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A) The work done by the electric field is zero
B) The work done by the electric field is
C) The work done by the electric field is
Explanation:
A)
The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).
The work done by a force is given by the equation
where
F is the magnitude of the force
d is the displacement of the particle
is the angle between the direction of the force and the direction of the displacement
In this problem, we have:
- The force is directed vertically upward (because the field is directed vertically upward)
- The charge moves to the right, so its displacement is to the right
This means that force and displacement are perpendicular to each other, so
and : therefore, the work done on the charge by the electric field is zero.
B)
In this case, the charge move upward (same direction as the electric field), so
and
Therefore, the work done by the electric force is
and we have:
is the magnitude of the electric force. Since
is the magnitude of the electric field
is the charge
The electric force is
The displacement of the particle is
d = 0.660 m
Therefore, the work done is
C)
In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is
Moreover, we have:
(electric force calculated in part b)
While the displacement of the charge is
d = 2.50 m
Therefore, we can now calculate the work done by the electric force:
And the work is negative because the electric force is opposite direction to the displacement of the charge.
Learn more about work and electric force:
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Answer:
Inlet :
Outlet:
Explanation:
1) Notation and important concepts
Flow of mass represent "the mass of a substance which passes per unit of time".
Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"
Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."
Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".
We know that the flow of mass is given by the following expression
, where
represent the flow rate and
the specific volume at the pressure and temperature given.
is the inlet area
pressure at the inlet area
temperature at the inlet area
is the outlet area
pressure at the outlet area
temperature at the outlet area
represent the flow of mass
If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is
and the entropy is
With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy
Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)
Our interest value would be given using interpolation like this:
2) Solution to the problem
Now since we have all the info required to solve the problem we can find the velocities on this way.
We know from the definition of flow of mass that , but since
we have this:
If we solve from the velocity v we have this:
(*)
And now we just need to replace the values into equation (*)
For the inlet case:
For the oulet case:
