Physics: 7. A 0.045-kg bullet traveling at a speed of 456 m/s is fired vertically into a sponge with a mass of 0.0060kg that is initially at rest. Assuming that the bullet is embedded in the sponge, how high would the combined bullet and sponge rise after the perfectly inelastic collision?

7. A 0.045-kg bullet traveling at a speed of 456 - id700454565886124, kenneth0125

Ответ:

maxi12312345

23.01.2022

Physics

Combined velocity is 402.35 m/s.

Explanation:

Given that,

Mass of the bullet, m = 0.045 kg

Initial speed of the bullet, v = 456 m/s

Mass of the sponge, m' = 0.006 kg

Initial speed of the sponge is 0 as it was at rest. The momentum remains conserved in this case. The bullet is embedded in the sponge, it is a case of inelastic collision. So,

$mv+m'v'=(m+m')V\\\\V=\dfrac{mv+m'v'}{(m+m')}\\\\V=\dfrac{0.045\times 456+0}{(0.045+0.006)}\\\\V=402.35\ m/s$

So, the combined bullet and sponge velocity after the perfectly inelastic collision is 402.35 m/s.

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Ответ:

issagirl05

10.08.2021

Physics

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Explanation:

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10. intergovernmental panel on climate change (IPCC).

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7. A 0.045-kg bullet traveling at a speed of 456 m/s is fired vertically into a sponge with a mass of 0.0060kg that is initially at rest. Assuming that the bullet is embedded in the sponge, how high would the combined bullet and sponge rise after the perfectly inelastic collision?

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