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.
vMaster0fPuppet
Messages
16
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
Thread 'Collision of a bullet on a rod-string system: query'

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...
View full post »
Thread 'A cylinder connected to a hanged mass'

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...
View full post »

Hot Threads

Recent Insights

Back
Top