Assuming atmospheric pressure to be 1.01 x 10^5 Pa and the density of sea water to be 1025 kg/m^3, what is the absolute pressure at a depth of 15.0m below the surface of the ocean?
Could you show me how to do this problem? Thank you.
P,abs = P + pgh
P = (1.01*10^5 Pa) + (1025 kg/m^3)*(9.8 m/s^2)(15 m)
P = 101000 + 150675
P = 251675 Pa
Check the significant figures.
Hope this helps~ `u`
Verify that the atmospheric pressure at sea level is 1.o1×10pa
Sure! But first, a joke:
Why don't scientists trust atoms?
Because they make up everything!
Now, let's dive into the question. When calculating the absolute pressure at a certain depth, we need to consider both atmospheric pressure and the pressure due to the depth in the ocean.
The pressure at a specific depth in a fluid is given by the equation:
P = P₀ + ρgh
Where:
P is the pressure at depth,
P₀ is the atmospheric pressure,
ρ is the density of the fluid,
g is the acceleration due to gravity, and
h is the depth.
Given:
P₀ = 1.01 x 10^5 Pa (atmospheric pressure),
ρ = 1025 kg/m^3 (density of sea water), and
h = 15.0 m (depth).
Now, let's calculate the absolute pressure:
P = P₀ + ρgh
P = 1.01 x 10^5 Pa + (1025 kg/m^3) x (9.8 m/s^2) x (15.0 m)
Calculating this value gives us the absolute pressure at a depth of 15.0 m below the surface of the ocean. Let me crunch some numbers for you!
Sure! To solve this problem, you need to consider the combined pressure due to the atmospheric pressure and the pressure exerted by the column of water above the given depth.
1. Start by finding the pressure due to the column of water. The pressure at any point within a 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 height of the column.
In this case, the height of the column of water is given as 15.0m, the density of sea water is 1025 kg/m^3, and the acceleration due to gravity is approximately 9.8 m/s^2.
So, the pressure due to the column of water can be calculated as:
P_water = ρgh
= 1025 kg/m^3 * 9.8 m/s^2 * 15.0 m
2. Next, find the absolute pressure by adding the pressure due to the column of water to the atmospheric pressure. The absolute pressure is the total pressure at a given point, including both the atmospheric pressure and the pressure due to any fluid above.
Absolute pressure = Atmospheric pressure + Pressure due to water column
= 1.01 x 10^5 Pa + P_water
3. Now, substitute the calculated value of P_water from Step 1 into the equation:
Absolute pressure = 1.01 x 10^5 Pa + (1025 kg/m^3 * 9.8 m/s^2 * 15.0 m)
4. Simplify the equation and compute the result to find the absolute pressure at 15.0m below the surface of the ocean.
I hope this helps!
Sure! To find the absolute pressure at a certain depth below the surface of the ocean, you can use the formula:
P = P0 + ρgh
Where:
P is the absolute pressure at the depth,
P0 is the atmospheric pressure,
ρ is the density of the sea water, and
h is the depth below the surface.
In this case, we are given:
P0 = 1.01 x 10^5 Pa (atmospheric pressure)
ρ = 1025 kg/m^3 (density of sea water)
h = 15.0 m (depth below the surface)
Now, let's substitute these values into the formula:
P = 1.01 x 10^5 Pa + (1025 kg/m^3) * (9.8 m/s^2) * (15.0 m)
First, let's calculate the second part of the equation:
(1025 kg/m^3) * (9.8 m/s^2) * (15.0 m) = 150,225 N/m^2
Now, let's substitute this value back into the formula:
P = 1.01 x 10^5 Pa + 150,225 N/m^2
To combine the units, we need to convert N/m^2 to Pa:
1 N/m^2 = 1 Pa
Therefore, we have:
P = 1.01 x 10^5 Pa + 150,225 Pa
Adding them together, we get:
P = 2.51 x 10^5 Pa
So, the absolute pressure at a depth of 15.0 m below the surface of the ocean is 2.51 x 10^5 Pa.