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The image shows a problem where the student is converting units in an equation involving pressure (Pa) and distance (mm and m). The equation starts with \( (68 \, \text{Pa} \cdot \text{mm}^2) \), and the student needs to convert it to \( ? \, \text{Pa} \cdot \text{m}^2 \).

To solve this, we need to convert millimeters squared (\(\text{mm}^2\)) to meters squared (\(\text{m}^2\)).

### Conversion from mm² to m²:
- \( 1 \, \text{mm} = 10^{-3} \, \text{m} \)
- Therefore, \( 1 \, \text{mm}^2 = (10^{-3} \, \text{m})^2 = 10^{-6} \, \text{m}^2 \)

### Solution:
The missing part of the equation should convert \(\text{mm}^2\) to \(\text{m}^2\). To convert from \(\text{mm}^2\) to \(\text{m}^2\), multiply by \(10^{-6}\).

So, the equation becomes:
\[
(68 \, \text{Pa} \cdot \text{mm}^2) \cdot 10^{-6} = ? \, \text{Pa} \cdot \text{m}^2
\]
Now, calculate:
\[
68 \cdot 10^{-6} = 6.8 \times 10^{-5}
\]

Therefore, the missing part of the equation is \( 10^{-6} \), and the answer is \( 6.8 \times 10^{-5} \, \text{Pa} \cdot \text{m}^2 \).

Would you like further details or have any questions? Here are some related questions to deepen your understanding:

1. How do you convert between different units of pressure?
2. What is the significance of converting \(\text{mm}^2\) to \(\text{m}^2\) in practical applications?
3. How does dimensional analysis help in checking unit consistency in equations?
4. How do you handle unit conversions in physics equations involving forces and areas?
5. What are common mistakes to avoid when performing unit conversions?

**Tip:** Always write down conversion factors explicitly to avoid mistakes when working through unit conversions!

A student sets up the following equation to convert a measurement: (68 Pa·mm²) × ? = ? Pa·m². Fill in the missing part of this equation.

Learn how to convert the pressure unit Pa·mm² to Pa·m² using dimensional analysis. The conversion involves multiplying by 10⁻⁶ due to the square of the millimeter to meter conversion. Suitable for students in Grades 8-10.

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