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Diano4ka-milaya [45]
2 years ago
14

A lead weight falls from a height of 6 m onto a muddy surface. It comes to rest after penetrating 0.4 cm into the surface. What

was the magnitude of the average acceleration during the impact? How long did it take to stop?
(Also can I have like a little explanation :))
Physics
1 answer:
WITCHER [35]2 years ago
8 0

The the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.

<h3>Average acceleration of the lead weight during the impact</h3>

The lead weight will fall under the influence of gravity with magnitude of 9.8 m/s².

<h3>Time of motion of the lead weight</h3>

t = √2h/g

where;

  • h is the total height of fall, h = 6 m + 0.4 cm = 6.004 m
  • g is acceleration due to gravity

t = √((2 x 6.004)/9.8)

t = 1.11 s

Thus, the the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

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Answer:

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\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{-25}-\frac{1}{u}=\frac{1}{\frac{400}{7}}\\\Rightarrow -\frac{1}{u}=\frac{7}{400}+\frac{1}{25}\\\Rightarrow \frac{1}{u}=-\frac{7}{400}-\frac{1}{25}\\\Rightarrow \frac{1}{u}=-0.0575\\\Rightarrow u=-17.4\ cm

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