# Answer to Question #71302 in Other Physics for Courtney

Question #71302

A rock makes a dynamometer show 1.04N, when it’s lowered into a cup of water the dynamometer shows 0.68N. What is the rock’s density?

Expert's answer

Given:

W_(in air)=1.04 N,

W_(in water)=0.68 N,

ρ=?

Buoyancy is the force acting opposite the direction of gravity that affects all objects submerged in a fluid.

Buoyancy = weight of displaced fluid.

The weight of the displaced fluid is directly proportional to the volume of the displaced fluid.

Archimedes’ Principle: The buoyant force on a submerged body equals the weight of the fluid it displaces:

BF =ρ_water gV_rock

The buoyant force is also equal to the weight of the object in the air minus the weight of the object in the fluid:

BF =W_(in air)-W_(in water)

The volume of rock is

V_rock=m/ρ=mg/ρg=W_(in air)/ρg

Equating the two expressions for the buoyant force we can solve for the density of rock:

ρ=(W_(in air) ρ_water)/(W_(in air)-W_(in water) )

Density of water is 1 g/cm3 or 1000 kg/m3.

Thus,

ρ=(1.04×1000)/(1.04-0.68)=2888.8 kg/m^3≈2890 kg/m^3≈2.89 g/cm^3

Answer: 2.89 g/cm^3

W_(in air)=1.04 N,

W_(in water)=0.68 N,

ρ=?

Buoyancy is the force acting opposite the direction of gravity that affects all objects submerged in a fluid.

Buoyancy = weight of displaced fluid.

The weight of the displaced fluid is directly proportional to the volume of the displaced fluid.

Archimedes’ Principle: The buoyant force on a submerged body equals the weight of the fluid it displaces:

BF =ρ_water gV_rock

The buoyant force is also equal to the weight of the object in the air minus the weight of the object in the fluid:

BF =W_(in air)-W_(in water)

The volume of rock is

V_rock=m/ρ=mg/ρg=W_(in air)/ρg

Equating the two expressions for the buoyant force we can solve for the density of rock:

ρ=(W_(in air) ρ_water)/(W_(in air)-W_(in water) )

Density of water is 1 g/cm3 or 1000 kg/m3.

Thus,

ρ=(1.04×1000)/(1.04-0.68)=2888.8 kg/m^3≈2890 kg/m^3≈2.89 g/cm^3

Answer: 2.89 g/cm^3

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment