21.09.2015 Views

Apostila Mec. Flu

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Aula 2<br />

Prof. MSc. Luiz Eduardo Miranda J. Rodrigues<br />

Solução do Exercício 1<br />

Massa Específica:<br />

ρ =<br />

ρ =<br />

m<br />

V<br />

1500<br />

2<br />

Peso Específico:<br />

γ = ρ ⋅<br />

g<br />

γ = 750 ⋅10<br />

Peso Específico Relativo:<br />

γ =<br />

r<br />

γ r<br />

=<br />

γ<br />

γ<br />

H<br />

2<br />

O<br />

7500<br />

10000<br />

ρ = 750<br />

kg/m³<br />

γ = 7500<br />

N/m³<br />

γ r<br />

= 0,75<br />

<strong>Mec</strong>ânica dos <strong>Flu</strong>idos

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