Excercise 3


"The Container Carryer" is a ship with, on earth, a mass of 1 025 000 kg. The area of the ships surface is 100 m². And it's height is 50m
1) How deep does the ship sink in the sea when it is not yet loaded?
2) How many containers can be loaded on the ship when it still has to be 5m above the surface of the sea?
(density sea-water = 1025kg/m³)
(one container ways 7 500 kg)




1)
How many seawater does the ship represents under the water?
V(seawater) = 1 025 000 kg / 1025 kg/m³ = 1000 m³
How deep is the ship under the waterline?
Ship under water => 1000 m³ / 100 m² = 10 m


2)
*M = unknown value = Mass of all the containers

Still 5m of the ship above water => 45m of the ship is under the surface of the water.(50m-5m)
*Unloaded: 10m below waterline.
*Loaded: 45m below waterline.
=> 45m-10m = 35m => The ship may sink 35m more into the water.

*height = 35m
*surface = 100m²
=> V(seawater)= 100m² * 35m = 3500m³

*V(seawater) = 3500m³
*density sea-water = 1025kg/m³
=> mass: M = 3500m³*1025kg/m³ = 3587500 kg

*one container weighs 7500kg
*we may load 3587500 kg on the ship
=> 3587500 kg/ 7500kg = 478.333.. => we can load 478 containers on the ship.