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Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Uniformity of Acc Lab
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???????????Using the 'Experiment-Specific Calculations' button of the data program, select 1, as you did in the preceding experiment, and respond with the information necessary to calculate the speed of the ball at the end of the ramp, based on the mean distance observed for your first set of 5 trials.
Repeat this process for each of the remaining 11 trials.
Report the resulting speeds in the box below, three speeds for each setup and ramp orientation, in the same order and the same format used in the preceding:
For the setup you skipped, just enter 'skipped'.
---------->>>>> ball speeds based on mean distances
Your answer (start in the next line):
14.52cm/0.2 s= 72.6 cm/s
18.9cm/0.2s= 94.5 cm/s
23.98cm/0.2 s= 119.9 cm/s
14.46 cm/0.2 s=72.3 cm/s
21.08cm/0.2 s=105.4 cm/s
24.98 cm/ 0.2 cm=124.9 cm/s
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I am not sure really how to get the time. I thought I did it right with 0.2 sec when I first send it in, however I do not see how I got it now, and it does not make sense.
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How far did the ball fall, and how did you use this to find the time of fall?
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After I find the time I can find the velocity, then do I find the acceleration?
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The horizontal velocity of the ball will be constant, starting at the instant it leaves the ramp.
It will not have had time to speed up or slow down between the instant it reaches the end of the ramp and the instant it leaves the ramp.
So its horizontal velocity during the fall is the same as its velocity at the end of the ramp.
Having the velocity at the end of the ramp, you can combine this with other information about its trip down the ramp to find the acceleration.
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#$&*
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
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I am not sure if I did this part right for the Lab for assignment 13
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Report in the first line below the data you previously obtained for the trial with the least average velocity.
Starting in the second line report your average angular velocity vs. midpoint clock time data for this trial. You will report a clock time and an average angular velocity for each 180-degree interval. Use comma-delimited format with one interval reported per line.
After reporting the requested information, explain how you obtained your midpoint clock times and your average angular velocities.
Your answer (start in the next line):
50, 1.187, 1.377, 2.579
85.4 degrees/sec, 1.377
I found the average angular velocity from my previous data (total degrees/total seconds), then the middle clock time was 1.377 from previous data as well.
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Am I on the right track?
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Your data report would indicate that the ball traveled the first 180 degrees in 1.187 sec, the next 180 degrees in 1.377 sec, and an additional 50 degrees in 2.579 sec.
It would follow that the strap took about 5 seconds to travel 410 degrees, which is consistent with your 85.4 degrees / second average angular velocity.
I believe the present exercise starts by asking you about the average angular velocity for the 180-degree intervals, as well as the 50-degree interval at the end, then asks for average angular velocity vs. midpoint clock time.
The exercise explains what is meant by midpoint clock time. Be sure you understand that part, as it is often misinterpreted.
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#$&*
Phy 121
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
QA 19 Revision
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Your solution:
15 degree incline= 255 degrees+15 degrees= 270 degrees
x component: 80 kg*cos(255 deg)=-56.6
y component: 80kg *sin (255 deg)=-56.6
?? not sure what to do next???
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The sine and cosine of 255 degrees are not the same (a sketch of this vector will show that one component is clearly much greater than the other). So you need to correct at least one of the components.
Which component acts parallel to the incline?
What force is present to oppose that force?
How big does that force have to be?
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x component: 80 kg*cos(255 deg)=-56.6
y component: 80kg *sin (255 deg)=-56.6
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I am still getting the components as the same in my calculator even though I know one is bigger. Help! :/
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The sine of 255 deg is between -.9 and -1.
The cosine is in the neighborhood of -.25.
What does your calculator give you for these values on the sine and cosine?
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