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Sample – Electronic Lesson
CopyrightProf J Pieter H van Wyk
© Direct-Science 2
VECTORS AND
SCALARS
© Direct-Science 3
VECTORS AND SCALARS: Resultant velocity
A plane flies at 550 km.h-1
from town A to town B
that is in a direction of 00
and 1200 km from town
A. During the flight the
plane experiences a wind
that blows at 150 km.h-1 in
a direction 900. The plane
has to land at B.
Complete a vector
diagram of the above
motion and indicate:
i) direction of wind
ii) direction in which
plane must aim to reach
town B
iii) resultant velocity of
plane.
© Direct-Science 4
VECTORS AND SCALARS: Resultant velocity
Town A
Town B
1200 km
Town B is 1200 km North of Town A
A wind is blowing at 150 km.h-1
in a direction 900.
If the plane aims at town B the wind will cause the plane to fly in a direction east of Town B.
A plane is at town A.
The plane would like to fly from A to B.
The plane will not reach town B.
150 km.h-1
150 km.h-1
150 km.h-1
© Direct-Science 5
VECTORS AND SCALARS: Resultant velocity
Town A
Town B
1200 km
In order to land at Bthe plane has to aim in a direction into the
wind.The plain must fly
into the wind.
150 km.h-1
150 km.h-1
150 km.h-1
While aiming into the wind the plane will fly
directly to Town B.
Aim in this direction
while flying into the wind
at550 km.h-1.
Speed relative to the wind
is 550 km.h-1.
The resultant velocity of the plane can now be calculated.
© Direct-Science 6
VECTORS AND SCALARS: Resultant velocity
Town A
Town B
1200 km
Complete a vector diagram.
Speed of plane relative to the wind:550 km.h-1
Speed of wind:150 km.h-1
Resultant Speed of plane relative to the ground:
Direction in which the plane must aim:
xSin x = opposite
hypotenuse= 150 km.h-1
550 km.h-1
= 0,27
X = 15,660 Direction:15,660 west of north
or 344,340
© Direct-Science 7
VECTORS AND SCALARS: Resultant velocity
Town A
Town B
1200 km
Complete a vector diagram.
Speed of plane relative to the wind:550 km.h-1
Speed of wind:150 km.h-1
Resultant Speed of plane relative to the ground:
x
Resultant Speed of plane:
[(550 km.h-1)2 – (150 km.h-1)2]1/2
529 km.h-1
© Direct-Science 8
VECTORS AND SCALARS: Resultant velocity
Town A
Town B
1200 km
Speed of plane relative to the wind:550 km.h-1
Speed of wind:150 km.h-1
Resultant Velocity of plane:
[(550 km.h-1)2 – (150 km.h-1)2]1/2
529 km.h-1, 00
15,660
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).
© Direct-Science 9
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.:
Calculate:b.) The electric field strength between the plates:
______________________________________________________
c.) Magnitude of the gravitational force acting on the oil drop:
______________________________________________________
d.) Magnitude of the Coulomb force acting on the oil drop:
______________________________________________________
ELECTRIC FIELD STRENGTH (E)
20 mm
- - - - - -
+ + + + + +
E = V /d = 4000 V / 0,02 M = 200 000 V.m-1
Fg = mg = 1,2 x 10-14 kg x 10 m.s-2 = 1,2 x 10-13 N
FE = QE = 8 X 10-19 C x 200 000 V.m-1 = 1,6 x 10-13 N
© Direct-Science 10
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
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 dropwill accelerate in direction of the resultant force
(upwards).
Oil drop
© Direct-Science 11
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
Downward gravitational force: 1,3.10-13N = F = mg
Oil drop
- - - - - -
+ + + + + +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.
© Direct-Science 12
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
Downward gravitational force: 1,3.10-13N = F = mg
Oil drop
- - - - - -
+ + + + + +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 13
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
Downward gravitational force: 1,3.10-13N = F = mg
Oil drop
- - - - - -
+ + + + + +
E = V/dIf 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.
© Direct-Science 14
CURRENT ELECTRICITY
© Direct-Science 15
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:
Symbol: ________(3)
Formula: ________(4)
Unit:_________(5)
Electric Current:
Symbol: ________(9)
Formula: ________(10)
Unit:___________(11)
Resistance:
Symbol: _______(6)
Formula: _______(7)
Unit:_________(8)
V
V = IRVolt
RI = V/R
Ohm
II = V/R
Ampere
© Direct-Science 16
CURRENT ELECTRICITY
Rtotal = 1,2 ohm + 2,4 ohm
= 3,6 ohm (12)
2 ohm
3 ohm
4 ohm
6 ohm
I1
I2
I4
I5
I3 V1V2 20V
1
R=
1
2 ohm+
1
3 ohm
1
R=
6 ohm
5
R
1=
6 ohm
5
1
R=
6 ohm
3 + 2
R = 1,2 ohm
1
R=
1
4 ohm+
1
6 ohm
1
R=
12 ohm
5
R
1=
12 ohm
5
1
R=
12 ohm
3 + 2
R = 2,4 ohm
I3 = Potential over external circuit
total external resistance
= 20 V / 3,6 ohm = 5,6 A (13)
© Direct-Science 17
CURRENT ELECTRICITY
V1 = I3 x 1,2 ohm
= 5,6 A x 1,2 ohm
= 6,72 V (14)
2 ohm
3 ohm
4 ohm
6 ohm
I1
I2
I4
I5
I3 V1V2 20V
1
R=
1
2 ohm+
1
3 ohm
1
R=
6 ohm
5
R
1=
6 ohm
5
1
R=
6 ohm
3 + 2
R = 1,2 ohm
© Direct-Science 18
CURRENT ELECTRICITY
V2 = I3 x 2,4 ohm
= 5,6 A x 2,4 ohm
= 13,44 V (15)
2 ohm
3 ohm
4 ohm
6 ohm
I1
I2
I4
I5
I3 V1V2 20V
1
R=
1
4 ohm+
1
6 ohm
1
R=
12 ohm
5
R
1=
12 ohm
5
1
R=
12 ohm
3 + 2
R = 2,4 ohm
© Direct-Science 19
CURRENT ELECTRICITY
I1 = V1 / 2 ohm
= 6,72 V / 2 Ohm
= 3,36 A (16)
2 ohm
3 ohm
4 ohm
6 ohm
I1
I2
I4
I5
I3 V1V2 20V
I4 = V2 / 4 ohm
= 13,44 V / 4 Ohm
= 3,36 A (17)
I4
I2 = V1 / 3 ohm
= 6,72 V / 3 Ohm
= 2,24 A (18)
I2
I5 = V1 / 2 ohm
= 13,44 V / 6 Ohm
= 2,24 A (19)
I5
© Direct-Science 20
CURRENT ELECTRICITY
Amount of Energy (W) conversion in a resistor: W = VIt W = I2Rt W = V2t / R
2 ohm
3 ohm
4 ohm
6 ohm
I1 = 3,36 A
V2 = 13,44 V 20V
Calculate the Energy conversion in the following Resistors in 1 minute:2 ohm: ________________________________________________(20)3 ohm: ________________________________________________(21)4 ohm: _________________________________________________(22)6 ohm: _________________________________________________(23)
I4 = 3,36 A
I2 = 2,24 A
I5 = 2,24 A
Time in seconds
I 3 =
5,6
A
V1= 6,72 V
W = VIt = 6,72 V x 3,36 A x 60 s = 1354,75 J
W = VIt = 6,72 V x 2,24 A x 60 s = 903,17 J
W = VIt = 13,44 V x 3,36 A x 60 s = 2709,5 J
W = VIt = 13,44 V x 2,24 A x 60 s = 1806,34 J
© Direct-Science 21
CURRENT ELECTRICITYPower in each resistor can be calculated:
P = VI P = I2R P = V2/R
2 ohm
3 ohm
4 ohm
6 ohm
I1 = 3,36 A
V2 = 13,44 V 20V
Calculate the Power in each of the following Resistors:2 ohm: ________________________________________________(20)3 ohm: ________________________________________________(21)4 ohm: _________________________________________________(22)6 ohm: _________________________________________________(23)
I4 = 3,36 A
I2 = 2,24 A
I5 = 2,24 A
I 3 =
5,6
A
V1= 6,72 V
P = VI = 6,72 V x 3,36 A = 22,58 W
P = VI = 6,72 V x 2,24 A = 15,05 W
P = VI = 13,44 V x 3,36 A = 45,16 W
P = VI = 13,44 V x 2,24 A = 30,11 W
© Direct-Science 22
CURRENT ELECTRICITY
Calculate:Total resistance of the circuit. (1)
2V
2V
2V
3 ohm 3 ohm
12 ohm
3 ohm
6 ohm
3 ohm
A
1
R=
1
3 ohm+
1
3 ohm
1
R=
3 ohm
2
R
1=
3 ohm
2
1
R=
3 ohm
1 + 1
R = 1,5 ohm
V
1 R =
12 ohm 7
R1
=12 ohm
7 R = 1,7 ohm
1
R=
1
12 ohm+
1
3 ohm+
1
6 ohm1
R=
12 ohm 1 + 4 + 2 (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
© Direct-Science 23
CURRENT ELECTRICITY
Calculate:Reading on voltmeter. (3)
2V
2V
2V
3 ohm 3 ohm
12 ohm
3 ohm
6 ohm
3 ohm
A
V
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.
V = IR = 0,97 A x 4,7 Ohm = 4,6 V
1,7 V
I = V / R = 1,7 V / 12 Ohm = 0,14 A
Calculate:Power in 6-ohm resistor. (5)P = V2 / R= (1,7 V)2 / 6 Ohm= 0,48 Watt
