Respuesta :
Answer:
Specific weight of the pop, [tex]w_{s} = 8619.45 N/m^{3}[/tex]
Density of the pop, [tex]\rho_{p} = 8790.76 kg/m^{3}[/tex]
[tex]g_{s} = 8.79076[/tex]
[tex]w_{w} = 9782.36 N/m^{3}[/tex]
Given:
Volume of pop, V = 360 mL = 0.36 L = [tex]0.36\times 10^{-3} m^{3}[/tex]
Mass of a can of pop , m = 0.369 kg
Weight of an empty can, W = 0.153 N
Solution:
Now, weight of a full can pop, W
W' = mg = [tex]0.369\times 9.8 = 3.616 N[/tex]
Now weight of the pop in can is given by:
w = W' - W = 3.616 - 0.513 = 3.103 N
Now,
The specific weight of the pop, [tex]w_{s} = \frac{weight of pop}{volume of pop}[/tex]
[tex]w_{s} = \frac{3.103}{0.36\times 10^{- 3}} = 8619.45 N/m^{3}[/tex]
Now, density of the pop:
[tex]\rho_{p} = \frac{w_{s}}{g}[/tex]
[tex]\rho_{p} = \frac{86149.45}{9.8} = 8790.76 kg/m^{3}[/tex]
Now,
Specific gravity, [tex]g_{s} = \frac{\rho_{p}}{density of water, \rho_{w}}[/tex]
where
[tex]g_{s} = \frac{8790.76}{1000} = 8.79076[/tex]
Now, for water at [tex]20^{\circ}c[/tex]:
Specific density of water = [tex]998.2 kg/m^{3}[/tex]
Specific gravity of water = [tex]0.998 kg/m^{3}[/tex]
Specific weight of water at [tex]20^{\circ}c[/tex]:
[tex]w_{w} = \rho_{20^{\circ}}\times g = 998.2\times 9.8 = 9782.36 N/m^{3}[/tex]
