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LiRa
3 months ago
11

Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move w

ith the wind, but getting them to be windborne requires the moss to shoot the spores upward. Some species do this by using a spore-containing capsule that dries out and shrinks. The pressure of the air trapped inside the capsule increases. At a certain point, the capsule pops, and a stream of spores is ejected upward at 3.6 m/s, reaching an ultimate height of 20 cm.
A)

What fraction of the initial kinetic energy is converted to the final potential energy?

Express your answer numerically.

UfKi =

B)

What happens to the "lost" energy?

Choose the correct answer

It has been transformed into thermal energy of the spores and surrounding air.
It has been transformed into thermal energy of the spores and potential energy of surrounding air.
It has been transformed into potential energy of the spores and kinetic energy of surrounding air.
It has been transformed into potential energy of the spores and surrounding air.
Physics
1 answer:
Keith_Richards [3.2K]3 months ago
8 0

Solution:

Em_{f} / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

      K = ½ m v²

      U = mgh

We assess the mechanical energy at two positions:

Initial. Lower

    Em₀ = K = ½ m v²

At its highest point

    Em_{f} = U = mg and

Now let's compute

    Em₀ = ½ m 3.6²

    Em₀ = m 6.48

    Em_{f} = m 9.8 × 0.2

    Em_{f} = m 1.96

Thus the energy lost is given by:

    Em_{f} / Em₀ = m 1.96 / m 6.48

   Em_{f} / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

This energy changes into thermal energy affecting the spores and air, since it cannot be regained.

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Early investigators (including Thomas Young) measured the thickness of wool fibers using diffraction. One early instrument used
ValentinkaMS [3465]

Answer:

w= 1.867\times10^{-2}

Explanation:

Provided:

fiber diameter d= 18 μm

screen distance D= 30 cm

wavelength λ= 560 nm

from this, we can determine the fringe width

w=\frac{2\lambda D}{d}

substituting the values yield

w=\frac{2\times560\times10^{-9}\times0.3}{18\times10^{-6}

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2 months ago
A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn
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The alteration in potential energy is  \Delta PE = - 3.8*10^{-16} \ J

In the query, it is stated that

  The intensity of the uniform electric field equals E = 950 \ N/C

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Typically, the force exerted on this electron is expressed mathematically as

     F = qE

Where F signifies the force and  q represents the charge of the electron, which is a fixed value of q = 1.60*10^{-19} \ C

    Thus  

      F = 950 * 1.60 **10^{-19}

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Generally, the work-energy theorem is mathematically framed as

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Where W denotes the work done on the electron by the electric field and  \Delta KE  is the change in kinetic energy

Additionally, work done on the electron can also be described as

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Where \Delta PE signifies the change  in  potential energy  

Thus  

        \Delta PE = - 3.8*10^{-16} \ J

               

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