A plastic boat with a 25 cm² square cross section floats in a liquid. One by one, you place 50-g masses inside the boat and measure how far the boat extends below the surface.
Your data are as follows:
Mass added, m(g) - Depth, d(cm)
50 - 2.9
100 - 5.0
150 - 6.6
200 - 8.6
Graphing either m versus d or d versus m gives a straight line. In the graph shown above, we chose to plot d on the vertical axis and m on the horizontal axis. From the equation for the line of best fit given, determine the density rho of the liquid.
Please Explain.

Respuesta :

Answer:

Explanation:

Archimedes principle states that the upward buoyant foce exrted on a body is equal to th wight o the liquid displaced.

Now, the buoyant force on the boat is given by:

[tex](m+m_b)g=V\rho g[/tex]

[tex]V[/tex] is the volum [tex]\rho[/tex] is the density [tex]m_b[/tex] is the mass of the boat and [tex]m[/tex] is the mass added to the boat.

[tex](m+m_b)g=(Sd)\rho[/tex]

[tex]S[/tex] is the surface area and [tex]d[/tex] is the depth.

[tex]m=Sd\rho - m_b...(1)[/tex]

The equation for thebest fit linis,

[tex]d=(0.374m/kg)m+0.11m[/tex]

Re-arrangethis equati for [tex]m[/tex]

[tex]m=\frac{d}{(0.374m/kg)}-\frac{0.11m}{0.374m/kg}...(2)[/tex]

From equations(1) and (2),

[tex]Sd\rho=\frac{d}{0.374m/kg}[/tex]

the density is,

[tex]\rho=\frac{1}{S(0.0374m/kg)}=\frac{1}{(25cm^2)(\frac{1m^2}{10^4cm^2})(0.374m/kg)}=1.069\times 10^3 kg/m^3[/tex]

Therefore, the density of the liquid is

[tex]\rho=1.07\times 10^3 kg/m^3[/tex]