Buoyant Force Contents: How to calculate Whiteboards.

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Buoyant Force Contents: How to calculate Whiteboards

Buoyant force is the weight of displaced water: ρ f = Density of fluid in kg m -3 V f = Volume of displace fluid in m 3 g = 9.81 N kg -1 Derive Demo: Float not float/Cartesian Diver/hot air balloons/hydrometer

Buoyant force is the weight of displaced water: ρ f = Density of fluid in kg m -3 V f = Volume of displace fluid in m 3 g = 9.81 N kg -1 What is the buoyant force on a 3.0 cm diameter air bubble under water? ρ H2O = 1.0E3 kg m N

ρ f = Density of fluid in kg m -3 V f = Volume of displace fluid in m 3 g = 9.81 N kg -1 What is the buoyant force on a 5.45 kg iron shot submerged in water? What is the weight of the shot in air, and what is its apparent weight submerged? ρ Fe = 7.8E3 kg m -3, ρ H2O = 1.0E3 kg m -3, ρ = m/V so V = m/ρ 6.85 N, 53.5 N, 46.6 N

ρ f = Density of fluid in kg m -3 V f = Volume of displace fluid in m 3 g = 9.81 N kg -1 The King’s crown has a mass of 14.7 kg, but appears to have a mass of only 13.4 kg when weighed when it is submerged in water. What is the density of the crown? Is it gold? ρ Au = 19.3E3 kg m -3, ρ H2O = 1.0E3 kg m -3, ρ = m/V so V = m/ρ 11.3x10 3 kg m -3, so no

??????? ρ f = Density of fluid in kg m -3 V f = Volume of displace fluid in m 3 g = 9.81 N kg -1 The King’s crown has a mass of 14.7 kg, but appears to have a mass of only 13.4 kg when weighed when it is submerged in water. What is the density of the crown? Is it gold? ρ Au = 19.3E3 kg m -3, ρ H2O = 1.0E3 kg m -3, ρ = m/V so V = m/ρ 11.3x10 3 kg m -3, so no

Buoyant Forces 11 | 2 | 323

50. N What is the buoyant force on a rectangular block of wood that measures 12x23x15 cm if it is submerged in the Dead Sea where the density of the water is 1240 kg m -3 ? (convert cm to m first) 1240*.12*.23*.15*9.81 = N

620 kg m -3 A 15x15x5.0 cm piece of wood floats in water (1000. kg m -3 ) face down in the water with the waterline 3.1 cm up the 5.0 cm side: What is its mass? What is its density? Volume of water displaced:.15*.15*.031 = 6.975E-4 m 3 mass of water displaced: 6.975E-4*1000 = kg mass of block = kg density of block = /(.15*.15*.05) = 620 kg m -3 btw: (3.1/5)*1000 = 620 amount submerged = % of fluid’s density object is

0.37 N A 5.0x4.0x4.0 cm piece of wood with a density of 530 kg m -3 is tied to the bottom of a pail of water (1000. kg m -3 ) with a string and held completely submerged. What is the tension on the string? mass of the block:.04*.04*.05*530 = kg weight of block: * N Buoyant force on the block:.04*.04*.05*1000*9.81 = N y Equilibrium of block: F = 0, F = N

5.7 cm A 25x25x10 cm block of iron (7.80x10 3 kg m -3 ) floats on mercury (13.6x10 3 kg m -3 ) If one of the 25x25 cm faces is down into the mercury, how far into the mercury does the block sink before coming to equilibrium? mass of iron block:.25*.25*.10*7800 = kg Volume of displaced mercury: 48.75/13,600 = 3.584E-3 m -3 depth into mercury:.25*.25*D = 3.584E-3 m -3 D = ≈ 5.7 cm. (BTW 7.8E3/13.6E3 =.5735…..)