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3 years ago

What happens to the convection fluid as it heats up? How might this relate to masses of heated air?

Physics

1 answer:

Hitman42 [59]3 years ago

8 0

Answer:

The convection process plays an important role in the liquid. Due to the increasing heat supply or high amount of temperature, the fluid gets heated up, as a result of which it becomes warm, less dense and eventually rises up forming convection cells.

In the interior of the earth, the hot molten rocks get heated up due to the heat supplied by the core of the earth. This makes the magma warm and less dense and rises upward forming convection currents in the mantle.

This convection process is similar to the convection cells that form in the atmosphere, where the hot, less dense air rises up in the atmosphere forming a low-pressure zone. This uprising air forms convection cells, in which the warm air rises and as it rises high in the atmosphere, the temperature becomes low, making the air cold and it eventually sinks.

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What is the thermal energy of an object?

Vedmedyk [2.9K]

The total kinetic and potential energies

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3 years ago

What is happening at the atomic level to give rise to the observed energy?

ziro4ka [17]

Here is the answer. What is happening at the atomic level to give rise to the observed energy is that the atomic level is affected by the movement of electrons so as to give rise to the observed energy. Hope this answers your question. Have a great day!

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3 years ago

A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w

Scilla [17]

Answer:

$v_f = 20 m/s$

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2$

now for pure rolling condition we will have

$v = R\omega$

also we have

$I = mR^2$

now we will have

$KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}$

$KE = mv^2$

now by work energy theorem we can say

$W = KE_f - KE_i$

$842 J = mv_f^2 - mv_i^2$

$842 = 3(v_f^2) - 3\times 11^2$

now solve for final speed

$v_f = 20 m/s$

3 0

2 years ago

Relative density has no unit .why ?

Klio2033 [76]

Answer:

Relative density is a ratio of the density of a certain material to the density of a reference material. Relative density has no unit, because the units of the material in question cancel out with the units of the reference material. For example, say a cooking oil’s density is 0.9 g/cm^3.

Explanation:

7 0

3 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

$b=2.898\cdot 10^{-3} m \cdot K$ is Wien's displacement constant

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

The Keck telescopes is actually a system of 2 telescopes in the Keck Observatory, located in Mauna kea, Hawai.

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.

7 0

3 years ago