Colonial Adventure Tours calculates the total price of a trip by adding the trip price plus other fees and multiplying the resul
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In order to understand a monomer let´s first see the structure of a polymer. As an example, in the first figure polyethylene (or polyethene) is shown. This polymer, like every other one, is composed of many repeated subunits, these subunits are called monomer. In the second figure, polyethylene's monomer is shown.
a) Potential in A: -2700 V
b) Potential difference: -26,800 V
c) Work:
Explanation:
a)
The electric potential at a distance r from a single-point charge is given by:
where
is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.
Charge 1 is
and is located at the origin (x=0, y=0)
Charge 2 is
and is located at (x=0, y = 0.40 m)
Point A is located at (x = 0.40 m, y = 0)
The distance of point A from charge 1 is
So the potential due to charge 2 is
The distance of point A from charge 2 is
So the potential due to charge 1 is
Therefore, the net potential at point A is
b)
Here we have to calculate the net potential at point B, located at
(x = 0.40 m, y = 0.30 m)
The distance of charge 1 from point B is
So the potential due to charge 1 at point B is
The distance of charge 2 from point B is
So the potential due to charge 2 at point B is
Therefore, the net potential at point B is
So the potential difference is
c)
The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by
where
q is the charge of the particle
is the potential difference
In this problem, we have:
is the charge of the electron
is the potential difference
Therefore, the work required on the electron is
Answer:
the graph and the answer can be found in the explanation section
Explanation:
Given:
Network rated voltage = 24 kV
Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi
Rn = 0.07 * 8 = 0.56 Ω
Xn = 0.5 * 8 = 4 Ω
If the alternator terminal voltage is equal to network rated voltage will have
Vt = 24 kV/√3 = 13.85 kV/phase
The alternative current is
The impedance Zn is
The voltage drop is
rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω
The effective armature resistance is
The induced voltage for leading power factor is
if cosθ = 0.5
if cosθ= 0.6
EF = 12790.8 V
if cosθ = 0.7
EF = 13731.05 V
if cosθ = 0.8
EF = 14741.6 V
if cosθ = 0.9
EF = 15809.02 V
if cosθ = 1
EF = 13975.6 V
The voltage regulation is
For each value:
if cosθ = 0.5
voltage regulation = -13.8%
if cosθ = 0.6
voltage regulation = -7.6%
if cosθ = 0.7
voltage regulation = -0.85%
if cosθ = 0.8
voltage regulation = 6.4%
if cosθ = 0.9
voltage regulation = 14%
if cosθ = 1
voltage regulation = 0.9%
the graph is shown in the attached image
for 10% of regulation the power factor is 0.81
