There is no image!?...was there meant to be something attached?
Answer:
Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:
where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is
Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:
Substituting this value into the previous formula and re-arranging it for Q, we find the charge:
Answer:
The<u> heat transfer </u>model showed convection.
In the convection model, the red water on the bottom of the beaker <u>is hot</u>
This means that the water at the bottom of the beaker was <u> less dense than </u>the water near the top of the beaker.
Explanation:
<em>Convection</em> is the transference of heat energy by the movement (translation) of the particles of fluid (liquids or gases).
When the water on the bottom of the beaker is heated, it expands and becomes less dense.
The water near the top of the beaker is cold which makes it denser than the water at the bottom of the beaker.
Thus, the hot water from the bottom of the beaker will ascend toward the top of the beaker, while the cold water on top will descend toward the bottom. As long, as there is a difference of temperature between the water on the bottom and on top of the beaker, there will be a continuous movement of the particles: cold particles from the top replace hot particles from the bottom that ascend, and when the cold particles are heated they will ascend and will be replaced by new cold particles. This continuous translation of hot and cold particles in fluids is the model of heat transfer by convection.
Answer:
ΔS= 10.66 J/K
Explanation:
Given that
For hot reservoir :
T₁= 500 K
For cold reservoir :
T₂= 300 K
Energy transfer ,Q= 8000 J
The entropy change for the Hot reservoir :
The entropy change for the cold reservoir :
The entropy for universe
ΔS=ΔS₁+ΔS₂
ΔS= - 16 + 26.66 J/K
ΔS= 10.66 J/K
