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Question Number 137641 by mey3nipaba last updated on 04/Apr/21

Answered by mr W last updated on 05/Apr/21

Commented by mey3nipaba last updated on 05/Apr/21

oh okay. i′m sorry. thanks though for the solution.

$${oh}\:{okay}.\:{i}'{m}\:{sorry}.\:{thanks}\:{though}\:{for}\:{the}\:{solution}. \\ $$

Commented by mr W last updated on 05/Apr/21

wood alone in liquid:  ρ_W V_W =ρ_L V_L   ρ_W =(V_L /V_W )ρ_L =((75)/(100))×1.2=0.9  wood with glas in liquid:  ρ_G V_G +ρ_W V_W =ρ_L (V_G +V_W )  (ρ_G −ρ_L )V_G =(ρ_L −ρ_W )V_W   V_G =(((ρ_L −ρ_W )V_W )/(ρ_G −ρ_L ))=((1.2−0.9)/(2.4−1.2))×100=25 cm^3   m_G =1×ρ_G V_G =1×2.4×25=60 g

$${wood}\:{alone}\:{in}\:{liquid}: \\ $$$$\rho_{{W}} {V}_{{W}} =\rho_{{L}} {V}_{{L}} \\ $$$$\rho_{{W}} =\frac{{V}_{{L}} }{{V}_{{W}} }\rho_{{L}} =\frac{\mathrm{75}}{\mathrm{100}}×\mathrm{1}.\mathrm{2}=\mathrm{0}.\mathrm{9} \\ $$$${wood}\:{with}\:{glas}\:{in}\:{liquid}: \\ $$$$\rho_{{G}} {V}_{{G}} +\rho_{{W}} {V}_{{W}} =\rho_{{L}} \left({V}_{{G}} +{V}_{{W}} \right) \\ $$$$\left(\rho_{{G}} −\rho_{{L}} \right){V}_{{G}} =\left(\rho_{{L}} −\rho_{{W}} \right){V}_{{W}} \\ $$$${V}_{{G}} =\frac{\left(\rho_{{L}} −\rho_{{W}} \right){V}_{{W}} }{\rho_{{G}} −\rho_{{L}} }=\frac{\mathrm{1}.\mathrm{2}−\mathrm{0}.\mathrm{9}}{\mathrm{2}.\mathrm{4}−\mathrm{1}.\mathrm{2}}×\mathrm{100}=\mathrm{25}\:{cm}^{\mathrm{3}} \\ $$$${m}_{{G}} =\mathrm{1}×\rho_{{G}} {V}_{{G}} =\mathrm{1}×\mathrm{2}.\mathrm{4}×\mathrm{25}=\mathrm{60}\:{g} \\ $$

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