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Tom [10]
3 years ago
9

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

of 30.0 m. During the collision at the bottom of the elevator shaft, a 86.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision

Physics
1 answer:
Elena L [17]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

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We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

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Here,

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Both distance are the same, then

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\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}

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 Rearranging to find the second wavelength

m_1 \lambda_1 = m_2 \lambda _2

\lambda_2 = \frac{m_1\lambda_1}{m_2}

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<h3>Answer Sentence: </h3><h2><em><u>By the time she had managed to open the door, the postman had already gone</u></em>.</h2>
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