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Refer to the topographic map and orthophoto map to answer the following questions: 2.1.1 How many kilometres would you travel by train from point U in block B2 on the topographic map to Harrismith? 2.1.2 Determine the true bearing of spot height 1096 in block B9 from the spot height 1101 in block C9 on the topographic map - NSC Geography - Question 2 - 2016 - Paper 2

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Refer-to-the-topographic-map-and-orthophoto-map-to-answer-the-following-questions:--2.1.1-How-many-kilometres-would-you-travel-by-train-from-point-U-in-block-B2-on-the-topographic-map-to-Harrismith?--2.1.2-Determine-the-true-bearing-of-spot-height-1096-in-block-B9-from-the-spot-height-1101-in-block-C9-on-the-topographic-map-NSC Geography-Question 2-2016-Paper 2.png

Refer to the topographic map and orthophoto map to answer the following questions: 2.1.1 How many kilometres would you travel by train from point U in block B2 on t... show full transcript

Worked Solution & Example Answer:Refer to the topographic map and orthophoto map to answer the following questions: 2.1.1 How many kilometres would you travel by train from point U in block B2 on the topographic map to Harrismith? 2.1.2 Determine the true bearing of spot height 1096 in block B9 from the spot height 1101 in block C9 on the topographic map - NSC Geography - Question 2 - 2016 - Paper 2

Step 1

2.1.1 How many kilometres would you travel by train from point U in block B2 on the topographic map to Harrismith?

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Answer

To determine the distance from point U in block B2 to Harrismith, use the scale provided on the topographic map. Measure the distance between the two points in centimetres and multiply by the scale factor to convert to kilometres. For instance, if the measured distance is 5 cm and the scale indicates that 1 cm = 2 km, then the journey would be:

Distance = 5 cm * 2 km/cm = 10 km.

Step 2

2.1.2 Determine the true bearing of spot height 1096 in block B9 from the spot height 1101 in block C9 on the topographic map.

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Answer

To find the true bearing, use a protractor on the map to measure the angle from the north to the line connecting the two spot heights. Ensure to account for any declination if provided on the map. For instance, if the angle measures 75 degrees, then the true bearing is 75°.

Step 3

2.2 Calculate the average gradient between 4 and 5 on the orthophoto map. Show ALL calculations.

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Answer

The average gradient can be calculated using the formula: Gradient=Vertical height (VH)Horizontal equivalent (HE)\text{Gradient} = \frac{Vertical\ height\ (VH)}{Horizontal\ equivalent\ (HE)}

If the vertical height between points 4 and 5 is 80 m and the horizontal distance is 400 m:

  1. Calculate VH: 80 m
  2. Calculate HE: 400 m
  3. Gradient = ( \frac{80\ m}{400\ m} = 0.2 ) or 20%.

Step 4

2.2.2 Describe the gradient you calculated in QUESTION 2.2.1.

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Answer

The gradient of 0.2 indicates a relatively gentle slope between points 4 and 5. This means for every 100 m of horizontal distance, there is an elevation change of 20 m, suggesting that the terrain is not steep and may be navigable.

Step 5

2.3 Is there intervisibility between points 4 and 5 on the orthophoto map? Give TWO reasons evident on the orthophoto map to support your answer.

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Answer

Yes, there is intervisibility between points 4 and 5:

  1. There are no obstructions between points 4 and 5, allowing a clear line of sight.
  2. Point 5 is at a higher elevation (1,080 m) than point 4 (1,015 m), which means that point 5 can be seen from point 4 due to the difference in altitude.

Step 6

2.4.1 What is the importance of drawing cross-sections?

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Answer

Drawing cross-sections is essential for several reasons:

  1. They assist in understanding gradient and slope types, providing a clear visualization of changes in elevation.
  2. Cross-sections are useful tools for determining intervisibility and can highlight areas of steep or gentle slopes, aiding in land-use planning.

Step 7

2.4.2 Calculate the vertical exaggeration of the cross-section between 4 and 5.

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Answer

To calculate the vertical exaggeration, use the formula: Vertical exaggeration=Vertical scale (VS)Horizontal scale (HS)\text{Vertical exaggeration} = \frac{Vertical\ scale\ (VS)}{Horizontal\ scale\ (HS)}

Given that the vertical scale is 1 cm = 20 m and if the horizontal scale is 1:10,000:

  • Convert the vertical scale into a number: 20 m = 20,000 cm
  • Calculate the horizontal scale: 1 cm = 10,000 cm

Therefore, Vertical exaggeration = ( \frac{20000}{10000} = 2 ) times.

Step 8

2.4.3 Suggest ONE reason why the vertical scale in a cross-section is exaggerated/made bigger.

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Answer

The vertical scale is often exaggerated to enhance clarity and make variations in height more apparent. This helps viewers easily interpret the slope and elevation changes, thus providing a clearer understanding of the terrain's profile.

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