Question

PART ONE

Answer 1

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

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

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

$D=806.46$ $kg/m^{3}$