Compression of a sphere underwater

[ultrapowerpie]

Homework Statement

A solid metal sphere of volume 1.05 m^3 is lowered to a depth in the ocean where the water pressure is equal to 1.38 x 10^7 N/m^2. The bulk modulus of the metal from which the sphere is made is 9.9 X 10^9 N/m^2

Given: The atmosphereic pressure is 1.013 10^5 Pa.

What is the change in the volume of the sphere? Give in m^3

Homework Equations

The equation is:

\Delta V = (-1/B)*Pressure*V

B= Bulk modulus
V= volume

The Attempt at a Solution

I really just have one question: what does the atmospheric pressure have to do with anything? I know that F/A (which is the original equation) is pressure, but can I just use the water pressure, or do I somehow have to incooperate the atmospheric pressure?

 

[asleight]

Homework Statement

A solid metal sphere of volume 1.05 m^3 is lowered to a depth in the ocean where the water pressure is equal to 1.38 x 10^7 N/m^2. The bulk modulus of the metal from which the sphere is made is 9.9 X 10^9 N/m^2

Given: The atmosphereic pressure is 1.013 10^5 Pa.

What is the change in the volume of the sphere? Give in m^3

Homework Equations

The equation is:

\Delta V = (-1/B)*Pressure*V

B= Bulk modulus
V= volume

The Attempt at a Solution

I really just have one question: what does the atmospheric pressure have to do with anything? I know that F/A (which is the original equation) is pressure, but can I just use the water pressure, or do I somehow have to incooperate the atmospheric pressure?

You probably have to use the change in pressure, \Delta\rho.

 

[ultrapowerpie]

Hmmm, that makes a little bit of sense, let me try that real fast...

Nope, problem says it's wrong. I also tried adding the two together, that's wrong. And multiplying them makes the Delta V larger then the total volume of the sphere. >.>

 

[mgb_phys]

Wether you should subtract the air presusre depends wether they mean that the total pressure is 1.38 x 10^7 ps or the pressure only due to seawater is 1.38 x 10^7pa. They almost certainly mean the total pressure in which case the atmospheric pressure is unnecessary.
But since the difference is much smaller than the accuracy you are given any figures for you can just ignore it.

 

[ultrapowerpie]

Ok, I don't know what is going on here, but the stupid online homework keeps saying it's wrong. I don't know what the heck I'm doing wrong, and I really wish that the system had more then just a 1% margin of error >.>

1/B= 1.01e-10

(1/B)*P= .00139

.00139*1.05= .00146 m^3

Yet, for some reason, it says it's wrong. >.>

 

[mgb_phys]

The simplest way is to just look at the units.
Bulk modulus has units of pressure.
The volume obviously has units of volume and the answer you want has units of volume.
You don't expect the volume of a metal ball to change very much so you are looking for a small number.

So it must be something like (bulk modulus/pressure)*volume.
(9.9 X 10^9 N/m^2/1.38 x 10^7 N/m^2) * 1.05 m^3 = 0.075 m^3 (check units balance)

Working it out this way is a lot more reliable than trying to remember the formula .

 

[ultrapowerpie]

Ok, when I tried doing it your way, I'm getting 753.26 m^3, not .075 m^3, did I do something wrong?

 

[mgb_phys]

After what I said about wanting a small number I typed them the wrong way up!
(1.38 x 10^7 N/m^2/9.9 X 10^9 N/m^2) * 1.05 m^3 = 0.0015 m^3

You always have to be careful with calculators and big/small numbers, so do an estimate
1.38 x 10^7 /9.9 X 10^9 = 1.38/9.9 x 10^(7-9) = (1.4/10) * 10^-2 = 0.14/100 = 0.0014