Answered: Can some one please help filling in the…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Can some one please help filling in the empty graph. The other one is to help fill in the empty one . The 6 things I'm listing aren't questions just directions to understand . 1. Correctly round all of your TE values in Trial 5 to a whole number. Correctly round all of your TE values in Trial # 7 to a one decimal place only. 2. Compare the 5 TE values for each trial. Choose the trial in which your 5 TE values are the most similar. This will be your CHOSEN trial . 3. Once you have decided on the trial with the closest 5 TE values, create a bar graph on the attached graph paper. I have labeled the X- axis for you. Notice that you will have a PE, KE, and TE bar for each point. 4. You will have to decide on the numbers for the Yaxis. Type numbers in the blank column on the left of the graph . 5. Remember to title your graph "Y - axis variable name vs X-axis variable name." That is not the actual title, just the template for you to use to write your title. 6. Please make your PE bars pink your KE bars green and your TE bars yellow

Point on
trajectory
Point 1
Point 2
Point 3
Point 4
Point 5
Point on
trajectory
Point 1
Point 2
Point 3
Position (m)
[height]
0.683
Point 4
Point 5
0.818
0.856
0.799
0.651
Data Table 3: Trajectory Points for Trial 7
0.589
Velocity
(m/s)
1.806
0.869
0.628
-0.092
0.575
0.428
-1.026
Position (m) Velocity
[height]
(m/s)
0.0454
1.733
-1.895
0.869
-0.063
-1.002
-1.709
Trial 5
PE (J)
PE=mgh
1.00401 J
1.20246 J
1.858 J
1.17459 J
0.956
Trial 7
PE (J)
PE=mgh
0.0668 J
0.8658 J
0.924 J
0.846 J
0.629 J
KE (J)
KE=1/2mv²
0.244 J
0.0566 J
0.0006348 J
0.0789 J
0.2699 J
KE (J)
KE=1/2mv²
0.225 J
0.0566 J
0.000297 J
0.075 J
0.219 J
TE (J)
TE=PE+KE
1.24801 J
1.259 J
1.2586 J
1.25348 J
1.2262 J
TE (J)
TE=PE+KE
0.3 J
0.9 J
0.9 J
0.9 J
0.8 J

PE
ΚΕ
Point 1
TE
PE KE
Point 2
TE
KE TE
Point 3
Types of Energy
PE
PE KE TE
Point4
PE
KE TE
Point 5

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