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ololo11 [35]
3 years ago
11

Un móvil que viaja a 40 m/s aplica los frenos y desacelera a razón de -0,25 m/seg 2 , ¿Cuánto tiempo le llevo

Physics
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

Where is a picture?

Explanation:

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After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.
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Answer:

Imp = 11.666\,\frac{kg\cdot m}{s}

Explanation:

Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

(0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (19\,m) = \frac{1}{2}\cdot (0.320\,kg)\cdot v_{A}^{2}

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The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

Imp = (0.32\,kg)\cdot [(17.153\,\frac{m}{s} )-(-19.304\,\frac{m}{s} )]

Imp = 11.666\,\frac{kg\cdot m}{s}

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