In this experiment we had a bucket full of water and had to determine the size of a hole that was drilled into the bottom of the bucket. We would not find the size by simply measuring it, but instead we would determine its size by using Bernoulli's equation.
We had the bucket filled up 3 inches high with water and timed how long it took for 16 oz. of water to run out of the bucket. We did this repeated this process 6 times.
Trial 1: 40.68 +/- 0.30s Trial 2: 41.16 +/- 0.30s Trial 3: 41.77 +/- 0.30s
Trial 4: 41.65 +/- 0.30s Trial 5: 42.94 +/- 0.30s Trial 6: 42.70s +/- 0.30s
We had two people with stopwatches time our experiment and averaged their times. The biggest variation between their times was 0.30s so we will use that as our uncertainty.
The volume emptied theoretically was 16oz/99.883 oz/ft^3 which is 0.016ft^3 which is the same as 453.07mL. The area of the drain hole is πr^2 = π(0.0025m)^2 = 1.963 * 10^-5m^2.
the height of the water 3.0 in. which is 0.0762m. The theoretical time it would take for it to empty is expressed by this equation. t = V/A(2gh)^1/2 when the values are put in we get that 19.71s
Percent Error: Trial 1: 51.5% Trial 2: 52.1% Trial 3: 52.8% Trial 4: 52.7% Trial 5: 54.1% Trial 6: 52.70%
.051 +/- .005
We had a drastic percent error through all of our trials. This was because when we measured our water in the bucket, we filled the bucket up to 3 in. We should have filled the bucket up 3 in. above the hole. This would speed the rate of the water flowing out of the bucket and would have given us a more accurate reading.
Using the formula above and solving for the radius we get r=(V/πt(2gh)^1/2) ^1/2
When we input our numbers we can determine that r = 0.00168m +/- 0.0008m,
so d is .00336m +/- 0.0008m
The given value for the diameter was 0.005m
Our percent error from calculating the radius was 48.83%. Our error is from reusing the same water from each experiment along with some of the water spilling. It is also from filling up the bucket to the incorrect volume.
The volume emptied theoretically was 16oz/99.883 oz/ft^3 which is 0.016ft^3 which is the same as 453.07mL. The area of the drain hole is πr^2 = π(0.0025m)^2 = 1.963 * 10^-5m^2.
the height of the water 3.0 in. which is 0.0762m. The theoretical time it would take for it to empty is expressed by this equation. t = V/A(2gh)^1/2 when the values are put in we get that 19.71s
Conclusion
Percent Error: Trial 1: 51.5% Trial 2: 52.1% Trial 3: 52.8% Trial 4: 52.7% Trial 5: 54.1% Trial 6: 52.70%
.051 +/- .005
We had a drastic percent error through all of our trials. This was because when we measured our water in the bucket, we filled the bucket up to 3 in. We should have filled the bucket up 3 in. above the hole. This would speed the rate of the water flowing out of the bucket and would have given us a more accurate reading.
Using the formula above and solving for the radius we get r=(V/πt(2gh)^1/2) ^1/2
When we input our numbers we can determine that r = 0.00168m +/- 0.0008m,
so d is .00336m +/- 0.0008m
The given value for the diameter was 0.005m
Our percent error from calculating the radius was 48.83%. Our error is from reusing the same water from each experiment along with some of the water spilling. It is also from filling up the bucket to the incorrect volume.
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