Your initial message (if any):
Is flow rate increasing, decreasing, etc.?
As time progresses the rate of flow would decreases. There is less volume or mass pushing down on the liquid in the bottom of the cylinder. The less amount to push down the slower the output.
Is the velocity of the water surface increasing, decreasing, etc.?
Velocity is how fast the objects position is changing. The initial velocity would therefore be zero. As the water flows the velocity of the buoy would increase. As the cylinder approaches empty the buoy's velocity would decrease and eventually stop again.
How would the velocity of the water surface, the velocity of the exiting water, the diameter of the cylinder and the diameter of the hole be interrelated?
The velocity of the water surface is based upon the water underneath the surface. The initial portion of exiting water will have the highest velocity because it has more volume pushing down on it. The diameter of the cylinder will affect how fast the water flows because it affect the amount of water. The smaller the diameter the faster the flow. If the hole of the cylinder is large water will flow faster. If the hole of the cylinder is small the water cannot leave in large amounts and therefore the exiting velocity is slower. The bigger the hole and the smaller the cylinder the faster the velocity. This velocity slows as it reaches emptiness.
Explain how we know that a change in velocity implies the action of a force:
The force must be exerted to do work. The change in velocity shows work by the water exiting the tube at an accelerated pace. Gravity is probably the force but I am not certain so I did not note this earlier. This might have more to do with the speed of exit and such than the volume.
Does the depth seem to be changing at a regular rate, at a faster and faster rate, or at a slower and slower rate
Assuming the pictures are taken at precise increments the water appears to be changing faster in depth as the water exits. The first to second picture has had less of a liquid loss compared to the liquid loss from the second to the third picture. The depth is changing at a faster and faster rate. This of course is going by the pictures.
What do you think a graph of depth vs. time would look like?
The independent variable or the time would be graphed along the x axis. The dependent variable or the depth would be graphed along the y axis. The line would start at the top left hand corner. As time increases the depth decreases. The line would decrease but a fast rate.
Does the horizontal distance (the distance to the right, ignoring the up and down distance) traveled by the stream increase or decrease as time goes on?
As time progresses the stream decreases in length.
Does this distance change at an increasing, decreasing or steady rate?
The distance changes in a increasing rate. As more time goes by the smaller and smaller the distance becomes.
What do you think a graph of this horizontal distance vs. time would look like?
The time is the independent variable graphed along the x axis. The dependent horizontal distance would be graphed along the y axis. The line would start at the top indicating a larger distance. As time progresses the horizontal distance decreases. The distance becomes smaller at a faster and faster rate.
The contents of TIMER program as you submitted them:
1 672.6953 672.6953
2 675.1016 1.40625
3 677.0469 1.945313
4 678.9766 1.929688
5 680.9375 1.960938
6 683.1094 2.171875
7 685.2891 2.179688
8 687.8203 2.53125
9 690.9375 3.117188
10 693.9844 3.046875
11 697.9609 3.976563
12 703.5625 5.601563
13 707.1328 3.570313
The vertical positions of the large marks as you reported them, relative to the center of the outflow hole
Tube to 30 mL = 2.2 cm
Tube to 50 mL = 4.4
Tube to 70 mL = 6.4
Tube to 90 mL = 8.4
Tube to 110 mL = 10.2
Tube to 130 mL = 12.1
Tube to 150 mL = 14.0
Tube to 170 mL = 15.9
Tube to 190 mL = 17.8
Tube to 210 mL = 19.7
Tube to 230 mL = 21.4
Tube to 250 mL = 23.2
As the cylinder is measured vertically from the bottom the 20 mL increments get smaller in size, the diameter must change.
Your table for depth (in cm) vs clock time (in seconds)
0, 23.2
1.40625,21.4
3.351563,19.7
5.281251,17.8
7.242189,15.9
9.414064,14.0
11.593752,12.1
14.125002,10.2
17.24219,8.4
20.289065,6.4
24.265628,4.4
29.876191,2.2
Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
The depth is changing at a slower and slower rate. The pictures did not depict this but I now see that the time for the water level to drop approximately two centimeters is longer after each interval.
Your description of your depth vs. t graph:
The independent x axis is the Clock time and the dependent variable is graphed along the y axis and it is Water Depth. The plots start in the upper left quadrant (#2) and decend at a slower and slower rate as the plots fall into the lower right quandrant (#4) in a diagonal fashion.
Your explanation and list of average average velocities:
Velocity is equal to the change in position divided by the change in time.
Velocity = Delta S/ delta T
V= 1.9 cm / 1.40625 sec
V= 1.351 cm/s
V= 1.7 cm / 1.945313 s
V= 0.8739 cm/s
V= 1.9 cm / 1.929688 s
V= 0.9846 cm/s
V= 1.9 cm / 1.960983 s
V= 0.9689 cm/s
V= 1.9 cm / 2.171875 s
V= 0.8748 cm/s
V= 1.9 cm / 2.179688 s
V= 0.8717 cm/s
V= 1.9 cm / 2.53125 s
V= 0.7506 cm/s
V= 1.8 cm / 3.117188 s
V= 0.5774 cm/s
V= 2 cm / 3.046875 s
V= 0.6564 cm/s
V= 2 cm / 3.976563 s
V= 0.5029 cm/s
V= 2.2 cm / 5.601563
V= 0.3927 cm/s
The midpoints of your time intervals and how you obtained them:
1.40625 --> 0.703125
1.945313 --> 0.9726565
1.929688 --> 0.964844
1.960983 --> 0.9804915
2.171875 --> 1.0859375
2.179688 --> 1.089844
2.53125 --> 1.265625
3.117188 --> 1.558594
3.046875 --> 1.5234375
3.976563 --> 1.9882815
5.601563 --> 2.8007815
Your table of average velocity of water surface vs. clock time:
1.351 cm/s,0.703125 s
0.8739 cm/s,0.9726565 s
0.9846 cm/s,0.964844 s
0.9869 cm/s,0.9804915 s
0.8748 cm/s,1.0859375 s
0.8717 cm/s,1.089844 s
0.7506 cm/s,1.265625 s
0.5774 cm/s,1.558594 s
0.6564 cm/s , 1.5234375 s
0.5029 cm/s, 1.9882815 s
0.3927 cm/s, 2.8007815 s
Your description of your graph of average velocity vs clock time:
The graph is decreasing with one slight peak at the the third and fourth point. This is probably an error. The graph slopes downward starting in the upper left quandrant and descending diagonally at a decreasing rate.
Your explanation of how acceleration values were obtained:
The average acceleration is determined by dividing the change in velocity by the change in time.
a= v2-v1 / t2- t1
a = 1.351 cm/s - 0.8730 cm/s / 0.9726 s - 0.703125 s
a = 1.7738 cm/s^2
a = 0.8730 cm/s - 0.9846 cm/s / 0.9726 s- 0.964844
a= -14.956 cm/s^2 ????
a= 0.9869 cm/s - 0.9846 cm/s / 0.9804915 s - 0.964844 s
a= .1469 cm/s^2
a= 0.9869 -0.8748 cm/s / 0.9804915 s - 1.0859375 s
a= -1.06 cm/s^2 ???
a= 0.8748 - 0.8717 cm/s / 1.0859375 - 1.089844 s
a= -0.7848 cm/s^2
a= 0.8717 -0.7506 cm/s / 1.089844 - 1.265625 s
a= -0.6856 cm/s^2
a= 0.7506 - 0.5774 cm/s / 1.265625 - 1.558594 s
a= -0.5912 cm/s^2
a= 0.5774 - 0.6564 cm/s / 1.558594 - 1.265625 s
a= -0.2696 cm/s^2
a= 0.6564 - 0.5029 cm/s / 1.5234375 - 1.988281 s
a= -0.1699 cm/s^2
a= 0.5029 -0.3927 cm/s / 1.9882815 - 2.8007815 s
a= -0.1356 cm/s^2
***THIS ISNT RIGHT****
please help me then I will answer the rest of the questions.
Your acceleration vs clock time table:
1.7738 cm/s^2, 0.703125 s
-14.956 cm/s^2, 0.9726565 s
.1469 cm/s^2, 0.964844 s
-1.06 cm/s^2, 0.9804915 s
-0.7848 cm/s^2, 1.0859375 s
-0.6856 cm/s^2, 1.089844 s
-0.5912 cm/s^2, 1.265625 s
-0.2696 cm/s^2, 1.558594 s
-0.1699 cm/s^2, 1.5234375 s
-0.1356 cm/s^2, 1.9882815 s
According to the evidence here, is acceleration increasing, decreasing, staying the same or is in not possible to tell?
This is inconclusive because I believe the acceleration is calculated incorrectly.
Please help me with this very last part Prof. Smith.
13:04:13 01-29-2006