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en/4.2.6.md
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| ### Statement | |||
| − | $4.2.6$ A piece of iron weighs $9.8 \mathrm{~N}$ in water. Determine its volume. The density of iron is $7.8 \cdot 10^3 \mathrm{m^3}$. | ||
| + | $4.2.6$ A piece of iron weighs $9.8 \mathrm{~N}$ in water. Determine its volume. The density of iron is $7.8 \cdot 10^3 \mathrm{~m^3}$. | ||
| ### Solution | |||
| Let's consider the following figure | |||
|  | |||
| The normal force $N$ is what the equipment measures as apparent weight. From Newton Second Law $$N = mg-F_A$$ $$N = \rho~g~V-\rho_w~g~V$$ $$N = (\rho -\rho_w)~g~V$$ and taking $\rho_w = 1000$ kg/m$^3$ $$\boxed{V = \frac{N}{(\rho -\rho_w)~g}=147 \rm{cm}^3}$$ | |||
| unchanged lines 5 | |||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $4.2.6$ A piece of iron weighs $9.8 \mathrm{~N}$ in water. Determine its volume. The density of iron is $7.8 \cdot 10^3 \mathrm{m^3}$. | $4.2.6$ A piece of iron weighs $9.8 \mathrm{~N}$ in water. Determine its volume. The density of iron is $7.8 \cdot 10^3 \mathrm{~m^3}$. | ||
| ### Solution | ### Solution | ||
| Let's consider the following figure | Let's consider the following figure | ||
|  |  | ||
| The normal force $N$ is what the equipment measures as apparent weight. From Newton Second Law $$N = mg-F_A$$ $$N = \rho~g~V-\rho_w~g~V$$ $$N = (\rho -\rho_w)~g~V$$ and taking $\rho_w = 1000$ kg/m$^3$ $$\boxed{V = \frac{N}{(\rho -\rho_w)~g}=147 \rm{cm}^3}$$ | The normal force $N$ is what the equipment measures as apparent weight. From Newton Second Law $$N = mg-F_A$$ $$N = \rho~g~V-\rho_w~g~V$$ $$N = (\rho -\rho_w)~g~V$$ and taking $\rho_w = 1000$ kg/m$^3$ $$\boxed{V = \frac{N}{(\rho -\rho_w)~g}=147 \rm{cm}^3}$$ | ||
| unchanged lines 5 | |||