Dynamometer

What is the volume of a body that stretches a dynamometer in air, on which it is suspended by a force of 2.5 N, and if it is immersed in alcohol with a density of 800 kg/m ^ 3, does it tension the dynamometer with a force of 1.3 N?

Correct answer:

V =  0.15 l

Step-by-step explanation:

$$\begin{gathered} F_1=2.5\text{ N } \\ {F}_2=1.3\text{ N } \\ \rho =800{\text{kg/m}}^3 \\ g=10{\text{m/s}}^2 \\ F=mg \\ m=V\rho \\ F=V\rho g \\ {V}_1=(F_1-F_2)\mathrm{/}(\rho \cdot g)=(2.5-1.3)\mathrm{/}(800\cdot 10)={\displaystyle \frac{3}{20000}}\doteq 0.0002{\text{m}}^3 \\ V=V_1\to \text{ l}=V_1\cdot 1000\text{ l}=0.0002\cdot 1000\text{ l}=0.15\text{ l} \end{gathered}$$

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