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3 years ago

The source of the earth’s magnetic field is found in the _____.

Physics

2 answers:

liraira [26]3 years ago

6 0

Answer:

Molten iron and metals in the earth's core

Explanation:

The core of earth consists of molten iron and metals which are continuously spinned due to the rotation of earth. This spinning of earth produces result equivalent to “passing of electric current through a coil to produce magnetic field”. The neutral particles in the core are charged and thus the flow of neutrally charged particle creates a magnetic field.

telo118 [61]3 years ago

5 0

The source of the earth's magnetic field is found in the ;LIQUID MAGNETIZED IRON IN THE CORE OF THE EARTH. The earth's magnetic field is believed to be generated very deep down in the earth's core as a result of molten iron which it contains.

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A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185N at

Vitek1552 [10]

Answer:

$2.75 m/s^2$

Explanation:

We can solve the problem by writing the equations of motion along the horizontal and vertical direction.

Along the horizontal direction we have:

$T cos \theta - \mu N = ma$ (1)

where

$T cos \theta$ is the horizontal component of the tension, where

T = 185 N is the tension in the rope

$\theta=25^{\circ}$ is the angle between the rope and the horizontal

$\mu N$ is the force of friction, where

$\mu=0.27$ is the coefficient of friction

N is the normal reaction of the floor on the box

m = 35.0 kg is the mass of the box

a is the acceleration

Along the vertical direction we have:

$N+T sin \theta-mg=0$ (2)

where

N is the normal force (upward direction)

$T sin \theta$ is the vertical component of the tension in the rope (upward direction)

$mg$ is the weight of the box (downward direction), where

m = 35.0 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

From eq.(2) we get:

$N=mg-T sin \theta$

And substituting into (1), we can find the acceleration:

$T cos \theta - \mu (mg-T sin \theta) = ma\\Tcos \theta -\mu mg + \mu T sin \theta = ma\\a=\frac{T cos \theta- \mu mg + \mu T sin \theta}{m}=\\=\frac{(185)(cos 25^{\circ})-(0.27)(35.0)(9.8)+(0.27)(185)(sin 25^{\circ})}{35.0}=2.75 m/s^2$

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3 years ago