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KiRa [710]
2 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]2 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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Complete Question

The complete question is shown on the first uploaded image

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Explanation:

In order to get a better understanding of this question let us explain some concepts

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We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

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Looking at the diagram the person is at equilibrium so

                 584 = Ha + Lg

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         So,   0.410 Ha  = 0.840 Lg

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A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial
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Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

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Explanation:

Time of flight of a projectile is given by the expression,

               t=\frac{2usin\theta}{g}

           Here θ = 30° and t = 20 s

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                         u = 169.91 m/s

                         a = 0 m/s²

                          t = 15 s

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                         u = 98.1 m/s

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