When a load of 0.97 106 N is placed on a battleship, the ship sinks only 3.0 cm in the water. Estimate the cross-sectional area of the ship at water level. (The density of sea water is 1.025 103 kg/m3.)
Why did the battleship go to the therapist? Because it had some serious sinking issues!
But don't worry, I'm here to help. To estimate the cross-sectional area of the ship at water level, we can use Archimedes' principle. According to Archimedes, the buoyant force experienced by an object immersed in a fluid is equal to the weight of the fluid displaced by that object.
The buoyant force can be calculated as the weight of the displaced water, which is equal to the density of the fluid multiplied by the volume of the displaced fluid. Since the ship sinks by 3.0 cm, we can consider this as the volume of water displaced.
Now, let's dive into the calculation!
Volume of displaced water = area of the ship at water level × distance sunk in the water
Volume of displaced water = Area × 3.0 cm
Now, we need to convert this volume into cubic meters to match the units of density. Since 1 cm = 0.01 m, we get:
Volume of displaced water = Area × (3.0 cm × 0.01 m/cm) = Area × 0.03 m
The buoyant force is equal to the weight of the displaced water, so we have:
Buoyant force = density of seawater × Volume of displaced water
0.97 × 10^6 N = (1.025 × 10^3 kg/m^3) × (Area × 0.03 m)
Now we can solve for the cross-sectional area of the ship at water level:
Area = (0.97 × 10^6 N) / [(1.025 × 10^3 kg/m^3) × (0.03 m)]
And voila! You have your estimation for the cross-sectional area of the ship at water level. Just be careful not to upset any sea creatures down there!
To estimate the cross-sectional area of the ship at water level, we can use Archimedes' principle.
Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
We are given the load placed on the battleship, which is 0.97 * 10^6 N, and the amount the ship sinks in the water, which is 3.0 cm.
First, let's convert the sinking depth from centimeters to meters:
3.0 cm = 0.03 m
Next, let's calculate the volume of water displaced by the ship:
Volume of water displaced = sinking depth * cross-sectional area of the ship at water level
Using Archimedes' principle, we know that the weight of the water displaced is equal to the load on the ship. Therefore,
Weight of water displaced = 0.97 * 10^6 N
The weight of the water displaced is equal to the density of sea water multiplied by the volume of water displaced and the acceleration due to gravity (g = 9.8 m/s^2).
Weight of water displaced = density of sea water * volume of water displaced * g
Putting all these together, we can calculate the cross-sectional area of the ship at water level.
density of sea water * volume of water displaced * g = 0.97 * 10^6 N
volume of water displaced = (0.97 * 10^6 N) / (density of sea water * g)
cross-sectional area of the ship at water level = volume of water displaced / sinking depth
Now, let's substitute the values given:
density of sea water = 1.025 * 10^3 kg/m^3
g = 9.8 m/s^2
sinking depth = 0.03 m
volume of water displaced = (0.97 * 10^6 N) / (1.025 * 10^3 kg/m^3 * 9.8 m/s^2)
volume of water displaced = 94.12 m^3
cross-sectional area of the ship at water level = 94.12 m^3 / 0.03 m
cross-sectional area of the ship at water level ≈ 3137.33 m^2
Therefore, the estimated cross-sectional area of the ship at water level is approximately 3137.33 square meters.
To find the cross-sectional area of the ship at water level, you can use the concept of buoyancy and Archimedes' principle. Here's how you can approach this problem:
1. Start by understanding the concept of buoyancy and Archimedes' principle. According to Archimedes' principle, an object immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object.
2. Calculate the volume of water displaced by the battleship using the information provided. Since the ship sinks only 3.0 cm in the water, this distance represents the height of the displaced water column.
- Convert the given value of 3.0 cm into meters by dividing it by 100 (1m = 100cm).
- Now, you have the height (h) of the water column displaced by the ship.
3. Use the density of sea water to find the mass of the displaced water using the formula:
- mass = density × volume
- Since density is given as 1.025 × 10^3 kg/m^3 and volume = area × height, you can rewrite the formula as:
mass = density × area × height
4. Find the weight of the displaced water using the formula:
- weight = mass × gravity
- Since the weight is given as 0.97 × 10^6 N and gravity is approximately 9.8 m/s^2, you can rewrite the formula as:
weight = density × area × height × gravity
5. Rearrange the formula from step 4 to solve for the cross-sectional area (area):
- area = weight / (density × height × gravity)
- Plug in the known values to calculate the area.
By following these steps, you can estimate the cross-sectional area of the ship at water level.
rho1*V1 = rho2*V2
where rho1 is the density of the object; V1 is the volume of the ship; rho2 is the density of the water; V2 is the volume of the water
mass of object = 0.97 106/g = 0.97 106/9.8
density of object = 0.97 106/(9.8*0.03*A)
where A is the area
(0.97 106/(9.8*0.03*A))*0.03*A = 1.025 103 * (0.03*A)
(0.97 106/(9.8*0.03*A))= 1.025 103
Solve for A, the area