The buoyancy of a hollow metal sphere with a weight of 1kg and a volume of 1.2dm is calculated

The buoyancy of a hollow metal sphere with a weight of 1kg and a volume of 1.2dm is calculated

F = liquid density x acceleration of gravity x volume of liquid
One kilogram receives 9.8 N of gravity
Limit method: let it sink completely in liquid and receive buoyancy of
(1) F kerosene = 800x9.8x1.2/1000 = 9.408 n 9.8 N, so the metal ball will float on the water surface because it is still floating on the surface
According to the mechanical balance: buoyancy = gravity = 9.8 n
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A metal ball is immersed in a container containing water and kerosene successively. The buoyancy it receives is F1 and F2 respectively. Compare their sizes

The problem is to use the same metal ball, that is, V row
So F1 = ρ water V row g F2 = ρ kerosene V row G
Because the density of water is higher than that of kerosene, F1 is higher than F2

A metal ball weighing 25 n is hung under the spring dynamometer and slowly immersed in the water. When it displaces 5 N of water, how many n is the buoyancy received? This is the indication of the spring dynamometer

The buoyancy is equal to the weight of the displaced water (Archimedes principle)
So the buoyancy is 5N and the indication of dynamometer is f = 25n-5n = 20n

How much buoyancy does a metal ball with a weight of 5N displace 4N when immersed in water?

The buoyancy received in the water is 4 N, and the buoyancy received in the mercury is 5 n. because the watermark density is larger than the metal ball, the metal ball returns to the mercury, and the buoyancy is equal to the gravity

When a solid metal ball with a weight of 10 N is immersed in the water, it displaces 500 cm of water. What is the buoyancy of the metal ball

It's very simple. The answer can be obtained directly from Archimedes' principle
F floating = g water = ρ water GV drainage = 1.0 × 10 ﹥ kg / m ﹥ 179; × 10N / kg × 0.0005m ﹥ 179; = 5N
Special tips:
Is there something wrong with the question raised by the building owner? The volume of boiled water should be 500cm, not 500cm;

A solid metal ball with a weight of 10N, when immersed in water, displaces 400g of water. Question: buoyancy of the metal ball?

F = g row = m row, g = 0.4KG * 10N / kg = 4N

When there is a metal block weighing 78.4n in air and 68.6N when it is immersed in water, what is the buoyancy, volume and capacity of the metal block? (g = 9.8N / kg)

(1) F floating = G-G '= 78.4n-68.6n = 9.8N (2) according to f floating = P liquid * g * V row, v = V row = f floating / P liquid * g = 9.8n/1.0 * 10 ^ 3kg / m ^ 3 * 9.8n/kg = 1.0 * 10 ^ 3M ^ 3 (3) P metal = m / v = g / g / v = 78.4n/9.8n/kg/1.0 * 10 ^ 3M ^ 3 = 8 * 10 ^ 3kg / m ^ 3

If there is a metal block weighing 78.4n in air and 68.6N in water, the buoyancy received by the metal block, the volume of the metal block and the weight of the metal block will be larger
Finally, what is the density of the metal block

The buoyancy is 78.4-68.6 = 9.8N
F = ρ VG
9.8N=1.0g/cm^3·V·9.8N/kg
1kg=1g/cm^3·V
V=1000cm^3
V=1dm^3

There is a metal ball hanging from the lower end of the spring dynamometer. The indication of the spring dynamometer is 10N. Now the ball is completely immersed in the water. Under 4N buoyancy, what is the volume of the metal ball

Buoyancy = density * g * V

A solid solid with a weight of 19.6n and a volume of 2dm3 is put into alcohol, water and mercury respectively to find out the buoyancy?

∵ g = mg = ρ VG, ∵ ρ solid = GVG = 19.6n2 × 10 − 3m3 × & nbsp; 9.8n/kg = 1 × 103kg / m3; ∵ ρ solid > ρ alcohol, ∵ solid sinking in alcohol, ∵ buoyancy of solid in alcohol: F1 = ρ alcohol V row g = ρ alcohol VG = 0.8 × 103kg / m3 × 9.8N / kg × 2 × 10-3m3 = 15.68