course phy201

I am sending this back I am not sure what lab this is, but it is not one that I have done. It is not mine. I will revise the testing hypothesis time interval and resubmit that one. :) thanks

Thanks for the information. I'll try to match this up.

1.568

1.633

1.508

1.563

1.594

1.789

1.718

1.765

1.785

1.781

The first five lines is the data for one domino of a ball rolling down the ramp in one direction, and the next five lines is in the opposite direction.

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1.258

1.094

1.172

1.153

1.223

1.226

1.164

1.252

1.143

1.201

The first five lines is the data for two dominoes of a ball rolling down the ramp in one direction, and the next five lines is in the opposite direction

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0.883

0.773

0.883

0.953

0.853

0.883

0.794

0.903

0.883

0.835

The first five lines is the data for three dominoes of a ball rolling down the ramp in one direction, and the next five lines is in the opposite direction.

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1.573, 0.04580

1.768, 0.02919

1.18, 0.06349

1.197, 0.04440

.869, 0.06504

.8596, 0.04440

The results indicated above are the mean and standard deviation of the five time trials for a ball rolling down a ramp. First two lines are for one domino in each direction, the next two lines are for two dominoes in each direction, and the last two lines are for three dominoes in each direction. The first number ineach line is the mean value and the second number is the standard deviation. These values were calculated using the data program.

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17.800 cm/s

15.837 cm/s

23.729 cm/s

23.392 cm/s

32.221 cm/s

32.573 cm/s

The above results for average velocity were obtained by dividing the change in position by the change in clock time. For example, the average velocity, which is the change in position with respect to clock time,for 1 domino from right to left is, vAve = 'ds / 'dt. With 'dt being the mean time for the five trials and 'ds is the displacement of the ball, we have vAve = 28 cm / 1.573 = 17.800 m/s.

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22.632 cm/s^2

17.915 cm/s^2

40.218 cm/s^2

39.084 cm/s^2

74.156 cm/s^2

75.787 cm/s^2

The results above were obtained by using the second fundamental equation of motion, df = d0 + v0t + 1/2at^2. We rearrange the terms to solve for acceleration, which is the change in velocity divided by the time that it takes to make that change. Before we rearrange the terms we can simplify it more by knowing that the initial position and initial velocity are both zero. When we rearrange the terms and use for example the trial with one domino from right to left, we then have, a = 'ds / (1/2 * t^2) = 28 cm / (1/2 * 1.573^2) = 22.632 cm/s^2. For these results the time that was used ws the mean time for the five trials, and the distance was the given 28 cm.

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20.274 cm/s^2

39.651 cm/s^2

74.972 cm/s^2

The results above were obtained by adding the average rate of change of velocity for each direction and dividing by two.

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20.068 cm/s^2

39.645 cm/s^2

74.965 cm/s^2

The values above were obtained by recalculating the average velocity based on the average mean time between the two directions. Once the average velocity was recalculated then the new rate of change of velocity was calculated.

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I would expect the two results to be the same or very close to the same.

The results that I calculated for the two mean times is very close to the previously obtained results.

I would expect them to be nearly the same because we are using the results of the experiment, but just averageing the different values.

The reason that they are not exactly the same is probably because rounding errors.

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.03, 20.068

.06, 39.645

.09, 74.965

The lines above indicate the ramp slope for the first number which is an unitless quantity, and the second number is the average rate of change of velocity expressed in cm/s^2. As we can tell from the results, as the slope increases so does the average rate of change of velocity.

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.01

-10.0

The horizontal intercept, which is the offset from the zero point on the vertical axis where the graph crosses the horizontal axis. This value is unitless.

The units of the vertical intercept in the second line are cm/s^2. This is the offset where the graph crosses the vertical axis from the zero point on the horizontal axis.

This tells us that the slope is quite steep, which means that the acceleration changes by a large value for each increase in ramp steepness.

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.10, 82.0

Made the estimate by extending the line and finding the .10 point on the horizontal axis and then finding the corresponding point on the vertical axis. The estimate is fairly accurate with maybe an error of +-.5.

** **

.09

82

911.11

The slope is telling us the average rate of change of acceleration with respect to the ramp slope.

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Very good. Ideally the slope will match the acceleration of gravity, as corrected for the fact that the ball is rotating about its axis as it rolls. You have a good result.

1.573, 1.768, 1.671

0.046, 0.029, .038

The first line above indicates the mean time for one domino in each direction and the average mean time. The second line above indicates the standard deviation for one domino in each direction and the average of the standard deiations. Finding the averages should give us more accurate results when calculating values.

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1.633, 1.709

The numbers above represent the boundaries for the mean +- the standard deviation.

** **

17.146, 20.999

16.384, 19.174

19.174, 20.999

The numbers in the first line represent the average velocity and acceleration for the minimum mean time value. The numbers on the second line indicate the average velocity and acceleration for the maximum mean time value. If everything is set up correctly based on the standard deviation the results should always be between these two values.

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1.18, 1.197, 1.1885

0.063, 0.044, 0.054

1.135, 1.243

24.670, 43.471

22.526, 36.245

36.245, 43.471

The first line indicates the mean times in each direction and the average mean time. The second line indicates the standard deviation for each direction and the average standard deviation. The third line is the minimum and maximum times based on the average mean and average standard deviation. The fourth and fifth lines are the maximum and minimum average velocity and acceleration based on the average mean and standard deviation. The sixth line is the minimum and maximum acceleration.

** **

0.869, 0.8596, 0.8643

0.065, 0.044, 0.055

0.809, 0.919

34.611, 85.564

30.468, 66.307

66.307, 85.564

The first line indicates the mean times in each direction and the average mean time. The second line indicates the standard deviation for each direction and the average standard deviation. The third line is the minimum and maximum times based on the average mean and average standard deviation. The fourth and fifth lines are the maximum and minimum average velocity and acceleration based on the average mean and standard deviation. The sixth line is the minimum and maximum acceleration.

** **

0.3, 19.174

0.6, 36.245

0.9, 66.307

0.3, 20.999

0.6, 43.471

0.9, 85.564

The numbers above indicate the lower and upper limits of acceleration based on the slope of the ramp. They show us that the acceleration increases as the slope increases and also that there is a range in which these values can fall.

** **

The graph does fit this description. The best fit line passes through the 0.6 and 0.9 slope values and is right on the 0.3 slope values. Overall it fits good through the vertical lines.

** **

77.5, 0.092, 842.39

The results above are the rise, run and slope for the steepest line. The coordinates use to calculate these values were, (0.10, 77.5) and (0.008, 0).

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68.5, .10, 685.00

The results above are the rise, run and slope for the steepest line. The coordinates use to calculate these values were, (0.10, 68.5) and (0, 0).

** **

2 Hours

** **

There appear to be 2 labs in this submission; if they aren't both yours then there's a glitch with the form, so please let me know.

See my notes on the quick-click means and standard deviations. I'll have to ask you to revise that one, since your means don't seem to correspond to your data. Questions are always welcome.

The second experiment looks good. I hope that's yours, but again let me know if not."

&#Good responses. See my notes and let me know if you have questions. &#