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A 950 kg cylindrical buoy floats vertically in seawater. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0 kg man stands on top of it.

Respuesta :

Numenius

Answer:

Additional depth is 0.13 m

Explanation:

mass of cylinder, M = 950 kg

Let the initial height inside water is h.

diameter = 0.9 m

radius, r = 0.45 m

mass of man, m = 80 kg

density of water = 1000 kg/m^3

Let the additional distance is y.

For initial stage:

Buoyant force = weight

Volume immersed x density of water x g = M g

3.14 x 0.45 x 0.45 x 1000 x h = 950

h = 1.49 m

Now :

3.14 x 0.45 x 0.45 x 1000 x ( h + y) = 950 + 80

950 +  635.85 y = 950 + 80

y = 0.13 m

onyebuchinnaji

The additional distance the buoy will sink is 0.126 m.

The given parameters:

  • Mass of the buoy, m = 950 kg
  • Diameter of the buoy, d = 0.9 m
  • Additional mass, = 80 kg
  • Density of water, ρ = 1000 kg/m³
  • Radius of the buoy, r = 0.45 m

The initial distance the buoy would sink is calculated as follows;

Buoyant force = weight

[tex]\rho Vg = mg\\\\\rho V = m\\\\\rho \pi r^2 h = m\\\\h = \frac{m}{\rho \pi r^2} \\\\h = \frac{950}{1000 \times \pi \times (0.45)^2} \\\\h = 1.493 \ m[/tex]

Let the additional distance the buoy will sink = d

The additional distance is calculated as follows;

[tex]h + d = \frac{950 + 80}{\rho \pi r^2} \\\\d = \frac{950 + 80}{\rho \pi r^2} - h\\\\d = \frac{1030}{1000 \times \pi \times 0.45^2} \ - \ 1.493\\\\d = 0.126 \ m[/tex]

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