Fred (mass 60 kg) is running with the football at a speed of 6.0 m/s when he is met head-on by brutus (mass 120 kg), who is moving at 4.0 m/s. brutus grabs fred in a tight grip, and they fall to the ground. which way do they slide and how far? the coefficient of kinetic friction between football uniforms and the ground is 0.30.

By applying the law of conservation of momentum, the distance Fred and Brutus would slide is equal to 0.0756 meters.

Given the following data:

To find the direction and how far (distance) Fred and Brutus would slide, we would use the law of conservation of momentum:

Where:

Making the subject of formula, we have:

Substituting the given parameters into the formula, we have;

Therefore, they would both slide in the direction Brutus is falling due to the negative sign.

To find the distance, we would solve for the acceleration of both players by applying Newton's Second Law of Motion:

The kinetic frictional force is equal to the sum of the net force acting on both players.

Now, we can determine the distance covered by using the third equation of motion;

Distance, S = 0.0756 meters

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  x = 0.0756 m

Fred moves in the direction where Brutus moves

Explanation:

This exercise is for the moment, we define the system as formed by the two players, for this system the forces in the clash are internal, so the moment is preserved

Initial. Before the crash

       p₀ = m v₀₁ - M v₀₂

Final. After the crash

        = (m + M) v

       p₀ =

       m v₀₁ –M v₀₂ = (m + M) v

       v = (m v₀₁ - M v₀₂) / (m + M)

Let's calculate

       v = (60 6 - 120 4) / (60 +120)

       v = - 120/180

       v = - 0.667 m / s

The negative sign indicates that the final speed is the direction where Brutus runs

Let's use Newton's second law to find the acceleration of the two players

       fr = (m + M) a

       fr = μ N

      N- W = 0

      N = (m + M) g

      μ (m + M) g = (m + M) a

      a = μ g

      a = 0.30 9.8

      a = 2.94 m / s²

We use kinematics to find the distance traveled, the final speed is zero

     v² = v₀² - 2 a x

     x = v₀² / 2 a

     x = 0.667² / (2 2.94)

     x = 0.0756 m

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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