# 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 1.013x10

^{5}P (Pascal= 1N/m

^{2}) 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__

Answer is Negative.

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 =*. If you have a square like I do the

**0.1342 m***Area = side*. If you have another shape find the area for that shape. So let's do that

^{2}*0.1342*.

^{2}=**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 P

_{atm}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 = ρ**. Now we need to find the

_{water}*g*h**h**so the formula becomes

**h = P/(ρ**(

_{H2O}*g)*Sheet 14*). If it was for mercury we would use the density for mercury which is

*ρ*. The formula for that would look something like this

_{Hg}= 13600*h = P/(ρ*.

_{Hg}*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/in

^{2}) = 6.89x10

^{3}N/m

^{2}) 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:

*P*. That's all it is asking.

_{atm}+P_{gauge}> P_{gauge}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 P

_{atm}+P

_{tire gauge}

P

_{abs}= P

_{atm}+P

_{tire gauge}

P

_{abs}= 101300+212487.6

P

_{abs}=

- 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 = ρ**. 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.

_{H2O}*g*hP = ρ

_{H2O}*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)

P

_{abs}= 17599.14 + 101300

P

_{abs}=

- 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

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

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

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

## question3

answer is positive

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

1.9564 + 1atm = 2.9564 as your answer

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

1.9564 + 1atm = 2.9564 as your answer

**guest111**- 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

The CD's answer was 2.901041

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

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

The CD's answer was 2.901041

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

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

Page

**1**of**1****Permissions in this forum:**

**cannot**reply to topics in this forum