question form

#$&*

Phy 122

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** Question Form_labelMessages **

Question 5 revisions

** **

** **

** **

question form

#$&*

Phy 122

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** Question Form_labelMessages **

Question 6

** **

** **

The answers below are from the questions you posed to me from the experiment raising atmospheric pressure I.

** **

How much additional pressure does it take to support a column of water 52 cm high?

&&&&&&

P = rho g y = 1000 kg / m^3 *9.8 m/sec^2 * 0.52 m = 5096 N/M^2 = 5096 Pascals.

&&&&&&&

This can be answered most simply with the relationships of Introductory Problem Set 5, Problem 1. It can also be answered using Bernoulli's equation.

*@

By what percent did the length of the air column change when you squeezed the bottle?

25%

Can you use the answers to these two questions to estimate atmospheric pressure?

&&&&&

Atmospheric pressure I guess would be similar to the 0.25 * 5096 = 1274 Pascals

@&

5096 Pa would be 25% of atmospheric pressure. Atmospheric pressure would not be 25% of 5096 Pa.

*@

&&&&&

@&

The pressure in the air column was originally atmospheric pressure. The pressure then changed by around 25%.

So the change in pressure would have been around 25% of atmospheric pressure.

*@

&&&&&

I reset up the experiment and the air column still changed by 15%

The air pressure would then change by 0.15 * 5096 = 764 Pascals.

That would then be added to the 5096 Pascals to get 5860 Pascals.

&&&&&

@&

Atmospheric pressure is greater than 25% of atmospheric pressure.

So if 25% of atmospheric pressure it 5000 Pascals, then atmospheric pressure must be greater than 5000 Pascals, not less.

&&&&&

@&

Your results are not in accordance with this note. If 25% of atmospheric pressure is 5000 Pa, then atmospheric pressure cannot be 1270 Pa. 25% of 1270 Pa is only about 300 Pa, not 5000 Pa.

*@

Revisions have been made.

&&&&&

*@

@&

You need answers to both of the preceding questions to answer this question. Check my notes on the first question.

*@

@&

See if my notes help. Either way, please submit another question form with either answers or additional questions.

*@

@&

See my note indicating why your results indicate that 25% of atmospheric pressure is about 5000 Pa.

Then see if you can revise your conclusion.

Your result will be less than accepted atmospheric pressure because the length of a confined air column will not change by 25% in response to the pressure increase required to raise water 50 cm. Most likely the air column wasn't completely confined, indicating perhaps a small leak in the capped end.

*@

&&&&&

I reset up the experiment and re-answered the question above at approximately 15%.

&&&&&

@&

Send me a revision of that one result, and/or additional questions. It should take you only a few minutes to complete this one minor revision (or if necessary to ask a question).

*@

@&

If 25% of atmospheric pressure is 5000 Pa, it does not follow that atmospheric pressure is 25% of 5000 Pa.

Check my notes and see if you can provide another revision. This time use #### to denote your insertions.

*@