total volume of ice = 100*1.1 = 110m^3
mass of ice = 917*110 = 100870 kg
that is the mass of water displaced (Archimedes)
so
volume of water displaced = 100870/1025
=98.41m^3
so
100m^2 * x = 98.41 where x is the draft
x = .984 m
so height out of water = 1.1-.984 = .116m
added mass of penguins = 27 n
mass of water displaced by penquin mass = 1025*100*.116
so
27 n = 1025*100*.116
n = 440
A rectangular block of ice 10 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 *
.116 OK
HELP: Use Archimedes' Priciple.
b) How many penguins of mass 27 kg each can stand on the ice block before they get their feet wet?
4 NO
HELP: Their feet will get wet if the ice block sinks below the water surface.
Just need b
1 answer
1 answer