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Natali [406]
4 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]4 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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Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.
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Answer:

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2

where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})} (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

I_c=\frac{1}{2}MR^2\\\\I_r=mR^2

Finally, by replacing in (1):

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

hope this helps!!

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<h3>Weight of the person in the elevator</h3>

The weight of the person in the elevator is calculated as follows;

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