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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 #$... prompt):
pressure=1g/cm^3*(980cm/s^2)*40cm
39200 dyne/cm^2
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• If the same squeeze causes the pressure in the air column of a pressure-indicating tube (one with a plug of water and a closed end) to increase by 5%, then what is the pressure of the atmosphere?
**** Your response (insert your response beginning in the next line; the next line is blankd and doesn't include the #$... prompt):
If atmospheric pressure is about 100000 pascal, then it would be 105000 pascal or 1.05 atm.
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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 #$... prompt):
1.5cm is 6.25% of a 24cm column. So, atmospheric pressure would be about 106250 kg/(m/s^2).
#$&* (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 with my eyes close several times as close as I could get to being the same squeeze. After each squeeze I opened my eyes to see how close I was to the previous squeezes. I continued this until they were fairly consistant. With an uncapped hose the water reached 50cm. With approximately the same squeezed on a capped hose the water plug was at 60cm high and it moved 2.5 cm. This is around a 4% change. Atmospheric pressure is around 100000 kg/(m/s2) so the pressure would have been about 104000 kg/(m/s2)
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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:
Weight is mass*force of gravity, which is around 1000 cm/s2 or 10 m/s2. 1cm^3*1000cm/s2= 1000 dyne. .001m^3*10m/s2= .01 newton
@&
That would be mass * acceleration of gravity.
mass * force doesn't give you a force, while, as you know, mass * acceleration does.
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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:
Density of water is about 1000 kg/m^3
1000 kg/m^3*(10m/s2)*1m= 10000 kg/(m/s2)= 10000 Pascal
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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:
Work= 40g*980cm/s2*50cm= 1960000 erg or 0.196 joules. The change in gravitational PE of the water is -196000erg or -0.196 joules
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196 000 ergs, as in your last answer, not 1 960 000 ergs as in your first. I understand that was just a typo.
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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:
300g*(4.18J/gc)*50*C= 62700 Joules heating water
.196 Joules raising water
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&#Your work looks very good. Let me know if you have any questions. &#