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DaniilM [7]
3 years ago
13

PART ONE

Physics
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

Density of the object is 9708.52 kg/m^3

Density of the oil is 806.46 kg/m^3

Explanation:

Given Information:

F_{air} =301N\\F_{water} =270N\\F_{oil} =276N

Part one: Find the density of the object

To find the volume of an object we use

V=\frac{F_{B} }{D*g}

Where F_{B} is the Buoyant force and D is the density of the object

The Buoyant force is given by

F_{B}=F_{air}-F_{water}

F_{B}=301-270=31N

The density of water is 1000 kg/m^3

Therefore Volume will be

V=31/1000*9.8=0.0031632 m^3

Mass of the object is given by

m=F_{air} /g

m=301/9.8=30.71 kg

Density of the object is given by

D=m/V

D=30.71/0.0031632=9708.52 kg/m^{3}

Part two: Find the density of the oil

The Buoyant force is given by

F_{B} =F_{air} -F_{oil}

F_{B}=301-276=25N

V=\frac{F_{B} }{D*g}

Rearranging above equation yields

D=\frac{F_{B} }{V*g}

D=\frac{25}{0.0031632*9.8}

D=806.46 kg/m^{3}

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