Answer:
Please see below as the answer is self-explanatory.
Explanation:
- The visible range extends roughly from 400 nm (violet) to 700 nm (red).
- Below the violet is the ultra-violet spectrum (with higher energy) and above red, we have the infra-red spectrum.
- The wavelengths in the range of 650 to 690 nm have red as the dominant color.
Answer:
by moving your hands up and down your we're creating transverse wave which travel in a right angle shape towards your friend
a) the number of protons is
more than the electrons
b) 
Explanation:
The net electric charge on the ball is

This electric charge is given by the algebraic sum of the charge of the protons and of the charge of the electrons.
The charge of one proton is:

While the charge of one electron is

So the net charge on the metal ball will be given by

where
is the number of protons
is the number of electrons
So we find:

This means that the number of protons is
more than the electrons.
b)
In this case, we want to make the ball neautral, so we have to remove a net charge of Q' such that the new charge is zero:

This implies that the charge that we must remove is

To do that (and to make the ball losing mass at the same time), we have to remove protons, since they have positive charge.
The number of protons that must be removed is:

The mass of one proton is

Therefore, the total mass that must be removed from the ball is

The magnitude and direction of the electric field at the position of this charge.3.33 N/C upward
An electric field is the electric force per unit of charge. It is assumed that the field's direction corresponds to the direction in which a positive test charge would experience force. The electric field is directed radially inward toward a negative point charge and radially outward from a positive charge.
Value of force F given = 10N
value of charge Q = 3 C
We know that E = F/Q
E = 10/3
= 3.33N
where charge is scalar quantity so the direction of force is the direction of electric field
Hence the magnitude and direction of the electric field at the position of this charge.3.33 N/C upward
Learn more about Electric field here
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