Answer:
Refraction
Explanation:
- Whenever there is a change in media through which light travels there is change in the speed of light due to which there is bending of light when there is change in the medium.
- When a ray of light passes from and optically denser medium to and optically rarer medium then it bends away from the normal to the interface and vice-versa.
As here the scuba driver is looking from an optically denser medium to an optically rarer medium the boy may appear nearer to the diver in position than is actually in the air.
Snell's law gives the relation between the angle of incidence and the angle of refraction as:

where
n = refractive index
i = angle of incidence
r = angle of refraction
Answer:
372,400 N
Explanation:
The volume of the column is ...
V = Bh = (2 m^2)(19 m) = 38 m^3
If we assume the density is 1000 kg/m^3, then the mass of the water is ...
M = ρV = (1000 kg/m^3)(38 m^3) = 38,000 kg
The force of gravity on that mass is ...
F = Mg = (38,000 kg)(9.8 m/s^2) = 372,400 N
The magnitude of a vertical electric field that will balance the weight of a plastic sphere of mass 2. 1 g that has been charged to -3. 0 NC is 6.86 ×
N/C.
An electric field is the physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field of a system of charged particles.
It is given that,
Mass of sphere, m = 2.1 g =0.0021kg
Charge,q = ₋3nC = ₋ 3 ₓ 
To balance the weight of a plastic sphere, we must determine the magnitude of the electric field. So,

a = g

E = 
E = 6860000 N/C
E = 6.86 ×
N/C
Hence, the magnitude of the electric field that balances its weight is 6.86 ×
N/C .
To know more about electric field refer to: brainly.com/question/8971780
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A) When a charge is moved in an electric field the work done (W) is calculated as charge*(change in potential). We can write W = q*V or V = W/q = 10/1 = 10V . This voltage is a difference in electric potential between 2 points within the field. If the charge is positive, and positive work is done upon it, then the final position is more positive than the original one.
<span>b) If a charge (Q) is released from rest and falls through a potential difference V, then its gain in energy (KE if no other force acts on the charged body) is q*V = 10J. This is the same as the work done in moving the charge to its new position in part (a), and is an example of the conservation of energy.</span>