Find pressure at depth of freshwater

  • Thread starter Thread starter cc2hende
  • Start date Start date
  • Tags Tags
    Depth Pressure
AI Thread Summary
To calculate the pressure at a depth of 100m while scuba diving, the equation P2 = P1 + pgd is used, where P1 is the atmospheric pressure at the surface. The density of freshwater is given as 1000 kg/m^3, and the air pressure at the lake surface is 1 atm or 1.013 x 10^5 N/m^2. The calculation results in P2 = 1,081,300 N/m^2, but confusion arises regarding the units of this answer. It's clarified that to convert the pressure to atmospheres, the result must be divided by the atmospheric pressure in N/m^2, ensuring consistent units throughout the calculation. Understanding the relationship between Newtons and other units is essential for accurate pressure measurement.
cc2hende
Messages
7
Reaction score
0

Homework Statement



You are scuba diving to reach a ship sunk at a depth of 100m. What is the pressure in atm at this depth? Density of fresh water is 1000kg/m^3 and air pressure at lake surface=1atm=1.013x10^5 N/m^2

Homework Equations



P2=P1+pgd
p=density

The Attempt at a Solution


P2=P1+pgd
P2=(1.013x10^5 N/m^2) + (1000kg/m^3)(9.8m/s^2)(100m)
P2=1081300
Its at this point that i get confused. I don't know what the units are for this answer.
 
Physics news on Phys.org
You have the correct answer for the pressure in N/m^2
If you are asked for the pressure in atmospheres then you must divide by atmospheric pressure in N/m^2
 
instead of 9.8 m/s^2 you should use 9.8 N/kg to get correct units

(kg/m^3)(N/kg)(m) will give you N/m^2
 
cc2hende said:

Homework Statement



You are scuba diving to reach a ship sunk at a depth of 100m. What is the pressure in atm at this depth? Density of fresh water is 1000kg/m^3 and air pressure at lake surface=1atm=1.013x10^5 N/m^2

Homework Equations



P2=P1+pgd
p=density

The Attempt at a Solution


P2=P1+pgd
P2=(1.013x10^5 N/m^2) + (1000kg/m^3)(9.8m/s^2)(100m)
P2=1081300
Its at this point that i get confused. I don't know what the units are for this answer.

Both terms in the equation have to have the same units. That is, the units of the two terms must be consistent. You already know the units of the first term. Just make sure that the units of the second term are consistent. It, of course, helps to know that Newtons are the same kg-m/sec2.
 
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...
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...
Back
Top