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An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600 kg . It is anchored to the sea bottom by a cable. The density of seawater is 1025 kg/m3. Part A What is the buoyant force on the chamber?

Respuesta :

Answer:

the buoyant force on the chamber is F = 7000460 N

Explanation:

the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.

Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume

mass of water displaced = density of seawater * volume displaced

m= d * V , V = 4/3π* Rext³

the buoyant force is the weight of this volume of seawater

F = m * g = d * 4/3π* Rext³ * g

replacing values

F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N

Note:

when occupied the tension force on the cable is

T = F buoyant - F weight of chamber  = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N

The buoyant force on the chamber is mathematically given as

F= 8.75*10^{5} N

What is the buoyant force on the chamber?

Question Parameter(s):

An undersea research chamber is spherical with an external diameter of 5.50 m .

The mass of the chamber, when occupied, is 87600 kg

Generally, the equation for the   is mathematically given as

[tex]V = \frac{4\pi (\frac{d}{2} )^{3}}{3}\\\\ V = \frac{4(3.14) (\frac{5.50}{2} )^{3}}{3}[/tex]

V = 87.07 m^{3}

In conclusion

F = d V g

F= (1025) (87.07) (9.8)

F= 8.75*10^{5} N

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