Răspunsuri

2014-10-25T20:18:58+03:00
p_{fier}=\dfrac{mg}{S}=\dfrac{\rho Vg}{S}\\ \\ \\ p_{apa}=\dfrac{\rho_0 Vg}{S}

Raportul celor doua presiuni se reduce la:

\dfrac{p_{fier}}{p_{apa}}=\dfrac{\rho}{\rho_0}

Date:
\rho=\text{densitatea fierului}\\ \rho_0=\text{densitatea apei}

Spor la calcul.