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course PHY 202
1/19 3
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 #$... prompt):
1000N/m^2 to 10,000N/cm^2 times 40 =400,000N/cm^2
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Consider:
A 1 cm cube with mass 1 gram weighs about 1000 dynes or .01 Newtons, and the pressure on its base is 1000 dynes / cm^2 or .01 N / cm^2, or 100 N / m^2.
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• If the same squeeze causes the pressure in the air column of a pressure-indicating tube (one with a plug of water and a closed end) to increase by 5%, then what is the pressure of the atmosphere?
**** Your response (insert your response beginning in the next line; the next line is blankd and doesn't include the #$... prompt):
1.05 atms
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• 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):
1.07 amts
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The pressure in the air column would be 1.07 atmospheres. That's an increase of .07 atmospheres over the original 1-atmosphere pressure.
What is the pressure of a 40 cm column of water?
According to the information given here that would be equal to .07 atmospheres.
What therefore is the pressure of 1 atmosphere?
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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:
We squeezed the bottle to get it to the top and then using the same squeezed (estimated) we got the air column to rise 10 cm. The original column was at 40. So we estimated our atm pressure to be 1.25.
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You need to base your result on the data you have indicated, as well as on the original length of the air column, and the change in its length.
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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:
The how much weight it takes to hold up one cm^3 of water, if you had a 10 cm stack of water, it would take 10 * 1000 dynes = 10,000 dynes
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What you say is correct, and that knowledge is useful, but it doesn't answer the question.
How do you get the weight of a given mass?
What is the mass of 1 cm^3 of water?
What therefore is 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:
because 100pa*100cm =10,000pa
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A pascal is a N / m^2. A pascal multiplied by a cm is a N * cm / m^2, which is not a N / m^2 and therefore not a Pascal.
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However, 100 Pa for every cm of height can be expressed as a Pa per cm of height, and when multiplied by 100 cm of height you get 100 Pa / cm * 100 cm = 10 000 Pa.
So, except for the units, you have explained how the 10 000 Pa result follows from the 100 Pa result.
Now can you explain how we get the 100 Pa result?
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The weight a one-centimeter cube of water is .01 Newton, as you were asked to show in a previous problem, and the area of its base is .0001 m^2. What therefore is the pressure at the base of that one-centimeter cube?
How does this result lead to an answer to the given question?
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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:
rising the level of water from 40 to 50 cm changes the amount of dynes to 40*1000=40,000 to 50*1,000=50,000 dynes
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40 cm^3 is the amount of water being raised, not a height. Height wouldn't be measured in cm^3, which is a unit of volume.
However you do have the right idea.
How much would 40 cm^3 of water weigh?
How much work would it therefore take to raise the water 50 cm?
Note that the weight of the water is in dynes, and when you multiply a force in dynes by a distance in cm you get dynes * cm = ergs.
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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:
4.18J*300*50=62,700J It is not specified how high the water was raised in cm or m, so I do not think I have enough information to tell how the gravitational PE because it could be at different highs and gravitational PE depends on height.
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I believe the problem states that the water is raised to a height of 50 cm. The amount of water raised in 40 cm^3. (see also the preceding problem)
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Considering that you had to miss Wednesday's class you've made a very good attempt on this document.
I've inserted a number of notes to help you clarify some of the calculations and concepts.
You are welcome to submit a revision, which would allow me to give you more help, should you need it.
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Be sure to include the entire document, including my notes. &#
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