The density of ice is 917 kg/m3, and the density of sea water is 1025 kg/m3. A swimming polar bear climbs onto a piece of floating ice that has a volume of 6.76 m3. What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?

Di + Db = Dsw

917 + Db = 1025 kg/m^3
Db = 108 kg/m^3 = Density of the bear.

Mass of bear = 6.76m^3*108kg/m^3=730 kg
Wt. of bear = 730kg * 9.8N/kg = 7154 N.

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To determine the weight of the heaviest bear that the ice can support without sinking completely beneath the water, we need to compare the buoyant force exerted by the water on the ice with the weight of the bear.

The buoyant force on the ice is equal to the weight of the water displaced by the submerged part of the ice. Since part of the ice is submerged, we can calculate the weight of the water displaced as follows:

Weight of water displaced = Volume of submerged ice * Density of water

The volume of submerged ice can be calculated using the difference in densities between ice and sea water:

Volume of submerged ice = Volume of ice * (Density of ice / Density of water)

Now, let's calculate the weight of the water displaced:

Weight of water displaced = 6.76 m3 * (917 kg/m3 / 1025 kg/m3)
Weight of water displaced ≈ 6.76 m3 * 0.8951 ≈ 6.05 kg

To ensure that the bear does not sink completely, the weight of the bear should be less than or equal to the weight of the water displaced. Hence, the weight of the heaviest bear that the ice can support is approximately 6.05 kg.