Answer to Question #71302 in Other Physics for Courtney
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?
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