Answer:
10604 square meters
Explanation:
0.23 cm = 0.0023m
Assume the carrier has a shape of a rectangular box. When the carrier sinks 0.0023m deeper into water, the extra volume submerged into water is the same as the extra water volume being replaced. This extra volume would add an additional buoyant force to counter balance the extra weight created by the 25000 kg fighter.
With g being constant, the mass of the extra water displaced is the same as the mass of the fighter airplane [tex] m_f[/tex]. And mass of the water displaced is its volume V times its density [tex]\rho[/tex]
[tex]V\rho = m_f[/tex]
[tex]1025V = 25000[/tex]
[tex]V = 25000/1025 = 24.39 m^3[/tex]
We assume the carrier has a shape of a rectangular box, so this extra displaced volume is the extra depth d = 0.0023 m times cross-section area A
[tex]V = dA[/tex]
[tex]24.39 = 0.0023A[/tex]
[tex]A = 24.39 / 0.0023 = 10604 m^2[/tex]
The cross-sectional area of the carrier is 1.06 x 10⁴ m².
The given parameters;
The pressure experienced by the carrier is calculated as follows;
[tex]P = \rho gh\\\\P = 1025 \times 9.8\times 0.0023\\\\P = 23.104 \ N/m^2[/tex]
The area of the carrier is calculated as follows;
[tex]P = \frac{F}{A} \\\\A = \frac{F}{P} \\\\A = \frac{mg}{P} = \frac{25000\times 9.8}{23.104} \\\\A = 1.06\times 10^4 \ m^2[/tex]
Thus, the cross-sectional area of the carrier is 1.06 x 10⁴ m².
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