Suppose a distant world with surface gravity of 7.12 m/s2 has an atmospheric pressure of 7.48 104 Pa at the surface.

(a) What force is exerted by the atmosphere on a disk-shaped region 2.00 m in radius at the surface of a methane ocean?
________N

(b) What is the weight of a 10.0-m deep cylindrical column of methane with radius 2.00 m? Note: The density of liquid methane is 415 kg/m3.
________N

(c) Calculate the pressure at a depth of 10.0 m in the methane ocean.
________Pa

To find the force exerted by the atmosphere on a disk-shaped region at the surface of a methane ocean, we can use the formula:

Force = Pressure x Area

(a) First, let's calculate the area of the disk-shaped region. The radius is given as 2.00 m, so the area is:

Area = π x (radius)^2
= π x (2.00 m)^2
= 4π m^2

Now, we can use the given atmospheric pressure to calculate the force:

Force = 7.48 x 10^4 Pa x 4π m^2
= (7.48 x 10^4 Pa) x (4π m^2)
≈ 9.4 x 10^5 N

Therefore, the force exerted by the atmosphere on the disk-shaped region is approximately 9.4 x 10^5 Newtons.

(b) To find the weight of the cylindrical column of methane, we can use the formula:

Weight = Mass x Gravity

The density of liquid methane is given as 415 kg/m^3. To find the mass, we need to calculate the volume of the cylindrical column first. The height or depth of the column is given as 10.0 m and the radius is given as 2.00 m.

Volume = Area of base x Height
= π x (radius)^2 x height
= π x (2.00 m)^2 x 10.0 m
= 40π m^3

Now, we can calculate the mass of the cylindrical column using the density:

Mass = Density x Volume
= 415 kg/m^3 x 40π m^3

Finally, we can find the weight:

Weight = Mass x Gravity
= (415 kg/m^3 x 40π m^3) x (7.12 m/s^2)

(c) To calculate the pressure at a depth of 10.0 m in the methane ocean, we can use the formula:

Pressure = Pressure at the surface + Density x Gravity x Depth

Given:
Pressure at the surface = 7.48 x 10^4 Pa
Density = 415 kg/m^3
Gravity = 7.12 m/s^2
Depth = 10.0 m

Now, substitute these values into the formula:

Pressure = (7.48 x 10^4 Pa) + (415 kg/m^3 x 7.12 m/s^2 x 10.0 m)

Calculate the result and you will find the pressure at a depth of 10.0 m in the methane ocean in Pascals.