Convection. It is the transfer of heat by the circulation and movement of the molecules that come from liquid or gas.
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v
= ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v = ( u + u ) / ( 1 + ( u×u / c²) )
v = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v = 1.904c / ( 1 + 0.906304 )
v = 1.904c / 1.906304
v = 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c
Well, with the light spectrum there technically is no middle color. Both green and yellow meet up in the middle at 560 nm (wavelength interval) and 540 THz (frequency interval).
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. 
, the amount of charge stored in this capacitor, will stay the same.
The formula 
relates the electric potential across a capacitor to:
- , the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential 
by changing the capacitance of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
