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pantera1 [17]
3 years ago
7

The sun heats earth's atmosphere unevenly. this causes convection currents to move in large circles in the atmosphere. what is t

he result of convection currents in the atmosphere?
Physics
2 answers:
topjm [15]3 years ago
6 0
The answer is wind. we all know that this is often the horizontal drive of air. All wind is created by the not level heating of layer, that arrays convection currents in the signal. Convection currents on an enormous<span> scale make global winds; convection currents on a minor scale cause local winds.

Explanation:


</span>Convection currents area unit is known<span> in Earth's mantle. Heated mantle material is shown rising from deep </span>within<span> the mantle, </span>whereas cooler<span> mantle material sinks, </span>making<span> a convection current. </span>it's<span> thought that </span>this sort<span> of current is </span>responsible for<span> the movements of the plates of </span>crust.

<span>Convection is </span>one in each of the key<span> processes in </span>making<span> our weather. Convection is one </span>explanation for<span> rising air in our atmosphere, </span>sometimes heat<span> air rises </span>higher than<span> cold air.</span>
<span>Convection </span>happens once the planet<span> is heated </span>inconsistently<span>.

The air over </span>the planet can<span> heat </span>in step with how briskly the planet under that<span> heats. That </span>a part of<span> the atmosphere that heats </span>quicker can<span> expand.</span>
vladimir2022 [97]3 years ago
3 0

The answer is wind. We know that this is the horizontal drive of air. All wind is produced by the not level heating of Earth's surface, which arrays convection currents in signal. Convection currents on a big scale make global winds; convection currents on a minor scale cause local winds.

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A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
4 years ago
Question 11
Katarina [22]
It depends on the graphics, color, structure all that stuff for me to believe if the image is real or virtual
Hope I helped you
4 0
3 years ago
Which material is the best heat insulator?<br><br> metal<br><br> wood<br><br> plastic<br><br> glass
OlgaM077 [116]
Of the materials listed wood is the best insulator. It would be the least hot if exposed to similar temperatures.
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3 years ago
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Find the mass of a 52.2N bucket.​
Goshia [24]

Answer:

m = 5.22 kg

Explanation:

The force acting on the bucket is 52.2 N.

We need to find the mass of the bucket.

The force acting on the bucket is given by :

F = mg

g is acceleration due to gravity

m is mass

m=\dfrac{F}{g}\\\\m=\dfrac{52.2}{10}\\\\=5.22\ kg

So, the mass of the bucket is 5.22 kg.

3 0
3 years ago
What happens if :<br> . The test charge is not tiny.
docker41 [41]

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

brainly.com/question/16737526

#SPJ9

3 0
2 years ago
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