Your work on flow experiment has been received. After the due date we will continue analysis of the results of this experiment, using your results and those obtained from other students.
i would expect the rate of flow to decrease as water flows from the cylinder.
i would expect the velocity of the water surface would decrease as the water flows out of the cylinder.
with the diameter of the cylinder and hole you can find the C.S. area for the cylinder and hole. then with the velocity of the exiting water you can find the velocity of the water surface. to find the velocity you take (V_exit times A_exit)/A_cylinder and that will find the velocity of the water surface.
because if the velocity is not changing then the water depth will moving at the same rate as the water exiting. and the C.S. area of the hole and the cylinder are not the same so that can not happen. a force is what makes the water move faster coming out of the hole.
I think the nature of the force is the pressure that the water has on the hole because the hole is so small and their is so much water wanting to come out of it.
the depth is changing a a slower and slower rate
the slope will decerease as the clock time increases. the graph will look like a y=1/x(squared) graph
the horizontal distance traveled by the stream decreases as time goes on
i think the distance changes at a decreasing rate
i think the graph will look like the depth vs. clock time graph. it will also look like a y=1/x(squared) graph
2.7 cm tube length
0
1.19
1.86
1.84
1.89
2.20
2.25
2.73
2.71
3.48
3.89
6.13
1.73
3.46
5.19
6.92
8.65
10.38
12.11
13.84
15.57
17.30
19.03
20.78
0,20.78
1.19,19.03
3.05,17.30
4.89,15.57
6.78,13.84
8.98,12.11
11.23,10.38
13.96,8.65
16.67,6.92
20.15,5.19
24.04,3.46
30.17,1.73
the depth is changing at a slower and slower rate
the graph is decreasing at an exponitional rate. like a y=1/x(squared)
I take (v2-v1)/(t2-t1) to find the average velocity
-1.47
-.93
-.94
-.92
-.78
-.77
-.63
-.64
-.50
-.44
-.28
i took the (t1+t2)/2
.595
1.525
1.85
1.865
2.045
2.225
2.49
2.72
3.095
3.685
5.01
.595,-1.47
1.525,-.93
1.85,-.94
1.865,-.92
2.045,-.78
2.225,-.77
2.49,-.63
2.72,-.64
3.095,-.50
3.685,-.44
5.01,-.28
if i use the negative number for the velocity then the graph is increase to zero and the slope is decreaseing as the clock time is incereasing. Some of the numbers are not in the line but spread out around a line that would best fit the points.
i took the (average velocity)/(clock time)
-2.47
-.61
-.51
-.49
-.38
-.35
-.25
-.24
-.16
-.12
-.06
.595,-2.47
1.525,-.61
1.85,-.51
1.865,-.49
2.045,-.38
2.225,-.35
2.49,-.25
2.72,-.24
3.095,-.16
3.685,-.12
5.01,-.06
if you dont include the first acceleration time then the data shows that the acceleration of the water surface is decreasing.
i think the acceleration of the water surface is actually decreasing as the data shows.