Wn the load is applied?
6. What is the total mass of the Earth’s atmosphere? (The radius of the Earth is
6.37x106m, and atmospheric pressure at the surface is 1.013x105N/m2.)
7. (a) Calculate the absolute pressure at an ocean depth of 1000m. Assume the
density of seawater is 1024kg/m3 and that the air above exerts a pressure of
101.3kPa. (b) At this depth, what force must the frame around a circular
submarine porthole having a diameter of 30.0cm exert to counterbalance the
force exerted by the water?
8. Calculate the buoyant force, if a steel ball of radius 6 cm is totally immersed
in water.
6. The total mass of the Earth's atmosphere can be calculated using the formula:
mass = density x volume
The density of air at sea level is approximately 1.225 kg/m3. The volume of the Earth's atmosphere can be approximated as the volume of a spherical shell with inner radius equal to the Earth's radius (6.37x10^6 m) and outer radius extending into the upper atmosphere. Assuming a thickness of 100km for the atmosphere, the outer radius would be 6.47x10^6 m.
So, the mass of the Earth's atmosphere would be:
mass = density x volume
mass = 1.225 kg/m^3 x (4/3) π (6.47x10^6)^3 - (4/3) π (6.37x10^6)^3
mass = 1.225 kg/m^3 x 7.24x10^19 m^3 - 1.19x10^20 m^3
mass = 8.86x10^19 kg - 1.46x10^20 kg
mass = -5.71x10^19 kg
Therefore, the total mass of the Earth's atmosphere is approximately 5.71x10^19 kg.
7. (a) To calculate the absolute pressure at an ocean depth of 1000m, you can use the formula:
P = P0 + ρgh
Where:
P = absolute pressure at depth (Pa)
P0 = atmospheric pressure at surface (1.013x10^5 N/m^2)
ρ = density of seawater (1024 kg/m^3)
g = acceleration due to gravity (9.81 m/s^2)
h = depth (1000m)
Plugging in the values:
P = 1.013x10^5 + 1024 x 9.81 x 1000
P = 1.013x10^5 + 1.001x10^7
P = 1.011x10^7 Pa
Therefore, the absolute pressure at an ocean depth of 1000m is approximately 1.011x10^7 Pa.
(b) To calculate the force exerted by the water on the submarine porthole, you can use the formula for pressure:
Pressure = Force / Area
The force exerted by the water on the porthole is equal to the absolute pressure at the depth multiplied by the area of the porthole:
Force = Pressure x Area
Force = 1.011x10^7 Pa x π(0.30/2)^2
Force ≈ 3551 N
Therefore, the force the frame around the submarine porthole must exert to counterbalance the force exerted by the water is approximately 3551 N.
8. To calculate the buoyant force acting on the steel ball immersed in water, you can use the formula for buoyant force:
Buoyant force = ρVg
Where:
ρ = density of water (1000 kg/m^3)
V = volume of the steel ball
g = acceleration due to gravity (9.81 m/s^2)
Assuming the density of steel is 7850 kg/m^3 and using the formula for the volume of a sphere (V = (4/3)πr^3), we can calculate the buoyant force:
V = (4/3)π(0.06)^3
V ≈ 0.00090478 m^3
Buoyant force = 1000 x 0.00090478 x 9.81
Buoyant force ≈ 8.86 N
Therefore, the buoyant force acting on the steel ball immersed in water is approximately 8.86 N.