Respuesta :
Answer:[tex]\begin{gathered} \text{Buoyant Force on the vessel= 7.40}\times10^5N \\ \text{Tension in the cable = 1.04}\times10^4N \end{gathered}[/tex]Explanations:
The mass of the vessel, m = 74400 kg
The external diameter, d = 5.20 m
Radius, r = 5.2 / 2 = 2.6 m
[tex]\begin{gathered} \text{Density of sea water, }\rho_w=1.025\times10^3\operatorname{kg}m^{-3}^{} \\ \rho_w=1025\operatorname{kg}m^{-3} \end{gathered}[/tex]The volume of the vessel is given by the volume of a sphere
[tex]\begin{gathered} V\text{ = }\frac{4}{3}\pi r^3 \\ V\text{ = }\frac{4}{3}\times3.142\times2.6^3 \\ V\text{ = }73.63m^3 \end{gathered}[/tex]a) The buoyant force (B) is given by the formula:
[tex]\begin{gathered} B\text{ = }\rho_wgV \\ B\text{ = 1025}\times9.8\times73.63 \\ B\text{ = }739613.35 \\ B\text{ = 740000 (to the nearest ten thousand)} \\ B\text{ = 7.40}\times10^5N \end{gathered}[/tex]b) The tension in the cable is given by the equation
[tex]T\text{= B - mg}[/tex]Where T represents the tension
T = 739613.35 - 74400(9.8)
T = 10493.35
[tex]T\text{ = 1.04}\times10^4N[/tex]