# Physical Sciences Grade 12 Paper 1 Gauteng Mock 2016 Memo

> Grade 12 Physical Sciences Paper 1 memorandum from the Gauteng Mock Exam 2016, 13 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2016afrgp-physics-fisika-memo-p1/7a6dc9d2-f5a6-4fe6-8e7a-a5072c74cbae
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: Gauteng Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 13

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Physical Sciences · Grade 12 · Gauteng Mock Exam · 2016. Memorandum, 13 pages. Read online or download the PDF.

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          PREPARATORY EXAMINATION
          VOORBEREIDENDE EKSAMEN
                               2016
                        MEMORANDUM

              PHYSICAL SCIENCES: PHYSICS P1 (10841)

                    FISIESE WETENSKAPPE V1 (10841)

13 pages / 13 bladsye

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                            GAUTENG DEPARTMENT OF EDUCATION
                          GAUTENG DEPARTEMENT VAN ONDERWYS
                                    PREPARATORY EXAMINATION
                                    VOORBEREIDENDE EKSAMEN

          PROOFREAD      DATE ___________                PHYSICAL SCIENCES: PHYSICS
          SIGNATURE      _________________               (First Paper)
          PROOFREAD      DATE ___________                FISIESE WETENSKAPPE: FISIKA
          SIGNATURE      _________________               (Eerste vraestel)

                                                         MEMORANDUM

QUESTION / VRAAG 1

1.1       D                                                                                         (2)

1.2       B                                                                                         (2)

1.3       A                                                                                         (2)

1.4       B                                                                                         (2)

1.5       C                                                                                         (2)

1.6       A                                                                                         (2)

1.7       A                                                                                         (2)

1.8       B                                                                                         (2)

1.9       D                                                                                         (2)

1.10      C                                                                                         (2)
                                                                                                     [20]

QUESTION / VRAAG 2

2.1       When a resultant/net force acts on an object, the object accelerates in the direction
          of the (net) force at an acceleration directly proportional to the force  and inversely
          proportional to the mass of the object. 

          Wanneer ’n resultante /netto krag op ’n voorwerp uitgeoefen word, versnel die
          voorwerp in die rigting van die (netto) krag met ’n versnelling direk eweredig aan die
          grootte van die krag  en omgekeerd eweredig aan die massa van die voorwerp.              (2)

                                                     2

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2.2                         180 N/F/Fapplied/applied force         Notes/Aantekeninge:
                                                                    • Correct arrow and label for each
                            180 N/F/F toegepas/toegepaste krag       force for one mark. / Korrekte pyle
                                                                      en benoeming vir elke krag vir een
  T/Force of Y on X                                                  punt.
                                                                    • Comparative lengths of arrows are
  T/Krag Y op X            Weight/mg/w/Fg/39,2 N                    not required. / Vergelykende lengte
                            Gewig/mg/w/Fg/39,2 N                     van pyle nie vereiste nie.

                                                                                                       (3)

2.3       2.3.1   For block X / Vir blok X:              Note/Aantekeninge:
                  F net = ma                               • If the system approach is used to
                  180 - w – T = ma                             first calculate acceleration and
                  180 - (4)(9,8) – T = 4a                     then acceleration is substituted to
                  140,8 – T = 4a ………………(i)                     obtain T:
                                                            Max. 2
                  For block Y/ Vir blok Y:                           4
                  F net = ma                                • Indien die stelsel benader word
                  T - w = ma                                   om eerstens versnelling te
                  T - (8)(9,8) = 8a                           bereken en dan die versnelling
                  -78,4 +T = 8a……………….(ii)                     gebruik om T te bereken:
                                                            Maks 2
                  281,6 – 2T = 8a                                    4
                   -78,4 + T = 8a
                    360 – 3T = 0
                  T = 120 N
                  T = 120 N upwards / opwaarts                                                        (4)

          2.3.2   POSITIVE MARKING FROM QUESTION 2.3.1 / POSITIEWE NASIEN VAN
                  VRAAG 2.3.1
                  OPTION / OPSIE 1          OPTION / OPSIE 2
                  -78,4 +T = 8a             140,8 – T = 4a
                  -78,4 + 120 = 8a         140,8 – 120 = 4a 
                             -2
                  a = 5,2 m∙s              a = 5,2 m∙s-2 

                  OPTION / OPSIE 3
                  281,6 -2T = 8a
                  281,6 + 220 = 8a 
                  a = 5,2 m∙s-2                                                                       (2)
                                                                                                      [11]

QUESTION / VRAAG 3

3.1       The object had an upward velocity when it was released.  / Die voorwerp het
          ’n opwaartse snelheid toe dit losgelaat is.
          OR/ OF
          The object continues with its state of motion in a straight line. / Die voorwerp
          beweeg voort in dieselfde rigting van beweging in ’n reguit lyn.
          OR/ OF
          The object has inertia. / die voorwerp het inersie.                                          (1)
                                                     3

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3.2       3.2.1   Upwards / opwaarts                                                                      (1)

          3.2.2   Downwards / afwaarts                                                                    (1)

3.3       3.3.1   Upward positive / opwaarts                 Downward positive/ afwaarts
                  positief                                   positief

                  vf = vi + a ∆ t                           vf = vi + a ∆ t 
                  0 = 110 - 9,8Δt                           0 = - 110 + 9,8Δt 
                  ∆ t = 11,22 s                             ∆ t = 11,22 s                                (3)

          3.3.2   OPTION 1
                  Upward positive / opwaarts positief               Upward negative / opwaarts
                                                                    negatief
                   Δy = v i Δt + 21 a∆t 2                          Δy = v i Δt + 21 a∆t 2 
                  -550  = 110Δt + ½(-9,8)Δt2                      550  = -110Δt + ½(9,8)Δt2 
                  4,9Δt2 -110 Δt – 550 = 0                          4,9Δt2 -110 Δt – 550 = 0
                       − ( −110 ) ± ( −110 )2 − 4( 4,9)( −550 )     Δt = 26,66 s or – 4,21 s
                  ∆t =                                              Δt = 26,66 s 
                                       2( 4,9)
                  Δt = 26,66 s or – 4,21 s
                  Δt = 26,66 s 

                  OPTION 2
                  POSITIVE MARKING FROM QUESTION 3.3.1 / POSITIEWE NASIEN VAN
                  VRAAG 3.3.1.

                  Time(point Q to R): 11,22 s                     Time(point Q to R): 11,22 s
                  Time(point R to Q): 11,22 s                     Time(point R to Q): 11,22 s
                  Time(point Q to ground):                        Time(point Q to ground):

                  Upward positive / opwaarts positief             Upward negative / opwaarts
                                                                  negatief
                   Δy = v i Δt + 21 a∆t 2                        Δy = v i Δt + 21 a∆t 2 
                  -550  = (-110)Δt + ½(-9,8)Δt2                 550  = (110)Δt + ½(9,8)Δt2 
                  4,9 Δt2 + 110 Δt – 550 = 0                      4,9 Δt2 + 110 Δt – 550 = 0
                       − 110 ± (110 )2 − 4( 4,9)( −550 )               − 110 ± (110 )2 − 4( 4,9)( −550 )
                  ∆t =                                            ∆t =
                                  2( 4,9)                                         2( 4,9)
                  Δt = 4,21 s                                     Δt = 4,21 s

                  Total time = 11,22 + 11,22 + 4,21               Total time = 11,22 + 11,22 + 4,21
                             = 26,65 s                                      = 26,65 s 
                                                                                                           (4)

                                                        4

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                          OPTION 3
                          POSITIVE MAKING FROM QUESTION 3.3.1 / POSITIEWE NASIEN VAN
                          VRAAG 3.3.1.
                          Upward positive / opwaarts positief Upward negative / Opwaarts
                                                               negatief
                          Height reached above point Q:
                          v f = v i + 2a∆y 2                   Height reached above point Q:
                             2      2

                                                               v f = v i + 2a∆y 2
                                                                  2      2
                          (0)2 = (110)2 + 2(-9,8) ∆y 2
                          ∴ ∆y 2 = 617,35 m                    (0)2 = (-110)2 + 2(9,8) ∆y 2
                                                               ∴ ∆y 2 = 617,35 m
                          Displacement from point R to ground:
                          ∆y = 550 + 617,35 = 1 167,35 m       Displacement from point R to
                           Δy = v i Δt + 21 a∆t 
                                               2               ground:
                                                               ∆y = 550 + 617,35 = 1 167,35 m
                          -1 167,35  = (0)Δt + ½(-9,8)Δt 
                                                         2

                          Δt = 15,435 s                         Δy = v i Δt + 21 a∆t 2 
                                                               1 167,35 = (0)Δt + ½(9,8)Δt2
                          Total time = 11,22 + 15,435          Δt = 15,435 s
                                        = 26,66 s 
                                                               Total time = 11,22 + 15,435
                                                                            = 26,66 s 

3.4       POSITIVE MARKING FROM QUESTION 3.3 / POSITIEWE NASIEN VAN VRAAG 3.3.

          Downward as positive / afwaarts as positief:

                                                         Marking criteria/nasienriglyne
                                                         Positive slope / positiewe helling   
                                                         Graph starts at v = -110 m·s-1

          v (m·s-1)
                                                                                              
                                                         Grafiek begin by v = -110 m·s-1
                                                         Graph intercepts x axis at 11,22 s
                                                                                              
                                                         Grafiek sny x axis by 11,22 s
                                                         Graph ends at 26,66 s
                                                                                              
                      0      11,22     26,66    t(s)    Grafiek eindig by 26,66 s

             -110
          Upward as positive / opwaarts as positief:

               110
                                                         Marking criteria
                      0      11,22     26,66     t(s)   Negative slope / negatiewe helling   
                                                         Graph starts at v = 110 m·s-1

          v (m·s-1)
                                                                                              
                                                         Grafiek begin by v = 110 m·s-1
                                                         Graph intercepts x axis at 11,22 s
                                                                                              
                                                         Grafiek sny x axis by 11,22 s
                                                         Graph ends at 26,66 s
                                                                                              
                                                         Grafiek eindig by 26,66 s

                                                                                                    (4)
                                                                                                   [14]

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QUESTION / VRAAG 4

4.1       The total (linear) momentum of a closed system remains constant/is conserved. 
          (2 marks or zero)

          Die totale (liniêre) momentum van ’n geslote sisteem bly konstant/bly behoue. 
          (2 punte or niks)                                                                    (2)

4.2       Σ pi = Σ pf 
          m 1 v 2 + m 2 v 2 = (m 1 + m 2 )v c
          (68 x 20) + (12 x 0) = (68 + 12)v c 
          ∴ v c = 17 m·s-1                                                                    (4)

4.3       POSITIVE MARKING FROM QUESTION 4.2 / POSITIEWE NASIEN VAN VRAAG
          4.2.
          OPTION / OPSIE 1
          (E p + E k ) i = (E p + E k ) f 
          (mgh + ½mv c 2) i = (mgh + ½ mv c 2) f
          0 + ½(80)(17)2  = (80)(9,8)h + 0 
          ∴ h = 14,75 m

          Distance up the incline, d/ Afstand opwaarts teen skuinsvlak
                   h
          sin θ =
                  ∆x
                    14,75
          sin25° =        
                     ∆x
          ∴ ∆ x = 34,89 m 

          OPTION / OPSIE 2
          W net = ΔE k 
          W w = E kf -E ki
                               1
          mgsin25°cos180° =      m(v 2f –v i2 )
                               2
                                               1
          (80)(9,8)sin25° ∆ xcos180° = (80)(02 - 172) 
                                               2
          Δx = 34,89 m 
                                                                                               (5)

4.4       Decreases / verminder 

          Friction is a non-conservative force/ opposes motion/removes kinetic energy from
          the system. 

          Wrywing is ‘n nie-konserwatiewe krag /opponeer beweging / verwyder kinetiese
          energie van die sisteem.                                                            (2)
                                                                                              [13]

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QUESTION / VRAAG 5

5.1       Zero/0 (J)                                                                                (1)

5.2
                     FN/N/Normal                 Force of the engine/F/FA
                     / Normaal                  Krag van enjin /F/FA       
                                     
          Frictional force/f/Ff
          Wrywingskrag /f/Ff 
                                         weight/w/mg/Fg/6 860 N
                                         gewig/w/mg/Fg/6 860 N
                                                                                                    (4)

5.3       5.3.1   The net/total work done on an object is equal to the change in the object’s
                  kinetic energy.  (2 marks or zero.)

                  Die netto/totale werk verrig op ’n voorwerp is gelyk aan die verandering in
                  die voorwerp se kinetiese energie.  (2 punte or niks.)                           (2)

          5.3.2   Marking guidelines / nasienriglyne
                  • Formula/ Formule: W net = ∆E k or W nc = ∆E p + ∆E k 
                  • Formula / Formule: W = F ∆ xcos θ 
                  • Substitution to calculate / Vervanging in formulef = µ k mgcos30° 
                  • Substitute to calculate / Vervanging om te bereken: W f 
                  • Substitute to calculate / Vervanging om te bereken: W w OR ΔE p 
                  • Substitute to calculate / Vervanging om te bereken: W F 
                  • Substitute to calculate / Vervanging om te bereken: ∆E k 
                  • Final answer / Finale antwoord: 11,59 m·s-1 

OPTION / OPSIE 1:
W net = ∆E k 
                                            W N + W f + W w + W F = ∆E k
           0 + f ∆ xcos 180° + w ∆ xcos 180° + F ∆ xcos 0°  = E kf - E ki
                                                                     1
µ k mgcos30° ∆ xcos 180° + mg ∆ xsin30°180° + F ∆ xcos 0° = m(v f 2 –v i 2)
                                                                     2
        0,32(700)(9,8)cos30°(70)cos 180° + (700)(9,8)sin30°(70)cos180°+ (6000)(70)cos 0° =
                                                                                     1
                                                                                       (700)(v 2f – 0 ) 
                                                                                     2
                                     5           5             5         2
                         -1,33 x 10 – 2,4 x 10 + 4,2 x 10 = 350v f
                                                              v f = 11,59 m·s-1 
OPTION / OPSIE 2:
                              W nc = ΔE p + ΔE k 
                         W f + W F = mgΔh + ½ m(v f 2 – v i 2)
   f ∆ xcos 180° + F ∆ xcos 0°  = mgΔh + ½ m(v f 2 – v i 2)
0,32(700)(9,8)cos30°(70)cos 180°  + (6000)(70)cos 0° =
                                                  (700)(9,8)(70sin30° – 0)  + ½ (700(v f 2 – 0) 
                                v f = 11,59m∙s-1                                                      (8)
                                                                                                     [15]
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QUESTION / VRAAG 6

6.1       6.1.1   It is the apparent change in frequency of a source when there is relative
                  motion between the source and the observer. 

                  Dit is die skynbare verandering in frekwensie van ’n bron wanneer daar ’n
                  relatiewe verandering in beweging tussen die bron en die waarnemer is.      (2)

          6.1.2   Away / Weg                                                                   (1)

          6.1.3          v ± vL
                  fL =          fs 
                         v ± vS
                          340
                  73 =             75 
                         340 + v s
                  v s = 9,32 m ⋅ s −1                                                          (4)

6.2       •   The absorption spectrum observed for elements on a distant star compared to
              spectrum of elements on the earth or sun. 
          •   Absorption lines in spectrum from star shifted to red end of the spectrum/ red
              shifted. 
          •   Wavelengths in the absorption spectrum lines have increased. 
          •   According to the Doppler effect the star is thus moving away from the earth. 

          •   The absorbsiespektrum waargeneem vir elemente van ’n verafgeleë ster in
              vergelyking met die spektrum van elemente nader aan die aarde of die son. 
          •   Absorbsielyne in die spektrum van ster verskuif na die rooi kant van die
              spektrum/ rooiverskuiwing. 
          •   Golflengte in the absorbsiespektrumlyne vermeerder. 
          •   Volgens die Doppler effek beweeg die ster dus verder van die aarde.              (4)

6.3       Ultrasound scans / ultraklankskandering
          Measuring rate of flow of blood (Blood-flow meter) / Meting van die tempo waarteen
          bloed vloei (bloedvloeimeter)                                                        (2)
                                                                                               [13]

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QUESTION / VRAAG 7

7.1       The force of attraction that two charges at rest exert on each other is directly
          proportional to the product of the two charges and
          inversely proportional to the square of the distance between their centres. 

          Die aantrekkingskrag wat twee ladings in rus op mekaar uitoefen is direk
          eweredig aan die produk van die twee ladings  en omgekeerd eweredig aan die
          vierkant van die afstand tussen die middelpunte.                                        (2)

7.2                                F12 

                                   Fnet
          F13 
                                                                                                   (2)

7.3               kQ1Q 2
          F12 =     2
                         
                   r12
                                       
                (9 × 10 9 )(2 × 10 − 5 )(2 × 10 − 4 )
              =
                              (0,5)2
              = 144 N, to the right / na regs

                  kQ1Q3
          F13 =     2
                   r13
                                       
                (9 × 10 9 )(2 × 10 − 5 )(2 × 10 − 4 )
              =
                              (0,4)2 
              = 225 N, downwards / afwaarts

          F net = (225 )2 + (144 )2
               = 267,13 N
                   225
          tan θ =       ∴ θ = 57,38°
                   144

          F net = 267,13 N,  147,38° / E57,38°S / 57,38° south of east / suid van oos            (7)
                                                                                                  [11]

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QUESTION / VRAAG 8

8.1       12 V                                                                                          (1)

8.2       8.2.1    OPTION / OPSIE 1                         OPTION 2 / OPSIE 2
                    1   1    1                                    R1R 2
                      =   +                                Rp =          
                   R p R1 R 2                                    R1 + R 2
                        1 1                                       (6)(10)
                       =  +                                    =         
                        6 10                                       6 + 10
                   Rp = 3,75 Ω                               Rp = 3,75 Ω

                   R external = 3,75 + 2  = 5, 75 Ω       R external = 3,75 + 2 = 5,75 Ω             (4)

          8.2.2    POSITIVE MARKING FROM QUESTION 8.2.1 / POSITIEWE NASIEN VAN
                   VRAAG 8.2.1.
                   OPTION / OPSIE 1           OPTION / OPSIE 2
                   V 1 = IR ext              V 1 = IR ext 
                           10                         10
                   ∴I =                      ∴I =         
                          5,75                       5,75
                       = 1,74 A                   = 1,74A

                   ε = IR ext + Ir                         V lost = Ir 
                                                            12 – 10 = 1,74r 
                   12 = (1,74)(5,75) + 1,74r 
                                                            r = 1,15 Ω 
                   ∴ r = 1,15 Ω                                                                         (5)

8.3       Increases / Verhoog 
          Closing S 2 effectively cuts off the 2 Ω resistor / is a short circuit. / Sluit van S 2 sny
          die 2 Ω weerstand effektief af 
          Total resistance decreases / Totale weerstand verminder.                                      (3)
                                                                                                        [13]

QUESTION / VRAAG 9

9.1       9.1.1    AC / Alternating current / Wisselstroom                                              (1)

          9.1.2    DC 
                   Due to the presence of the split ring commutator, current always exits
                   through the one brush and then through the other brush. 

                   Agv die teenwoordigheid van die splitringkommutator, is daar altyd stroom
                   wat deur een borsel beweeg en dan deur die ander borsel. 
                                                                                                         (2)

9.2       9.2.1    AC generator has slip rings  whilst a DC generator has a split ring
                   commutator. 

                   AC generator het slipringe,  terwyl die DC generator splitringkommutator
                   het.                                                                                 (2)

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          9.2.2   An AC voltage can be stepped up. / ’n AC stroom kan verhoging ondergaan. 
                  Increase in voltage and decrease in current. / Verhoging in potensiaalverskil
                  en verlaging in stroom. 
                  Loss in power/P = I2R decreases. / Verlies aan drywing/P = I2R verminder     (3)

9.3       9.3.1   OPTION / OPSIE 1                    OPTION / OPSIE 2
                      1                                  cycles
                  f=                                  f=
                     T                                    time
                       1                                  3,5
                   =                                  =      
                     0,02                                0,07
                   = 50 Hz                            = 50 Hz                                 (2)

          9.3.2   OPTION 1                            OPTION 2
                         V                                   V
                  Imax = max                         Vrms = max 
                          R                                    2
                         200                                 200 
                       =                                  =
                          20                                   2
                       = 10 A                              = 141,42 V

                  P ave = ½(V max I max )                  Vrms
                       = ½(200 x 10 )                Irms =
                       = 1000 W (1 kW)                      R
                                                            141,42
                  OPTION / OPSIE 3                        =
                                                              20
                         V                                = 7,07 A
                  Vrms = max 
                           2
                         200                          Pave = I rms
                                                               2
                       =                                          R
                           2                               = (7,07)220
                       = 141,42 V                          = 999,70 W
                                                      OR
                          2                           Pave = Vrms I rms 
                        Vrms
                  Pave =                                  = (141,42)(7,07) 
                         R
                                                           = 999,84 W 
                        (141,42)2 
                      =
                            20                        Accept range / Aanvaar:
                      = 999,98 W 
                                                          999,70 to 1000 W                      (5)
                                                                                               [15]

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QUESTION / VRAAG 10

10.1   Minimum energy required to release (photo) electrons from a metal surface.

       Minimum energie nodig om (foto)elektrone uit ’n metaaloppervlak vry te stel.                  (2)

10.2   OPTION / OPSIE 1                               OPTION / OPSIE 2
       W o = hf                                         c
                                                      f= 
           =
             hc                                         λ
              λ                                            3 × 10 8
             (6,63 × 10 −34 )(3 × 10 8 )              =              
           =                                             500 × 10 − 9
                   500 × 10 − 9                       = 6 × 1014 Hz
           = 3,978 x 10-19J 
                                                      Wo = hf 
                                                      = (6,63 × 10 − 34 )(6 × 1014 ) 
                                                      = 3,98 × 10 −19 J                               (5)

10.3   10.3.1     Remains the same / Bly dieselfde                                                    (1)

       10.3.2     Remains the same / Bly dieselfde                                                    (1)

10.4   POSITIVE MARKING FROM QUESTION 10.2 / POSITIEWE NASIEN VAN VRAAG
       10.2.
       OPTION / OPSIE 1                                       OPTION / OPSIE 2
       E = Wo + Ek(max)                                          c
                                                              f=
       hc                                                         λ
            = W o + E k(max)
        λ                                                           3 × 10 8
                                                               =                 
                                                                  400 × 10 − 9
                      
       (6,63 × 10 −34 )(3 × 10 8 )                            = 7,5 × 1014 Hz
                        −9
                                   = 3,978 x 10-19 + E k(max)
              400 × 10 
                                                              E = hf
       E k(max) = 9,95 x 10-20 J 
                                                              = (6,63 × 10 − 34 )(7,5 × 1014 ) 
                                                           = 4,97 × 10 −19 J
       Range/ aanvaar:
        9,93 x 10-20 J to 9,95 x 10-20 J                   E = Wo + Ek(max) 
                                                           4,97 x 10-19 = 3,98 x 10-19 + E k(max) 
                                                           E k(max) = 9,93 x 10-20 J                  (5)
                                                                                                      [14]

                                                 12

### Page 13

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   4                                                                                                    10841/16

QUESTION / VRAAG 11

11.1   The energy transferred to / work done on  each coulomb (of charge) / per C
       charge  passing through the battery.

       Die lading oorgedra / arbeid verrig op elke coulomb (lading) / per C lading  wat
       deur die battery beweeg.                                                                                   (2)

11.2                          8 − 0,5
                            3,                       Note/Aantekening:
       Gradient/helling =             = 0,37 (V-1)
                            9−0                       • Accept any correct values from graph
                                                        to calculate the gradient.
                                                       • Aanvaar enige korrekte waarde vanaf
                                                         grafiek om helling te bereken.                           (3)

11.3    1
           
       emf                                                                                                        (1)

11.4   11.4.1   POSITIVE MARKING FROM QUESTION 11.2 / POSITIEWE NASIEN
                VAN VRAAG 11.2.
                OPTION /OPSIE 1         OPTION / OPSIE 2
                   1                    emf = 2(0) + 2r
                      = 0,37 
                 emf                    emf = 0,26(9) + 0,26r
                ε (emf) = 2,70 V       0 = - 2,34 + 1,74r
                                        r = 1,34 Ω
                                        ε = 2(0) + 2r
                                          = 2(1,34)
                Range/Aanvaar:            = 2,67 V 
                  2,67 – 2,70 V                                                                                   (2)

       11.4.2   POSITIVE MARKING FROM QUESTION 11.2 AND 11.3.1 / POSITIEWE
                NASIEN VAN VRAAG 11.2 EN 11.3.1.
                OPTION / OPSIE 1                OPTION / OPSIE 2
                Emf = I(R + r)                  Any two equation using values from the
                               1     1       r  graph/ Enige twee vergelykings waar
                y = mx + c:      =      R+      waardes van grafiek gebruik is:
                               I emf        emf
                                         r
                y int ercept / afsnit =         emf = 2(0) + 2r 
                                        emf
                                                emf = 0,26(9) + 0,26r 
                           r
                0,5  =                        0 = - 2,34 + 1,74r
                         2,70                   r = 1,34 Ω 
                R = 1,35 Ω 

                Range/ aanvaar:
                 1,34 - 1,35 Ω
                                                                                                                  (3)
                                                                                                                 [11]

                                                                                        GRAND TOTAL: 150
                                                          (s) only. If you have received this email in error p

                                                     13

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