Hydrolic Pressure and Bulk Modulus

AI Thread Summary
To determine the pressure required to compress a solid copper cube from an edge length of 87.0 cm to 85.7 cm, the bulk modulus of copper, given as 1.4 x 10^11 N/m², is used in the calculations. The initial and final volumes of the cube are calculated, yielding a change in volume (deltaV) of 0.03 m³. The pressure is then calculated using the formula p = (B x deltaV)/V, resulting in an initial incorrect answer. The issue was resolved by correctly calculating deltaV as the difference between the cubes' volumes using precise values. Ultimately, the calculations were confirmed to be correct when significant figures were properly considered.
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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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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.
 
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.
 
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.
 

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