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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

A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy

vladimir2022 [97]

Answer:

a)KE=878.8 J

b)W=2636.4 J

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

We know that kinetic energy KE is given as follows

$KE=\dfrac{1}{2}mu^2$

m=mass

u=velocity

Now by putting the values in the above equation we get

$KE=\dfrac{1}{2}\times 65\times 5.2^2\ J$

KE=878.8 J

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Now by putting the values in the above equation we get

$W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J$

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

8 0

3 years ago

In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t

Vitek1552 [10]

Answer:

So the ratio will be $\frac{T_L}{T_H}=-0.171$

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So $Q=450j$

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency $\eta =\frac{W}{Q}=\frac{290}{450}=0.6444$

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine $=\frac{0.6444}{0.55}=1.171$

Efficiency of Carnot engine is $\eta =1-\frac{T_L}{T_H}$

$1.171 =1-\frac{T_L}{T_H}$

$\frac{T_L}{T_H}=-0.171$

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3 years ago

Read 2 more answers

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85+5=90 :DD
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4 years ago

The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0

nirvana33 [79]

(a) $7.25\cdot 10^{-8}m$

Wien's displacement law is summarized by the equation

$\lambda = \frac{b}{T}$

where

$\lambda$ is the peak wavelength

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T is the absolute temperature at the surface of the star

For an O-type star, we have

T = 40,000 K

Therefore, its peak wavelength is

$\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m$

(b) Ultraviolet

We can answer this part by looking at the wavelength range of the different parts of the electromagnetic spectrum:

gamma rays

X-rays 1 nm - 1 pm

ultraviolet 380 nm - 1 nm

visible light 750 nm - 380 nm

infrared $25 \mu m - 750 nm$

microwaves 1 mm - $25 \mu m$

radio waves > 1 mm

The peak wavelength of this star is

$\lambda=7.25\cdot 10^{-8}m=72.5 nm$

Therefore, it falls in the ultraviolet region.

(c) No

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The two telescopes, thanks to several instruments, are able to detect much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

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3 years ago