\times 10^{14}\ Hz$

So, the frequency of visible red light is $4.41\times 10^{14}\ Hz$ .

3 0

3 years ago

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla

lesantik [10]

Answer:

When they are connected in series

The 50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

The power rating of the first bulb is $P_1 = 100 \ W$

The power rating of the second bulb is $P_2 = 50 \ W$

Generally the power rating of the first bulb is mathematically represented as

$P_1 = V^2 R$

Where $V$ is the normal household voltage which is constant for both bulbs

$R_1 = \frac{V^2}{P_1 }$

substituting values

$R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

$R_2 = \frac{V^2}{50}$

From the above calculation we see that

$R_2 > R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 1 = I^2_1 R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 2 = I^2_2 R_2$

Now given that they are connected in series which implies that the same current flow through them so

$I_1^2 = I_2^2$

This means that

$P \ \alpha \ R$

So when they are connected in series

$P_2 > P_1$

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0

3 years ago

To do the same amount of work in less time you need to

Dima020 [189]

Answer:

requires more power

Explanation:

5 0

3 years ago

In the diagram, q1 = +6.39*10^_9 C and

ivanzaharov [21]

Answer:

The electric potential will be "259.695 volt".

Explanation:

In the given question, the figure is not provided. Below is the attached figure given.

Given:

$q_1=6.39\times 10^{-9} \ C$

$q_2=3.22\times 10^{-9} \ C$

$AP=(0.150+0.250)$

$=0.40 \ m$

$BP=0.25 \ m$

Now,

At point P, the electric potential will be:

⇒ $V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}$

By putting values, we get

⇒ $=9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]$

⇒ $=259.695 \ Volt$

4 0

3 years ago

In a device like the one shown below, the cylinder is allowed to fall a distance of 300 m. As a result, the temperature of the w

Delvig [45]

Answer:

Increase in the temperature of water would be 0.9 degree C

Explanation:

As we know by energy conservation

Change in the gravitational potential energy of the cylinder = increase in the thermal energy of the water

Here we know that the gravitational potential energy of the cylinder is given as

$U = mgh$

here we have

h = 300 m

now we can say

$Mc\Delta T = (m \times 9.8 \times 300)$

now if the cylinder falls from height h = 100 m

then we have

$Mc\Delta T' = (m \times 9.8 \times 100)$

now from above two equations

$\frac{\Delta T'}{\Delta T} = \frac{100}{300}$

$\Delta T' = 2.7 \times \frac{1}{3} = 0.9 Degree C$

8 0

3 years ago