The pressure at 10 below the surface of the ocean is about2.00×105. Which of the following statements istrue?…
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Question “The pressure at 10 below the surface of the ocean is about2.00×105. Which of the following statements istrue?…”
The pressure at 10
below the surface of the ocean is
about2.00×105.
Which of the following statements
istrue?
The weight of a column of seawater 1
in cross section and 10
high is about 2.00×105.
The weight of a column of seawater 1
in cross section and 10
high plus the weight of a column of air with thesame cross section
extending up to the top of the atmosphere isabout
2.00×105.
The weight of 1
of seawater at 10
below the surface of the ocean is
about2.00×105.
The density of seawater is
about2.00×105 times the density of air at sealevel.
Now consider he pressure at 20
below the surface of the ocean.
This pressure is
twice that at a depth of 10 .
the same as that at a depth of 10 .
equal to that at a depth of 10
plus the weight of a column of seawater 1
in cross section and 10
high.
equal to the weight of a column of seawater
1
in cross section and 20
high.
The pressure at 10 m below the surface of the ocean is about2.00�10^5 Pa. Which of the following statements istrue? The weight of a column of seawater 1 m^2 in cross section and 10 m high is about 2.00�10^5 N. The weight of a column of seawater 1m^2 in cross section and 10 m high plus the weight of a column of air with thesame cross section extending up to the top of the atmosphere isabout 2.00�10^5 N The weight of 1 m^3 of seawater at 10 m below the surface of the ocean is about2.00�10^5 N.Now consider he pressure at 20 m below the surface of the ocean.The density of seawater is about2.00�10^5 times the density of air at sealevel. This pressure is twice that at a depth of 10 m.the same as that at a depth of 10 m. equal to that at a depth of 10 m plus the weight of a column of seawater 1 m^2 in cross section and 10 m high. equal to the weight of a column of seawater 1m^2 in cross section and 20 m high
Answer
Concepts and Reason
This problem can be solved by combining variation in pressure and depth within a fluid.
To calculate the total pressure, first use relation between atmospheric pressure and static liquid pressure
To calculate the force, use the relationship between area and total pressure.
Fundamentals
The container can be used to measure the fluid’s depth.
The expression for the object’s weight is:
W refers to the object’s weight, M refers to the object’s mass, and G indicates the acceleration caused by gravity.
The volume of the object is
A refers to the area of an object, and H refers to the fluid’s depth.
For the force, expression is:
F here is the force.
The static fluid pressure can be expressed as:
Here
refers to the static fluid pressure.
To get the total pressure, the container must be exposed to the atmosphere.
The expression for total pressure is:
represents the total pressure, and
the atmospheric pressure.
(1)
The expression for total pressure is:
represents the total pressure at #media_tag_10$ under the ocean’s surface.
Consider the pressure at 20 meters below the ocean’s surface. The pressure at 20m will equal the pressure at 10m plus the pressure at the 10m top 10m below.
This means that the pressure at a depth
plus the weight of a column seawater in cross-section and high must equal the
.
Part 1: Ans
A column of seawater tag_25$ with a cross section of
high and a column containing the same cross section but extending to the top is about .
Conclusion
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