Question

If a 1-megaton nuclear weapon is exploded at ground level, the peak overpressure (that is, the pressure increase above normal atmospheric pressure) will be 0.2 atm at a distance of 6 km. Atmospheric pressure is 1.013 × 105 Pa. What force due to such an explosion will be exerted on the side of a house with dimensions 5.69 m × 13.8 m? Answer in units of N.

Answer 1

To calculate the force exerted by the nuclear explosion on the side of the house, we need to follow these steps:

1. **Calculate the peak overpressure in pascals:**
The peak overpressure is given as 0.2 atm. First, we convert this pressure into pascals (Pa):

\[
0.2 \, \text{atm} = 0.2 \times 1.013 \times 10^5 \, \text{Pa}
\]

\[
= 0.2 \times 101300 \, \text{Pa}
\]

\[
= 20260 \, \text{Pa}
\]

2. **Calculate the area of the house:**
The dimensions of the house are given as 5.69 m × 13.8 m. So, the area \( A \) can be calculated as:

\[
A = \text{length} \times \text{height} = 5.69 \, \text{m} \times 13.8 \, \text{m}
\]

\[
A \approx 78.582 \, \text{m}^2
\]

3. **Calculate the force exerted on the side of the house:**
The force \( F \) exerted by the overpressure can be calculated using the formula:

\[
F = P \times A
\]

where \( P \) is the peak overpressure in pascals and \( A \) is the area in square meters. Substituting the values:

\[
F = 20260 \, \text{Pa} \times 78.582 \, \text{m}^2
\]

\[
F \approx 1596805.52 \, \text{N}
\]

4. **Final answer:**
Rounding to three significant figures, the force exerted on the side of the house will be approximately:

\[
\boxed{1597000} \, \text{N}
\]