Question

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? Express your answer to three significant figures and include the appropriate units.Part B What is the tension in the cable? Express your answer to two significant figures and include the appropriate units.

Answer 1

A)

`8.75*10^(5) N`

B)

`1.7*10^(4) N`

Step-by-step explanation:

`d` = diameter of the spherical chamber = 5.50 m

Volume of the spherical chamber is given as

`V = (4\pi ((d)/(2) )^(3))/(3)\\ V = (4(3.14) ((5.50)/(2) )^(3))/(3)\\V = 87.07 m^(3)`

`\rho` = density of the seawater = 1025 kg m⁻³

Part A)

Force of buoyancy by seawater on the chamber is given as

`F_(b) = \rho V g \\F_(b) = (1025) (87.07) (9.8)\\F_(b) = 8.75*10^(5) N`

Part B)

`T` = tension force in the cable in upward direction

`m` = mass of the chamber = 87600 kg

Weight of the chamber is given as

`F_(g) = mg \\`

Using equilibrium of force in the vertical direction, we have

`T + F_(g) = F_(b)\\T + m g = F_(b)\\T + (87600) (9.8) = 8.75*10^(5)\\T = 16520 N \\T = 1.7*10^(4) N`