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course Phy 122
10:21 PM1/21/2014
1. Go to the page more_communication.htm and do as instructed there. This should not take you long.
2. Go to the document Preliminary Q-A-Self_Critique and do as instructed. This should take no more than 15 minutes or so.
3. Copy the questions in the indented section below into a text editor (e.g, notepad for Windows or whatever text editor is available on the device you use).
Answer the questions. I suggest you work out your answers using pencil and paper, then insert a brief synopsis of your work the copy in your text editor.
Submit your completed document using the Submit Work Form.
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1. By squeezing the bottle I raise a column of water 40 cm.
• What pressure is required to support that column?
**** Your response (insert your response beginning in the next line; the next line is blankd and doesn't include the #$... prompt):
3920 N/m^2
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• If the same squeeze causes the pressure in the air column of a pressure-indicating tube (one with a plug of water and a closed end) to increase by 5%, then what is the pressure of the atmosphere?
**** Your response (insert your response beginning in the next line; the next line is blankd and doesn't include the #$... prompt):
4116 N/m^2
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You don't show how you got this, but 5% pf 4116 N/m^2 is not 3920 N/m^2.
3920 N/m^2 should be 5% of atmospheric pressure, to the extent your data are correct. Using this information, you in fact do get a pretty good approximation of accepted atmospheric pressure.
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#$&* (Note that your response was to go into 'the next line'; your response will therefore be inserted before this line, not after. This is obvious when you're looking at the form, but if you've copied the form into a text editor it might be less obvious. Hence this note.)
• If the air column in the pressure-indicating tube has original length 24 cm, and the same squeeze that raised water 40 cm caused the length of the air column to change by 1.5 cm, then what is atmospheric pressure?
**** Your response (insert your response beginning in the next line; the next line is blankd and doesn't include the #$... prompt):
1.039 atm
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1.039 atm is 1.039 times atmospheric pressure, and is not equal to atmospheric pressure.
It is in any case unclear where 1.039 came from.
By what percent did the pressure in the air column change? This is directly related to the percent by which its volume, and hence its length, changed.
If you set the change in pressure, in atmospheres, equal to the 3920 N/m^2 pressure of the 40 cm water column, you will be able to solve to get the pressure of the atmosphere in N/m^2.
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#$&* (Note that your response was to go into 'the next line'; your response will therefore be inserted before this line, not after. This is obvious when you're looking at the form, but if you've copied the form into a text editor it might be less obvious. Hence this note.)
2. What were your data for the experiment done in class today, and what did you get for atmospheric pressure. Explain how conducted the experiment and how you obtained your result for the pressure.
**** Your response:
2.3 N/m^2, Data obtained from water experiment done in class
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The question asked for your data, how the experiment was conducted, and how you obtained your result for pressure.
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3. The weight of 1 cm^3 of water is about 1000 dynes or .01 Newton. Explain how these results were obtained.
**** Your response:
It's the mass unit per volume unit
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That would be density, which is 1 gram / cm^3 or 1000 kg / m^3.
You know from this that the mass of 1 cm^3 of water is 1 gram. How did we determine that a 1 gram mass has weight 1000 dynes or .01 Newton?
Given a mass, how do we determine its weight?
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4. A column of water requires a pressure at its base of 100 Pa for every cm of height, or 10 000 Pa for every meter of height. Explain how these results were obtained.
**** Your response:
It's the amount of pressure per unit of height
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I believe we divided the weight of a column of 1 cm^3 water "cubes" by the area of the base on which they rested. Do you recall the details of how this was worked out?
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5. How much work does gravity do on 40 cm^3 of water as it is raised to a height of 50 cm? What therefore is the change in the gravitational PE of this water?
**** Your response:
0.196 J
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This is correct.
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6. Approximately 4.18 Joules of thermal energy are required to raise the temperature of 1 gram of water by 1 degree Celsius. If the temperature of 300 grams of water is raised from 20 Celsius to 70 Celsius, then poured over a bottle in order to raise 40 cm^3 of water to a height of 50 cm, how does the thermal energy required to raise the temperature of the 300 grams of water compare to the change in the gravitational potential energy of the raised water?
**** Your response:
Thermal energy is 0.5007 J more
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You don't indicate how you got this.
However each gram of water requires over 4 Joules for every Celsius degree, and there are many grams of water, compounded by many Celsius degrees of temperature change.
In other words, 1 gram of water raised by 1 Celsius would required over 4 Joules. 300 grams of water would require 300 times this, and the 50 Celsius rise would require many times that result.
The correct result is between 50 000 and 100 000 Joules.
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See Introductory Problem Set Set 5, Problem 2. Solve the problem without looking at the solution, or give it your best try. Then look at the solution and the Generalized Solution. Be sure you can solve a problem of this nature on your own.
See also:
Set 5, Problem 10. You probably won't be able to solve this problem without information given in the solution, which introduces the fairly simple concept of a temperature gradient. However by understanding the solution you will hopefully understand everything. We can of course discuss this concept in class, after you've had a chance to think about it.
Set 5, Problem 12. In a manner similar to the preceding, this problem also introduces a fairly straightforward concept.
You should know from previous science courses that P V = n R T, where P is the pressure, V the volume, n the number of moles of gas and T the temperature of a confined gas. If n and T are constant (i.e., if the number of moles and the temperature are both constant, as might be the case if you squeeze a capped bottle), the product P V must be constant. This means that if P goes up, V must go down, and vice versa. PV = constant is also very specific about just how P and V vary. For example, if V doubles, it follows that to keep the product PV unchanged, P must become half as great. Just as halving is the reciprocal of doubling (i.e., 1/2 is the reciprocal of 2), if V changes by any factor, P must change by the reciprocal factor. If you wanted to double the pressure in that container, you would have the squeeze it down to half its volume.
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You're on the right track, but you are missing some details. Check my notes and see if you can make the necessary modifications.
We will of course discuss this in class tomorrow.
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