TBR physics chp 7 (fluids and solids) pg 59

[athockey7]

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hey guys

i am really really struggling for some reason with the examples on pg 59......
i have NO IDEA what is going on even though I have wasted over an hour trying to figure these out.. there is some concept or link that I am missing..

if anyone could explain those (or even 1 of those) pressure problems I would greatly appreciate it.

thanks

this is the problem btw.

At a specific depth in a swimming pool, a barometer measures the total pressure to be twice that of atmospheric pressure. if the barometer is now submerged to a depth that is twice its initial depth, by how much does the total pressure increase?

A by 50%
B by 100%
C by 200%
D by 300%

answer in white: A

 

[SOOyC52H1Sof]

Since the pressure at the initial depth is twice atmospheric pressure, the pressure from the water is equal to 1 atm. The air constitutes the other 1 atm at STP (Combining to give you 2 atm).

If the inital depth is doubled, the pressure from the water is now 2 atm. Adding in your 1 atm from STP, you now have a total pressure of 3 atm.

So pressure increase is now:
[(final-initial)/initial x 100%]

(3-2)/2 x 100% = 50%

 

[keikoblue2]

The two questions are asking about the change in pressure as depth changes.
Total pressure = atmospheric pressure + (rho x g x h) <-- also know as gauge pressure (pressure relative to atmospheric pressure. Since pressure gauges usually read 0 for/ignore atmospheric pressure, we can think of gauge pressure as the pressure pertaining to a certain depth).

Imagine you're at the bottom of a 5m pool. If the pool is empty, only atmospheric pressure will be pushing down on you. However, if the pool is filled, you have 5m of water pushing down on top of you (a certain pressure value), in addition to the atmospheric pressure. The gauge pressure is thus the pressure of the water pushing down on you. You're experiencing more pressure in the second scenario because there are more things weighing down on top of you.

For the next question (7.2b), it's asking for the pressure change again. When there's a problem involving change, there's an initial situation and a final situation. The initial situation is that the swimmer is at the bottom of the pool (doesn't matter how deep because at the end, he rises to the surface). The gauge pressure (rho x g x h) is 0.5 atm. The total pressure for the initial situation is that the swimmer has the water pressure and atmospheric pressure pushing down on him. Total Pressure = Patm + (rho x g x h) = 1 + 0.5 = 1.5 atm.

In the final situation, the swimmer rises to the surface. Now, he doesn't have a column of water on top of him, just air, so only atmospheric pressure is pushing down on him. Total pressure = Patm + (rho x g h) = 1 + 0 = 1 atm.

By how much does the total pressure change = what's the percent increase or decrease? The pressure got smaller so it must be a decrease in pressure, eliminating choices A and B. Plug into the eqn: (final - initial)/initial -> (1 - 1.5)/1.5 = -0.5/1.5 = -1/3 = -33%. The negative sign means the pressure decreased by 33%.

 

[cheetoh]

Your equation is the pressure-to-depth relation

P(total) = P(atm) + p * g * h

Gauge pressure is directly proportional to height; double the height --> double the gauge pressure.

2 = 1 + 1 (Given)
X = 1 + 2 (Find new pressure total) X = 3

3/2 = 150% --> 50% increase