A balloon filled with helium with a density of 0.18 kg / m ^ 3 has a buoyancy of 6321 n near the ground and an air density of 1.29 kg / m ^ 3. The volume of the balloon is________ M ^ 3

A balloon filled with helium with a density of 0.18 kg / m ^ 3 has a buoyancy of 6321 n near the ground and an air density of 1.29 kg / m ^ 3. The volume of the balloon is________ M ^ 3

According to Archimedes principle: F floating = g row = mg = PV empty g V empty = f floating / PG = 6321 / (1.29x9.8) = 500 cubic meters
Because V space = V helium, V helium = 500 cubic meters

Using the density of air and your own weight, you can estimate the buoyancy in the air. Human is 50kg, and ρ human is slightly larger than ρ water

The density of human body is a little higher than that of water, which is usually calculated in the same way as that of water. According to ρ = m / V, v = m / ρ water = 50 / 1000 = 0.05 (M & # 179;) because under standard conditions, the density of air is ρ air = 1.29kg/m & # 179;, g = 10N / kg, according to Archimedes' buoyancy law

How much air buoyancy does the average adult suffer? (air density t = 1.29Kg / m ^ 3)

The mass of human is generally 60kg, and the average density is about 1.0 * 1000kg / m3
So the volume of human is about 0.06m cubic
Substituting formula: F = PGV
=1.29kg/m cubic * 9.8n/kg * 0.06m cubic
=0.75852=0.76N

The air density is 1.29kg/m3. Can you estimate the buoyancy of a person with a mass of 50kg in the air? (it is estimated that the density of people is similar to that of water)

The volume of helium is 100 cubic meters, and the density of air is 1.29kg/cubic meter. What is the buoyancy?

The density of salt water is 1.1 × 1000kg / m3 for seed selection with salt water. Now 500 cubic centimeters of salt water is prepared, and the weight is 600kg. How much salt or water do you need

Let the volume of added water be x ml, then: the mass of added water is x g  after adding water, The mass of brine M = 600g + XG = (600 + x) g, the volume of brine v = 500cm3 + xcm3 = (500 + x) cm3. If the density of brine after adding water is 1.1g/cm3, the equation (600 + x) g / (500 + x) cm3 = 1.1g/cm3 can be obtained The solution is x = 500 ｛ adding water 500ml (0.5kg) can make 1.1 × 10 cubic / m3 brine. A: adding water 500ml (0.5kg) can make 1.1 × 10 cubic / m3 brine

Salt water with density of 1.1 × 10 ~ (- 179); kg / M ~ (- 179); is needed for seed selection. 500ml salt water is prepared, and its mass is 0.6kg. If the salt water is unqualified, should salt or water be added, and how many kg?

600 △ 500 = 1.2g/cm3 = 1.2 × 10 & # 179; kg / M & # 179; > 1.1 × 10 & # 179; kg / M & # 179;, so it is not qualified. Water should be added. If x g of water is added, it is equivalent to x ml. the equation: (600 + x) / (500 + x) = 1.1 is given, and the solution is x = 500 g

The volume of a bucket is 24 cubic decimeters, the bottom area is 7.5 square decimeters, and there is a hole 0.7 decimeters away from the bucket mouth. Now the bucket can hold water at most______ Kg. (1 kg per cubic decimeter of water)

Now this bucket can hold water: 1 × (24-7.5 × 0.7), = 1 × (24-5.25), = 18.75 (kg); answer: now this bucket can hold water up to 18.75 kg. So the answer is: 18.75

The volume of a cylindrical bucket is 56.52 cubic decimeters, and the bottom diameter measured from the inside is 4 decimeters. Now the bucket is filled with 9 / 7 barrels of water, and the depth of water can be calculated

Depth = 56.52 ÷ (2 × 2 × 3.14) × 9 / 7 = 5.7857142857143 decimeters ≈ 6 decimeters

The volume of this cylindrical bucket is 62.8 cubic decimeters, the bottom area is 12.56 square decimeters, and there is a hole 0.5 decimeters away from the bucket mouth
How many cubic decimeters of water can you hold at most?

62.8÷12.56=5dm
(5-0.5)×12.56=56.52dm³