(a) The force exerted by the atmosphere on a disk-shaped region 2.00 m in radius at the surface of a methane ocean is 6.56 x 10^5 N.
We can use the formula:
Force = Pressure x Area
The pressure is given as 8.28 x 10^4 Pa, and the area of the disk-shaped region is πr^2 = π(2.00 m)^2 = 12.57 m^2.
Substituting the given values gives:
Force = (8.28 x 10^4 Pa) x (12.57 m^2) = 6.56 x 10^5 N
Therefore, the force exerted by the atmosphere on the disk-shaped region is 6.56 x 10^5 N.
(b) The weight of a 10.0-m deep cylindrical column of methane with radius 2.00 m is 3.25 x 10^6 N.
The volume of the cylindrical column is given by:
Volume = πr^2h = π(2.00 m)^2(10.0 m) = 125.66 m^3
The mass of the cylindrical column is:
Mass = Volume x Density = (125.66 m^3) x (415 kg/m^3) = 52170.9 kg
The weight of the cylindrical column is:
Weight = Mass x Gravity = (52170.9 kg) x (5.20 m/s^2) = 3.25 x 10^6 N
Therefore, the weight of the cylindrical column is 3.25 x 10^6 N.
(c) The pressure at a depth of 10.0 m in the methane ocean is 4.14 x 10^5 Pa.
The pressure at a depth h in a liquid is given by:
P = ρgh
where ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth of the liquid.
Substituting the given values gives:
P = (415 kg/m^3) x (5.20 m/s^2) x (10.0 m) = 4.14 x 10^5 Pa
Therefore, the pressure at a depth of 10.0 m in the methane ocean is 4.14 x 10^5 Pa.