(Q&A) Physics question

Q:-
A rectangular block of ice 6 m on each side and 1.1 m thick floats in sea water. The density of the sea water is 1025 kg/m3. The density of ice is 917 kg/m3.



a) How high does the top of the ice block float above the water level?

m
5.3678 NO

HELP: Use Archimedes' Priciple.
HELP: The buoyant force of the water must balance the weight of the ice.

b) How many penguins of mass 24 kg each can stand on the ice block before they get their feet wet?

A:-
a. The fraction of the ice that is submerged in water is: 917/1025.
Thus, the height submerged is: (917/1025)*1.1.
Thus, the distance from the top of the ice floats to the water is: 1.1-(submerged height)

b. #penguins = (1025-917)/24