A hollow copper ball with a volume of 1 cubic decimeter is hung on a spring dynamometer and immersed in water. The indication is the buoyancy of the copper ball in 5 / 6 of the air? The gravity? The volume of the hollow part? The density of copper ball is 9000kg / m3, g = 10N / kg (2,

A hollow copper ball with a volume of 1 cubic decimeter is hung on a spring dynamometer and immersed in water. The indication is the buoyancy of the copper ball in 5 / 6 of the air? The gravity? The volume of the hollow part? The density of copper ball is 9000kg / m3, g = 10N / kg (2,

1. F floating = ρ V row, g = 10 & # 179;, kg / M & # 179; × 1 × 10 - (# 179; M & # 179; × 10N / kg = 10n2. Because the indication in the submerged water is 5 / 6 of the air, and the buoyancy is equal to G minus the apparent weight (the indication of the spring dynamometer), G-G × 5 / 6 = f floating solution, g = 60n3. G actual = mg = ρ V actual g brings V actual = 2 / 3 = 0

A copper block is hung on the spring dynamometer. When the copper block is immersed in water, the indication on the spring dynamometer indicates () a, buoyancy B and weight of the copper block
A copper block is hung on the spring dynamometer. When the copper block is immersed in water, the indication on the spring dynamometer indicates the ()
A. Buoyancy B, gravity C, difference between gravity and buoyancy D, sum of gravity and buoyancy

Answer [C]
Because the buoyancy of an object is: F = go-g ',
The "gravity indicator" when immersed in water is g '= go-f floating

A solid steel ball weighing 17.8n is hung on the spring dynamometer. After immersion in water, the indication of the spring dynamometer is 15.8n, and the buoyancy of the copper ball is 15.8n______ N. What is the volume of the copper ball______ Cm3. (G is 10N / kg)

When a copper ball with a weight of 4.5n and a volume of 0.5dm and 179; is immersed in water and let go, how much buoyancy does the copper ball bear when it is still?

Answer: 4.5n
Analysis:
Suppose that the copper ball will be submerged:
F floating = g row = ρ water GV row, V row = 0.5dm ^ 3 = 0.0005m ^ 3
＝1000×10×0.0005
＝5N
5N ＞ 4.5n, buoyancy is greater than gravity, the ball will float up, and finally become floating
That is: F = g = 4.5n
I don't know what to ask