When a load of 1(10)^6 N is placed on a ship, the ship only sinks 2.5 cm into the ocean (sea water). Estimate the cross-sectional area of the ship at water level.
How amusing - you happened upon a Naval Architect.
increased volume of ship in the water = Area * .025 meters
mass of water displaced = density of seawater * that volume
density of seawater is about 1029 kg/m^3
so
mass of water displaced = 1029 * .025 * A
Buoyancy force due to that water = mass * 9.81 m/s^2 gravity
so
additional buoyant force up
= 9.81*1029 * .025 * A
so Archimedes says
1,000,000 = 9.81 * 1029 * .025 * A
To estimate the cross-sectional area of the ship at water level, we can use Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Given:
- Load on the ship (force) = 1(10)^6 N
- Sinking depth = 2.5 cm
To calculate the cross-sectional area of the ship, we need to determine the volume of water displaced by the ship. However, we are only given the sinking depth, which is not sufficient to directly calculate the volume.
We can make an assumption that the ship sinks uniformly to its full immersion depth. Based on this assumption, we can calculate the volume of water displaced using the sinking depth and then solve for the cross-sectional area.
Let's convert the sinking depth from centimeters to meters:
Sinking depth = 2.5 cm = 2.5/100 = 0.025 m
The volume of water displaced (V) is given by the formula:
V = sinking depth * cross-sectional area
Since the ship sinks uniformly, the immersed volume will be a cylinder with a height equal to the sinking depth. Thus, the volume V can be expressed as:
V = π * (cross-sectional area) * (sinking depth)
Using Archimedes' principle, the buoyant force is equal to the weight of the displaced water:
Buoyant force = weight of water displaced
Buoyant force = Load on the ship = 1(10)^6 N (given)
We know that the density of seawater is approximately 1025 kg/m^3. The weight of the water displaced can be calculated using the formula:
Weight of water displaced = density of water * volume of water displaced
Weight of water displaced = 1025 kg/m^3 * V
Setting the buoyant force equal to the weight of water displaced, we have:
1(10)^6 N = 1025 kg/m^3 * V
Rearranging the equation to solve for V, we get:
V = (1(10)^6 N) / (1025 kg/m^3)
Now we can plug in the known values to calculate the volume of water displaced:
V = (1(10)^6 N) / (1025 kg/m^3)
V ≈ 976.1 m^3
Finally, we can solve for the cross-sectional area of the ship at water level:
V = π * (cross-sectional area) * (sinking depth)
976.1 m^3 = π * (cross-sectional area) * 0.025 m
Rearranging the equation to solve for the cross-sectional area, we get:
(cross-sectional area) = 976.1 m^3 / (π * 0.025 m)
(cross-sectional area) ≈ 12418.91 m^2
Therefore, the estimated cross-sectional area of the ship at water level is approximately 12418.91 square meters.
To estimate the cross-sectional area of the ship at water level, we can use Archimedes' principle, which states that the buoyant force experienced by a submerged object is equal to the weight of the fluid it displaces.
Let's break down the solution step by step:
1. We are given the weight of the load placed on the ship, which is 1(10)^6 N (1 million newtons).
2. We are also given the sinking depth of the ship, which is 2.5 cm (0.025 m).
3. According to Archimedes' principle, the buoyant force acting on the ship is equal to its weight. Therefore, the buoyant force can be calculated as 1(10)^6 N.
4. The buoyant force is equal to the weight of the fluid displaced by the ship. Since the fluid in this case is sea water, we need to determine the density of sea water.
5. The average density of seawater is approximately 1025 kg/m^3. This means that for every cubic meter of seawater displaced by the ship, it experiences a buoyant force equivalent to 1025 kg.
6. To find the cross-sectional area of the ship at water level, we'll use the formula:
Area = Buoyant force / (Density of seawater * Acceleration due to gravity)
Area = 1(10)^6 N / (1025 kg/m^3 * 9.8 m/s^2)
7. Plugging the values into the formula:
Area = (1 * 10^6) / (1025 * 9.8) m^2
8. Simplifying the equation:
Area = 101.63 m^2
Therefore, the estimated cross-sectional area of the ship at water level is approximately 101.63 square meters.