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KiRa [710]
4 years ago
7

A professional boxer hits his opponent with a 1025 N horizontal blow that lasts 0.150 s. The opponent's total body mass is 116 k

g and the blow strikes him near his center of mass and while he is motionless in midair. Determine the following.
(a) impulse the boxer imparts to his opponent by this blow
kg · m/s
(b) the opponent's final velocity after the blow
m/s
(c) Calculate the recoil velocity of the opponent's 5.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer's body.
m/s
Physics
1 answer:
mylen [45]4 years ago
5 0

Answer:

The impulse is  I  =  153.8 \ N \cdot s

The opponents velocity is  v=  1.33 m/s

The opponents head recoils velocity  v_r = 30.8 \ m/s

Explanation:

From the question we are told that

    The force of the blow is  F=  1025 \ N

    The duration of the blow is  t =  0.150

      The mass of the opponent is  m_o  =  116 \ kg

       The mass of the opponents head is  m_h  = 5 \ kg

The impulse the boxer imparts is mathematically represented as

         I  =  F *  t

substituting values

         I  =  1025 * 0.150

          I  =  153.8 \ N \cdot s

The impulse can also be mathematically evaluated as

         I  =  m_o * v

substituting values

          153.8  =  116 * v

          v=  \frac{153.8}{116}

          v=  1.33 m/s

The recoil velocity is mathematically represented as  

              v_r =  \frac{I}{m_h}

substituting values

                v_r =  \frac{153.8}{5}

                v_r = 30.8 \ m/s

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During the last shot of the game, the basketball goes from rest to 15 m/s and reaches the backboard in 0.41 s. What was the acce
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How long must a simple pendulum be if it is to make exactly one swing per four seconds? (That is, one complete oscillation takes
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9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
mojhsa [17]

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

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Hope it helps :)

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