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3 years ago

The field lines around one end of a bar magnet are shown

Physics

2 answers:

Vitek1552 [10]3 years ago

8 0

<h3>C ⇒ IT IS A SOUTH POLE BECAUSE THE FIELD LINES ENTER THE MAGNET AT THIS END</h3>

In a magnet, the lines (forces) always enter through the <em>south </em>pole and leave through the <em>north</em> pole.

I hope that helps ya'll.

- sincerelynini

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

Vlad1618 [11]3 years ago

7 0

Answer:

Explanation:

Field lines are lines of forces around a bar magnet. The show the direction of force field in a magnet.

Usually, around a bar magnet, the field lines originates and spreads out from the north pole.

Then they converge and enter through the south pole.

Therefore, we can make our choice by inspecting the given diagram. If the lines enters through the pole, it is the south pole.

Where they originate or leave is the north pole.

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Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

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3 years ago

A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?

slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

Ec is the kinetic energy, which is measured in Joules (J).
m is the mass measured in kilograms (kg).
v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

m= 0.500 kg
v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

Finally, the bullet has a kinetic energy of 156.25 J.

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