A research submarine is designed to operate 3 km below the ocean surface. If then interior pressure is 1 atm, what is the total pressure on a 15cm diameter window and force on window.

The average density of seawater is 1028 kg/m3

To calculate the total pressure on the window of the research submarine, we need to consider the pressure due to the depth of water and the atmospheric pressure.

1. Pressure due to the depth of water:
The pressure due to the depth of the water is given by the formula:

Pressure = Density × Gravitational acceleration × Depth

Given:
Density of seawater = 1028 kg/m^3
Gravitational acceleration = 9.8 m/s^2 (approximately)
Depth = 3 km = 3000 m (since 1 km = 1000 m)

Pressure = 1028 kg/m^3 × 9.8 m/s^2 × 3000 m
= 30,264,000 Pa

2. Atmospheric pressure:
Atmospheric pressure at sea level is generally defined as 1 atm, which is approximately equal to 101,325 Pa.

Therefore, the total pressure on the window is the sum of the pressure due to the depth of water and the atmospheric pressure:

Total Pressure = Pressure due to depth + Atmospheric pressure
= 30,264,000 Pa + 101,325 Pa
= 30,365,325 Pa

To find the force on the window, we can use the formula:

Force = Pressure × Area

Given:
Diameter of the window = 15 cm = 0.15 m (since 1 cm = 0.01 m)
Radius of the window (r) = Diameter / 2 = 0.075 m
Area of the window = π × r^2 = 3.14 × (0.075)^2

Force = 30,365,325 Pa × 3.14 × (0.075)^2
≈ 53,930 N

So, the total pressure on the 15 cm diameter window is approximately 30,365,325 Pa, and the force on the window is approximately 53,930 Newtons.

To determine the total pressure on the window of the research submarine, we need to take into account the interior pressure (1 atm) and the pressure exerted by the column of water above it.

The pressure exerted by a column of fluid is given by the formula:

P = ρgh

where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid.

First, we need to calculate the pressure exerted by the column of water above the submarine at a depth of 3 km.

Given:
Density of seawater (ρ) = 1028 kg/m³
Depth (h) = 3000 m
Acceleration due to gravity (g) = 9.8 m/s² (approximate value)

Using the formula above, we can calculate the pressure exerted by the water column:

P = (1028 kg/m³) * (9.8 m/s²) * (3000 m)
P ≈ 30,258,000 Pa

Now, to find the total pressure on the window, we need to add the interior pressure (1 atm) to the pressure exerted by the water column:

Total pressure = Interior pressure + Pressure exerted by water column

Converting 1 atm to Pascal:
1 atm ≈ 101,325 Pa

Total pressure = 101,325 Pa + 30,258,000 Pa
Total pressure ≈ 30,359,325 Pa

To find the force exerted on the window, we need to use the formula:

Force = Pressure * Area

Given:
Diameter of the window = 15 cm

First, we need to convert the diameter to radius:
Radius = Diameter / 2 = 15 cm / 2 = 7.5 cm = 0.075 m

Now we can calculate the force on the window:

Force = Total pressure * Area
Area = π * (radius)² = 3.14 * (0.075 m)²

Substituting the values:

Force = 30,359,325 Pa * 3.14 * (0.075 m)²

Calculating this value will give you the force exerted on the 15 cm diameter window by the surrounding water pressure.

pressure above=weightwater+weightatmosphere

=density*height+1atm
=1028kg/m^3*g*3000m + 1atm
figure that out.

then, pressure on glass is above -1atm, so force on glass is that pressure times area, or area= pi r^2=PI*.075^2