flow experiment_data

Your initial message (if any):

Is flow rate increasing, decreasing, etc.?

As the water level decreases the rate of flow from the cylinder will decrease.

Is the velocity of the water surface increasing, decreasing, etc.?

The velocity of the surface would decrease as the water level decreased thus making the velocity of the buoy decrease.

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 faster the velocity of the water surface the faster the velocity of the exiting water -- if the water level is decreasing at a rapid rate then the water exiting the cylinder will be moving more rapidly. Since v=sqrt(2gh),we know that the height of the cylinder will affect the velocity of the exiting water. We learned in class that the pressure of the water on the surface and the water exiting the cylinder have the same pressure and using Bernoulli's Equation we know that if the pressure is constant then if velocity increases depth decreases.

Explain how we know that a change in velocity implies the action of a force:

Pressure equals force/area so force equals P*area. From Bernoulli's Equation if we have altitude, velocity and pressure - if velocity changes then either pressure or altitude must remain constant. If the depth remains constant then when velocity increases then pressure must decrease. V=sqrt(2gh) so we know that gravity plays a major part in accelerating the water from inside the cylinder to the outside of the outflow hole.

Does the depth seem to be changing at a regular rate, at a faster and faster rate, or at a slower and slower rate

The depth of the water changes faster at the top of the cylinder and as the water level decreases the depth changes slower and slower.

What do you think a graph of depth vs. time would look like?

Depth decreases as the time increases so the graph will be curved. the graph decreases on the y axis and approaches zero on the x axis.

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 goes on the distance traveled horizontally by the stream decreases.

Does this distance change at an increasing, decreasing or steady rate?

This distance changes at a fairly steady rate but decreases at the end.

What do you think a graph of this horizontal distance vs. time would look like?

A graph of the horizontal distance vs. time would be fairly linear and decrease on the y axis as it increases on the x axis.

The contents of TIMER program as you submitted them:

Our timer results are saved on a computer in class.

The vertical positions of the large marks as you reported them, relative to the center of the outflow hole

1.5 cm to 30 mL mark

3.5 cm to 50 mL mark

5.4 cm to 70 mL mark

7.3 cm to 90 mL mark

9.3 cm to 110mL mark

11.1cm to 130mL mark

13.0cm to 150mL mark

14.9cm to 170mL mark

16.7cm to 190mL mark

18.5cm to 210mL mark

20.4cm to 230mL mark

22.2cm to 250mL mark

Your table for depth (in cm) vs clock time (in seconds)

0, 22.2

1.86, 20.4

4.07, 18.5

6.17, 16.7

8.43, 14.9

10.88, 13.0

13.59, 11.1

16.58, 9.3

19.84, 7.3

24.14, 5.4

29.29, 3.5

38.55, 1.5

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.

Your description of your depth vs. t graph:

my graph of depth vs. clocktime is curved and starts at 250 mL on the y axis and decreases as the time increases to 38.55.

Your explanation and list of average average velocities:

velocity is the average rate of change of position with respect to clocktime and in this case is mL/sec. velocity = change in position/change in clocktime or final mL - initial mL/change in t.

230-250mL = -20mL/1.86s = -10.75 mL/sec

210-230mL = -20mL/2.21s = -9.050 mL/sec

190-210mL = -20mL/2.10s = -9.523 mL/sec

170-190mL = -20mL/2.26s = -8.849 mL/sec

150-170mL = -20mL/2.45s = -8.163 mL/sec

130-150mL = -20mL/2.71s = -7.380 mL/sec

110-130mL = -20mL/2.99s = -6.689 mL/sec

90-110mL = -20mL/3.26s = -6.135 mL/sec

70-90mL = -20mL/4.30s = -4.651 mL/sec

50-70mL = -20mL/5.15s = -3.883 mL/sec

30-50mL = -20mL/9.26s = -2.159 mL/sec

The midpoints of your time intervals and how you obtained them:

we used the timer program to obtain our time intervals - I am not sure what you are asking for here.

Your table of average velocity of water surface vs. clock time:

1.86s, -10.75mL/sec

2.21, -9.050

2.10, -9.523

2.26, -8.849

2.45, -8.163

2.71, -7.380

2.99, -6.689

3.26, -6.135

4.30, -4.651

5.15, -3.883

9.26, -2.159

Your description of your graph of average velocity vs clock time:

my graph of average velocity vs. clocktime is curved, with velocity decreasing as the time increases.

Your explanation of how acceleration values were obtained:

acceleration is the average rate of change of velocity with respect to clocktime. avg accel = Vf-V0/'dt. I am not sure if 'dt is the time we got from the timer or the difference in the times between the two intervals. If you could clarify this for me I will complete the form.

Your acceleration vs clock time table:

According to the evidence here, is acceleration increasing, decreasing, staying the same or is in not possible to tell?

13:04:20 01-29-2006