We can model the situation by considering the rail as a segment of a circle with a small rise in the middle. Let’s use geometry to estimate how high the rail will rise due to the expansion.

Given:

• Initial length of the rail $L_0 = 2000 \, \text{m}$ (since 2 kilometers = 2000 meters)
• Expansion of the rail $\Delta L = 0.19 \, \text{m}$ (since 19 centimeters = 0.19 meters)

Now, let's consider the rail as forming a circular arc after expanding, with the chord being the original length $L_0$, and the arc length being the expanded length $L = L_0 + \Delta L$.

Step 1: Find the radius of the circle.

The relationship between the arc length $L$, the chord length $L_0$, and the radius $R$ is approximated by: $L = R \theta$ where $\theta$ is the central angle in radians.

For small expansions, the angle $\theta$ can be approximated as: $\theta \approx \sqrt{\frac{6 \Delta L}{L_0}}$ Substituting the given values: $\theta \approx \sqrt{\frac{6 \times 0.19}{2000}} = \sqrt{0.00057}$ $\theta \approx 0.02385 \, \text{radians}$

Step 2: Calculate the rise of the center.

The height $h$ (the rise of the center of the rail) is related to the radius and angle by: $h \approx \frac{L_0}{2} \sin\left(\frac{\theta}{2}\right)$ For small angles, $\sin\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}$. Thus: $h \approx \frac{L_0}{2} \times \frac{\theta}{2} = 1000 \times \frac{0.02385}{2}$ $h \approx 1000 \times 0.01193 = 11.93 \, \text{meters}$

Final Answer:

The center of the rail would rise approximately 11.9 meters above the ground.

Would you like further details or have any questions?

Here are five related questions to think about:

1. How would the rise change if the rail expanded by 25 centimeters instead of 19?
2. What is the relationship between the expansion length and the height of the rise?
3. How would the result differ for a 5-kilometer-long rail?
4. What other factors, besides temperature, might affect rail expansion?
5. How does the geometry of arcs help in calculating similar problems?

Tip: When working with small angles, approximations like $\sin \theta \approx \theta$ for small $\theta$ can simplify calculations without much loss in accuracy.