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Sagot :
To determine the pressure at the crush depth of 100 meters in seawater, we need to calculate both the pressure induced by the water column and the atmospheric pressure above the sea surface. Here's a detailed step-by-step solution:
1. Calculate the pressure due to the water column:
The pressure exerted by a column of fluid is given by the formula:
[tex]\[ P = \rho gh \][/tex]
where:
- [tex]\( \rho \)[/tex] is the density of the fluid (seawater in this case), which is [tex]\( 1025 \, \text{kg/m}^3 \)[/tex].
- [tex]\( g \)[/tex] is the acceleration due to gravity, [tex]\( 9.81 \, \text{m/s}^2 \)[/tex].
- [tex]\( h \)[/tex] is the depth of the fluid, which is [tex]\( 100 \, \text{m} \)[/tex].
Substituting the given values:
[tex]\[ P = 1025 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 100 \, \text{m} \][/tex]
[tex]\[ P = 1005525 \, \text{Pa} \][/tex]
So, the pressure exerted by the water column is [tex]\( 1005525 \, \text{Pa} \)[/tex].
2. Convert the water pressure from Pascal to atmospheres:
The conversion factor between Pascal and atmospheres is:
[tex]\[ 1 \, \text{atm} = 1.01325 \times 10^5 \, \text{Pa} \][/tex]
To find the pressure in atmospheres:
[tex]\[ \text{Pressure in atm} = \frac{1005525 \, \text{Pa}}{1.01325 \times 10^5 \, \text{Pa/atm}} \][/tex]
[tex]\[ \text{Pressure in atm} \approx 9.9238 \, \text{atm} \][/tex]
3. Add the atmospheric pressure at the surface:
The pressure at the surface due to the atmosphere is [tex]\( 1 \, \text{atm} \)[/tex].
Therefore, the total pressure at the crush depth is:
[tex]\[ \text{Total pressure} = 9.9238 \, \text{atm} + 1 \, \text{atm} \][/tex]
[tex]\[ \text{Total pressure} \approx 10.9238 \, \text{atm} \][/tex]
4. Determine the correct answer choice:
The total pressure at the crush depth, including both the water pressure and the atmospheric pressure, is approximately [tex]\( 10.9 \, \text{atm} \)[/tex].
Thus, the correct answer is:
D. 10.9 atm
1. Calculate the pressure due to the water column:
The pressure exerted by a column of fluid is given by the formula:
[tex]\[ P = \rho gh \][/tex]
where:
- [tex]\( \rho \)[/tex] is the density of the fluid (seawater in this case), which is [tex]\( 1025 \, \text{kg/m}^3 \)[/tex].
- [tex]\( g \)[/tex] is the acceleration due to gravity, [tex]\( 9.81 \, \text{m/s}^2 \)[/tex].
- [tex]\( h \)[/tex] is the depth of the fluid, which is [tex]\( 100 \, \text{m} \)[/tex].
Substituting the given values:
[tex]\[ P = 1025 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 100 \, \text{m} \][/tex]
[tex]\[ P = 1005525 \, \text{Pa} \][/tex]
So, the pressure exerted by the water column is [tex]\( 1005525 \, \text{Pa} \)[/tex].
2. Convert the water pressure from Pascal to atmospheres:
The conversion factor between Pascal and atmospheres is:
[tex]\[ 1 \, \text{atm} = 1.01325 \times 10^5 \, \text{Pa} \][/tex]
To find the pressure in atmospheres:
[tex]\[ \text{Pressure in atm} = \frac{1005525 \, \text{Pa}}{1.01325 \times 10^5 \, \text{Pa/atm}} \][/tex]
[tex]\[ \text{Pressure in atm} \approx 9.9238 \, \text{atm} \][/tex]
3. Add the atmospheric pressure at the surface:
The pressure at the surface due to the atmosphere is [tex]\( 1 \, \text{atm} \)[/tex].
Therefore, the total pressure at the crush depth is:
[tex]\[ \text{Total pressure} = 9.9238 \, \text{atm} + 1 \, \text{atm} \][/tex]
[tex]\[ \text{Total pressure} \approx 10.9238 \, \text{atm} \][/tex]
4. Determine the correct answer choice:
The total pressure at the crush depth, including both the water pressure and the atmospheric pressure, is approximately [tex]\( 10.9 \, \text{atm} \)[/tex].
Thus, the correct answer is:
D. 10.9 atm
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