[tex]dane:\\h = 12 \ cm = 0,12 \ m\\\rho_{l} = 910\frac{kg}{m^{3}} \ - \ gestosc \ lodu\\\rho_{w}=1000\frac{kg}{m^{3}} \ - \ gestosc \ wody\\szukane:\\d = ? \ - \ grubosc \ kry[/tex]
Rozwiązanie
Ciało pływa całkowicie lub częściowo zanurzone w wodzie, gdy siła wyporu jest równa sile grawitacji:
Fw = Fg
[tex]F_{w} = \rho_{w}\cdot V_{z}\cdot g\\\\V_{z} = S\cdot(d-h), \ zatem\\\\F_{w} = \rho_{w}\cdot S\cdot (d-h)\cdot g\\-----------\\\\Masa \ kry:\\m = \rho_{l}\cdot V=\rho_{l}\cdot S\cdot d\\\\F_{g} = \rho_{l}\cdot S\cdot d\cdot g\\--------\\\\F_{w} = F_{g}\\\\\rho_{w}\cdot S\cdot(d-h)\cdot g = \rho_{l}\cdot S\cdot d\cdot g \ \ /:(S\cdot g)\\\\\rho_{w}\cdot(d-h) = \rho_{l}\cdot d\\\\\rho_{w}\cdot d - \rho_{w}\cdot h = \rho_{l}\cdot d\\\\\rho_{w}\cdot d - \rho_{l}\cdot d = \rho_{w}\cdot h[/tex]
[tex]d(\rho_{w}-\rho_{l}) = \rho_{w}\cdot h \ \ /:(\rho_{w}-\rho_{l})\\\\d = \frac{\rho_{w}\cdot h}{\rho_{w}-\rho_{l}}\\\\d = \frac{1000\frac{kg}{m^{3}}\cdot0,12 \ m}{1000\frac{kg}{m^{3}}-910\frac{kg}{m^{3}}} = \frac{120}{90} \ m\\\\d = 1,(3) \ m\approx1,33 \ m[/tex]
Odp. Grubość kry wynosi ≈ 1,33 m.
