If a scuba diver descends too quickly into the sea, the internal pressure on each eardrum remains at atmospheric pressure while the external pressure increases due to the increased water depth. At sufficient depths, the difference between the external and internal pressures can rupture an
eardrum. Eardrums can rupture when the pressure difference is as little as 35 kPa. What is the depth at which this pressure difference could occur? The density of seawater is 1025 kg/m3.
delta p = rho g delta h
35*10^3 = 1025 *9.81 * h
h = 3.48 meters
Oh, the depth of underwater surprises! Let's dive into this problem with a splash of humor! So, to find the depth where the eardrum may rupture, we'll have to channel our inner mermaids and do a little math.
Now, let's use the equation of pressure, which is given by P = ρgh, where P is the pressure, ρ is the density, g is the acceleration due to gravity, and h is the depth. But hold your breath, here's the twist - we need to convert 35 kPa to Pa, which is 35,000 Pa.
Alright, let's do some calculations. Our density, as specified, is 1025 kg/m3. Assuming you're not diving on another planet, we'll take gravity as 9.8 m/s2. Using these values, we can rearrange the equation to solve for depth:
h = (P) / (ρg)
Plugging in the values, we get:
h = (35,000 Pa) / (1025 kg/m3 * 9.8 m/s2)
After those tedious calculations, the answer is... drumroll, please... approximately 3.6 meters!
So, if a scuba diver descends to a depth greater than 3.6 meters, they better be warned that their eardrums might just burst into an underwater symphony! Just remember to ascend slowly and keep your ears safe and sound. Happy diving, my aquatic friend!
To calculate the depth at which the pressure difference could occur, you can use the equation for hydrostatic pressure:
Pressure = Density × Gravity × Height
First, convert the pressure difference from kPa to Pa:
35 kPa = 35,000 Pa
The density of seawater is given as 1025 kg/m^3, and the acceleration due to gravity is approximately 9.8 m/s^2.
Let's assume that the depth is represented by "h."
The external pressure at this depth is given by:
P_external = (Density of seawater) × (Gravity) × (Depth)
P_external = 1025 kg/m^3 × 9.8 m/s^2 × h
Since the pressure on the eardrum is the atmospheric pressure, it is 101,325 Pa.
The equation for the pressure difference can be written as follows:
Pressure difference = P_external - P_internal
35,000 Pa = (1025 kg/m^3 × 9.8 m/s^2 × h) - 101,325 Pa
Now, let's solve for the depth (h):
35,000 Pa + 101,325 Pa = 1025 kg/m^3 × 9.8 m/s^2 × h
136,325 Pa = 10,045 kg/m^3 × h
Divide both sides of the equation by 10,045 kg/m^3 to isolate h:
h = 136,325 Pa / (1025 kg/m^3 × 9.8 m/s^2)
h ≈ 13.55 meters
Therefore, the depth at which the pressure difference of 35 kPa could occur is approximately 13.55 meters.
To find the depth at which this pressure difference could occur, we can use the concept of hydrostatic pressure.
The hydrostatic pressure in a fluid increases with depth. It can be calculated using the equation:
P = ρ*g*h
Where:
P is the pressure,
ρ is the density of the fluid (seawater in this case),
g is the acceleration due to gravity, and
h is the depth.
In this scenario, we want to find the depth at which the pressure difference is 35 kPa. We can assume that the atmospheric pressure at the surface is 101.3 kPa.
So, the external pressure at the depth we're looking for will be the sum of the atmospheric pressure and the pressure difference:
External Pressure = Atmospheric Pressure + Pressure Difference
External Pressure = 101.3 kPa + 35 kPa
Now, we need to convert the pressure from kilopascals (kPa) to pascals (Pa) for consistency:
External Pressure = (101.3 kPa + 35 kPa) * 1000 Pa/kPa
External Pressure = 136.3 kPa * 1000 Pa/kPa
External Pressure = 136,300 Pa
We have the external pressure, and we know the density of seawater is 1025 kg/m³. We can rearrange the formula to solve for depth:
h = P / (ρ*g)
h = 136,300 Pa / (1025 kg/m³ * 9.8 m/s²)
Now, let's calculate the depth:
h ≈ 13.31 meters
Therefore, at a depth of approximately 13.31 meters, the pressure difference could be sufficient to rupture an eardrum if a scuba diver descends too quickly.