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3 years ago

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Una gota de lluvia (m = 3.54 x 10-5 Kg.) cae verticalmente a velocidad constante bajo la influencia de la gravedad y la resisten

cia del aire. Después de que la gota ha descendido 121 m, ¿cuál es el trabajo (mJ) realizado por la gravedad?

1 answer:

konstantin123 [22]3 years ago

4 0

Answer:

Hola no speak Espanol

Explanation:

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A) +5 J

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Explanation:

A)

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

$\Delta U=-5 J$ is the loss in potential energy of the system

$\Delta K=+6 J$ is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

B)

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

$E=K+U$

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

$\Delta E= \Delta K + \Delta U$

In this problem:

$\Delta K=+6 J$

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So, the change in mechanical energy is:

$\Delta E=+6+(-5)=+1 J$

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

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Products must equal which in a chemical reaction?

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Answer:

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Explanation:

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v = Final velocity

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Earth

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -32\times t^2\\\Rightarrow s=96t-16t^2\ ft$

Differentiating the first equation with respect to time we get

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Equating with zero

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Equating with zero

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Applying the time taken to the above equations, we get

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