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Natali [406]
3 years ago
9

A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the high

est point, as a wave passes. If the ripples decreaseto 4.7 cm, by what factor does thebug's maximum KE change?
Physics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω²     .................1

and kinetic energy is  directly proportional to square of the amplitude.

so

\frac{KE2}{KE1} =  \frac{A2^2}{A1^2}      .............2

\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}

\frac{KE2}{KE1} = 0.52284

so factor that bug maximum KE change is 0.52284

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irina [24]

The angular momentum is defined as,

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L_i=mvl

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At the end with the collition we have

L_f=(I_b+I_s)\omega

Substituting

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Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

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Substituting

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Answer:

The  value  is  \lambda   = 1.329 *10^{-9} \  m

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