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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

Oil having a density of 930 kg/m3 floats on water. A rectangular block of wood 4.00 cm high and with a density of 960 kg/m3 floats partly in the oil and partly in the water. The oil completely covers the block. How far below the interface between the two liquids is the bottom of the block?

To determine
The distance of the bottom of the block below the interface between the two liquids.

Explanation
The buoyant force on the floating block must be equal to the weight of the block that is the density of the oil is ρoil=[A(dx)]g+ρwater[Ax]g=ρwood(Ad)g where A is the surface area of the top or bottom of the rectangular block and solving this equation, we can calculate the distance as x=(ρwoodρoil/ρwaterρoil)d .

Given info: The density of oil is 930kg/m3 , the height of the block is 4.00cm , the density of the wood is 960kg/m3 , and the density of water is 1.00×103kg/m3 .

The formula for the distance of the bottom of the block below the interface between the two liquids is,

x=(ρwoodρoilρwaterρoil)d

  • ρwood is density of wood

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