Hydrolic Pressure and Bulk Modulus

In summary, the question asks for the amount of pressure needed to reduce the edge length of a solid copper cube from 87.0 cm to 85.7 cm. Using the equations for volume and pressure, the calculation was done with the bulk modulus of copper as 1.4 x 10^11 N/m^2. The correct answer was obtained by using the correct number of significant figures and calculating deltaV as .870^3-.857^3.
  • #1
vMaster0fPuppet
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Homework Statement



A solid copper cube has an edge length of 87.0 cm. How much pressure must be applied to the cube to reduce the edge length to 85.7 cm? The bulk modulus of copper is 1.4 multiplied by 1011 N/m2.

Homework Equations



Vcube= s^3

F/A=(B x deltaV)/V
p=F/A
p=(B x deltaV)/V

The Attempt at a Solution

V= (.870m)^3= .659m^3
Vn= (.857m)^3= .629m^3
deltaV= V-Vn= .03m^3

p=(1.4e^10N/m^2)x(.03m^3)/(.659m^3)= 6.37e^9N/m^2This is apparently not the answer?
 
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  • #2
vMaster0fPuppet said:

Homework Statement



A solid copper cube has an edge length of 87.0 cm. How much pressure must be applied to the cube to reduce the edge length to 85.7 cm? The bulk modulus of copper is 1.4 multiplied by 1011 N/m2.

Homework Equations



Vcube= s^3

F/A=(B x deltaV)/V
p=F/A
p=(B x deltaV)/V

The Attempt at a Solution

V= (.870m)^3= .659m^3
Vn= (.857m)^3= .629m^3
deltaV= V-Vn= .03m^3

p=(1.4e^10N/m^2)x(.03m^3)/(.659m^3)= 6.37e^9N/m^2This is apparently not the answer?

Is that 1.4 x 1010 Pa or 1.4 x 1011 Pa.
I think think it is 1.4 x 1011 Pa for copper as you originally gave in the problem.
 
  • #3
Its 10^11, and I did the calculations with 10^11.

Thanks anyway, I actually figured the problem out. When I used .870^3-.857^3 for deltaV the calculations worked just fine. I didn't realize that making the calculation easier by breaking it up would cause such a problem if I used the correct number of significant figures.
 
  • #4
vMaster0fPuppet said:
Its 10^11, and I did the calculations with 10^11.

Thanks anyway, I actually figured the problem out. When I used .870^3-.857^3 for deltaV the calculations worked just fine. I didn't realize that making the calculation easier by breaking it up would cause such a problem if I used the correct number of significant figures.

OK then rightio. I didn't see a problem with your method.
 

1. What is hydrolic pressure?

Hydrolic pressure is the force per unit area exerted by a liquid on its container or any surface it comes into contact with. It is typically measured in units of pressure, such as pounds per square inch (psi) or pascals (Pa).

2. How is hydrolic pressure calculated?

Hydrolic pressure is calculated by dividing the force applied by the area over which it is applied. The formula for hydrolic pressure is P = F/A, where P is the pressure, F is the force, and A is the area.

3. What is bulk modulus?

Bulk modulus is a measure of a substance's resistance to compression when subjected to hydrolic pressure. It is a constant that relates the change in volume of a substance to the change in pressure applied to it. Bulk modulus is typically measured in units of pressure, such as psi or Pa.

4. How is bulk modulus related to hydrolic pressure?

Bulk modulus and hydrolic pressure are directly related. As hydrolic pressure increases, the substance's bulk modulus also increases, meaning it becomes more difficult to compress. Conversely, as hydrolic pressure decreases, the substance's bulk modulus decreases, making it easier to compress.

5. What are some practical applications of hydrolic pressure and bulk modulus?

Hydrolic pressure and bulk modulus have many practical applications, including in hydraulic systems, such as car brakes and heavy machinery, where the pressure of a liquid is used to produce motion or force. They are also important in determining the stability of structures, such as dams and bridges, as well as in the study of fluid dynamics and geology.

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