The answer is A. the fields lines never cross, if you bring another magnet near it, the lines work just compress
Answer: The result of "the upper bound of the density" does not go on the denominator.
So simplified, no. The answer is no.
The answers to your questions are as written below:
- The objects that represents a negatively charged particle is : Image B
- The object that represents a positively charged molecule is : Image A
- The object that represents an uncharged molecule is : Image C
- The object the will not move when in an electric fied is : Image C
<h3>Different types of charges molecules</h3>
A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.
A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.
Hence we can concude that the answers to your questions are as listed above.
Learn more about electric charges :brainly.com/question/857179
#SPJ4
attached below is the missing image
In one of the most amazing coincidences in all of science,
the part of the electromagnetic spectrum that's visible to the
human eye is called "visible light".
Visible light is not 'divided' into anything. We mention the names
to seven of the colors in visible light. But all of the thousands of
OTHER colors that we can see are in there too, even though we
don't bother to list their names when we buzz through the rainbow
in the third grade.
Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

And the distance traveled downwards is:

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

Replacing into the first equation:

Rationalizing:

Let's call v1 the final speed of the package dropped from a height H. Thus:

Let v2 be the final speed of the package dropped from a height 4H. Thus:

Taking out the square root of 4:

Dividing v2/v1 we can compare the final speeds:

Simplifying:

The final speed of the second package is twice as much as the final speed of the first package.