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The anchor points are \( 25 \, \text{m} \) from the base of the tower. ### Part D

分類: physics 更新時間: 2026-02-02

題目

Turn 1 Question

transmission towers
imagine that you are an engineer. you are designing a radio transmission
tower that will be 100 m high. it will be supported by cables, called guy
wires, as follows:

  • the guy wires will be anchored on the ground at three equally spaced

locations in a circle around the tower.

  • the distance of the anchor points from the base of the tower will be

one quarter the height of the tower.

  • there will be five guy wires from each anchor point. the guy wires

will be attached to five points on the tower at 20 m intervals, going up.
? how much guy wire do you need to support the tower?
a. draw a diagram as if you were looking down at the base of the
tower. show the location of the three anchor points. explain
how you know that the anchor points are equally spaced around
the tower, and that they are all the same distance from the base
of the tower.
b. how far are the anchor points from the base of the tower?
c. draw a side view of the tower, showing all the guy wires from
one anchor point.
d. calculate the total amount of guy wire needed. each guy wire
requires 2 m at each end to attach it to the tower and the base.
round your answer to the closest metre. show all your work.
image of transmission tower with guy wires
task checklist

  • did you draw your

diagrams accurately?

  • did you show all the

steps you used?

  • did you explain your

thinking?

  • did you check your

answers?

Turn 2 Question

Why is it 3 and not 4x351.09

Turn 3 Question

Yep but wheeze is a and c

Turn 4 Question

Can you draw it for me

解題步驟

  1. Understand the question
    Turn 1 Question

    transmission towers
    imagine that you are an engineer. you are designing a radio transmission
    tower that will be 100 m high. it will be supported by cables, called guy
    wires, as follows:

    • the guy wires will be anchored on the ground at three equally spaced

    locations in a circle around the tower.

    • the distance of the anchor points from the base of the tower will be

    one quarter the height of the tower.

    • there will be five guy wires from each anchor point. the guy wires

    will be attached to five points on the tower at 20 m intervals, going up.
    ? how much guy wire do you need to support the tower?
    a. draw a diagram as if you were looking down at the base of the
    tower. show the location of the three anchor points. explain
    how you know that the anchor points are equally spaced around
    the tower, and that they are all the same distance from the base
    of the tower.
    b. how far are the anchor points from the base of the tower?
    c. draw a side view of the tower, showing all the guy wires from
    one anchor point.
    d. calculate the total amount of guy wire needed. each guy wire
    requires 2 m at each end to attach it to the tower and the base.
    round your answer to the closest metre. show all your work.
    image of transmission tower with guy wires
    task checklist

    • did you draw your

    diagrams accurately?

    • did you show all the

    steps you used?

    • did you explain your

    thinking?

    • did you check your

    answers?

    Turn 2 Question

    Why is it 3 and not 4x351.09

    Turn 3 Question

    Yep but wheeze is a and c

    Turn 4 Question

    Can you draw it for me

  2. Response
    Turn 1 Answer
    Part B
  3. Explanation

    Step1: Identify tower height

    Tower height \( h = 100 \, \text{m} \).

    Step2: Calculate anchor distance

    Anchor distance is \( \frac{1}{4} \) of tower height:
    \( \text{Distance} = \frac{1}{4} \times 100 = 25 \, \text{m} \).

  4. Explanation

    Step1: Identify attachment length

    Each guy wire needs \( 2 \, \text{m} \) at each end, so total attachment per wire: \( 2 + 2 = 4 \, \text{m} \).

    Step2: Find tower attachment heights

    The five attachment points are at \( 20 \, \text{m} \) intervals: \( 20 \, \text{m}, 40 \, \text{m}, 60 \, \text{m}, 80 \, \text{m}, 100 \, \text{m} \) (height from base).

    Step3: Calculate length of each guy wire

    Using Pythagoras (\( \text{Length} = \sqrt{\text{height}^2 + 25^2} + 4 \)):

    • For \( 20 \, \text{m} \): \( \sqrt{20^2 + 25^2} + 4 \approx \sqrt{400 + 625} + 4 \approx \sqrt{1025} + 4 \approx 32.02 + 4 = 36.02 \, \text{m} \)
    • For \( 40 \, \text{m} \): \( \sqrt{40^2 + 25^2} + 4 \approx \sqrt{1600 + 625} + 4 \approx \sqrt{2225} + 4 \approx 47.17 + 4 = 51.17 \, \text{m} \)
    • For \( 60 \, \text{m} \): \( \sqrt{60^2 + 25^2} + 4 \approx \sqrt{3600 + 625} + 4 \approx \sqrt{4225} + 4 = 65 + 4 = 69 \, \text{m} \)
    • For \( 80 \, \text{m} \): \( \sqrt{80^2 + 25^2} + 4 \approx \sqrt{6400 + 625} + 4 \approx \sqrt{7025} + 4 \approx 83.82 + 4 = 87.82 \, \text{m} \)
    • For \( 100 \, \text{m} \): \( \sqrt{100^2 + 25^2} + 4 \approx \sqrt{10000 + 625} + 4 \approx \sqrt{10625} + 4 \approx 103.08 + 4 = 107.08 \, \text{m} \)

    Step4: Sum lengths of five wires

    Total length for one anchor:
    \( 36.02 + 51.17 + 69 + 87.82 + 107.08 \approx 351.09 \, \text{m} \)

    Step5: Total for three anchor points

    Total guy wire: \( 3 \times 351.09 \approx 1053.27 \, \text{m} \).
    Round to nearest metre: \( 1053 \, \text{m} \).

  5. Final answer

    The anchor points are \( 25 \, \text{m} \) from the base of the tower.

    Part D

答案

Response

Turn 1 Answer
Part B

Explanation

Step1: Identify tower height

Tower height \( h = 100 \, \text{m} \).

Step2: Calculate anchor distance

Anchor distance is \( \frac{1}{4} \) of tower height:
\( \text{Distance} = \frac{1}{4} \times 100 = 25 \, \text{m} \).

Answer

The anchor points are \( 25 \, \text{m} \) from the base of the tower.

Part D

Explanation

Step1: Identify attachment length

Each guy wire needs \( 2 \, \text{m} \) at each end, so total attachment per wire: \( 2 + 2 = 4 \, \text{m} \).

Step2: Find tower attachment heights

The five attachment points are at \( 20 \, \text{m} \) intervals: \( 20 \, \text{m}, 40 \, \text{m}, 60 \, \text{m}, 80 \, \text{m}, 100 \, \text{m} \) (height from base).

Step3: Calculate length of each guy wire

Using Pythagoras (\( \text{Length} = \sqrt{\text{height}^2 + 25^2} + 4 \)):

  • For \( 20 \, \text{m} \): \( \sqrt{20^2 + 25^2} + 4 \approx \sqrt{400 + 625} + 4 \approx \sqrt{1025} + 4 \approx 32.02 + 4 = 36.02 \, \text{m} \)
  • For \( 40 \, \text{m} \): \( \sqrt{40^2 + 25^2} + 4 \approx \sqrt{1600 + 625} + 4 \approx \sqrt{2225} + 4 \approx 47.17 + 4 = 51.17 \, \text{m} \)
  • For \( 60 \, \text{m} \): \( \sqrt{60^2 + 25^2} + 4 \approx \sqrt{3600 + 625} + 4 \approx \sqrt{4225} + 4 = 65 + 4 = 69 \, \text{m} \)
  • For \( 80 \, \text{m} \): \( \sqrt{80^2 + 25^2} + 4 \approx \sqrt{6400 + 625} + 4 \approx \sqrt{7025} + 4 \approx 83.82 + 4 = 87.82 \, \text{m} \)
  • For \( 100 \, \text{m} \): \( \sqrt{100^2 + 25^2} + 4 \approx \sqrt{10000 + 625} + 4 \approx \sqrt{10625} + 4 \approx 103.08 + 4 = 107.08 \, \text{m} \)

Step4: Sum lengths of five wires

Total length for one anchor:
\( 36.02 + 51.17 + 69 + 87.82 + 107.08 \approx 351.09 \, \text{m} \)

Step5: Total for three anchor points

Total guy wire: \( 3 \times 351.09 \approx 1053.27 \, \text{m} \).
Round to nearest metre: \( 1053 \, \text{m} \).

Answer

The total amount of guy wire needed is approximately \( \boldsymbol{1053} \) metres.

Turn 2 Answer

Oh good question! The problem says there are three equally spaced anchor points around the tower, right? Each anchor point has five guy wires, so we only multiply by 3 (one set for each anchor) instead of 4. There's no fourth anchor point mentioned—just the three in a circle around the base of the tower. That makes sense? 😊

Turn 3 Answer

Oh right, let's break those down super simply!

For Part A (top-down diagram):
Imagine drawing a big circle (that's the area around the tower base). Put a dot right in the center—that's the tower itself. Then draw three dots evenly spaced around the circle, so the angle between each pair of dots from the center is 120 degrees (since 360/3 = 120). Each of those outer dots is an anchor point, and they're all the same distance (25m) from the center tower dot. That's what "equally spaced in a circle" means!

For Part C (side view):
Draw a straight vertical line—that's the 100m tower. Mark 5 points along it at 20m, 40m, 60m, 80m, and 100m from the bottom. Then draw a dot 25m to the side of the tower's base (that's one anchor point). Now draw 5 straight lines connecting that side dot to each of the 5 marked points on the tower. That's all the guy wires from one anchor, shown in side view!

Turn 4 Answer

I wish I could draw directly, but I can describe them so clearly you’ll be able to sketch them in 2 seconds!

Part A (Top-Down View):
  1. Grab a piece of paper, draw a small dot in the middle → that’s the tower base.
  2. Draw a big circle around the dot (doesn’t need to be perfect!).
  3. Mark 3 dots on the circle: one at the top, one at the bottom-right, one at the bottom-left. (Make sure they’re spread out evenly—like slicing a pizza into 3 equal slices!)
  4. Label the middle dot "Tower Base" and the 3 outer dots "Anchor 1, 2, 3." Add a note: "All anchors 25m from tower, 120° apart (360°/3 = 120°)"
Part C (Side View):
  1. Draw a tall straight vertical line → label it "100m Tower."
  2. Make 5 small marks along the line: label the bottom mark "0m (Base)," then the next ones up "20m, 40m, 60m, 80m, 100m (Top)."
  3. Draw a dot 2-3 cm to the left of the 0m mark → label it "Anchor Point (25m from tower)."
  4. Draw 5 straight lines connecting the anchor dot to each of the 5 marks on the tower. That’s all 5 guy wires from one anchor!

Question Analysis

Subject natural science
Sub Subject engineering
Education Level high school
Difficulty unspecified
Question Type with image, multiple part question (a - d)
Multi Question Yes
Question Count 4
Analysis Status completed
Analyzed At 2026-02-02T12:43:51

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Can you draw it for me

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natural scienceengineeringwith image, multiple part question (a - d)high schoolliked-answerturns-4

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