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3 years ago

Which step is not part of a normal convection cycle?

Physics

2 answers:

meriva3 years ago

7 0

Answer:

B. Warmed air rises away from the surface.

Explanation:

It can be said that the convection cycle responsible for the dynamics of the lithospheric plates occurs not only in the nucleus, but also in the mantle and is transferred as a result transfers determined on the Earth's surface.

In this sense, we know that when the presence of a hot fluid occurs, the substances swell, which results in a reduction in density and then, this increases, since it is less dense than the medium where it is inserted. Thus, convection currents result from the heat flow that causes the hotter material, therefore less dense, to be propelled to the surface.

In summary, we can say that the normal convection cycle occurs when a hot fluid expands, it reduces density and rises because it is less dense than the surrounding environment. Convection currents are the result of the heat flow that makes the hotter material less dense and ascends the surface. This phenomenon occurs in the nucleus and in the mantle, and as a result it transfers certain heat from the interior to the Earth's surface.

Varvara68 [4.7K]3 years ago

4 0

Answer:

Air flows from a high-pressure area to a low-pressure area.

Explanation:

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3 years ago

At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f

jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp = change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s, 0.65 kg m/s)

The magnitude of this vector is calculated as follows:

$|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s$

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

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3 years ago