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The problem asks us to convert the rate \( R = 0.0709 \, \text{kPa} \cdot \text{km}^{-1} \) to units of \(\text{kJ} \cdot \text{m}^{-4}\).

Here's how to approach it:

1. **Convert kPa to kJ:**  
   \[ 1 \, \text{kPa} = 1 \, \text{kN/m}^2 = \frac{1 \, \text{kJ}}{\text{m}^3} \]  
   So, \( \text{kPa} = \text{kJ} \cdot \text{m}^{-3} \).

   Applying this to \( R = 0.0709 \, \text{kPa} \cdot \text{km}^{-1} \):
   \[ R = 0.0709 \, (\text{kJ} \cdot \text{m}^{-3}) \cdot \text{km}^{-1} \]

2. **Convert km\(^{-1}\) to m\(^{-4}\):**
   Since \( 1 \, \text{km} = 1000 \, \text{m} \), we have:
   \[ \text{km}^{-1} = (1000 \, \text{m})^{-1} = 10^{-3} \, \text{m}^{-1} \]

   Substituting this into \( R \):
   \[ R = 0.0709 \cdot 10^{-3} \, \text{kJ} \cdot \text{m}^{-4} \]

3. **Simplify the expression:**
   \[ R = 0.0000709 \, \text{kJ} \cdot \text{m}^{-4} \]

So, the answer is:
\[
R = 0.0000709 \, \text{kJ} \cdot \text{m}^{-4}
\]

Would you like further explanation on any step?

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**Relative Questions:**

1. How do you convert units involving exponents?
2. What are other examples of unit conversions in physics?
3. How does understanding unit conversions help in real-world applications?
4. Why is it important to convert between different units of measurement accurately?
5. What is the relationship between kPa and kJ in terms of dimensional analysis?

**Tip:** When converting units, remember to apply the conversion factor to the exponent if it’s raised to a power.

A NASA spacecraft measures the rate R at which atmospheric pressure on Mars decreases with altitude. The result at a certain altitude is: R = 0.0709 kPa·km⁻¹. Convert R to kJ·m⁻⁴.

Learn how to convert the rate at which atmospheric pressure changes on Mars from units of kPa·km⁻¹ to kJ·m⁻⁴. This detailed step-by-step guide demonstrates unit conversion techniques, including transforming between kilometers and meters and using dimensional analysis. Suitable for high school students in Grades 10-12.

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