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3 years ago

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In the diagram, which is demonstrated by arrow 1? A. liquid water evaporating from the surface to the atmosphere B. water vapor

condensing into liquid water in the atmosphere C. liquid water moving from the atmosphere to the surface D. liquid water moving along the surface

1 answer:

Shkiper50 [21]3 years ago

6 0

I think D. liquid water moving along the surfac

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Consider this row in the periodic table of elements. Which element(s) are most likely to conduct electricity?

Sidana [21]

A) lithium and beryllium

Explanation:

From the given row on the periodic table, only lithium and beryllium will conduct electricity.

What makes a substance able to conduct electricity?

The presence of free mobile electrons and in some, ions allows them to carry electric currents.

Metals are generally known to be good conductors of heat and electricity. This is because, metals have a large pool of electrons i.e free mobile electrons. They are electropositive with a large size and readily release their electrons for conduction.

Lithium and Beryllium are in the metallic block on the periodic table.

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Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$

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- We can decrease the radius to 1/4 of its original value:

$r'=\frac{1}{4}r$

So the new acceleration is

$a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a$

so, the acceleration has increased by a factor 4 again.

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OlgaM077 [116]

The Answer Is Lowering the amount of reactants

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