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hoa [83]
2 years ago
5

PLS SOMONE PLS FILL OUT I WILL GIVE BRAINLIEST HURRYYY

Chemistry
1 answer:
Soloha48 [4]2 years ago
8 0

Answer:

ano yarnnn????? dko maintindihan

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Two billiard balls having masses of 0.2 kg and 0.15kg approach each other. The first ball having a velocity of 2m/s hit the seco
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Answer: The velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.

Explanation:

Given: m_{1} = 0.2 kg,      m_{2} = 0.15 kg

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Formula used is as follows.

m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}

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Substitute the values into above formula as follows.

m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}\\0.2 kg \times 2 m/s + 0.15 kg \times 0 m/s = 0.2 kg \times v'_{1} + 0.15 kg \times 1.5 m/s\\0.4 kg m/s + 0 = 0.2v'_{1} + 0.225 kg m/s\\0.2v'_{1} kg = 0.175 kg m/s\\v'_{1} = \frac{0.175 kg m/s}{0.2 kg}\\= 0.875 m/s

Thus, we can conclude that the velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.

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