How much pressure is exerted on a submarine at a depth of 8.50 km in the Pacific Ocean? (density of sea water = 1.025 x 10^3 kg/m3, P0 = 1.01 x 10^5 Pa)
8.6*106pa
Well, that submarine probably feels like it's under a lot of pressure at a depth of 8.50 km in the Pacific Ocean. In terms of numbers, the pressure can be calculated using the formula P = P0 + ρgh, where P is the pressure, P0 is the atmospheric pressure at the surface, ρ is the density of sea water, g is the acceleration due to gravity, and h is the depth.
So, substituting the given values into the formula, we have:
P = 1.01 x 10^5 Pa + (1.025 x 10^3 kg/m^3)(9.8 m/s^2)(8.50 x 10^3 m)
Doing the calculations, it comes out to be a whopping pressure of roughly 1.93 x 10^8 Pa!
I guess you could say that at that depth, the submarine is being squeezed like a clown car packed with way too many clowns. Stay cool down there!
To calculate the pressure exerted on a submarine at a depth of 8.50 km in the Pacific Ocean, you can use the hydrostatic pressure formula:
P = P0 + ρgh
Where:
P is the pressure at depth,
P0 is the atmospheric pressure at the surface (1.01 x 10^5 Pa),
ρ is the density of sea water (1.025 x 10^3 kg/m^3),
g is the acceleration due to gravity (9.8 m/s^2),
and h is the depth.
Let's plug in the values and calculate the pressure:
P = (1.01 x 10^5 Pa) + (1.025 x 10^3 kg/m^3) * (9.8 m/s^2) * (8.50 km)
Since 1 km = 1000 m, we need to convert the depth from kilometers to meters:
8.50 km * (1000 m/km) = 8500 m
Plugging in the values:
P = (1.01 x 10^5 Pa) + (1.025 x 10^3 kg/m^3) * (9.8 m/s^2) * (8500 m)
P ≈ 1.01 x 10^5 Pa + 8.45 x 10^7 Pa
P ≈ 8.46 x 10^7 Pa
Therefore, the pressure exerted on the submarine at a depth of 8.50 km in the Pacific Ocean is approximately 8.46 x 10^7 Pa.
To calculate the pressure exerted on a submarine at a certain depth in the ocean, we can use the concept of hydrostatic pressure, which is based on the density of the fluid and the height of the fluid column above the object.
The formula to calculate the hydrostatic pressure is:
P = P0 + ρgh
Where:
P is the total pressure exerted (in this case, the pressure on the submarine),
P0 is the atmospheric pressure at sea level (in this case, 1.01 x 10^5 Pa),
ρ is the density of the fluid (in this case, sea water density which is 1.025 x 10^3 kg/m^3),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
and h is the depth of the object (in this case, 8.50 km).
Now, let's plug in the given values into the equation:
P = 1.01 x 10^5 Pa + (1.025 x 10^3 kg/m^3)(9.8 m/s^2)(8.50 x 10^3 m)
First, multiply the density (1.025 x 10^3 kg/m^3) by the acceleration due to gravity (9.8 m/s^2), and then multiply that by the depth (8.50 km converted to meters by multiplying by 10^3), and finally add that value to the atmospheric pressure.
Calculating the equation, we get:
P ≈ 1.01 x 10^5 Pa + 8.45 x 10^7 Pa
Thus, the pressure exerted on the submarine at a depth of 8.50 km in the Pacific Ocean is approximately 8.45 x 10^7 Pa.