Calculating Water Level Rise in Low Pressure Zone | Atmospheric Pressure Effect

  • Thread starter Thread starter qspartan570
  • Start date Start date
  • Tags Tags
    Water Water level
AI Thread Summary
The discussion revolves around calculating the water level rise in a low-pressure zone at sea, where the pressure at the center is 64 mm-Hg lower than surrounding areas. The initial calculation converted this pressure difference to Pascals, resulting in 8505.27 Pa. Using the hydrostatic pressure equation, the calculated height of the water column was found to be 9.47 m. However, it was noted that the pressure difference alone suffices for the calculation, and the correct expected answer is 0.85 m, indicating a misunderstanding in the initial approach. The key takeaway emphasizes the importance of using the correct density for seawater in such calculations.
qspartan570
Messages
14
Reaction score
0

Homework Statement



Suppose that a zone of low atmospheric pressure (a "low") is at some place on the surface of the sea. The pressure at the center of the "low" is 64 mm-Hg less than the pressure at a large distance from the center. By how much will this cause the water level to rise at the center?



Homework Equations


p-p0= -p(density)gy



The Attempt at a Solution



First I converted 64 mm-Hg to Pa using 1.01X10^5 Pa/760 mm-Hg and I got 8505.27 Pa

p-p0=-p(density)gy
101325 Pa-8505.27 Pa=1000X9.8Xh
h=92819.73/9800=9.47 m

According to the book, the answer is 0.85 m
 
Physics news on Phys.org
qspartan570 said:

Homework Statement



Suppose that a zone of low atmospheric pressure (a "low") is at some place on the surface of the sea. The pressure at the center of the "low" is 64 mm-Hg less than the pressure at a large distance from the center. By how much will this cause the water level to rise at the center?



Homework Equations


p-p0= -p(density)gy



The Attempt at a Solution



First I converted 64 mm-Hg to Pa using 1.01X10^5 Pa/760 mm-Hg and I got 8505.27 Pa

p-p0=-p(density)gy
101325 Pa-8505.27 Pa=1000X9.8Xh
h=92819.73/9800=9.47 m

According to the book, the answer is 0.85 m

The pressure value that you've been given is already a pressure difference, so there's no need to invoke atmospheric pressure. All you need is the height of a column of water (seawater, which has a slightly different density than pure water!) that will produce that pressure at its base.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging 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