A 15-kilogram wagon originally at rest on a level road is pulled by a rope that exerts a constant horizontal force of 45 newton's, If the frictional forces on the wagon total 15 newton's, what will be its acceleration ? * 2.0 〖m/s〗^2
4.0 〖m/s〗^2
15 〖m/s〗^2
30 〖m/s〗^2
3.0 〖m/s〗^2

A. 2 m/

Explanation:

Mass of wagon = 15 kg

Horizontal force = 45 N

frictional force = 15 N

It is good to note that part of the exerted horizontal force would be used to overcome the frictional force on the wagon.

So that, total force exerted on the wagon can be determined by;

total force = horizontal force - frictional force

                 = 45 - 15

                 = 30 N

From Newton's second law of motion, we have;

F = ma

Where: F is the force applied on an object, m is its mass, and a is the acceleration of the object.

Thus,

30 = 15 x a

a =

  = 2 m/

The acceleration of the wagon is 2 m/.

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.