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ale4655 [162]
3 years ago
15

Two football players with mass 75 kg and 100 kg run directly toward each other with speeds of 6 m/s and 8 m/s respectively. if t

hey grab each other as they collide, the combined speed of the two players just after the collision would be:
Physics
1 answer:
3241004551 [841]3 years ago
6 0

<u><em>heyaaaaa</em></u>

<u><em>Momentum before Pb = momentum after Pa</em></u>

Pb = 75*6 - 100*8 = -350kgm/s = Pa = (75+100)V where V is the velocity of the combined mass of the two players after the collision.

<u><em>Velocity has magnitude (speed) and direction. V = -350/175 = -2m/s </em></u>

So the two players are moving at 2m/s in the direction the 100kg player was moving before the collision.

<em><u>I arbitrarily chose the direction of the smaller player as positive so the opposite direction (of the larger player) had to be negative. </u></em>

hope it helpssss!!!!!!

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7 0
2 years ago
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3 years ago
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Answer:

a) t = 1.75 s

b) x = 31.5 m

Explanation:

a) The time at which Tom should drop the net can be found using the following equation:

y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = 0

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0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2}

t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s

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b) We can find the distance at which is Jerry when Tom drops the net as follows:

v = \frac{x}{t}

x = v*t = 18 m/s*1.75 m = 31.5 m

Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.

I hope it helps you!                                                                    

3 0
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Answer:The amount if mass of all substances before or after a chemical change are equal

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