91Ó°ÊÓ

Understanding pressure difference is crucial in physics, especially when calculating the effects of varying pressures on surfaces. In the given exercise, the house experiences different pressures inside and outside due to the nuclear explosion. The pressure outside is 2.8 atm, while inside it's 1.0 atm. The first step to understanding the effect is finding the pressure difference, \[ \Delta P = P_{\text{outside}} - P_{\text{inside}} = 2.8 \, \text{atm} - 1.0 \, \text{atm} = 1.8 \, \text{atm} \]This difference is essential as it will directly impact the forces acting on structures. To work in scientific units, this difference in atmospheres must be converted to Pascals (Pa). Remember:

• 1 atm = 101325 Pa
• Hence, \(1.8 \, \text{atm} \) is converted to \(1.8 \, \text{atm} \times 101325 \, \text{Pa/atm} = 182385 \, \text{Pa} \)

This pressure difference in Pascals is now ready to be used in further calculations.

Calculating the net force on a surface is necessary to understand the physical impact of pressure differences. In the given problem, after determining the pressure difference as 182385 Pa, we can calculate the force on the front of the house.The net force is calculated using the formula:\[ F = \Delta P \times A \]By substituting the known values:

• Pressure Difference: \(182385 \, \text{Pa} \)
• Area \( A \)

This tells us how the pressure acts on a surface. Once the values are multiplied, the calculated force is:\[ F = 182385 \, \text{Pa} \times 49.95 \, \text{m}^2 = 9113325.75 \, \text{N} \]This force is what presses against the house's front due to the external and internal pressure disparity. The final answer might require rounding to account for significant figures from the original measurements. Hence, it might often be represented as \(9.11 \times 10^6 \, \text{N} \). This net force indicates the strength of the pressure acting on the structure.

The area of a surface is a basic but vital component in physics when dealing with forces and pressures. In this problem, the front area of the house needs to be determined to proceed with force calculations. With given dimensions, height and width can be easily used to find the area:\[ A = \text{Height} \times \text{Width} \]Substituting the given values:

• Height: 3.33 m
• Width: 15.0 m

The calculation becomes:\[ A = 3.33 \, \text{m} \times 15.0 \, \text{m} = 49.95 \, \text{m}^2 \]This is the total area the pressure is acting upon. Having a correct surface area is crucial because it scales the total force based on how much of the structure is exposed to pressure. This calculated area ensures we factor in the entire exposed surface for accurate force computations.