# Find in Newtons the equilibrium reading of the top spring scale.A 1.19 kg beaker containing 1.86 kg of oil with a density of 916 kg/m^3 rests on a scale. A 2.20 kg block of iron is suspended from a...

Find in Newtons the equilibrium reading of the top spring scale.

A 1.19 kg beaker containing 1.86 kg of oil with a density of 916 kg/m^3 rests on a scale. A 2.20 kg block of iron is suspended from a spring scale and completely submerged in the oil. (The density of iron is 7.86*10^3 kg/m^3.)

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A 1.19 kg beaker containing 1.86 kg of oil with density 916 kg/m^3 is resting on a scale. A 2.2 kg block of iron is suspended from a spring scale and completely submerged in the oil. The density of iron is equal to 7.86*10^3 kg/m^3. The volume of a 2.2 kg block of iron is 2.2/7.86*10^3 = 2.79*10^-4 m^3. The mass of 2.79*10^-4 m^3 of oil is 2.79*10^-4*916 = 0.2563 kg.

When the block of iron is submerged in the oil, the weight of the iron block is decreased by an amount equal to the weight of the oil displaced. The weight of the oil displaced is 0.2563*9.8

The spring scale from which the iron block is suspended initially shows displays a weight of 2.2*9.8 = 21.56 N. After the block is submerged in the oil the weight displayed is equal to 21.56 - 0.2563*9.8 = 19.05 N