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Sample – Electronic Lesson
Copyright
Prof J Pieter H van Wyk
VECTORS
AND
SCALARS
© Direct-Science
2
VECTORS AND SCALARS: Resultant velocity
A plane flies at 550 km.h-1
Complete a vector
from town A to town B that
diagram of the above
is in a direction of 00 and
motion and indicate:
1200 km from town A.
i) direction of wind
During the flight the plane
ii) direction in which
experiences a wind that
plane must aim to reach
blows at 150 km.h-1 in a
town B
direction 900. The plane
iii) resultant velocity of
has to land at B.
plane.
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3
VECTORS AND SCALARS: Resultant velocity
Town B
150 km.h-1
If the plane aims at town B the wind will
cause the plane to fly in a
direction east of Town B.
Town B is 1200 km North of
Town A
A plane is at town A.
150 km.h-1
The plane would like to
fly from A to B.
A wind is blowing at 150 km.h-1
in a direction 900.
1200 km
The plane will not reach town B.
150 km.h-1
Town A
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4
VECTORS AND SCALARS: Resultant velocity
150 km.h-1
Aim in this
direction
while flying
into the wind
at
550 km.h-1. 150 km.h-1
Speed relative
to the wind
is 550 km.h-1.
150 km.h-1
Town B
In order to land at B
the plane has to aim
in a direction into the
wind.
The plain must fly
into the wind.
While aiming into the
wind the plane will fly
1200 km
directly to Town B.
The resultant
velocity of the plane
can now be calculated.
Town A
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5
VECTORS AND SCALARS: Resultant velocity
Speed of wind:150 km.h-1
Town B
Complete a vector
diagram.
1200 km
Direction in which the
plane must aim:
Speed of plane
relative to the wind:
550 km.h-1
x
X = 15,660
Direction:15,660 west of north
or 344,340
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Resultant Speed of
plane relative to the
ground:
Sin x = opposite
hypotenuse
= 150 km.h-1
550 km.h-1
= 0,27
Town A
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VECTORS AND SCALARS: Resultant velocity
Speed of wind:150 km.h-1
Town B
Complete a vector
diagram.
1200 km
Speed of plane
relative to the wind:
550 km.h-1
x
Resultant Speed of
plane relative to the
ground:
Resultant Speed of
plane:
[(550 km.h-1)2 –
(150 km.h-1)2]1/2
529 km.h-1
Town A
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7
VECTORS AND SCALARS: Resultant velocity
Speed of wind:150 km.h-1
Town B
1200 km
Speed of plane
relative to the wind:
550 km.h-1
15,660
Resultant Velocity
of plane:
[(550 km.h-1)2 –
(150 km.h-1)2]1/2
529 km.h-1, 00
How long does it take
the plane to reach
town B?
t = s/v
1200 km / 529 km.h-1
2,26 h (136 min).
Town A
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8
ELECTRIC FIELD STRENGTH (E)
6. In a Millikan-tipe experiment a positively charged oil drop was
placed between two horizontal plates, 20mm apart as indicated in the
diagram. The potential difference across the plates is 4000V. The
mass of the oil drop is 1,2 x10-14 kg and it has a charge of 8x10-19 C.
a.) Draw the electric field patterns between the plates.:
-
-
-
-
-
20 mm
+
+
+ + + +
Calculate:
b.) The electric field strength between the plates:
E = V /d = 4000 V / 0,02 M = 200 000 V.m-1
______________________________________________________
c.) Magnitude of the gravitational force acting on the oil drop:
Fg = mg = 1,2 x 10-14 kg x 10 m.s-2 = 1,2 x 10-13 N
______________________________________________________
d.) Magnitude of the Coulomb force acting on the oil drop:
FE = QE = 8 X 10-19 C x 200 000 V.m-1 = 1,6 x 10-13 N
______________________________________________________
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ELECTRIC FIELD STRENGTH (E)
e.) The oil drop is being observed with a microscope.
Explain the behavior of the oil drop:
Two forces
are acting
on the oil
drop.
Upward electrostatic force: 1,6.10-13N
Oil drop
Downward gravitational force: 1,3.10-13N
Upward force is stronger than downward force.
Oil drop experiences an upward resultant force (FRES)
and according to Newton’s second Law the oil drop
will accelerate in direction of the resultant force
(upwards).
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ELECTRIC FIELD STRENGTH (E)
f.) Describe two methods that can be applied to keep the oil
drop stationary:
-
-
-
-
-
-
Upward electrostatic force: 1,6.10-13N = F= QE
Oil drop
Downward gravitational force: 1,3.10-13N = F = mg
+
+
+ + + +
To keep the object stationary the forces must be in equilibrium.
The following are possible ways to balance these forces:
(i) Increase the downward force – mg
(i) m - mass of the particle is fixed and is not allowed to be changed.
(ii) g – constant at 10 m.s-2 and cannot be changed.
Downward force cannot be changed.
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ELECTRIC FIELD STRENGTH (E)
f.) Describe two methods that can be applied to keep the oil
drop stationary:
-
-
-
-
-
-
Upward electrostatic force: 1,6.10-13N = F= QE
Oil drop
Downward gravitational force: 1,3.10-13N = F = mg
+
+
+ + + +
To keep the object stationary the forces must be in equilibrium.
The following are possible ways to balance these forces:
(i) Decrease the upward force – QE
(i) Q – charge on the particle is fixed and is not allowed to be changed.
(ii) E – electric field strength is determined by the distance (d) between
the parallel plates as well as the potential difference over
these plates (V) : E = V/d
© Direct-Science
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ELECTRIC FIELD STRENGTH (E)
f.) Describe two methods that can be applied to keep the oil
drop stationary:
- - - - - Upward electrostatic force: 1,6.10-13N = F= QE
Oil drop
Downward gravitational force: 1,3.10-13N = F = mg
+
+
+ + + +
E = V/d
If V is decreased (without changing d), E will decrease and upward
force will decrease.
If d is increased (without changing V), E will decrease and upward
force will decrease.
To keep particle stationary:
(i) Decrease potential difference over plates.
(ii) Increase distance between plates.
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13
CURRENT
ELECTRICITY
© Direct-Science
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CURRENT ELECTRICITY
State Ohm’s Law:(1)
The current (I) in a metallic conductor is
directly proportional to the potential
difference (V) across its ends, provided
that the temperature remains constant.
Ohm’s Law Equation: (2)
V  I (V = IR)
Potential difference:
Resistance:
Electric Current:
Symbol: ________(3)
V
Symbol: _______(6)
R
Symbol: ________(9)
I
Formula: ________(4)
V = IR
Formula: I_______(7)
= V/R
Formula: ________(10)
I = V/R
Unit:_________(5)
Volt
Unit:_________(8)
Ohm
Unit:___________(11)
Ampere
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CURRENT ELECTRICITY
2 ohm
I3
3 ohm
I1
I2
V1
V2
4 ohm
6 ohm
I4
I5
Rtotal = 1,2 ohm + 2,4 ohm
= 3,6 ohm (12)
I3 = Potential over external circuit
total external resistance
= 20 V / 3,6 ohm = 5,6 A (13)
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20V
1
1
1
=
+
R
3 ohm
2 ohm
1
= 3 + 2
R
6 ohm
1
5
=
R
6 ohm
R
6 ohm
=
1
5
R
=
1,2 ohm
1
1
1
=
+
R
6 ohm
4 ohm
1
= 3 + 2
R
12 ohm
1
5
=
R
12 ohm
R
12 ohm
=
1
5
R
=
2,4 ohm
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CURRENT ELECTRICITY
2 ohm
I3
3 ohm
I1
I2
V1
V2
4 ohm
6 ohm
20V
I4
1
1
1
=
+
R
3 ohm
2 ohm
1
= 3 + 2
R
6 ohm
1
5
=
R
6 ohm
R
6 ohm
=
1
5
R
I5
=
1,2 ohm
V1 = I3 x 1,2 ohm
= 5,6 A x 1,2 ohm
= 6,72 V (14)
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17
CURRENT ELECTRICITY
2 ohm
I3
3 ohm
I1
I2
V1
V2
4 ohm
6 ohm
I4
I5
V2 = I3 x 2,4 ohm
= 5,6 A x 2,4 ohm
= 13,44 V (15)
© Direct-Science
20V
1
1
1
=
+
R
6 ohm
4 ohm
1
= 3 + 2
R
12 ohm
1
5
=
R
12 ohm
R
12 ohm
=
1
5
R
=
2,4 ohm
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CURRENT ELECTRICITY
2 ohm
I3
3 ohm
V1
I5
4 ohm
6 ohm
= 6,72 V / 2 Ohm
I2
I2
= 3,36 A (16)
V2
I4
I1 = V1 / 2 ohm
I1
20V
I4
I5
I4 = V2 / 4 ohm
= 13,44 V / 4 Ohm
= 3,36 A (17)
I5 = V1 / 2 ohm
= 13,44 V / 6 Ohm
= 2,24 A (19)
I2 = V1 / 3 ohm
= 6,72 V / 3 Ohm
= 2,24 A (18)
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I3 = 5,6 A
CURRENT ELECTRICITY
2 ohm
I1 = 3,36 A
3 ohm
I2 = 2,24 A
V1= 6,72 V
V2 = 13,44 V
Amount of Energy (W)
conversion in a resistor:
W = VIt
W = I2Rt
W = V2t / R
20V
I4 = 3,36 A 4 ohm
Time in seconds
I5 = 2,24 A 6 ohm
Calculate the Energy conversion in the following Resistors in 1 minute:
2 ohm:
W = VIt = 6,72 V x 3,36 A x 60 s = 1354,75 J
________________________________________________(20)
3 ohm:
W = VIt = 6,72 V x 2,24 A x 60 s = 903,17 J
________________________________________________(21)
4 ohm:
W = VIt = 13,44 V x 3,36 A x 60 s = 2709,5 J
_________________________________________________(22)
6 ohm:
_________________________________________________(23)
W = VIt = 13,44 V x 2,24 A x 60 s = 1806,34 J
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CURRENT ELECTRICITY
I3 = 5,6 A
2 ohm
3 ohm
I1 = 3,36 A
P = VI
P = I2R
P = V2/R
I2 = 2,24 A
V1= 6,72 V
V2 = 13,44 V
Power in each resistor
can be calculated:
20V
I4 = 3,36 A 4 ohm
I5 = 2,24 A 6 ohm
Calculate the Power in each of the following Resistors:
2 ohm:
P = VI = 6,72 V x 3,36 A = 22,58 W
________________________________________________(20)
3 ohm:
P = VI = 6,72 V x 2,24 A = 15,05 W
________________________________________________(21)
4 ohm:
P = VI = 13,44 V x 3,36 A = 45,16 W
_________________________________________________(22)
6 ohm:
_________________________________________________(23)
P = VI = 13,44 V x 2,24 A = 30,11 W
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21
CURRENT ELECTRICITY
2V 2V
1
1
1
=
+
R
3 ohm
3 ohm
1
= 1 + 1
R
3 ohm
1
2
=
R
3 ohm
R
3 ohm
=
1
2
=
R
1
=
R
1
R
1
R
R
1
R
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2V
3 ohm
3 ohm 3 ohm
+
1
3 ohm
12 ohm
1 + 4 + 2
=
12 ohm
7
=
12 ohm
12 ohm
=
7
=
1,7 ohm
V
12 ohm
3 ohm
1,5 ohm
1
A
6 ohm
+
1
6 ohm
Calculate:
Total resistance of the circuit. (1)
(1,5 + 1,7 + 3) Ohm
= 6.2 Ohm
Calculate:
Reading on ammeter. (2)
I = V/R = 6 V / 6.2 Ohm = 0,97 A
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CURRENT ELECTRICITY
Calculate:
Reading on voltmeter. (3)
V = IR = 0,97 A x 4,7 Ohm
= 4,6 V
2V 2V
2V
3 ohm
3 ohm 3 ohm
Voltmeter reading
reflects the potential
difference over the
section:
Calculate:
Current through 12 ohm resistor. (4)
Potential difference over 12 ohm resistor
= 1,7 ohm x 0,97 amp = 1,7 V.
I = V / R = 1,7 V / 12 Ohm
= 0,14 A
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A
V
12 ohm
3 ohm
6 ohm
1,7 V
Calculate:
Power in 6-ohm resistor. (5)
P = V2 / R
= (1,7 V)2 / 6 Ohm
= 0,48 Watt
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