1、铜块密度 ρ = 8.9 kg/dm^3,水密度 ρ1 = 1 kg/dm^3铜块一半浸入水中时,排开水的体积 V = (m/ρ)/2 = (8.9/8.9)/2 = 0.5dm^3;铜块所受的浮力 = 排开水的重力;即 F = Vρ1g = 1 * 0.5 * 9.8 = 4.9N;2、酒精密度 ρ2 = 0.8 kg/dm^3;铜块浸没在酒精中,排开酒精的体积V =m/ρ = 8.9/8.9 = 1dm^3;铜块所受的浮力 =排开酒精的重力;即F =Vρ2g = 1 * 0.8 * 9.8 =7.84N 。3、350×0.03 =35×0.3 3 5 X 0. 3———— 10. 5