suppose a block of lead of mass 0.4kg at temperature of 95°C is dropped in a 2 kg of water originally at 20°C in a container. wh
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Answer:
the position of the wood below the interface of the two liquids is 2.39 cm.
Explanation:
Given;
density of oil, = 926 kg/m³
density of the wood, = 974 kg/m³
density of water, = 1000 kg/m³
height of the wood, h = 3.69 cm
Based on the density of the wood, it will position across the two liquids.
let the position of the wood below the interface of the two liquids = x
Let the wood be in equilibrium position;
Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.