# Physical Sciences Grade 10 P1 Gauteng June 2022 Memo (Afr)

> Grade 10 Physical Sciences Paper 1 memorandum from the Gauteng June 2022 (Afrikaans), 6 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr10-ps-p1-afr-june-2022-possible-answers/a57313d8-9200-43ed-adcd-ea349c66cc01
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 10
- Document type: Memorandum
- Year: 2022
- Exam period: Gauteng June
- Paper: 1
- Curriculum: CAPS
- Pages: 6

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## About this document

Physical Sciences · Grade 10 · Gauteng June · 2022. Memorandum, 6 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

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            PROVINSIALE EKSAMEN
                    JUNIE 2022
                     GRAAD 10
                NASIENRIGLYNE
            FISIESE WETENSKAPPE (VRAESTEL 1)

6 bladsye

### Page 2

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                                                                FISIESE WETENSKAPPE
                                         NASIENRIGLYNE
                                                                (VRAESTEL 1)   GRAAD 10

VRAAG 1

1.1   B ✓✓

1.2   C ✓✓

1.3   B ✓✓

1.4   B ✓✓

1.5   C ✓✓

1.6   A ✓✓

1.7   D ✓✓

1.8   D ✓✓
                                                                            (2 x 8)   [16]

VRAAG 2

2.1   2.1.1   Skalaar is fisiese hoeveelheid met slegs grootte. ✓✓                     (2)

      2.1.2                              4,5 cm ✓

                                                    6,75 cm ✓
                           R = 8,11 cm ✓

                                                                                       (3)

      2.1.3   Fnet 2 = X 2 + Y 2 ✓

              Fnet = (4,5) 2 + (6,75) 2 ✓✓

              Fnet = 8,11 cm ✓                                                         (4)

      2.1.4   Ja ✓, die antwoorde moet dieselfde wees, ongeag die metode wat
              gebruik word.✓                                                           (2)

2.2   2.2.1   Afstand = 7,5 km ✓                                                       (1)

      2.2.2   Verplasing = 4,5 – 3 = 1,5 km ✓, Noord ✓                                 (2)
                                                                                      [14]

                                             2

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                                                              FISIESE WETENSKAPPE
                                        NASIENRIGLYNE
                                                              (VRAESTEL 1)   GRAAD 10

VRAAG 3

3.1   Die totale meganiese energie bly konstant in ŉ geïsoleerde sisteem. ✓✓

      OF

      Die som van gravitasise potensiële energie en kinetiese energie aan die bokant
      is gelyk aan die som van gravitasie potensiële energie en kinetiese energie aan
      die onderkant in die afwesigheid van wrywing. ✓✓                                   (2)

3.2   Ep = m.g.h ✓

           = 112 x 9,8 x 10 ✓

           = 10 976 J ✓                                                                  (3)

3.3   EMbokant = EMonderkant ✓

      m.g.h + .m.v2 = m.g.h + .m.V2

      (112 x 9,8 x 10) + 0 ✓ = (112 x 9,8 x 8) +   x 112 x V2 ✓

      10 976 = 8 780,8 + 56.V2

       V = 6,26 m·s-1 ✓                                                                  (4)

3.4   EMtop = EMbottom ✓

      m.g.h + .m.v2 = m.g.h + .m.V2

      (112 x 9,8 x 10) + 0 ✓ = 0 +    x 112 x V2 ✓

      10976 = 56·V2

       V = 14 m·s-1 ✓                                                                    (4)

3.5   Bly dieselfde. ✓ Totale meganiese energie word bewaar. Geen wrywing nie. ✓         (2)
                                                                                        [15]

                                              3

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                                                                FISIESE WETENSKAPPE
                                                NASIENRIGLYNE
                                                                (VRAESTEL 1)   GRAAD 10

VRAAG 4

4.1   Elke lading in die heelal bestaan uit heelgetal veelvoud van die elektronlading. ✓✓    (2)

4.2
      n=      ✓

      n=                     ✓

        = 5,125 x 1010 elektrone ✓                                                           (3)

4.3   Aantrekkingskrag✓                                                                      (1)

      4.4.1   S tot T ✓                                                                      (1)

      4.4.2   Die netto lading van ŉ geïsoleerde sisteem bly konstant tydens enige
              fisiese proses. ✓✓                                                             (2)

      4.4.3              Q T + QS
              Q nuut =            ✓
                             2

                ( +5,4  10 −9 ) + ( −8,2  10 −9 )
              =                                     ✓
                                 2

              = − 1,4  10 −9 C ✓                                                            (3)

4.5   Statiese elektrisiteit word baie beter gevorm wanneer die lug droog is of die
      humiditeit laag is. Wanneer die lug vogtig is, kan watermolekules op die
      oppervlak van verskeie materiale versamel. Dit kan die opbou van elektriese
      ladings voorkom. ✓✓                                                                    (2)
                                                                                            [14]

                                                        4

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                                                             FISIESE WETENSKAPPE
                                             NASIENRIGLYNE
                                                             (VRAESTEL 1)   GRAAD 10

VRAAG 5

5.1   Maksimum energie gegee aan elke coulomb lading wat deur battery gaan. ✓✓      (2)

5.2   Parallel ✓                                                                    (1)

5.3   5.3.1
                                    ✓

                  1    1   1
                     =   +   ✓
                  Rp   24 16

                  Rp = 9,6  ✓                                                     (3)

      5.3.2            V
                  I=     ✓
                       R

                       12
                  I=       ✓
                       9,6

                  I = 1,25 A✓                                                       (3)

5.4   5.4.1       Rs = R1 + R2

                  Rs = 24 + 16 ✓

                               Rs = 40  ✓                                          (2)

      5.4.2            V
                  I=
                       R

                      12
                  =      ✓
                      40

                  = 0,3 A ✓                                                         (2)

5.5        Q
      I=      ✓
           Δt

               Q
      0,3 =       ✓
              120

      Q = 36 C ✓                                                                   (3)

5.6   Stroombaan 1 ✓, om weerstande in parallel te verbind, verminder die totale
      weerstand van die stroombaan ✓, dus verhoog die totale stroom wat deur die
      stroombaan gaan/verhoog die helderheid van die gloeilamp. ✓                   (3)
                                                                                   [19]

                                                5

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                                                              FISIESE WETENSKAPPE
                                              NASIENRIGLYNE
                                                              (VRAESTEL 1)   GRAAD 10

VRAAG 6

6.1   Tempo van verandering in snelheid. OF Verandering in snelheid per sekonde. ✓✓    (2)

6.2   77 km .h −1
                   1 000 ✓ = 21,39 m·s-1 ✓
        3 600                                                                          (2)

6.3   V f = Vi + at ✓

      V f = 0 + (1,5)(6,5) ✓✓

      Vf = 9,75 m·s-1 ✓                                                                (4)

6.4   X = Vi t + 1 2 at 2 ✓

      X = (0)(6,5) + 1 2 (1,5)(6,5)2 ✓✓

      X = 31,69 m ✓                                                                   (4)

6.5         V f + Vi 
      X =          t ✓
             2 

            21,39 + 0 
      X =            (6,5) ✓✓
                2     

      X = 69,52 m ✓                                                                   (4)

6.6   Nee, ✓ MOTOR A beweeg met 2 keer die spoed van MOTOR B. ✓                        (2)

6.7   X = Vi t + 1 2 at 2 ✓

      25 = (21,39 )(2,5) + 1 2 a(2,5) ✓✓
                                  2

      a = −9,112 m.s −2 ✓ (vertraging)                                                 (4)
                                                                                      [22]

                                                                       TOTAAL:        100

                                                 6

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