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

[ScienceGeek24]

Homework Statement

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.

Homework Equations

For part a) p=P₀+ρGd

1 atm= 1.013*10⁵ Pa

ρ_{sea water}= 1.03*10³ kg/m³

For part b) Ʃy= F _buoyant-mg-T=0

F _buoyant=ρVg

ρ= m/V

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 Help?

 

[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.

 

[ScienceGeek24]

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