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a block of wood of density 0.6gcm-3 weighing 3.06N in air, floats freely in a liquid of density 0.9gcm-3. Calculate the volume of the portion immersed​

User NicoHood
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Answer:

M = ρ V to calculate the volume of the wood displaced

W = M g where M is the mass of the wood

W = ρ V g where W is the weight of the wood

ρ = .6 g/cm^3 = .6 (g / cm^3 */1000 g /kg / (1.00E-6 cm^3 / m^3)

ρ = 600 kg / m^3 density of wood

V = 3.06 / (600 * 9.80) m^3 = 5.20E-4 m^3 volume of wood

.9 g/cm^3 = .9 * 1/1000 * 10^6 = 900 kg / m^3 density of liquid

Vw = 3.06 / (900 * 9.80) = 3.47E-4 m^3 volume of liquid displaced

Weight of liquid displaced = Weight of wood

Note density of wood is 2/3 that of liquid so 2/3 the volume of the wood should be displaced

User Sashko
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