Physics
08.09.2022 04:42
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An object weighs 79.1 n in air. when it is suspended from a force scale and completely

An object weighs 79.1 n in air. when it is suspended from a force scale and completely immersed in water the scale reads 21.8 n. determine the density of the object. (b)when the object is immersed in oil, the force scale reads 48.4 n. calculate the density of the oil.
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a1supernut37owozkj
a1supernut37owozkj
4,9(19 marks)

(a). The density of the object is 1382 kg/m³.

(b). The density of the oil is 536.4 kg/m³.

Explanation:

Given that,

Weight in air = 79.1 N

Weight in water = 21.8 N

Weight in oil = 48.4 N

We need to calculate the volume of object

Using formula of buoyant force

F_{b}=W_{air}=W_{water}

F_{b}=79.1-21.8

F_{b}=57.3\ N

F_{b}=\rho g h

Put the value into the formula

57.3=1000\times V\times 9.8

V=\dfrac{57.3}{1000\times9.8}

V=5.84\times10^{-3}\ m^3

We need to calculate the density

Using formula of buoyant force

F_{b}=\rho Vg

79.1=\rho\times5.84\times10^{-3}\times9.8

\rho=\dfrac{79.1}{5.84\times10^{-3}\times9.8}

\rho=1382\ kg/m^3

The density of the object is 1382 kg/m³.

(b). We need to calculate the volume of object

Using formula of buoyant force

F_{b}=W_{air}=W_{oil}

F_{b}=79.1-48.4

F_{b}=30.7\ N

We need to calculate the density

Using formula of buoyant force

F_{b}=\rho_{oil} Vg

30.7=\rho_{oil}\times5.84\times10^{-3}\times9.8

\rho_{oil}=\dfrac{30.7}{5.84\times10^{-3}\times9.8}

\rho=536.4\ kg/m^3

The density of the oil is 536.4 kg/m³.

Hence, This is the required solution.

AshlynPlayz45
AshlynPlayz45
4,9(93 marks)
So i took the quiz and i got it right but i didn't any work but i got 466 so yeah sorry for not being able to explain it any better.

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