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nata0808 [166]
3 years ago
10

) A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the magnitude of the change i

n the momentum of the ball during the rebound?
Physics
1 answer:
deff fn [24]3 years ago
4 0

Answer:

The magnitude of the change in the momentum of the ball during the rebound is 4.4 kg-m/s.

Explanation:

Given that,

Mass of the ball, m = 0.1 kg

Initial speed of the ball, u = 25 m/s

Final speed of the ball, v = -19 m/s (the ball rebounds so it will be negative)

We need to find the magnitude of the change in the momentum of the ball during the rebound. The change in momentum of the object is equal to the difference of final and initial momentum.

\Delta p=m(v-u)

\Delta p=0.1\ kg\times (-19-25)\ m/s

\Delta p=-4.4\ kg-m/s

or

|\Delta p|=4.4\ kg-m/s

So, the magnitude of the change in the momentum of the ball during the rebound is 4.4 kg-m/s. Hence, this is the required solution.

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Answer:

17.7 m/s

Explanation:

Given:

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La moneda comienza a caer, pero la moneda esta inmersa en una sustancia, el aire. El aire comienza a aplicar una resistencia al movimiento de la moneda, y esta resistencia incremente a medida que la velocidad de la moneda incremente. Llega un punto en el que esta nueva fuerza es igual a la fuerza gravitatoria, y en sentido opuesto, lo que causa que la fuerza neta sea 0, y que la moneda caiga a velocidad constante hasta que esta impacta con el suelo.

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También se puede analizar el caso en el que, como la fuerza gravitatoria decrece con el radio al cuadrado, a medida que la moneda cae, la fuerza gravitatoria incrementa. El tema es que en para estas dimensiones, ese cambio en la fuerza gravitacional es imperceptible,

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marshall27 [118]

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Explanation:

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