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11.1 Bestudeer FIGUUR 11.1 hieronder en bereken die ware lengte van AC - NSC Mechanical Technology Welding and Metalwork - Question 11 - 2021 - Paper 1

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Question 11

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11.1 Bestudeer FIGUUR 11.1 hieronder en bereken die ware lengte van AC. Rond die antwoord tot die naaste heelgetal af. FIGUUR 11.1 11.2 Identifiseer die ontwikkel... show full transcript

Worked Solution & Example Answer:11.1 Bestudeer FIGUUR 11.1 hieronder en bereken die ware lengte van AC - NSC Mechanical Technology Welding and Metalwork - Question 11 - 2021 - Paper 1

Step 1

Ware lengte van AC:

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Answer

Om die ware lengte van AC te bereken, gebruik ons die Pythagoras-theorema:

AC2=AB2+BC2AC^2 = AB^2 + BC^2

Hier, gegee dat:

  • BC = \frac{90 - 50}{2} = 20 , \text{mm}
  • AB = 50 , \text{mm}

Nou kan ons die lengtes invul:

AC2=502+202AC^2 = 50^2 + 20^2

Dus,

AC2=2500+400AC^2 = 2500 + 400 AC2=2900AC^2 = 2900

Daarom is:

AC=290053,85mm54mmAC = \sqrt{2900} ≈ 53,85 \, \text{mm} \approx 54 \, \text{mm}

Step 2

Identifiseer die ontwikkeling in FIGUUR 11.2:

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Answer

Die ontwikkeling in FIGUUR 11.2 is 'n vierkantige/ reghoekige transformatormodel of oorgangstuk.

Step 3

Ware lengte na reghoek A-1:

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Answer

Om die ware lengte A-1 te bereken, gebruik ons die Pythagoras-theorema weer:

A1=2002+1302+5002A-1 = \sqrt{200^2 + 130^2 + 500^2}

Dus,

A1=40000+16900+250000=306900553,99mm554mmA-1 = \sqrt{40000 + 16900 + 250000} = \sqrt{306900} ≈ 553,99 \, \text{mm} \approx 554 \, \text{mm}

Step 4

Ware lengte na reghoek C-2:

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Answer

Die lengtes vir C-2 kan weer bereken word met behulp van die Pythagoras-theorema:

C2=4702+2002+5002C-2 = \sqrt{470^2 + 200^2 + 500^2}

Hier is:

C2=220900+40000+250000=510900714,77mm715mmC-2 = \sqrt{220900 + 40000 + 250000} = \sqrt{510900} ≈ 714,77 \, \text{mm} \approx 715 \, \text{mm}

Step 5

Identifiseer die geutbak in FIGUUR 11.4.1:

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Answer

Die geutbak wat in FIGUUR 11.4.1 getoon word, is 'n vierkantige geutbak.

Step 6

Identifiseer die geutbak in FIGUUR 11.4.2:

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Answer

In FIGUUR 11.4.2 is die geutbak 'n reghoekige geutbak.

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