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1. The figure below shows two boats, of which one is sunk and the other is afloat. If we consider the floating boat as boat 1 and the sunken boat as boat 2, then we can see that boat 1 has air within it and water around it, whereas boat 2 has water around it as well as inside it… Read TextPreview

- Subject: Other
- Type: Math Problem
- Level: High School
- Pages: 2 (500 words)
- Downloads: 0
- Author: barrowseddie

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Thus as boat 1 is displacing more volume of water than boat 2, it has a higher buoyancy force subjected upon it. Clearly, this is evident from the fact that boat 1 is afloat whereas boat two is sunken. 2. The volume of the given ice cube of 10cm dimensions is 1000cm3. It is given that when the ice cube is floating in pure water, 1cm of it is sticking up above the surface of the water. This shows that the weight of the water displaced by the 9cm thick portion of water is equivalent to the weight of the ice cube. The 9cm thick portion has a volume of 9cm*10cm*10cm i.e. 900cm3. The weight of 900cm3 of pure water is equivalent to 1000cm3 of ice. If we suppose the density of liquid water to be 1g/cm3, then we have 900g of liquid water equivalent to 1000cm3 of ice. This gives us a density of ice as 0.9g/cm3, which is approximately equal to the real value of density of ice (~0.91g/cm3). When the excess 1cm thick part is cut off, the remaining portion of ice has a volume of 900cm3. The weight of 900cm3 of ice is 0.9g/cm3*900cm3 i.e. 810g. Now, for liquid water, the volume of water that needs to be displaced to have the same weight as that of the ice is 810cm3 (810cm3*1g/cm3 = 810g). This means that only 810cm3 of the total 900cm3 of the remaining ice will get immersed in water. The rest 90cm3 volume will remain above the water level.
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