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docker41 [41]
3 years ago
11

PLEASE HELP!!! 30 POINTS

Physics
1 answer:
Tasya [4]3 years ago
6 0

The impulse and the action and reaction  law emit explain the sensation of the foot when kicking two different objects the answer in each case

a) kick the ball  the sensation is not painful

b) kick the wall the feeling is painful

Newton's second law establishes a linear relationship between force and mass

            F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration

Newton's third law or action and reaction law stable that the force acts on a pairs, the kicking of an object with a force F the object responds with a force of equal magnitude but opposite direction on us.

From Newton's second law we can derive a relationship between the impulse and the variation of the momentum

            I = ∫ F dt = Δp

Where I is the impulse, t the time and Δp the variation of the momentum

The impulse is related to the sensation that the body receives due to the variation of the amount of movement, in this case we have two situations

a) Kick a soccer ball

The mass of the ball is small so the force applied by the foot creates a large acceleration without the foot having a significant decrease in foot speed. Consequently the impulse applied by the ball on the foot is small, the sensation is not painful

b) Kick a brick wall

In this case, the mass of the wall is high and when kicked it does not acquire any movement, consequently the reaction of the wall creates a high impulse on the foot that causes its speed to quickly decrease to zero, the physical sensation of this reduction in quantity movement is painful.

In conclusion, the impulse and the law of action and reaction allow us to find the answer in each case.

a) kick the ball feeling painless

b) kick the wall painful feeling

learn more about the action and reaction law and impulse here:

brainly.com/question/6677486

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                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

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- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

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