Explanation:
Here are some of the ways that energy can change (transform) from one type to another:
The Sun transforms nuclear energy into heat and light energy.
Our bodies convert chemical energy in our food into mechanical energy for us to move.
An electric fan transforms electrical energy into kinetic energy.
the 2 main categories of nonmetals are REACTIVE NONMETALS an NOBLE GAS.
hope it helps
We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

From the question we are told
Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e
<h3>An E
xpression for the <em>magnitude </em>of charge moved</h3>
Generally the equation for the <em>magnitude </em>of charge moved, Q is mathematically given as

Therefore
An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

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Answer:
Explanation:
Given
Both cars mass is m
and solving problem in Vertical and horizontal direction
considering + y and +x to be positive and u be the final velocity of system
Conserving Momentum in Vertical direction

------1
Conserving momentum in x direction
-----2
squaring and adding 1 &2




The distance between Mars and the Sun in the scale model would be 1140 m
Explanation:
In this scale model, we have:
represents an actual distance of

The actual distance between Mars and the Sun is 228 million km, therefore

On the scale model, this would corresponds to a distance of
.
Therefore, we can write the following proportion:

And solving for
, we find:

Learn more about distance:
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