1. Sample – Electronic Lesson Copyright Prof J Pieter H van Wyk

2. © Direct-Science 2 VECTORS AND SCALARS

3. 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 0 0 and 1200 km from town A. During the flight the plane experiences a wind that blows at 150 km.h -1 in a direction 90 0. 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.

4. © 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 90 0. 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

5. © Direct-Science 5 VECTORS AND SCALARS: Resultant velocity Town A Town B 1200 km In order to land at B the plane has to aim in a direction into the wind. The plain must fly into the wind. 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 at 550 km.h -1. Speed relative to the wind is 550 km.h -1. The resultant velocity of the plane can now be calculated.

6. © 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: x Sin x = opposite hypotenuse = 150 km.h -1 550 km.h -1 = 0,27 X = 15,66 0 Direction:15,66 0 west of north or 344,34 0

7. © 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

8. © 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, 0 0 15,66 0 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).

9. © 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) ------ ++++++ E = V /d = 4000 V / 0,02 M = 200 000 V.m -1 F g = mg = 1,2 x 10 -14 kg x 10 m.s -2 = 1,2 x 10 -13 N F E = QE = 8 X 10 -19 C x 200 000 V.m -1 = 1,6 x 10 -13 N

10. © 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 -13 N Downward gravitational force: 1,3.10 -13 N Upward force is stronger than downward force. Oil drop experiences an upward resultant force (F RES ) and according to Newton’s second Law the oil drop will accelerate in direction of the resultant force (upwards). Oil drop

11. © 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 -13 N = F= QE Downward gravitational force: 1,3.10 -13 N = 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.

12. © 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 -13 N = F= QE Downward gravitational force: 1,3.10 -13 N = 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

13. © 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 -13 N = F= QE Downward gravitational force: 1,3.10 -13 N = F = mg Oil drop ------ ++++++ 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.

14. © Direct-Science 14 CURRENT ELECTRICITY

15. © 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 = IR Volt R I = V/R Ohm I I = V/R Ampere

16. © Direct-Science 16 CURRENT ELECTRICITY R total = 1,2 ohm + 2,4 ohm = 3,6 ohm (12) 2 ohm 3 ohm 4 ohm 6 ohm I1I1 I2I2 I4I4 I5I5 I3I3 V1V1 V2V2 20V 1 R = 1 2 ohm + 1 3 ohm 1 R = 6 ohm 5 R 1 = 5 1 R = 3 + 2 R =1,2 ohm 1 R = 1 4 ohm + 1 6 ohm 1 R = 12 ohm 5 R 1 = 5 1 R = 3 + 2 R =2,4 ohm I 3 = Potential over external circuit total external resistance = 20 V / 3,6 ohm = 5,6 A (13)

17. © Direct-Science 17 CURRENT ELECTRICITY V 1 = I 3 x 1,2 ohm = 5,6 A x 1,2 ohm = 6,72 V (14) 2 ohm 3 ohm 4 ohm 6 ohm I1I1 I2I2 I4I4 I5I5 I3I3 V1V1 V2V2 20V 1 R = 1 2 ohm + 1 3 ohm 1 R = 6 ohm 5 R 1 = 5 1 R = 3 + 2 R =1,2 ohm

18. © Direct-Science 18 CURRENT ELECTRICITY V 2 = I 3 x 2,4 ohm = 5,6 A x 2,4 ohm = 13,44 V (15) 2 ohm 3 ohm 4 ohm 6 ohm I1I1 I2I2 I4I4 I5I5 I3I3 V1V1 V2V2 20V 1 R = 1 4 ohm + 1 6 ohm 1 R = 12 ohm 5 R 1 = 5 1 R = 3 + 2 R =2,4 ohm

19. © Direct-Science 19 CURRENT ELECTRICITY I 1 = V 1 / 2 ohm = 6,72 V / 2 Ohm = 3,36 A (16) 2 ohm 3 ohm 4 ohm 6 ohm I1I1 I2I2 I4I4 I5I5 I3I3 V1V1 V2V2 20V I 4 = V 2 / 4 ohm = 13,44 V / 4 Ohm = 3,36 A (17) I4I4 I 2 = V 1 / 3 ohm = 6,72 V / 3 Ohm = 2,24 A (18) I2I2 I 5 = V 1 / 2 ohm = 13,44 V / 6 Ohm = 2,24 A (19) I5I5

20. © Direct-Science 20 CURRENT ELECTRICITY Amount of Energy (W) conversion in a resistor: W = VIt W = I 2 Rt W = V 2 t / R 2 ohm 3 ohm 4 ohm 6 ohm I 1 = 3,36 A V 2 = 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) I 4 = 3,36 A I 2 = 2,24 A I 5 = 2,24 A Time in seconds I 3 = 5,6 A V 1 = 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

21. © Direct-Science 21 CURRENT ELECTRICITY Power in each resistor can be calculated: P = VI P = I 2 R P = V 2 /R 2 ohm 3 ohm 4 ohm 6 ohm I 1 = 3,36 A V 2 = 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) I 4 = 3,36 A I 2 = 2,24 A I 5 = 2,24 A I 3 = 5,6 A V 1 = 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

22. © Direct-Science 22 CURRENT ELECTRICITY Calculate: Total resistance of the circuit. (1) 2V 3 ohm 12 ohm 3 ohm 6 ohm 3 ohm A 1 R = 1 + 1 1 R = 2 R 1 = 2 1 R = 1 + 1 R =1,5 ohm V 1 R = 12 ohm 7 R 1 = 7 R =1,7 ohm 1 R = 1 12 ohm + 1 3 ohm + 1 6 ohm 1 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

23. © Direct-Science 23 CURRENT ELECTRICITY Calculate: Reading on voltmeter. (3) 2V 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 = V 2 / R = (1,7 V) 2 / 6 Ohm = 0,48 Watt

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