It’s distance for my answer but tell me if anyone tells u this answer thanks
Depends on the sandwich, PB&J requires peanut butter and jelly and a butter knife and a piece of bread.
Answer:
Electrons are teeny tiny magnets. They have a north and a south pole, too, and spin around an axis. This spinning results in a very tiny but extremely significant magnetic field. Every electron has one of two possible orientations for its axis.In most materials, atoms are arranged in such a way that the magnetic orientation of one electron cancels out the orientation of another. Iron and other ferromagnetic substances, though, are different (ferrummeans iron in Latin). Their atomic makeup is such that smaller groups of atoms band together into areas called domains, in which all the electrons have the same magnetic orientation. Below is an applet that shows you how these domains respond to an outside magnetic field.
Explanation:
Hello there.
<span>If we increase the force applied to an object and all other factors remain the same that amount of work will
</span><span>C. Increase
</span>
Answer:


Explanation:
The Newton's law in this case is:

Here,
is the air temperture, C and k are constants.
We have
in
So:

And we have
in
, So:

Now, we have:

Applying (1) for
:

Applying (1) for
:
