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course Phy 202
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 #$... prompt):
P=(rho)(gravity)(height)
P=(1000kg/m3)(9.8m/s2)(0.4m) = 3920 N/m2
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• If the same squeeze causes the pressure in the air column of a pressure-indicating tube (one with a plug of water and a closed end) to increase by 5%, then what is the pressure of the atmosphere?
**** Your response (insert your response beginning in the next line; the next line is blankd and doesn't include the #$... prompt):
3920 x 0.05 = 196, so 3920 - 196 = 3724 N/m2 = P or 0.0368 atm
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In this situation the 3920 N/m^2 would be 5% of atmospheric pressure.
So .05 P_atm = 3920 Pa. You would solve this equation to get P_atm.
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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):
Air column changes by 6.25%. Protocol similar to above problem. P = 3675 N/m2 or 0.0363 atm
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This should be set up in a manner similar to the preceding (see the preceding note).
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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:
I squeezed the bottle with the end of the tube capped. The water raised 15cm of the total 80cm of the tube. That was about a 19% change. Therefore, we assumed atmospheric pressure was about 1.2 atm.
Also in pascals, I calculated 10,000 Pa / 0.19 = 52631.6 originially.
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You need to also use the height to which the same squeeze raised the water with the vertical tube uncapped.
Then you would be able to calculate the excess pressure in Pascals, based on that height, and set it equal to .19 of atmopheric pressure, as in my preceding notes.
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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:
Mass = 1g x 1kg/1000g = 0.001kkgx 10m/s2 = 0.01 N
9.8m/s2 x 100cm/1m = 980g x cm/s2 ~1000 dynes
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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:
P = (1000kg/m3)(9.8m/s2)(0.01m) = 98 N/m2 ~ 100 Pa
P = (1000kg/m3)(9.8m/s2)(1m) = 98000 N/m2 ~ 10,000 Pa
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Right calculations, but be sure you understand the units.
(1000kg/m^3)(9.8m/s^2)(1m) = 9800 kg / (m s^2).
This is the same as 9800 (kg m / s^2) / m^2, as you can verify, and this is the same as 9800 N / m^2 or 9800 Pa.
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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:
deltaPE = weight x deltaY so
deltaPE = (0.4N)(0.5m) = 0.2 Nxm
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Good.
Note on notation: use * for multiplication, not x.
Then note as well that N * m = Joules.
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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:
(4.18 Joules / 1g x 1 C) x (300g x 50 C) = 62,700 Joules
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Very good, but check my notes on the atmospheric pressure situation. You didn't quite set that one up correctly.
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