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 answer these questions, we will use some basic principles of physics:
(a) The force exerted by the atmosphere on the disk-shaped region can be calculated using the formula:
Force = Pressure x Area
Here, the pressure is given as 7.48 x 10^4 Pa, and the area of the disk-shaped region can be calculated as:
Area = π x r^2
where r is the radius of the disk, given as 2.00 m. Plugging in the values, we can find the force:
Area = π x (2.00 m)^2 = 12.57 m^2
Force = 7.48 x 10^4 Pa x 12.57 m^2 = 9.38 x 10^5 N
Therefore, the force exerted by the atmosphere on the disk-shaped region is approximately 9.38 x 10^5 N.
(b) The weight can be calculated using the formula:
Weight = Density x Volume x Gravity
Here, the density of liquid methane is given as 415 kg/m^3, and the volume of the cylindrical column can be calculated as:
Volume = π x r^2 x h
where r is the radius of the column (2.00 m) and h is the height of the column (10.0 m). Plugging in the values, we can find the weight:
Volume = π x (2.00 m)^2 x (10.0 m) = 125.66 m^3
Weight = 415 kg/m^3 x 125.66 m^3 x 7.12 m/s^2 = 370,136.9 N
Therefore, the weight of the cylindrical column of methane is approximately 370,136.9 N.
(c) The pressure at a depth in a fluid can be calculated using the formula:
Pressure = Pressure at Surface + (Density x Gravity x Depth)
Here, the pressure at the surface is given as 7.48 x 10^4 Pa, the density of methane is 415 kg/m^3, the gravity on the distant world is 7.12 m/s^2, and the depth is given as 10.0 m. Plugging in the values, we can find the pressure:
Pressure = 7.48 x 10^4 Pa + (415 kg/m^3 x 7.12 m/s^2 x 10.0 m) = 6.03 x 10^4 Pa
Therefore, the pressure at a depth of 10.0 m in the methane ocean is approximately 6.03 x 10^4 Pa.