The volume of the hollow aluminum ball is 30 cubic centimeter, the mass is 27 grams, and the total mass is 43 grams after filling with a certain liquid. The density of the liquid is 2.7 * 10

The volume of the hollow aluminum ball is 30 cubic centimeter, the mass is 27 grams, and the total mass is 43 grams after filling with a certain liquid. The density of the liquid is 2.7 * 10

V al = m Al / ρ al = 27g / 2.7g/cm & # 179; = 10cm & # 179;
V liquid = V ball - V aluminum = 20cm & # 179;
M liquid = m total - M aluminum = 43g-27g = 16g
ρ liquid = m liquid / V liquid = 16g / 20cm & # 179; = 0.8g/cm & # 179; = 0.8x10 & # 179; kg / M & # 179;

The volume of the hollow aluminum ball is 30cm3, the mass is 27g, and the total mass is 43g after filling with a certain liquid?

(1) 7 × 103kg / m3 = 2.7g/cm3, from ρ = MV, the volume of 27g aluminum: V aluminum = m ρ aluminum = 27g2.7g/cm3 = 10cm3, the volume of hollow part: V air = V ball-v aluminum = 30cm3-10cm3 = 20cm3; (2) the mass of liquid filled: m liquid = m total '- M = 43g-27g = 16g, the volume of liquid:

The volume of an aluminum ball is 30 cubic centimeters, and its mass is 54 grams. If so, the volume of the hollow part can be calculated, and the density of iron is 2700kg / cubic meter

Assuming that the sphere is not hollow, the mass of the sphere is m = ρ v = 2.7 × 30 = 81g
The quality given in the question is only 54G, indicating that the ball is hollow
Assuming that the volume of the hollow part is V1, then: 54 = 81 - ρ v1
The solution is: V1 = 10cm & # 179;