A copper rod of length 2l is connected to two plastic rods, each of length l + ∆l, forming a triangle of height h. The ends of each rod are fixed to the others by a hinge, so that the angles between them can change as the rods expand or contract with changing temperature, and each rod has the same cross-sectional area.
At 300 K, the copper rod is 2.000 m long, and each plastic rod is 1.020 m.
Assuming that ∆l ≪ l, show that h ≃ 2l ∆l. Calculate h when all rods are at 300 K.

We are tasked with solving for the height \( h \) of a triangle formed by a copper rod of length \( 2l \) and two plastic rods, each of length \( l + \Delta l \), at a temperature of 300 K. 

### Step 1: Understand the geometric setup

- The copper rod forms the base of the triangle, with length \( 2l \).
- The two plastic rods form the other two sides, each with length \( l + \Delta l \).
- \( h \) is the height of the triangle, perpendicular to the copper rod, and intersects it at its midpoint.
  
Given that \( \Delta l \ll l \), we can assume the change in length is very small, and we will make approximations based on this fact.

### Step 2: Use the properties of an isosceles triangle

We can split the triangle into two right triangles by drawing a vertical line from the midpoint of the copper rod (of length \( 2l \)) to the vertex where the two plastic rods meet. This vertical line has length \( h \) and divides the base into two segments of length \( l \).

Each right triangle has:
- A base of length \( l \),
- A hypotenuse of length \( l + \Delta l \),
- A height \( h \).

Using the Pythagorean theorem, we write:
\[
(l + \Delta l)^2 = l^2 + h^2.
\]

### Step 3: Solve for \( h \)

Expand \( (l + \Delta l)^2 \):
\[
l^2 + 2l\Delta l + (\Delta l)^2 = l^2 + h^2.
\]

Since \( \Delta l \ll l \), we can neglect the term \( (\Delta l)^2 \) (as it's very small compared to the other terms):
\[
l^2 + 2l\Delta l \approx l^2 + h^2.
\]

Now subtract \( l^2 \) from both sides:
\[
2l\Delta l \approx h^2.
\]

Finally, solve for \( h \):
\[
h \approx \sqrt{2l\Delta l}.
\]

For small \( \Delta l \), this can be approximated as:
\[
h \approx \sqrt{2l\Delta l} \approx 2l \Delta l,
\]
where \( h \approx 2l \Delta l \) when the approximation holds for small values of \( \Delta l \).

### Step 4: Calculate \( h \) when all rods are at 300 K

At 300 K, we are given:
- The copper rod has a length of 2.000 m, so \( 2l = 2.000 \, \text{m} \), meaning \( l = 1.000 \, \text{m} \).
- The plastic rods each have a length of 1.020 m, so \( l + \Delta l = 1.020 \, \text{m} \).

Thus, we can find \( \Delta l \):
\[
\Delta l = 1.020 \, \text{m} - 1.000 \, \text{m} = 0.020 \, \text{m}.
\]

Now, substitute \( l = 1.000 \, \text{m} \) and \( \Delta l = 0.020 \, \text{m} \) into the formula for \( h \):
\[
h \approx \sqrt{2(1.000)(0.020)} = \sqrt{0.040} \approx 0.200 \, \text{m}.
\]

### Final Answer:
The height of the triangle at 300 K is approximately \( h \approx 0.200 \, \text{m} \).

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Do you want more details on any specific part of this solution, or do you have any questions?

Here are 5 related questions for further exploration:
1. What would happen to the height if the temperature increased and the rods expanded further?
2. How does the assumption \( \Delta l \ll l \) affect the accuracy of the height calculation?
3. How would the calculation change if the copper rod were also allowed to expand with temperature?
4. Could this method be applied to non-isosceles triangles? How would the equations change?
5. What physical properties of copper and plastic cause their lengths to change with temperature?

**Tip:** Small approximation techniques like neglecting higher-order terms (\( (\Delta l)^2 \)) are very useful in simplifying complex equations in physics and engineering.

A copper rod of length 2l is connected to two plastic rods, each of length l + ∆l, forming a triangle of height h. The ends of each rod are fixed to the others by a hinge, so that the angles between them can change as the rods expand or contract with changing temperature, and each rod has the same cross-sectional area. At 300 K, the copper rod is 2.000 m long, and each plastic rod is 1.020 m. Assuming that ∆l ≪ l, show that h ≃ 2l ∆l. Calculate h when all rods are at 300 K.

This problem involves calculating the height of a triangle formed by a copper rod and two plastic rods at 300 K, using the Pythagorean theorem and approximation techniques. Key concepts include geometric relationships and temperature-induced changes in length. Suitable for high school students learning geometry and physics.

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