Sure, let's complete the temperature conversion table and solve the problem related to the expansion of the steel railroad track.
Firstly, let's fill out the missing row in the table:
1. Given: K = 405
2. We need to find the equivalent in Celsius and Fahrenheit.
To convert from Kelvin (K) to Celsius (°C), we use the formula:
[tex]\[ \text{Temperature in } °C = K - 273.15 \][/tex]
Given [tex]\( K = 405 \)[/tex]:
[tex]\[ °C = 405 - 273.15 = 131.85 °C \][/tex]
To convert from Celsius (°C) to Fahrenheit (°F), we use the formula:
[tex]\[ °F = (°C \times \frac{9}{5}) + 32 \][/tex]
Given [tex]\( °C = 131.85 \)[/tex]:
[tex]\[ °F = (131.85 \times \frac{9}{5}) + 32 = 269.33°F \][/tex]
Thus, the completed table is:
[tex]\[
\begin{tabular}{|c|c|c|}
\hline
${ }^{\circ} C$ & ${ }^{\circ} F$ & K \\
\hline
89 & 192.2 & 362.15 \\
\hline
13.38 & 56 & 286.48 \\
\hline
131.85 & 269.33 & 405 \\
\hline
\end{tabular}
\][/tex]
Now let's address the problem of the steel railroad track:
3. A 10,000 meter steel railroad track with a coefficient of linear expansion of [tex]$12 \times 10^{-6}$[/tex] per degree Celsius expands because of a temperature increase. If the temperature increases by 15°C, calculate the amount of expansion in meters.
The formula for linear expansion is:
[tex]\[ \Delta L = L_0 \alpha \Delta T \][/tex]
where:
- [tex]\(\Delta L\)[/tex] is the change in length (meters),
- [tex]\(L_0\)[/tex] is the original length (10,000 meters),
- [tex]\(\alpha\)[/tex] is the coefficient of linear expansion ([tex]\(12 \times 10^{-6} \text{ per }^{\circ} C\)[/tex]),
- [tex]\(\Delta T\)[/tex] is the change in temperature (15°C).
Substituting the values into the formula, we get:
[tex]\[ \Delta L = 10,000 \times 12 \times 10^{-6} \times 15 \][/tex]
Calculating the expansion:
[tex]\[ \Delta L = 10,000 \times 12 \times 10^{-6} \times 15 \][/tex]
[tex]\[ \Delta L = 10,000 \times 0.000012 \times 15 \][/tex]
[tex]\[ \Delta L = 0.12 \times 15 \][/tex]
[tex]\[ \Delta L = 1.8 \text{ meters} \][/tex]
Therefore, the steel railroad track will expand by 1.8 meters with a 15°C temperature increase.
