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Aleonysh [2.5K]
2 years ago
5

Which processes transfer energy from the core to the photosphere?

Physics
2 answers:
iragen [17]2 years ago
5 0

Answer:

C, radiation then convection

Explanation:

got it correct on edgen (the sun's energy quiz)

Cerrena [4.2K]2 years ago
3 0

Answer:

radiation, then convection

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Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

Thus for the exercise we will have

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

6 0
3 years ago
What are two ways an engineer can build a car in order for it to accelerate faster
Ket [755]

Explanation:

Take F=ma

a = F/m

For a higher, F higher or m lower

Means higher horse power for engine or lower mass for the car

4 0
2 years ago
What type of wave borders the violet end
wariber [46]
Violet light is at the end of the visible light section of the electromagnetic spectrum. Ultraviolet rays are directly next to violet rays on the EM Spectrum.
3 0
3 years ago
A tin can collapses if all air inside it is taken out why
Veseljchak [2.6K]

That only happens when the tin can is IN air.

In the familiar, comfy part of Earth's atmosphere where we live, the normal pressure of air is around 14.6 pounds on every square inch of everything. That's a big part of the reason why we're built with bodies that generate that same amount of pressure on the INSIDE pressing OUT. That way, we always have the same pressure pushing in both directions, so we know that we won't get crushed or blow up like balloons.

But we have to be careful with our bodies or other things when they're in places where the atmospheric pressure on the outside is NOT normal.

-- When a deep-sea diver goes hundreds of feet down in the ocean, and the pressure of the water is much GREATER than normal air.

-- When an astronaut has to go outside ... where there's NO air ... and fix something on the International Space Station.

When the pressure on the outside becomes very unusual, we have to wear special suits to protect our bodies from the unusual conditions.

The tin can in the story is a lot like our bodies. As long as it has air inside and air outside, the pressure is the same in both directions, so there's no particular force trying to deform the can. But ...

-- If you seal the can with the air inside it, take the can into a vacuum chamber, and pump the air out of the vacuum chamber, then the can only has pressure inside. It'll expand, and eventually spring a little hole in the metal, and all the air inside will blow out.

-- If you take all the air OUT of the can (so the can is REALLY 'empty'), then the pressure on it is all from the outside. In that situation, the can simply collapses, because there's nothing inside to provide pressure in the outward direction.

One more little thing to think about:

When you want some toothpaste to come drizzling out of the tube onto your brush, what do you do ? Do you perhaps squeeze the tube, and increase the pressure on the outside ?

4 0
3 years ago
Name the quantity which is measured by the area occupied below the velocity time graph
kiruha [24]

If you mark off a beginning time and ending time on the graph,
then the area under the part of the graph between those limits
is the distance covered during that period of time.

3 0
3 years ago
Read 2 more answers
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