a solid of density 5000 kg per metre cube 0.5 kgf in here it is completely immersed in water of density 1000 kilogram per metre cube calculate the Apparent weight of this solid in water
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50000kg weight of solid
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50000kg is the weight of the solid
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Density ρ = 5000 kg/m³ Weight = 0.5 kg f in air
mass M = 0.5 kg
Volume V = M / ρ = 0.5 / 5000 = 10⁻⁴ m³
mass of water displaced = V * density of water
= 10⁻⁴ * 1000 = 0.1 kg
Apparent weight = wt in air - wt of water displaced
= 0.5 kg f - 0.1 kg f
= 0.4 kg f
mass M = 0.5 kg
Volume V = M / ρ = 0.5 / 5000 = 10⁻⁴ m³
mass of water displaced = V * density of water
= 10⁻⁴ * 1000 = 0.1 kg
Apparent weight = wt in air - wt of water displaced
= 0.5 kg f - 0.1 kg f
= 0.4 kg f
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yes but can you pls solve it clearly
We have,
Density of solid, d = 5000 kg/m3
Weight in air, W = 0.5 kgf = 0.5g N
Density of water, q = 1000 kg/m3
Let, V be the volume of the solid.
So, W = Vdg
=> 0.5g = V × 5000 × g
=> V = 10-4 m-3
So, Buoyant force is, B = Vqg = 10-4 × 1000 × g = 0.1 g N = 0.1 kgf
Thus, apparent weight = W – B = 0.5 – 0.1 = 0.4 kgf
Density of solid, d = 5000 kg/m3
Weight in air, W = 0.5 kgf = 0.5g N
Density of water, q = 1000 kg/m3
Let, V be the volume of the solid.
So, W = Vdg
=> 0.5g = V × 5000 × g
=> V = 10-4 m-3
So, Buoyant force is, B = Vqg = 10-4 × 1000 × g = 0.1 g N = 0.1 kgf
Thus, apparent weight = W – B = 0.5 – 0.1 = 0.4 kgf
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