Solving SV(z) = ∫ ρ(z)gdz ~ ρgz

  • Thread starter g0ggs123
  • Start date
The ~ sign means that the integral is approximately the same as the ρgz expression.In summary, the conversation was about using a formula for SV(z) and applying p = density and z = depth. The formula is given as ∫ ρ(z)gdz ~ ρgz and the attempt at a solution involved manipulating this formula to get (ρ(z)(z)g))/2 and potentially (ρ(ρ)(z)(z)(z)g(g)))/2 if ~ means "multiple". The ~ sign in this context means "approximately equal to".
  • #1
g0ggs123
22
0

Homework Statement


I just need to use this formula and then apply p = density and I z= depth

Homework Equations



SV(z) = ∫ ρ(z)gdz ~ ρgz

The Attempt at a Solution


(ρ(z)(z)g))/2 and if ~ ρgz is multiple it would be (ρ(z)(z)g))/2 * ρgz which is (ρ(ρ)(z)(z)(z)g(g)))/2

Thanks for your help in advance,

Also not sure what ~ means in this instance??

G
 
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  • #2
Is there a question here?

This whole post is incredibly vague.
 
  • #3
g0ggs123 said:

Homework Statement


I just need to use this formula and then apply p = density and I z= depth

Homework Equations



SV(z) = ∫ ρ(z)gdz ~ ρgz

The Attempt at a Solution


(ρ(z)(z)g))/2 and if ~ ρgz is multiple it would be (ρ(z)(z)g))/2 * ρgz which is (ρ(ρ)(z)(z)(z)g(g)))/2

Thanks for your help in advance,

Also not sure what ~ means in this instance??

G
I agree with post 2, you are not making much sense.
However, the ~ sign means "approximately equal to". If ρ is substantially constant over the range of z then it can be taken out of the integral sign and you get ρgz.
 

What is the meaning of the equation "Solving SV(z) = ∫ ρ(z)gdz ~ ρgz"?

The equation represents the solution for calculating the specific volume (SV) of a substance at a given depth (z) in a fluid with a varying density (ρ) and gravitational acceleration (g). The integral sign (∫) indicates that the equation is solved using integration.

How is the equation "Solving SV(z) = ∫ ρ(z)gdz ~ ρgz" used in scientific research?

This equation is commonly used in studies that involve fluid dynamics, such as oceanography, meteorology, and geology. It allows scientists to calculate the specific volume of a fluid at different depths, which can provide valuable insights into the behavior and movement of fluids in various environments.

What are the units for the different variables in the equation "Solving SV(z) = ∫ ρ(z)gdz ~ ρgz"?

The specific volume (SV) is measured in cubic meters per kilogram (m3/kg). The depth (z) is typically measured in meters (m). The density (ρ) is measured in kilograms per cubic meter (kg/m3). The gravitational acceleration (g) is measured in meters per second squared (m/s2).

Can the equation "Solving SV(z) = ∫ ρ(z)gdz ~ ρgz" be applied to all types of fluids?

Yes, this equation can be applied to any type of fluid, including liquids and gases. However, it is important to note that the density (ρ) and gravitational acceleration (g) may vary depending on the type of fluid being studied.

Are there any limitations to using the equation "Solving SV(z) = ∫ ρ(z)gdz ~ ρgz"?

Like any equation, there are certain limitations to its use. This equation assumes that the fluid being studied is incompressible, and that there are no external forces acting on the fluid. In reality, these conditions may not always be true, so the results obtained from this equation should be interpreted with caution.

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