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dlinn [17]
4 years ago
8

1.Many people make the mistake of believing that when a larger defensive player collides with a wide receiver who is making a ca

tch in the air, the larger defender must be exerting a bigger force on the smaller receiver, leading to a greater chance of injury. This, however, is not true. Explain why and which of Newton's 3 laws supports your answer.
Physics
2 answers:
denis-greek [22]4 years ago
8 0

Answer:

See explanation

Explanation:

Solution:-

- larger defensive player with mass (M) collides at velocity (vi) with a wide receiver with mass (m) and velocity (vf) who is making a catch in the air. The larger defensive player exerts a force on the wide receiver.

- The force exerted by the larger defensive player on the wide receiver must follow the "Newton's 3rd Law". The law states that for every action (Force) there must be an equal but opposite reaction.

- So according to the Newton's Law the interaction between larger defensive player and wide receiver has a pair of equal but opposite forces. Hence, both larger defensive player and wide receiver experience the same magnitude of force.

- However, the probability of injury is higher for the smaller receiver than larger defensive player because of the principle of conservation of momentum.

- Assuming.

                                      M > m

                                      vi > vf

- The principle of conservation of momentum is valid because there are no external forces acting on the system consisting of two bodies i.e (  larger defensive player and wide receiver ).

- So for momentum to be conserved:

                       Initial momentum = Final momentum

                       M*vi + m*vf = Final momentum

- We know that after collision the larger defensive player usually comes to a halt while the wide receiver springs into air. Assuming the velocity of larger player to be zero after collision and wide receiver to spring in air with velocity V. Then we have:

                       m*V = M*vi + m*vf

                       V = (M/m)(vi) + vf

- We see that the velocity of wide receiver is significantly higher than that increases the probability of injury for wide receiver.    

lidiya [134]4 years ago
3 0

Answer: To every action there is an equal and opposite reaction. So as the defensive player collides with the receiver it feels an equal and opposite impact also.

Newton Third law of Motion

Explanation:

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude
matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, v_t = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

v_t = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, v_t = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4)  The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

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U

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I am 99999.999999999% sure that it is his second law of motion. 

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