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kondor19780726 [428]
3 years ago
11

An underwater air bubble has an excess inside pressure of 13 pa. what is the excess pressure inside an air bubble with twice the

radius?
Physics
1 answer:
Evgesh-ka [11]3 years ago
6 0
Pressure = 2 * pi * sigma * radius where sigma is the pressure constant of the medium or the material.

Thus, Pressure/Radius = constant

Based on this:
(pressure*radius) in case 1 = (pressure*radius) in case 2
(13 * R) = (P * 2R)
P = 13/2 = 6.5 pa
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(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

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The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

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W=F_f d cos \theta

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d = 4.5 m is the displacement

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W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

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As before, the work done by any force on the crate is

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The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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