In quantum mechanics, a central concept is that both matter and <u>energy</u> are alternate forms of the same entity and therefore both exhibit dual characteristics of particles and of <u>waves</u>.
Matter can be defined as anything that has mass and is able to occupy space.
Thus, any physical object or substance that is found on Earth is typically composed of matter.
Similarly, energy is highly affected by the mass of a any physical object or substance just like matter,
Hence, both energy and matter are known to be made up of atoms and as a result of this fact, exhibit dual characteristics of particles and of waves.
A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.
In conclusion, this central concept makes it easier for us to better understand the behavior of tiny particles such as electrons.
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Answer:
5.7 x 10^12 C
Explanation:
Let the charge on earth and moon is q.
mass of earth, Me = 5.972 x 10^24 kg
mass of moon, Mm = 7.35 x 10^22 kg
Let d be the distance between earth and moon.
the gravitational force between them is

The electrostatic force between them is

According to the question
1 % of Fg = Fe



q = 5.7 x 10^12 C
Thus, the charge on earth and the moon is 5.7 x 10^12 C.
<h3><u>Answer and Explanation</u>;</h3>
- input force refers to the force exerted on a machine, also known as the effort, while the output force is the force machines produce or the Load. The ratio of output force to input force gives the mechanical advantage of a simple machine
- <em><u>The output force exerted by the rake must be less than the input force because one has to use force while raking. The force used to move the rake is the input force. </u></em>
- <em><u>The rake is not going to be able to convert all of the input force into output force, the force the rake applies to move the leaves, because of friction.</u></em>
Answer:

Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e (
)
with this expression we can find the closest approach distance (r)