解：
(1) 已知銅球質(zhì)量?$m_{銅}=356g，$?根據(jù)?$G = mg，$?可得銅球所含銅的重力?$G_{銅}=m_{銅}g = 0.356kg\times10N/kg = 3.56N。$?
(2) 空心部分注滿水后，水的質(zhì)量?$m_{水}=m_{總}(cāng)-m_{銅}=706g - 356g = 350g。$?
由?$\rho=\frac{m}{V}$?可得，空心部分的體積?$V_{空}=V_{水}=\frac{m_{水}}{\rho_{水}}=\frac{350g}{1g/cm^{3}} = 350cm^{3}。$?
(3) 銅的體積?$V_{銅}=\frac{m_{銅}}{\rho_{銅}}=\frac{356g}{8.9g/cm^{3}} = 40cm^{3}，$?則空心銅球的總體積?$V_{總}(cāng)=V_{銅}+V_{空}=40cm^{3}+350cm^{3}=390cm^{3}。$?
(4) 注滿某種液體后總重力?$G_{總}(cāng)' = 8.81N，$?則總質(zhì)量?$m_{總}(cāng)'=\frac{G_{總}(cāng)'}{g}=\frac{8.81N}{10N/kg}=0.881kg = 881g。$?
液體的質(zhì)量?$m_{液}=m_{總}(cāng)'-m_{銅}=881g - 356g = 525g，$?液體體積?$V_{液}=V_{空}=350cm^{3}。$?
所以液體的密度?$\rho_{液}=\frac{m_{液}}{V_{液}}=\frac{525g}{350cm^{3}} = 1.5g/cm^{3}。$?
答：
(1) 該銅球所含銅的重力是?$3.56N；$?
(2) 該銅球空心部分的體積是?$350cm^{3}；$?
(3) 該空心銅球的總體積是?$390cm^{3}；$?
(4) 注入液體的密度是?$1.5g/cm^{3}。$?
