https://r2.erweima.ai/mathbot/image/1725697974273rju9ptlb.jpg

The expression in the image is:

\[
3.0 \times 10^7 \, \text{KN/m}^2 \times 2 \times \frac{1}{12} \times 0.3 \, \text{m} \times \left(0.7 \, \text{m}\right) \frac{3}{6.6} \, \text{m}
\]

Let's calculate it step by step.

1. Convert \(3.0 \times 10^7 \, \text{KN/m}^2\) to N/m² (since \(1 \, \text{KN} = 10^3 \, \text{N}\)):
   \[
   3.0 \times 10^7 \, \text{KN/m}^2 = 3.0 \times 10^{10} \, \text{N/m}^2
   \]
   
2. Simplify the rest of the expression.

I'll compute the result now.The result of the expression is approximately \( 4.77 \times 10^8 \, \text{N/m}^2 \).

Would you like more details on the steps or a breakdown of the units? 

Here are some related questions to consider:
1. How does unit conversion from \( \text{KN/m}^2 \) to \( \text{N/m}^2 \) work?
2. What is the significance of multiplying by \( \frac{1}{12} \)?
3. How do ratios affect physical calculations, like \( \frac{3}{6.6} \)?
4. What real-world scenarios involve such large pressures or forces?
5. How do factors like area or length affect calculations in mechanics?

**Tip:** Always ensure that units are consistent throughout the calculation to avoid mistakes.

3.0 × 10^7 KN/m² × 2 × 1/12 × 0.3 m × (0.7 m) × 3/6.6 m

Learn how to simplify a complex multiplication involving pressure in KN/m², unit conversion, and other factors with a detailed solution. This problem covers key physics concepts and mathematical calculations suitable for students in Grades 10-12.

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