A 130000 tonne iceberg is roughly the shape of a cube Assumi

A 130000 tonne iceberg is roughly the shape of a cube. Assuming sea water density of 1025 kg/m6 and ice density of 917 kg/m3 how high does the iceberg rise above the water level (m)? please show all work

Solution

By dividing the density of ice and water, we get the % of ice under water,

Thus ht under water= (917/1025)*ht

Thus, ht over water= 0.11* ht

Volume of cube= mass/ density

= 1300000*10³/917

= 1417666.303

Side of cube= 112m

Height above water= 112* .11

= 12.3mts