The largest ship in Christopher Columbus' "fleet" was the Santa Maria, which displaced about 100 tons
(1 ton = 2,000 pounds) of seawater when fully loaded. Assuming the average density of seawater surrounding
the ship was 1.024 g mL-1, calculate the total volume of seawater (in m3) displaced by the ship's hull.
1.024 g/cm^3 = 1.024 kg/dm^3 = 1024 kg/m^3
1 lb = .4536 kg
find kg in 100 tons (200000 lb)
divide by sea water density to find m^3
"about 100 tons" means an answer with one significant figure
volume * density = mass = weight/9.81
so,
v * 1.024g/mL * 10^6ml/m^3 * 1kg/1000g = 100ton * 2000lb/ton * 1kg/2.2lb
v * 1.024*10^3 kg/m^3 = 2*10^5/2.2 kg
v = 88.778 m^3
To calculate the total volume of seawater displaced by the ship's hull, we can use the formula:
Volume = Mass / Density
First, let's convert the displacement of the ship from tons to pounds:
Displacement in pounds = 100 tons * 2,000 pounds/ton = 200,000 pounds
Next, we need to convert the mass from pounds to grams:
Mass in grams = 200,000 pounds * 453.592 grams/pound ≈ 90,718,400 grams
Now, we can plug the values into the formula to calculate the volume:
Volume = 90,718,400 grams / (1.024 g/ml) ≈ 88,723,437.5 ml
Finally, let's convert the volume from milliliters to cubic meters:
Volume in m^3 = 88,723,437.5 ml * (1 L / 1,000 ml) * (1 m^3 / 1,000 L) ≈ 88.7 m^3
Therefore, the total volume of seawater displaced by the Santa Maria's hull is approximately 88.7 m^3.