What is the Relationship Between Pressure and Depth in the Ocean?

  • Thread starter Thread starter mantillab
  • Start date Start date
  • Tags Tags
    Ocean Pressure
AI Thread Summary
Pressure in fluids increases with depth, and at 20m below the ocean's surface, the pressure is not the same as at 10m. The correct calculation incorporates both the surface pressure and the weight of the seawater column above the point of measurement. Answer choice (c) is deemed correct as it aligns with the equation p = p_0 + pgh, accounting for both surface pressure and additional depth. Choice (d) is incorrect because it neglects the surface pressure. Understanding this relationship clarifies the concept of pressure in ocean depths.
mantillab
Messages
27
Reaction score
0

Homework Statement



I'm trying to understand the concept of pressure in fluids.

The pressure at 10m below the surface of the ocean is about 2.00×10^5 Pa.
Now consider he pressure at 20m below the surface of the ocean.
This pressure is:
a) twice that at a depth of 10m.
b) the same as that at a depth of 10m.
c) equal to that at a depth of 10m
plus the weight of a column of seawater 1m^2 in cross section and 10m high.
d) equal to the weight of a column of seawater 1m^2 in cross section and 20m high.

Homework Equations



p=p_0 +pgh

The Attempt at a Solution



Since pressure increases with depth, the pressure at 20m won't be the same as that at 10m. The pressure at a point 20m under the water is equal to the weight of the water above it and the pressure at the surface.

I don't think that (a) is the answer, because that would not take into account the pressure at the surface which isn't doubled. Answer choice (b) isn't correct because the pressure changes as the depth/height increases.

Answer choice (c) seems like it may be correct because it takes into account the pressure at the surface (p_0) and adds the extra weight of the increased depth, but would this conform to the equation: p=p_0 +pgh?

Answer choice (d) also seems like a possibility, except that it doesn't take into account the pressure at the surface at all and only refers to the weight of the sea water.

Any thoughts?
 
Physics news on Phys.org
mantillab said:
[
Answer choice (c) seems like it may be correct because it takes into account the pressure at the surface (p_0) and adds the extra weight of the increased depth, but would this conform to the equation: p=p_0 +pgh?

I don't get it...why won't it conform?
 
I guess I was confused by the way the problem was worded. But now that I think about it, it makes perfect sense: the pressure above (p_0) can be set equal to the pressure of the 10m and the surface of the water, plus (+) the weight of the next 10m (pgh). Thanks for the reality check!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top