course phy 202
This is a reply to 1/29/07
Here is the data from the experiement the other day that I emailed you but failed to copy and paste. Hope it turns out ok! Coffee Stirrer Length
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11.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
19 5 5
14.4 10 5
11.3 15 5
8.8 20 5
6.1 25 5
4.3 30 5
2.9 35 5
2.1 40 5
1.5 45 5
1.3 50 5
1 55 5
Coffee Stirrer Length
11.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
18.1 5 5
14.3 10 5
11.4 15 5
8.4 20 5
6.4 25 5
4.4 30 5
2.8 35 5
1.7 40 5
1.3 45 5
0.4 50 5
Coffee Stirrer Length
10.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
18.2 5 5
14.5 10 5
11 15 5
8.6 20 5
6.3 25 5
4 30 5
2.8 35 5
2 40 5
1.5 45 5
0.6 50 5
Coffee Stirrer Length
9.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
17.9 5 5
14.3 10 5
11.5 15 5
8.1 20 5
6 25 5
4.2 30 5
2.6 35 5
1.8 40 5
1.4 45 5
Coffee Stirrer Length
8.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
17.4 5 5
13.5 10 5
9.8 15 5
7.2 20 5
4.6 25 5
3.2 30 5
2.1 35 5
1.5 40 5
1.1 45 5
Coffee Stirrer Length
7.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
16.5 5 5
13.3 10 5
9.7 15 5
6.8 20 5
4.7 25 5
3.1 30 5
2.1 35 5
1.4 40 5
1.2 45 5
Coffee Stirrer Length
6.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
17.2 5 5
13.5 10 5
10.1 15 5
7.1 20 5
4.5 25 5
3 30 5
2 35 5
1.4 40 5
Coffee Stirrer Length
5.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
16.8 5 5
13.9 10 5
10.3 15 5
7.6 20 5
5 25 5
3.2 30 5
2.3 35 5
1.6 40 5
Coffee Stirrer Length
4.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
17.3 5 5
13.4 10 5
10 15 5
7.1 20 5
4.4 25 5
3.2 30 5
2.1 35 5
1.5 40 5
Coffee Stirrer Length
3.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
17.3 5 5
12.9 10 5
9.5 15 5
6.5 20 5
4.2 25 5
2.7 30 5
1.8 35 5
1.5 40 5
Coffee Stirrer Length
2.1cm Altitude with respect to out flow of hole Clock time Time interval
22.5 0 5
16.8 5 5
12.5 10 5
8.3 15 5
5.4 20 5
3.3 25 5
2 30 5
1.8 35 5
1.3 40 5
""
This looks like good data.
You now need to analyze the data as we did in class. Begin by determining the volume change, exit velocity, PE change and KE change for a single interval of your choice.
** Have you analyzed this data? **
course phy202
Here is the data from the experiement friday.
1st Trial
14.7 68.5
2nd Trial
Change in X Change in Y
13.9 110.3
3rd Trial
Change in X Change in Y
14.7 70.1
** You need to define a little more clearly what x and y mean so 100 years from now somebody can look back at this data and marvel at our ingenuity.
From your note below it appears that your reported 'change in x' is not actually a change in any quantity. My best guess is that your first column is the length of the air in the tube, and the second is the height of the supported water column. **
The intial length of the air in the tube was 9.7cm. The inital length of the air in the tube was 16.2. To get the change in x, I took 16.2-1.5 because that was the change in x for the first trial. I then repeated that for each trial using the changes 1.5, 2.3, and 1.5. (I forgot to put this in my table.)
Each change in y had to have 23 added to it because that was the amount that the water had to travel from out of the bottle to the top of the tube where I was actually measuring from.
Therefore each y actually started out at 45.4, 87.3, and 47.8 respectively.
The procedure was that we had a bottle filled with water and it was sealed. The only exit was the two tubes we were measuring. The y tube went straight up away from the bottle and the x tube went perpendicular to the bottle. Someone would squeeze the bottle while two other people measured the movement of the water from both the x and the y tubes. A harder squeeze would constitute a greater change in x and y. This is because the pressure on the water in the bottle increased. This forced is to go out. ""
So I know that length*cross sectional area=volume but I cannot remember the height of the cylinder. The density of water is 1000kg.m^3. Density is equal to mass over volume. So lets assume that the length of the cylinder is .25 meters and the c.s. area is .0025m^2. So .0025m^2*.25m=.000625m^3 which is the volume. And then you would take the 1000kg/m^3=mass/volume
1000kg/m^3*.000625m^3=mass =.625kg= 625grams. Would I now use the formula pv=nrt?"
** The c.s. area is the area of a circle 1/8 inch in diameter. That is not .0025 m^2, which is about the cross-sectional area of a child's wrist.
Start by using Bernoullil's Equation, not PV = n R T, to figure out the pressure change from the water surface to the top of the rising tube.
Then calculate the pressure in the air column. That one uses PV = n R T. I see that you have done this part correctly in your next submission. **