density of ice 917 kg/m^3, density of sea water is 1025 kg/m^3, a piece of ice floating has a volume of 5.2 m^3, what is the weight of the heaviest bear the ice can support without sinking?
what is (1027-917)kg*5.2?
1025 is the density of sea water and 917 is the density of ice and 5,2 is the volume of the ice floating on the sea water
To calculate the weight of the heaviest bear that the ice can support without sinking, we need to compare the weight of the bear with the buoyant force exerted by the water on the ice.
The buoyant force is equal to the weight of the water displaced by the immersed portion of the ice. Since the ice is floating, its weight is equal to the buoyant force.
First, let's calculate the weight of the ice:
Weight of the ice = Density of ice * Volume of the ice * Acceleration due to gravity
Weight of the ice = 917 kg/m^3 * 5.2 m^3 * 9.8 m/s^2
Weight of the ice = 47822.48 N
Now, let's find the weight of the bear that the ice can support without sinking. We'll assume the bear weighs more than the ice, but we don't know its exact weight yet. Let's call the weight of the bear W.
Weight of the bear = W
Since the ice and the bear are floating, the weight of the bear is balanced by the buoyant force. Therefore:
Buoyant force = Weight of the bear
The buoyant force is equal to the weight of the water displaced by the immersed portion of the ice. The displaced volume of water is equal to the volume of the ice.
Buoyant force = Density of water * Volume of the ice * Acceleration due to gravity
Buoyant force = 1025 kg/m^3 * 5.2 m^3 * 9.8 m/s^2
Buoyant force = 51188 N
Since the buoyant force is equal to the weight of the bear, we can set up the following equation:
51188 N = W
Therefore, the heaviest bear the ice can support without sinking weighs approximately 51188 Newtons.