There is a hollow copper ball with a volume of 30 cubic centimeters and a mass of 178 kg. If the hollow copper ball is filled with aluminum, what is the mass of aluminum? Density of aluminum: 2.7 x 10 cubic kg / M & # 179; The density of copper is 8.9 x 10 cubic kg / M and 179;

There is a hollow copper ball with a volume of 30 cubic centimeters and a mass of 178 kg. If the hollow copper ball is filled with aluminum, what is the mass of aluminum? Density of aluminum: 2.7 x 10 cubic kg / M & # 179; The density of copper is 8.9 x 10 cubic kg / M and 179;


27g



The volume is 30 cubic centimeter, the mass is 178 grams of copper ball is solid or hollow, in its hollow part filled with aluminum, aluminum


What's the density? You want to know, right? The definition of density is: the mass per unit volume in the material. For example, one cup of water, one cup of sand and one cup of sand, which is heavy? Of course, heavy sand filling, because the density of sand is greater than that of water. In the calculation process, we introduced the definition of density in order to quantify and calculate



Salt water with density of 1.1 × 103kg / m3 is needed for seed selection with salt water. At present, 500 cm3 salt water is prepared, and its mass is 600 g. does such salt water meet the requirements: if it does not meet the requirements, do you need to add salt or water? How much should I add?


Suppose the density of brine is ρ, then the density of brine is: ρ = MV = 600g500cm3 = 1.2g/cm3 = 1.2 × 103kg / m3, ∵ ρ > ρ 0 = 1.1 × 103kg / m3, ∵ the prepared brine does not meet the requirements, the density of brine is too high, and water needs to be added to reduce the density; suppose the mass of water to be added is △ m, then M total = m + △ m, while △ M = ρ water △ V, △ v = △ m ρ water, ∵ vtotal = V + △ v = V + △ m ρ water, from ρ 0 = M Total V: 1.1 × 103kg / m3 = m + △ MV + △ v = m + △ MV + △ m ρ water = 0.6kg + △ M500 × 10 − 6m3 + △ M1.0 × 103kg / m3, the solution is: △ M = 0.5kg = 500g