# Ch 11.1 Help

## Ch 11.1 Help

Question 1
Calculate the force exerted by the atmospheric pressure on an area of your body which is a square 13.42 cm on a side. Use 1.013x105 P (Pascal= 1N/m2) for the atmospheric pressure (see sheet 3,7). Indicate with a positive (negative) sign whether the force is small compared to (similar to) your body weight.

How To Solve

We need to use the formula P = F/A where P is the pressure, F is the Force, and A is the Area the force is acting on. Since we are looking for the Force we have to rearrange our formula which becomes F = P * A.

First we have to do some conversions, since neither the Area is given and it is not in S.I. Units.
Our cm should be in m so let's convert that: 13.42 cm *1000 = 0.1342 m. If you have a square like I do the Area = side2. If you have another shape find the area for that shape. So let's do that 0.13422 = 0.01800964.

Now we have the pieces for the formula. A = 0.01800964 and the pressure is that for the atmosphere, which is P = 101300, and just insert them into the formula.
F = 101300 * 0.01800964
F =
Spoiler:
1824.376532

Question 2
What is the height of a water column which produces a pressure equal to 1.0083 times the atmospheric pressure (see sheet 4,5,14) ? Indicate with a positive (negative) sign whether this height is smaller (larger) for Mercury.

How To Solve
Answer is Positive. Since the density of Mercury is larger (and multiplying it by gravity makes it larger), dividing it into the Patm would create a smaller number.

It is not exactly clear here where all our information is coming from. So let's think that through. We need the height of a fluid and we know that we can find the pressure by using the formula P = ρgh. The pressure is equal to ρ which is the density, g which is gravity and h and that's our height. Water is our fluid so we need to use the density of water. I'm not too sure if the CD gives it too clearly but it's actually pretty easy to remember it's just 1000g/cm (I believe are the correct units, but it is 1000). Anyway think of our formula as follows P = ρwater*g*h. Now we need to find the h so the formula becomes h = P/(ρH2O*g) (Sheet 14). If it was for mercury we would use the density for mercury which is ρHg = 13600. The formula for that would look something like this h = P/(ρHg*g).

Now our pressure is NOT not at sea level, or standard. My pressure (given in my equation) is 1.0083 times that of the standard pressure. So I'm just going to take the standard pressure which we know from Question 1 is 101300 and multiply it. 101300*1.0083 = 102140.79. This will be the pressure (P) in our equation.

h = 102140.79/(1000*9.81)
h = 102140.79 / 9810
h =
Spoiler:
10.4119052

Question 3
The gauge pressure in a car tire is 30.84 PSI (1 PSI (pounds per square inch, lb/in2) = 6.89x103N/m2) What is the absolute pressure in the tire in atmospheres (see sheet 7) ? Indicate with a positive (negative) sign whether the absolute pressure is always (never) larger than the gauge pressure.

How To Solve
Answer is Positive. Since we always have to add the atmospheric pressure on top of the gauge pressure it will always be larger. Look at it this way: Patm +Pgauge > Pgauge. That's all it is asking.

Again our units are not in S.I. so we have to convert. They give us the conversion here by telling us 1 PSI = 6890 atm. The pressure in my tire is 30.84 PSI and in atm that's 30.84 * 6890 = 212487.6 atm.

Our absolute pressure is just the pressure in the tire plus the atmospheric pressure so Patm+Ptire gauge
Pabs = Patm+Ptire gauge
Pabs = 101300+212487.6
Pabs =
Spoiler:
313787.6

Question 4

The container shown is open on the top and filled with water. The water height x shown is 1.794 m. What is the absolute pressure at point A (see sheet 10)? Indicate with a negative (positive) sign whether the pressure values measured at any point on the bottom of the container are all the same (are larger on the left end than on the right end of the bottom line).

How To Solve
Answer is Negative. The pressure is related to the height of the water regardless where the bottom point is, that is pressure is equal horizontally across at any vertical point. Does that make sense? [if it does not, please ask]

This question uses a lot of information we dealt with in the previous questions. Again we are finding the pressure relation to the height as we did in Question 2 so we can use the same formula P = ρH2O*g*h. Use the height given. The height in the smaller box, that is the area completely filled with water has no affect for what we are looking for.

P = ρH2O*g*h
P = 1000*9.81*1.794
P =
Spoiler:
17599.14

Since we need the absolute pressure we just add the standard pressure of 101300 to the pressure at the point we just calculated. This is because there is another force acting and we are at standard pressure (or it is rather assumed)

Pabs = 17599.14 + 101300
Pabs =
Spoiler:
118899.14

[w00t! Got a 4/4 the first shot. First time I got that]

Guest01

Posts : 133
Join date : 2008-09-19

## ques 3

I keep getting Number 3 wrong

The gauge pressure in a car tire is 30.97 PSI (1 PSI (pounds per square inch, lb/in2) = 6.89x103N/m2) What is the absolute pressure in the tire in atmospheres (see sheet 7) ? Indicate with a positive (negative) sign whether the absolute pressure is always (never) larger than the gauge pressure.

I did 30.97* 6890 = 213383.3
Pabs = 101300 + 213383.3 = 314683.3

and i put it as positive

Kathleen
Guest

## question 3

number 3 should be negative and the solution given is incorrect it should be a much smaller number according to the practice quiz on the c-d.

guest17
Guest

## Re: Ch 11.1 Help

That's interesting because MapleTA has mine marked correct for both the True/False question and the calculations

Guest01

Posts : 133
Join date : 2008-09-19

## Re: Ch 11.1 Help

Wait isnt 10_3 due tommrow and 11_1 on Wednesday??

j man
Guest

## Re: Ch 11.1 Help

Question number three isnt converted correctly... 1 psi is equal to .06805 atm. The question asks for the answer in atms and the atmospheric pressure in atms is only 1

guesto
Guest

## Question 3

Can someone please help with number 3 and give a step by step calculation for this I have tried it on my own and the method on here but nothing seems to work? I would appreciate some help.

angelbab
Guest

## Re: Ch 11.1 Help

yes tomorrow 11/10 ch 10_3 is due and 11_1 is due on wed.

guest17
Guest

## Re: Ch 11.1 Help

We are just trying to get ahead since physics labs start up again this week, and we don't want to have too much work!

kathleen
Guest

## question 3

Did anyone figure out how to do question 3? I am having a lot of trouble. If someone has figured it out could you please help?

sedwards
Guest

## Q3

Any ideas on Q3 guys?

hwilson
Guest

## question3

take your PSI # mine was 28.75 PSI * .06805 atm = 1.9564

guest111
Guest

## Re: Ch 11.1 Help

thank you! that works for ques 3 !

k
Guest

student
Guest

## info on #3

I was not comfortable with the accuracy of my calculation using the posts (The conversion # given was not being used). It was not accurately reflecting the answer given by the CD so I started to play with some numbers:
Looking on sheet 7, P_atm = 1.013x10^5
I then: (6.89x10^3)/(1.013x10^5) = 0.068016
My PSI was 27.95
27.95x0.068016 = 1.90105
1.90105 + 1 = 2.90105
Doing it by the posts I got 2.902. It has been my experience that if it is not accurate to the CD's answer something is not 100% correct.
Just thought you all might want to know

super mo
Guest

## Question 3 solved

After playing around with number 3, this is what I did, and it worked

First of all, the answer on maple TA is positive, in the Cd it is negative (possibly an error)

Well, this is what I did:

I had 32.09 PSI so I multiplied it by (6.89*10^3)= 221100.1 (gauge pressure)

I then added this to atmospheric pressure (1.013*10^5)= 322400.1 (absolute pressure)

I then divided this sum by (1.013*10^5) to get the absolute pressure in atmospheres.

answer= 3.182627 (and it is POSITIVE)

It worked for me, I hope it helps you all. Good Luck!

-B

B
Guest