A10.0-kg box starts at rest on a level floor. an external, horizontal force of 2.00 × 102 n is applied to the box for a distance of 4.00 m. if the coefficient of kinetic friction between the box and the floor is 0.440, how fast is the block moving at the end of the 4.00 m? assume it starts from rest.

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

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