A 1.0 m cubed object floats in fresh water with 20% of it above the waterline. What does the object weigh out of the water?
See http://www.jiskha.com/display.cgi?id=1290346564 for explanation of Archimedes principle of floatation.
Basically an object floating in a fluid displaces a volume of fluid equal to its own weight.
So this 1 m³ object has 80% of it submerged. How much weighs the water it displaced? And its own weight equals?
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To determine the weight of the object out of the water, we need to consider the concept of buoyancy. When an object is immersed in a fluid, it experiences an upward force called buoyant force that is equal to the weight of the fluid displaced by the object.
In this case, the object is partially submerged, with 20% of its volume above the waterline. Therefore, 80% of its volume is submerged in water.
We know the density of fresh water is approximately 1000 kg/m^3. Since the object floats, its average density must be equal to or less than the density of water.
To find the weight of the object out of the water, we can use the following steps:
Step 1: Calculate the volume of the submerged portion:
Volume of submerged portion = 80% x 1.0 m^3 = 0.8 m^3
Step 2: Calculate the weight of the displaced water:
Weight of displaced water = Volume x Density x Gravity
Weight of displaced water = 0.8 m^3 x 1000 kg/m^3 x 9.8 m/s^2
Step 3: Subtract the weight of the displaced water from the weight of the object:
Weight of object out of water = Weight of object - Weight of displaced water
It's important to note that we need the weight of the object to continue with these calculations. If you provide the weight of the object, I can provide you with the weight of the object out of water.
To find the weight of the object out of the water, we need to understand the concept of buoyancy and apply Archimedes' principle.
Archimedes' principle states that an object submerged in a fluid experiences a buoyant force equal to the weight of the fluid it displaces. This buoyant force opposes the weight of the object, causing it to float or sink.
In this case, we're dealing with fresh water, which has a density of approximately 1000 kg/m³.
First, let's calculate the volume of the object that is submerged in the water. Since 20% of the object is above the waterline, it means that 80% of it is submerged. Using the total volume of 1.0 m³, we can find the submerged volume:
Submerged Volume = 0.80 * 1.0 m³ = 0.80 m³
Next, using the density of water, we can calculate the weight of the water displaced by the submerged volume:
Weight of water displaced = Density of water * Submerged Volume * Acceleration due to gravity
Weight of water displaced = 1000 kg/m³ * 0.80 m³ * 9.8 m/s²
Weight of water displaced = 7840 N
According to Archimedes' principle, the weight of the object out of the water will be equal to the weight of the water displaced. Therefore, the object will weigh 7840 N out of the water.