# What is the pressure at 80 m below the surface of the ocean.

1. Apr 1, 2012

### ScienceGeek24

1. The problem statement, all variables and given/known data

a) A sailor uses a rope to lower an iron ball of radius 30 cm a to a depth of 80 m below the surface of the ocean. What is the pressure (in atm) at that depth? b) Find the tension in the rope in the previous problem. The density of iron is 7.86 *103 Kg/m3.

2. Relevant equations

For part a) p=P0+ρGd

1 atm= 1.013*105 Pa

ρsea water= 1.03*103 kg/m3

For part b) Ʃy= F buoyant-mg-T=0

F buoyant=ρVg

ρ= m/V

3. The attempt at a solution

In part a) p=(1.013*105 Pa)+(1.03*103 kg/m3)(9.8 m/s2)(80m)= 9.08*105 Pa

9.08*105 Pa / 1.013*105 Pa = 8.97 atm

However, the answer is 8.74 atm what did I do wrong??

In part b) I solved for T=F buoyant-mg

F buoyant= (7.86 *103 )(4π(0.3m)3/3)(9.8m/s2)= 8.70*103 N*m

For mg I used m=ρV = (7.86 *103 )(4π(0.3m)3/3)= 8.70*103 N*m

So T= 8.70*103 N*m - 8.70*103 N*m =0 ?????? The answer for part (b) is 7600 N :uhh: Help????

Last edited: Apr 1, 2012
2. Apr 1, 2012

### Nessdude14

You used 1030 kg/m3 for the density of the seawater. Could they have wanted you to use the density of water with no salt (1000 kg/m3)? The density of seawater varies quite a bit, depending on how much salt is in that part of the ocean. If they didn't specify the density of the water other than saying that it's "the ocean," I would guess they want you to use the standard density for water.

3. Apr 2, 2012

### ScienceGeek24

You were right! THANKS MAN!! I could solve both of them! thanks again!