10 Pages

Keeping It Connected

Course: MAS 361, Fall 2009
School: Syracuse
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Word Count: 3107

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Analyzing A1. Data Objective: To analyze each of the nine variables associated with keeping it connected with family members by using graphical and numerical descriptive techniques. A2. Our team constructed a questionnaire containing nine questions. These questions all pertained to how much time people spend trying to remain connected with their families, and what their cumulative GPAs were. Our groups target of...

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Analyzing A1. Data Objective: To analyze each of the nine variables associated with keeping it connected with family members by using graphical and numerical descriptive techniques. A2. Our team constructed a questionnaire containing nine questions. These questions all pertained to how much time people spend trying to remain connected with their families, and what their cumulative GPAs were. Our groups target of interest is Syracuse University college undergraduates. The sample that we used represents all Syracuse University undergraduates because the participants were randomly selected and were all from various states around the country. With the results from our questionnaire we will be able to formulate an idea about how trying to stay connected may, or may not, affect our academics. A3. According to Figure 1A most students spend approximately .5 hours per week talking on the phone with their family members. Based on our observation of the graph, the distribution is right skewed. In addition, the numerical descriptive data featured in Table 1 shows that on average students spend .66 hours (40 minutes) talking on the phone speaking with family members and the standard deviation is .3417. This low standard deviation tells us that the data is close the mean and therefore the data is reliable. Figure 2A displays the number of visits home for each one of our participants. According to this graph, most students visit home 3 or more times. Based on our observation of the graph, the distribution is left-skewed. In addition, Table 2 shows us that the average students travels home about 1.96 times per semester. The table also tells us what the standard deviation is, which we can use to see how reliable the data is. When interpreting Figure 3A, we can see that most of our participants talked to their family via email and instant messaging for less than two hours per week. The graphs distribution is rightskewed. Furthermore, Table 2 shows us that the average student emails and instant messages their family for about 1.63 hours per week (97.8 minutes). Figure 4A displays the cumulative GPAs of our participants. According to the dotplot, we can see that most of our participants have a cumulative GPA of 3.5. The distribution of this graph is left-skewed. Table 4 provides us with even more information. The table tells us that the average participant had a cumulative GPA of 3.2662. The low standard deviation shows us that this data is reliable. Figure 5A graphically illustrates the distribution of how much time (in hours) is spent on homework while on family visits. Most SU students spend zero hours on homework while at home. Numerical descriptive data tells us that the mean is .89 hours and the standard dev is 1.233. This means that on average, most students spend .89 hours or about 53 minutes on homework while on family visits at home. A standard deviation of 1.233 can be interpreted as a relatively close distribution of the data to the mean. The dotplot of how many family members a student keep in constant communication (Fig 6A) is left skewed and it shows that most people keep in contact with three or more family members. Descriptive numerical analysis (Table 6), illustrates that the mean is 2.64 and standard deviation is .6579. Figure 7A is right-skewed, and tells us how many hours away the participants live from Syracuse University. We can observe that most participants live less than 6 hours away. Table 7 indicates that the average student lives 5.244 hour away (313.44 minutes). Figure 8A tells us how many classes students skipped when they visited home last semester. Most of the students said that they missed zero classes when visiting home. Also, the graphs distribution is right-skewed telling us that the mean is greater than the median. Table 9 shows us that the average student misses about .36 classes per semester. Analysis of Figure 9A indicates most students do not spend any time on special events such as birthdays and holidays. The numerical descriptive analysis in Table 9 tells us that the mean of this variable is 2.64 the standard deviation is 3.770. A4. There were no qualitative variables used in our project A5. Summary This section of the project was all about taking the results that we got from our questionnaire and displaying them in both graphical and numerical formats. For each variable we created graphs (nine graphs in all). We used both dotplots and histograms, since they were the most relevant for displaying our data. For each variable we also found the mean and the standard deviation. With the mean we could come up with an on average number for each of our variables which would give us an idea of our target audience and the ways that are B1. Effects on Academic Success Objective: To determine whether certain kinds of communication with family members can have noticeable effects on academic success through observation of the relationships between independent (X) and dependent variables (Y). B2. Our dependent variables include cumulative GPA and study time. These variables will help us determine overall academic success by observing how each dependent variable responds to the independent or explanatory variables. B3. The independent/explanatory variables used in this statistical analysis include amount of time spent speaking on the phone with family members, the amount of time spent through internet communications such as emails, instant messaging, Facebook, and/or any other social networking sites, and the number of times a student visits home per semester. The amount of time a student spends on the phone relates to how well a students performance is because the time they spend on the phone takes away from studying or class time. Another factor that would take away from a students availability to study or attend class is the amount of time he or she spends on the computer contacting family members through instant messaging, emails, and Facebook or any of the other social networking sites. Finally, the most important factor that would be expected to affect student performance is the number of times a student goes home throughout the semester. From our initial survey, we found that students barely spend any time on school work while visiting with their families and often times some students will even skip classes to be able to catch the grey hound bus or a ride to the airport. So the more frequently a student decides to go home, most likely the less time they will spend on studying or reading for their next weeks classes. B4. Figure B1 relates the minutes a student speaks on the phone to his or her family with his or her GPA. Because the correlation coefficient is so near 0 and not -1 or 1, this means that a students phone time is a very weak predictor of GPA as well as proving that there is a positive relationship. Only 0.9% of GPA is predicted by the number of minutes a student uses the phone for family time. Figure B2 compares the amount of time a student spends on the internet contacting family to his or her GPA. The correlation coefficient proves that this is actually a negative weak relationship and the coefficient of determinate proves that a mere 0.08% of GPA is estimated by internet usage. Figure B3 demonstrates the number of visits home a student makes and how it predicts his or her GPA. The correlation coefficient being so close to 0 proves that this is a negative and weak relationship. The coefficient of determinate states that 1.41% of GPA is predicted by the number of visits a student makes home. Figure B4 compares the number of visits home to the amount of time a student spends studying in this scatter plot. The relationship is negative and weak as the correlation coefficient finds it to be near 0. The number of minutes put towards study time is found to be 1.23% based on the number of visits home. Figure B5 relates internet usage to the amount of time a student spends studying. The correlation coefficient is near 0, meaning that this relationship is weak but positive. The amount of time spent on the internet is estimated to determine 3.47% of the time spent on study. Figure B6 relates the amount of time a student spends on the phone with family with the amount of time he or she studies. The correlation coefficient proves the relationship to be negative and weak. The coefficient of determinate proves the amount of a time a student spends on the phone to only predict 0.42% of the time he or she spends studying. B5. Regression Analysis Regression Analysis: GPA versus Phone Time (mins) GPA = 3.00 + 0.00367 Phone Time (mins) S = 0.778345 R-Sq = 0.9% R-Sq(adj) = 0.0% This regression model indicates that every minute increase in phone usage the GPA will increase by .00367. If someone were not to talk on the phone at all they would have a 3.0 GPA. The coefficient determinant 0% indicates that this model is not a good determinant at all. Regression Analysis: GPA versus Phone Time (mins), Internet Time (mins) GPA = 3.01 + 0.00360 Phone Time (mins) - 0.000079 Internet Time (mins) S = 0.786493 R-Sq = 1.0% R-Sq(adj) = 0.0% This regression model indicates that with every minute increase spent on the internet to keep in touch with your family the GPA will drop by 0.000079 of a point. With every minute increase spent keeping in touch with your family on the phone the GPA will increase by 0.00360 of a point. If someone were to not do any of these their GPA would be 3.01 The coefficient determinant is 0%, which indicates that this model is not a good determinant. Regression Analysis: GPA versus Phone Time (, Internet Tim, No. of Home GPA = 3.17 + 0.00933 Phone Time (mins) - 0.000167 Internet Time (mins) - 0.197 No. of Home Visits S = 0.775960 R-Sq = 5.7% R-Sq(adj) = 0.0% This regression model indicates that every time someone visits home there GPA decreases by 0.197 of a point. With every minute increase spent on the internet to keep in touch with their family the GPA will drop by 0.000167 of a point. Every minute increase spent talking on the phone with family will increase the GPA by 0.00933 of a point. If someone were not to do any of these they would start off with a 3.17 GPA. The coefficient determinant is 0%, which indicates it is not a good regression model. Regression Analysis: Study Time (mins) versus Phone Time (mins) Study Time (mins) = 55.0 - 0.273 Phone Time (mins) S = 66.1259 R-Sq = 0.7% R-Sq(adj) = 0.0. This regression model indicates that every minute increase spent on the phone keeping in touch with your family will decrease study time by 0.273. If you were to not spend any time on the phone you would study 55 minutes. The coefficient determinant is 0% which indicates that it is not a good indicator. Regression Analysis: Study Time (versus Phone Time), Internet Tim Study Time (mins) = 43.7 - 0.196 Phone Time (mins) + 0.0788 Internet Time = (mins) S 65.7690 R-Sq = 3.8% R-Sq(adj) = 0.0%. This regression model indicates that with every minute increase spent on the internet the time spent study will increase by 0.0788 of a minute. With every minute increase spent on the phone time spent studying will decrease by 0.196 of a minute. If someone were not to spend any time on the internet or phone there study time would be 43.7 minutes. The coefficient determinant is 0% which indicates that it is not a good deciding factor. Regression Analysis: Study Time (versus Phone Time), Internet Tim, ... Study Time (mins) = 47.6 - 0.057 Phone Time (mins) + 0.0766 Internet Time (mins)- 4.8 No. of Home Visits S = 66.3473 R-Sq = 4.2% R-Sq(adj) = 0.0%. This regression model indicates that with every home visit increase study time will decrease by 4.8 minutes. With every minute increase spent on the internet keeping in touch study time will decrease by 0.0766 of a minute. With every minute increase spent talking on the phone, study time will decrease by 0.057 of a minute. If someone was not engaging i there study time would start off as 47.6 minutes. The coefficient determinant is 0% which indicates it is not a good determining factor. B6. Summary According to our findings there is not a strong correlation between GPA and how often you stay connected with your family. There is not a strong correlation between the amount of time you study and how much you stay connected with your family either. This information would be useful to campus universities because they will know that students are not affected by how much they stay in contact with their family. Whether they live close to home and like to visit often or live across the country and like to talk on the phone to relatives or remain in contact through the internet, their GPAs and study hours will not be affected. Appendix Appendix A Dotplotofnumberofvisitshome Dotplotoftimespentonphone( hours) 0 0.0 0.2 0.4 0.6 0.8 1.0 t imespent onphone( hours) 1.2 1 2 numberofvisit shome 3 1.4 Figure 1A Figure 2A HistogramofGPA 18 DotplotoffacebookandAI Mtime 16 14 Fr e que ncy 12 10 8 6 0.0 1.4 2.8 4.2 5.6 facebookandAIMt ime 7.0 8.4 4 9.8 2 0 Figure 3A 2.0 2.5 3.0 GPA 3.5 4.0 Figure 4A HistogramoftimespentonHWduringvisits 25 Dotplotofhowmanyclosefamilymembers Fr e que ncy 20 15 10 5 0 0.0 Figure 5A 1.2 2.4 3.6 t imespent onHWduringvisit s 4.8 6.0 0 Figure 6A 1 2 howmanyclosefamilymembers 3 Dotplotofclassesskippedforvisits Dotplotofhowfarawayfromhome 0 6 12 18 24 howfarawayfromhome 30 36 42 0 Figure 7A 1 classesskippedforvisit s 2 Figure 8A Dotplotoftimespentforfamevents 0 3 6 9 12 t imespent forfamevent s 15 18 Figure 9A Table 1: Descriptive Statistics: time spent talking on phone Variable N N* Mean StDev Minimum Q1 time spent talking on ph 50 0 0.0483 0.3417 0.0000 0.5000 Variable Maximum time spent talking on ph 1.5000 Mean Table 3: Descriptive Statistics: facebook and AIM time SE 0.6600 Median Q3 0.5000 0.6250 Table 2 : Descriptive Statistics: number of visits home Variable N N* Mean Mean StDev Minimum Q1 Median number of visits home~ 50 1 1.960 0.148 1.049 0.000 1.000 2.000 Variable number of visits home~ Q3 3.000 Maximum 3.000 SE Variable N N* Mean Mean StDev Minimum Q1 Median facebook and AIM time~ 49 0 1.630 0.345 2.415 0.000 0.097 1.000 Variable facebook and AIM time~ Q3 2.000 SE Maximum 10.000 Table 4 Descriptive Statistics: GPA Variable N N* Mean SE Mean Minimum Q1 Median Q3 GPA~ 50 0 3.2662 0.0609 2.0000 3.0750 3.3900 3.5625 Variable GPA~ StDev 0.4307 Maximum 3.9600 Table 5: Descriptive Statistics: time spent on HW during visits Variable N N* Mean Mean StDev Minimum Q1 Median time spent on HW during 50 0 0.890 0.174 1.233 0.000 0.000 0.416 Variable time spent on HW during Q3 2.000 SE Residual Error 48 29.0794 Total 49 29.3571 Unusual Observations Maximum 6.000 Table 6: Descriptive Statistics: how many close family members Variable N N* Mean StDev Minimum Q1 how many close family me 50 4 0.0980 0.6928 0.0000 2.0000 Variable Maximum how many close family me 3.0000 Mean SE 2.6400 Median Q3 3.0000 3.0000 Table 7: Descriptive Statistics: how far away from home Variable N N* Mean Mean StDev Minimum Q1 Median how far away from home 50 0 5.244 0.990 6.998 0.167 2.000 4.000 Variable how far away from home Q3 5.000 SE Maximum 42.000 Table 8 Descriptive Statistics: classes skipped for visits Variable N N* Mean StDev Minimum Q1 classes skipped for visi 50 2 0.0937 0.6627 0.0000 0.0000 Variable Maximum classes skipped for visi 2.0000 Mean SE 0.3600 Median Q3 0.0000 1.0000 Phone Time Obs (mins) St Resid 31 30.0 -4.04R 35 30.0 -4.04R 46 90.0 -0.73 X 47 90.0 0.31 X 48 90.0 0.66 X 49 90.0 -0.18 X 50 90.0 -0.09 X 0.6058 GPA Fit SE Fit Residual 0.000 3.106 0.122 -3.106 0.000 3.106 0.122 -3.106 2.800 3.326 0.295 -0.526 3.550 3.326 0.295 0.224 3.800 3.326 0.295 0.474 3.200 3.326 0.295 -0.126 3.260 3.326 0.295 -0.066 R denotes an observation with a large standardized residual. X denotes an observation whose X value gives it large leverage. Regression Analysis: GPA versus Phone Time (mins), Internet Time (mins) The regression equation is GPA = 3.01 + 0.00360 Phone Time (mins) 0.000079 Internet Time (mins) Predictor T P Constant 11.24 0.000 Phone Time (mins) 0.65 0.519 Internet Time (mins) -0.10 0.918 Coef SE Coef 3.0071 0.2675 0.003596 0.005531 -0.0000786 0.0007637 Table 9 Descriptive Statistics: time spent for fam events S = 0.786493 Variable N N* Mean Mean StDev Minimum Q1 Median time spent for fam event 50 0 2.460 0.533 3.770 0.000 0.000 1.000 Analysis of Variance Source DF SS P Regression 2 0.2843 0.796 Residual Error 47 29.0728 Total 49 29.3571 SE Variable Q3 Maximum time spent for fam event 4.000 20.000 Regression Analysis: GPA versus Phone Time (mins) Source Phone Time (mins) Internet Time (mins) Unusual Observations The regression equation is GPA = 3.00 + 0.00367 Phone Time (mins) Predictor P Constant 0.000 Phone Time (mins) 0.502 S = 0.778345 Coef SE Coef T 2.9958 0.2414 12.41 0.003672 0.005424 0.68 R-Sq = 0.9% Analysis of Variance Source DF SS P Regression 1 0.2777 0.502 R-Sq(adj) = 0.0% MS F 0.2777 0.46 R-Sq = 1.0% Phone Time Obs (mins) St Resid 13 30.0 -0.27 X 31 30.0 -4.00R 35 30.0 -4.04R DF 1 1 R-Sq(adj) = 0.0% MS F 0.1421 0.23 0.6186 Seq SS 0.2777 0.0066 GPA Fit SE Fit Residual 2.900 3.049 0.567 -0.149 0.000 3.104 0.125 -3.104 0.000 3.115 0.151 -3.115 R denotes an observation with a large standardized residual. X denotes an observation whose X value gives it large leverage. S = 66.1259 R-Sq = 0.7% Analysis of Variance Regression Analysis: GPA versus Phone Time (, Internet Tim, No. of Home Source DF SS MS F Regression 1 1532 1532 0.35 Residual Error 48 209886 4373 Total 49 211418 Unusual Observations Phone Time Study Time Obs (mins) (mins) Fit SE Fit Residual St Resid 6 30.0 180.00 46.82 10.35 133.18 2.04R 13 30.0 180.00 46.82 10.35 133.18 2.04R 14 30.0 180.00 46.82 10.35 133.18 2.04R 26 30.0 180.00 46.82 10.35 133.18 2.04R 33 30.0 240.00 46.82 10.35 193.18 2.96R 46 90.0 0.00 30.45 25.04 -30.45 -0.50 X 47 90.0 120.00 30.45 25.04 89.55 1.46 X 48 90.0 30.00 30.45 25.04 -0.45 -0.01 X 49 90.0 0.00 30.45 25.04 -30.45 -0.50 X 50 90.0 0.00 30.45 25.04 -30.45 -0.50 X The regression equation is GPA = 3.17 + 0.00933 Phone Time (mins) 0.000167 Internet Time (mins) - 0.197 No. of Home Visits Predictor T P Constant 11.14 0.000 Phone Time (mins) 1.40 0.167 Internet Time (mins) -0.22 0.826 No. of Home Visits -1.51 0.138 S = 0.775960 Coef SE Coef 3.1675 0.2844 0.009326 0.006645 -0.0001672 0.0007557 -0.1969 0.1303 R-Sq = 5.7% R-Sq(adj) = 0.0% Analysis of Variance Source DF SS P Regression 3 1.6599 0.439 Residual Error 46 27.6972 Total 49 29.3571 Source Phone Time (mins) Internet Time (mins) No. of Home Visits Unusual Observations Phone Time Obs (mins) St Resid 13 30.0 -0.39 X 31 30.0 -3.96R 35 30.0 -3.85R DF 1 1 1 MS F 0.5533 0.92 0.6021 P 0.557 R denotes an observation with a large standardized residual. X denotes an observation whose X value gives it large leverage. Seq SS 0.2777 0.0066 1.3756 GPA Fit SE Fit Residual 2.900 3.110 0.561 -0.210 0.000 3.030 0.132 -3.030 0.000 2.857 0.227 -2.857 R denotes an observation with a large standardized residual. X denotes an observation whose X value gives it large leverage. Regression Analysis: Study Time (mins) versus Phone Time (mins) The regression equation is Study Time (mins) = 55.0 - 0.273 Phone Time (mins) Predictor P Constant 0.010 Phone Time (mins) 0.557 R-Sq(adj) = 0.0% Coef SE Coef T 55.00 20.50 2.68 -0.2727 0.4608 -0.59 Regression Analysis: Study Time ( versus Phone Time (, Internet Tim The regression equation is Study Time (mins) = 43.7 - 0.196 Phone Time (mins) + 0.0788 Internet Time (mins) Predictor T P Constant 0.057 Phone Time (mins) 0.674 Internet Time (mins) 0.223 S = 65.7690 Coef SE Coef 43.67 22.37 1.95 -0.1960 0.4625 -0.42 0.07879 0.06386 1.23 R-Sq = 3.8% R-Sq(adj) = 0.0% Analysis of Variance Source Regression Residual Error Total DF 2 47 49 SS 8117 203301 211418 Source Phone Time (mins) Internet Time (mins) Unusual Observations Phone DF 1 1 MS 4058 4326 Seq SS 1532 6585 F 0.94 P 0.399 Time Study Time Obs (mins) (mins) Residual St Resid 6 30.0 180.00 142.21 2.20R 13 30.0 180.00 76.02 1.67 X 14 30.0 180.00 136.69 2.11R 26 30.0 180.00 137.48 2.12R 33 30.0 240.00 202.21 3.13R Fit SE Fit 37.79 12.63 103.98 47.46 43.31 10.68 42.52 10.86 Source Phone Time (mins) Internet Time (mins) No. of Home Visits 37.79 12.63 Unusual Observations R denotes an observation with a large standardized residual. X denotes an observation whose X value gives it large leverage. Regression Analysis: Study Time ( versus Phone Time (, Internet Tim, ... The regression equation is Study Time (mins) = 47.6 - 0.057 Phone Time (mins) + 0.0766 Internet Time (mins) - 4.8 No. of Home Visits Predictor Coef SE Coef T P Constant 47.56 24.32 1.96 0.057 Phone Time (mins) -0.0569 0.5681 -0.10 0.921 Internet Time (mins) 0.07664 0.06462 1.19 0.242 No. of Home Visits -4.78 11.14 -0.43 0.670 S = 66.3473 R-Sq = 4.2% R-Sq(adj) = 0.0% Analysis of Variance Source Regression Residual Error Total DF 3 46 49 SS 8928 202490 211418 DF 1 1 1 Phone Time Study Time Obs (mins) (mins) Residual St Resid 6 30.0 180.00 134.14 2.15R 13 30.0 180.00 74.54 1.63 X 14 30.0 180.00 133.56 2.05R 26 30.0 180.00 139.10 2.13R 33 30.0 240.00 208.48 3.29R MS 2976 4402 F 0.68 P 0.571 Seq SS 1532 6585 811 Fit SE Fit 45.86 22.71 105.46 48.00 46.44 13.02 40.90 11.59 31.52 19.38 R denotes an observation with a large standardized residual. X denotes an observation whose X value gives it large leverage.
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UNF - PHYSICS - 4360C
3.1117 2 mx = 02317 = and 2 = 2222222 r = 2 = 4 mx + 3 mx +(a)Amax =(b)=2d = 2 2 =25 24 d =522A15 2e Td =3.12F2m d r = 2 =121ln 2 = f d ln 2Td1 d = ( 2 2 ) 2(a)122So, = ( + )2d1111 2 2 2 ln 2 2 2f = fd
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21Td d 1 2=== 1 + 2 2 2 d 4 n TFor large n,Td1 1+ 2 28 nT13.14((a) r = 22122)22 = 2 2 = 0.70721 1 22 1 4(b) Q = d === 0.86622 2 2 2 ( 2 )222(c) tan = 2= 2 2 =2 43122() 2 = tan 1 = 146.331222
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3.16(b) Q =2 2d=222 =1,LC=R2L2 1 R 2 LC 4 L L 1Q== 2 R R C 42 2L L R(c) Q = = C =RR 2 3.17Fext = F sin t = Im F ei t and x ( t ) is the imaginary part of the solution to:mx + cx + kx = F ei ti.e.i t x ( t ) = Im Ae
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= tan 1 ( c 2m )m ( 2 2 ) c + kUsing sin 2 + cos 2 = 1 ,2F22= m ( 2 2 ) c + k + 2 ( c 2m )2AFA=cfw_m ( ) c + k + ( c 2m ) 2222122and x ( t ) = Ae t cos ( t ) + the transient term.3.19l A2 T 21 g8(a)for A =(b)412l1.041g,
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1 Tf ( t ) = c + 2T f ( t ) cos ( n t ) dt cos n tn T2T1+ 2T f ( t ) sin ( n t ) dt sin n t ,n T21Tn = 1, 2, . . ., and c = 2T f ( t ) dtT 2Now, due to the equality of terms in n :2 Tf ( t ) = c + 2T f ( t ) cos ( n t ) dt cos n tn T2T2
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3.22In steady state, x ( t ) = An e (i n t n )nAn =Fnm1( 2 n 2 2 )2 + 4 2 n 2 2 24F,n = 1, 3, 5, . . .andNow Fn =n9 2Q = 100 so 2 40, 00024F1A1 =1m22229 22( 9 ) + 4 40000 FA1 2m 24F1A3 =13m2 2 22 9 ( 9 2 9 2 ) +
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3.24The equation of motion is F ( x ) = x x 3 = mx . For simplicity, let m=1. Then(a)(b)x = x x 3 . This is equivalent to the two first order equations x = y andy = x x3The equilibrium points are defined byx x 3 = x (1 x ) (1 + x ) = 0Thus, the p
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3.25 + sin = 0d 2 cos = 0dt 22Integrating:2= cos 0)0T = 4 2 = 2 ( cos cosord1 2 ( cos cos ) 2Time for pendulum to swing from = 0 to = isNowsubstitute sin =sinsin2so =and after some algebra 2at = 2cos = 1 2 sinand use the iden
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CHAPTER 4GENERAL MOTION OF A PARTICLEIN THREE DIMENSIONS-Note to instructors there is a typo in equation 4.3.14. The range of the projectile is v 2 sin 2v 2 sin 2 2 NOT . 0R=x= 0gg-4.1V V VjkxyzF = c iyz + xz + kxyj(a) F = V = i()(b)
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er1 F = 2r sin r kr n4.3e r0e r sin = 0 conservative0(a)i F =xxyjycx 2k= k ( 2cx x )zz3jycxzy2kzxy2cx x = 01c=2(b)i F =xzya l so4.4(a) x cx = i 2 2 +yyx cx 2 2 =0yycz z+=0y2 y2 1 1 + k cz + z j2y
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v2 = v2 2( + + )m122v = v 2 ( + + ) m(b)v2 2( + + ) = 0m122v = ( + + ) m(c)mx = Fx = Vxmx = Vmy = = 2 yyVmz = = 3 z 2z4.5(a)jF = ix + ydr = idx + dyjon the path x = y :(1,1)(0,0 )11110000F dr = Fx dx + Fy
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(1,1)(and, with x = 1on this path1,0 )(1,1)(0,0 )1F dr = dy = 10F dr = 0 + 1 = 1F is not conservative.4.6From Example 2.3.2, V ( z ) = mgre2( re + z )zV ( z ) = mgre 1 + re From Appendix D,(1 + x )11= 1 x + x2 + z z2V ( z ) = mg
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h=re 1 2 2re v 2zre g22h=re re2v 21 z22greFrom Appendix D,( h, z1(1 + x ) 2 = 1 +h=x x2 +28re re v 2zv4++ z +2 2 2 g 4 grehre )v 2z v 2z 1 +2 g 2 gre 1v2 v2 From Example 2.3.2, h =1 2 g 2 gre v2 v2 1And with (1 x ) 1
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sin = gbv2cos 2 = 1 sin 2 =v 4 g 2b 2v4gb 2 v 4 g 2b 2 gb 2 v 2+= 2+v22 gv 22v2gMeasured from the ground,gb 2 v 2hmax = b + 2 +2v2ghmax = gb The mud leaves the wheel at = sin 1 2 v4.8x = R cos so t =andx = v x t = ( v cos ) tR
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Now sin = cos = cos + 4 2 4 2 2 4 2 2v 2 Rmax =cos 2 + 2g cos 4 2Again using Appendix B, cos 2 = cos 2 sin 2 = 2 cos 2 1v2 2v 2 1 1Rmax =cos + + =cos 2 + + 1222g cos 2 2 g cos Using cos + = sin ,22vRmax =(1 sin )g (1
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2sin 2 =2gh= 1 22csc v01gh 1 + 2 = v0 1 + ghv0 2gh Finally csc 2 = 2 1 + 2 v0 (b)Solving for Rmax Rmax =hhh==2sin 1 2 sin 1 2 csc2 Substituting for csc 2 and solving v2gh Rmax = 0 1 + 2 g v0 4.10 We can again use the results
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zzmaxv0h0h1x0xx1RWe have reversed the trajectory so that h0 (= 9.8 ft), and x0 , the height and range withinwhich Mickey can catch the ball represent the starting point of the trajectory. h1 (=3.28ft) is the height of the ball when Mickey strik
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u zmax = h1oru 2 zmax ( 2 ) = 2 zmax ( 2 .0475 ) = 3.9 zmax . This result is the correct one 3.9 zmax= 0.821 = 39.40x1Now solve for x0 using a relation identical to (4) Thus, tan =h0 = x0( x tan )tan 024 zmaxAgain we obtain a quadratic expres
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(a) The maximum height occurs ator1v cos sin = gT2v21H=cos 2 sin 2 2g2or atdz=0dt1v cos sin 2T=gmaximum at fixed dH=0ddH v 2 11212= 2 cos sin cos cos sin sin = 0d 2 g 222Using the above trigonometric identities, we get1
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projection on the xy-plane make an angle with the x-axis:x = s sin cos ,and Fx = Fr sin cos = mxy = s si n si n ,and Fy = Fr sin sin = myz = s cos ,and Fz = mg + Fr cos = mzSince Fr = c2 s = c2 ( x + y 2 + z 2 ) , the differential equations of moti
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2 x z 8x z 2 +g3g 2z = v sin and 2 x z = v 2 sin 2 :xmax =Forxmax =v 2 sin 2 4v 3 sin 2 sin +g3g 24.16x = A cos ( t + ) ,xx = A sin ( t + )from x = 0 ,from x = A ,y = B cos ( t + ) ,y =0x = A cos ty = B sin ( t + )12121 2kB = ky +
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V= 4 2 mxyy = B cos ( 2 t + )my = V= 9 2 mzzz = C cos ( 3 t + )mz = Since x = y = z = 0 at t = 0 , = = = 2x = A cos t = A sin t2x = A cos tSince v 2 = x 2 + y 2 + z 2 and x = y = z ,vx== A3vA=3vx=sin t3y = B sin 2 t ,y = 2 B c
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tmin =4.182=2Equation 4.4.15 istan 2 =2 AB cos A2 B 2Transforming the coordinate axes xyz to the new axes xyz by a rotation aboutthe z-axis through an angle given, from Section 1.8:x = x cos + y sin ,y = x sin + y cosor,x = x cos y sin , and
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1r1 = ( d x ) + y 2 + z 2 22 2x x2 + y2 + z 2 2+r = ( d 2 dx + x + y + z ) = d 1 dd21nFrom Appendix D, (1 + x ) = 1 + nx + n ( n 1) x 2 + 22 2x x + y2 + z2 1 r1 = d 1 + + 1d2 2 d 2 2 2 12222 22222222 2 x 2x3 2x x + y + z x
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zkB(F = q E+vBEjiv0())jv B = ix + y + kz kB = iyB xBjF = iqyB + q ( E xB )jymx = Fx = qyBqBxx =ymxqBmy = Fy = qE qxB = qE qB x +my22qE qBx qB eE eBx eB y=+ y y=mm mmm meEeBy +2 y = +x ,=mm1 eEy = 2 + x + A co
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hbh mv 2 mgmgR = mg = h ( b 2h ) = b ( 3h b )bbbthe particle leaves the side of the sphere when R = 0bbh = , i.e.,above the central plane33cos =4.2212mv + mgh = 02at the bottom of the loop, h = b12mv = mgb ,2v = 2 gbsohbmv 2
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CHAPTER 5NONINERTIAL REFERENCE SYSTEMS5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net forceacting on him is zero. The scale exerts an upward force, N , whose value is equal to thescale reading - the weight, W, of
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gg = g A = g ij10For small oscillations of a simple pendulum:1T = 2g2gg = g + = 1.005 g 10 2T = 25.5 (a)11= 1.9951.005 ggf = mg is the frictional force acting on theA0box, sof mA0 = ma(b)(a)f( a is the acceleration of the box re
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= i R sin t + R cos t x + i yjjx = y R sin ty = x + R cos there i = 1 !(c) Let u = x + iyu = x + iy = y R sin t i x + iR cos t= yi u =+i x u + i u = R sin t + iR cos t = i ReitTry a solution of the formu = Ae i t + Beitu = i Ae i t + iBeit
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A, A are the accelerations of the asteroid and the Earth in the fixed, inertial frameof reference.1st : examine:A A r= R R ( R R )= ( ) R = ( 2 2 ) Rnote: = k , = kThus:a = ( 2 2 ) R 2 vTherefore: jyix + = ( 2 2 ) iR cos ( ) t R sin ( ) t 2 x +
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= k , with = 0r = ber , so ( r ) = b 2 er v = ve , so v = ver From eqn 5.2.14,a = a + 2 v + ( r ) and putting in terms from abovev 2 2 v b 2ar = bFor no slipping F s mg , so a s gv 2+ 2 v + b 2 s gb2vm + 2 bvm + b 2 2 b s g = 0vm = b 2b 2
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5.10(See Example 5.3.3)m 2 x = mxx ( t ) = Ae t + Be tx ( t ) = Ae t Be tBoundary Conditions:lx2 ( 0) = = A + B2x ( 0 ) = 0 = ( A B )A=l4B=lx ( t ) = cosh t2lllx (T ) = + = cosh T222(a)(b) cosh T = 2lx ( t ) = sinh t2when the bea
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v = zv yi + zvx yvx k jj( v )horiz = zvyi + zvx ( v )horiz112222= ( z2v y + z2vx ) 2 = z ( vx + v y ) 2 = zvFcor = 2m v(F )cor5.13horiz= 2m ( v )horiz = 2m z v , independent of the direction of v .From Example 5.4.1 11 8h 3 2 = xh
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da = r + r + r + 2 r + 2 r dt rot()() a = r + ( r ) + 2 ( r ) + ( r ) + ( r ) + r + r Now( r )is to and r . Let this define a direction n : r = r nSince n , ( r ) is in the plane defined by and r and ( r ) = n r = r .Since ( r ) ( r ) = 2
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1z ( t ) = gt 2 + v t sin + v t 2 cos cos = 022v sin 2v sin t=org 2 v cos cos gWe have ignored the second term in the denominatorsince v would have to beimpossibly large for the value of that term to approach the magnitude gFor example, for = 4
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(jHence: a = a 2 v = 3 2 x i 2 k ix + ySo5.19)ja = ix + y = 3 2 xi + 2 yi 2 xj2x 2 y 3 x = 0y + 2 x = 0(mr = qE + q v BEquation 5.2.14)r = r + r + 2 v + ( r )v = v + r Equation 5.2.13q =B so = 02mqmr q B v B ( r ) = qE + q ( v + r
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Collecting terms and neglecting terms in 2 :gg x + x cos t + y + y sin t = 0llgg x + x sin t y + y cos t = 0ll24hourssin 24T== 73.7 hourssin 195.21T=5.22Choose a coordinate system with the origin at the center of the wheel, the x and
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r = 0Vb VtVt VtVtVt Vt k sin cos + V sin i cos j k sin cosbbbbbb22V Vt Vt V Vt Vt+ k cos 2 j cos ( r ) = k sin 2+ i sinbb bbb22VV ( r ) = k nbSince the origin of the coordinate system is traveling in a circle of r
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CHAPTER 6GRAVITATIONAL AND CENTRAL FORCES6.14m = V = rs331 3m 3rs = 4 F=Gmm( 2rs )222Gm 2 4 3 G 4 3 43== m4 3m 4 3 2FFGm 2 4 3 3==mW mg4g 3 121F 6.672 1011 N m 2 kg 2 4 11.35 g cm 3 1 kg 106 cm3 3=3 (1 kg ) 3W4 9.8
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The gravitational force on the particle is due only to the mass of the earth that isinside the particles instantaneous displacement from the center of the earth, r.The net effect of the mass of the earth outside r is zero (See Problem 6.2).4M = r334
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6.56.6GMm mv 2GMv2 ==so2rrrfor a circular orbit r, v is constant.2 rT=v4 2 r 2 4 2 3T2 =r r3=2vGM2 rvFrom Example 6.5.3, the speed of a satellite in circular orbit is (a) T =1 gR 2 2v= e r3T=2 r 21g 2 Re1 T 2 gRe2 3r =