jata and Results Mass, m (kg) _0. 02%kg Radius of path, r (m)0,(om Number of revolutions, B 20 Hanging Mass (kg) Trial 1 Time Trial 2 Time Trial 3 Time Trial 4 Time Trail 5 Time (s) (s) (s) (s) (s) 9.89 50 60 70 11.93 9.38 9.48 11.78 11.,68 10.11 10.43 9.56 9.72 9.26 9.58 8.42 9.41 9.70 80 9,25 8.19 9.02 8.96 8.60 9.18 90 9.32 9.16 9.25 Table 1: Experimental Data 8.90 Linear Speed Centripetal Average Total Force Linear Speed Squared Hanging Time for one Mass Time revolution Fe= Mg tavg T = tavg/B v = 2ar/T v? %3D %3D ((m/s)) 46.24 (kg) (N) (s) (s) (m/s) 0.5539 6.80 7.79 8.08 490 11.078 9. 670 50 60 588 0.4835 60.68 9.318 0.4659 65.28 70 80 90 686 784 8.804 0.4402 8.56 73.27 8.162 Table 2: Experimental Results 882 0.4081 9.23 85.19 Centripetal force =78.4 or 156.8 any # between Linear Speed 5.4o6 or 10.912 Squared any # between analysis 1. Why should the tape not touch the bottom of the tube while m is swinging? The tape touching the bottom of the tube would affect the Swinging mis velocity 2. Why should "Linear Speed Squared vs. The Centripetal Force" be plotted instead of "The Centripetal Force vs. Linear Speed Squared" If the graph was the centrcipetal force vs. linear Speed squared then the centripetal force would be x +linear Speed Squared would be y Showing the affected version of measure of linear Speed squared by the centripetal force is affectedby lincar speed squared. The y axis depends on the x axis. 3. Why should v and not v be plotted? Plotting v? instead of v Shows a transformed values. Acceleration is ac = Straighter line using linearly 4. The experimental mass = (the radius) / (the slope of the trend line). What is the experimental mass? o.6/ 5. Compare the mass found using the triple beam balance and the mass found using the slope. What is the % error? [Assume that the mass found using the triple beam balance is the correct value.]
jata and Results Mass, m (kg) _0. 02%kg Radius of path, r (m)0,(om Number of revolutions, B 20 Hanging Mass (kg) Trial 1 Time Trial 2 Time Trial 3 Time Trial 4 Time Trail 5 Time (s) (s) (s) (s) (s) 9.89 50 60 70 11.93 9.38 9.48 11.78 11.,68 10.11 10.43 9.56 9.72 9.26 9.58 8.42 9.41 9.70 80 9,25 8.19 9.02 8.96 8.60 9.18 90 9.32 9.16 9.25 Table 1: Experimental Data 8.90 Linear Speed Centripetal Average Total Force Linear Speed Squared Hanging Time for one Mass Time revolution Fe= Mg tavg T = tavg/B v = 2ar/T v? %3D %3D ((m/s)) 46.24 (kg) (N) (s) (s) (m/s) 0.5539 6.80 7.79 8.08 490 11.078 9. 670 50 60 588 0.4835 60.68 9.318 0.4659 65.28 70 80 90 686 784 8.804 0.4402 8.56 73.27 8.162 Table 2: Experimental Results 882 0.4081 9.23 85.19 Centripetal force =78.4 or 156.8 any # between Linear Speed 5.4o6 or 10.912 Squared any # between analysis 1. Why should the tape not touch the bottom of the tube while m is swinging? The tape touching the bottom of the tube would affect the Swinging mis velocity 2. Why should "Linear Speed Squared vs. The Centripetal Force" be plotted instead of "The Centripetal Force vs. Linear Speed Squared" If the graph was the centrcipetal force vs. linear Speed squared then the centripetal force would be x +linear Speed Squared would be y Showing the affected version of measure of linear Speed squared by the centripetal force is affectedby lincar speed squared. The y axis depends on the x axis. 3. Why should v and not v be plotted? Plotting v? instead of v Shows a transformed values. Acceleration is ac = Straighter line using linearly 4. The experimental mass = (the radius) / (the slope of the trend line). What is the experimental mass? o.6/ 5. Compare the mass found using the triple beam balance and the mass found using the slope. What is the % error? [Assume that the mass found using the triple beam balance is the correct value.]
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So I have my data but the problem I'm having is these two questions I'm not understanding what she's asking in these two questions.
Transcribed Image Text:jata and Results
Mass, m (kg) _0. 02%kg
Radius of path, r (m)0,(om
Number of revolutions, B 20
Hanging Mass
(kg)
Trial 1 Time
Trial 2 Time
Trial 3 Time
Trial 4 Time
Trail 5 Time
(s)
(s)
(s)
(s)
(s)
9.89
50
60
70
11.93
9.38
9.48
11.78
11.,68
10.11
10.43
9.56
9.72
9.26
9.58
8.42
9.41
9.70
80
9,25
8.19
9.02
8.96
8.60
9.18
90
9.32
9.16
9.25
Table 1: Experimental Data
8.90
Linear Speed
Centripetal Average Total
Force
Linear Speed
Squared
Hanging
Time for one
Mass
Time
revolution
Fe= Mg
tavg
T = tavg/B
v = 2ar/T
v?
%3D
%3D
((m/s))
46.24
(kg)
(N)
(s)
(s)
(m/s)
0.5539
6.80
7.79
8.08
490
11.078
9. 670
50
60
588
0.4835
60.68
9.318
0.4659
65.28
70
80
90
686
784
8.804
0.4402
8.56
73.27
8.162
Table 2: Experimental Results
882
0.4081
9.23
85.19
Centripetal force =78.4 or 156.8
any # between
Linear Speed 5.4o6 or 10.912
Squared
any # between
![analysis
1. Why should the tape not touch the bottom of the tube while m is swinging?
The tape touching the bottom of the tube would affect the Swinging
mis velocity
2. Why should "Linear Speed Squared vs. The Centripetal Force" be plotted instead of "The
Centripetal Force vs. Linear Speed Squared"
If the graph was the centrcipetal force vs. linear Speed squared then the
centripetal force would be x +linear Speed Squared would be y
Showing the affected version of measure of linear Speed squared by the
centripetal force is affectedby lincar speed squared. The y axis depends
on the x axis.
3. Why should v and not v be plotted?
Plotting v? instead of v Shows a
transformed values. Acceleration is ac =
Straighter line using linearly
4. The experimental mass = (the radius) / (the slope of the trend line). What is the experimental
mass?
o.6/
5. Compare the mass found using the triple beam balance and the mass found using the slope.
What is the % error? [Assume that the mass found using the triple beam balance is the correct
value.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb70fd92-421b-40d2-853c-67ceb5502be8%2F39f9c44b-521b-4dbc-b432-a58f8c71c21f%2Fqbf6bwe.jpeg&w=3840&q=75)
Transcribed Image Text:analysis
1. Why should the tape not touch the bottom of the tube while m is swinging?
The tape touching the bottom of the tube would affect the Swinging
mis velocity
2. Why should "Linear Speed Squared vs. The Centripetal Force" be plotted instead of "The
Centripetal Force vs. Linear Speed Squared"
If the graph was the centrcipetal force vs. linear Speed squared then the
centripetal force would be x +linear Speed Squared would be y
Showing the affected version of measure of linear Speed squared by the
centripetal force is affectedby lincar speed squared. The y axis depends
on the x axis.
3. Why should v and not v be plotted?
Plotting v? instead of v Shows a
transformed values. Acceleration is ac =
Straighter line using linearly
4. The experimental mass = (the radius) / (the slope of the trend line). What is the experimental
mass?
o.6/
5. Compare the mass found using the triple beam balance and the mass found using the slope.
What is the % error? [Assume that the mass found using the triple beam balance is the correct
value.]
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