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choli [55]
3 years ago
14

A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

erward? PLEASE HELP
Physics
2 answers:
Shtirlitz [24]3 years ago
7 0

Answer:

.345

Explanation:

pentagon [3]3 years ago
3 0

Answer:

2.877 m/s

Explanation:

According to the laws of conservation of linear momentum,

the momentum of the moving objects before impact is equal to the momentum of the objects after impact (Assuming no external forces were applied)

Let both players are tackled and moving in V velocity

  • M and m - masses of the players
  • U and  u -  velocities of them respectively (both velocities are towards east direction )

momentum before impact = momentum after impact

                          →MU + →mu  = →(M+ m )v

 91.5  * 2.73 + 63.5 * 3.09 =  (91.5 + 63.5) * V

                                       →V = 2.877 m/s (To East)

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The guy below is wrong!


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What is the electric field at a location b → = &lt;-0.5, -0.4, 0&gt; m, due to a particle with charge +2 nC located at the origi
Naddika [18.5K]

Answer:

E = \frac{8.99 x10^{9} \frac{Nm^2}{C^2} * 2x10^{-9} C}{(0.64m)^2} =43.85N

Explanation:

For this case we assume that we want to find the electrical field at the point P as we can see on the figure attached.

The electrical field wormula is given by:

E = \frac{K Q}{d^2}

Where r is the distance from the point and the charge. On this case we can use the Pythagoras theorem and we got:

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