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Ann [662]
2 years ago
11

A football kick returner catches the ball just as a player from the opposing team dives to tackle him. At the time of impact, th

e returner’s momentum is 0 kg-m/s, and the diving player’s momentum is 130 kg-m/s. What will the magnitude of the total momentum be just after the collision?
DOES ANYONE HAVE THE ANSWERS FOR "PHYSICS A SEMESTER EXAM" IN CONNEXUS?
Physics
1 answer:
pochemuha2 years ago
6 0

The total momentum of the players after collision is 130 kgm/s.

The given parameters:

  • <em>Initial momentum of the returner, </em>P_i_1<em> = 0 kgm/s</em>
  • <em>The initial momentum of the diving player, </em>P_i_2<em> = 130 kgm/s</em>

The total momentum of the players after collision is determined by applying the principle of conservation of linear momentum as follows;

P_f = P_i_1 + P_i_2\\\\P_f = 0 + 130\\\\P_f = 130 \ kgm/s

Thus, the total momentum of the players after collision is 130 kgm/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

3 0
4 years ago
Pick the answer that would make the following statement true.
Darina [25.2K]

Explanation:

option A is the correct answer, if the gravitational acceleration is taken 10m/s²(rounding of 9.8/ms²).

hope this helps you.

4 0
3 years ago
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