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5. Suppose the system described in question 4 has a string with a lincar density μ-2.95 x 104 kg/m. What is the tension T, in the string for one loops. Show your work.

5. Suppose the system described in question 4 has a string with a lincar density μ-2.95 x 104 kg/m. What is the tension T, in the string for one loops. Show your work.

Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter3: Vectors
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Hello, based on page 1 can you do ONLY question number 5 (on page 2)
Data Table n 2 3 4 5 6 7 8 μ= 0.00031 fexp M (g) kg/m L= Slope= 260 2 240 9 105 5 65 2 50 9 9 35 = 58.4 H₂ 2m M (kg) 0.260 Ks 0240 kg Vo-00031 x 0.973 0.105 kg 0.65 kg 10.50 ks 0.35 Kg cm= 200 m M 9.81 T, Mg (N) 2 (cm) 2.548 2.352 1.029 6.37 4.90 3.43 Calculation Table height = 19 x 0.27 = 0.973 / base 17 x 0.31 √T, (VN) 200.com 1.596 2 133 cm 133 1.533 100 cm 800 cm Slope= 0.973 / fexp= 58.4 H₂ λ (m) 222 =2 670 cm 2x2 Page 1 I 1. Using at least 75% of the graph paper, make a graph with 2 on Y axis and √T, on the X axis. 2. Draw a best fit line (a line that doesn't hit any data points but rather goes in between data points) and calculate the slope by picking two points exactly on the best fit line, do not use data points to calculate slope. Show your calculation to get credit. Record the slope of the best fit line in Data Table. 0.8 0.67 570m 0.57 3. According to equation (4), à should be linearly proportional to √Ts with Equate your value obtained for slope to this value in Data Table as fexp- 1.014 0.798 0.7 0.585 % Error= 2.67% as a slope. 1 and solve for the unknown frequency f. Record S√μ
Transcribed Image Text:Data Table n 2 3 4 5 6 7 8 μ= 0.00031 fexp M (g) kg/m L= Slope= 260 2 240 9 105 5 65 2 50 9 9 35 = 58.4 H₂ 2m M (kg) 0.260 Ks 0240 kg Vo-00031 x 0.973 0.105 kg 0.65 kg 10.50 ks 0.35 Kg cm= 200 m M 9.81 T, Mg (N) 2 (cm) 2.548 2.352 1.029 6.37 4.90 3.43 Calculation Table height = 19 x 0.27 = 0.973 / base 17 x 0.31 √T, (VN) 200.com 1.596 2 133 cm 133 1.533 100 cm 800 cm Slope= 0.973 / fexp= 58.4 H₂ λ (m) 222 =2 670 cm 2x2 Page 1 I 1. Using at least 75% of the graph paper, make a graph with 2 on Y axis and √T, on the X axis. 2. Draw a best fit line (a line that doesn't hit any data points but rather goes in between data points) and calculate the slope by picking two points exactly on the best fit line, do not use data points to calculate slope. Show your calculation to get credit. Record the slope of the best fit line in Data Table. 0.8 0.67 570m 0.57 3. According to equation (4), à should be linearly proportional to √Ts with Equate your value obtained for slope to this value in Data Table as fexp- 1.014 0.798 0.7 0.585 % Error= 2.67% as a slope. 1 and solve for the unknown frequency f. Record S√μ
4. Calculate and record the percentage error of fexp compared to the known value (standard) of the frequency f-60 Hz. F=IF-F|X100 P 160-58-41 x 100 = 2.67% 60 5. Suppose the system described in question 4 has a string with a linear density µ-2.95 x 104 kg/m. What is the tension T, in the string for one loops. Show your work. ↑ Answer all questions: 1. Was the experiment accurate in finding the experimental value for the frequency? State clearly the basis for your answer. not accurate, since the experimental value obtained i has 2.67 % error from the actual valve 2. Was the experiment of finding the frequency précised? What would your need to answer this question? Explain your reasoning. the result is precise because the values obtained are cumpled around the Some mean value Page 2
Transcribed Image Text:4. Calculate and record the percentage error of fexp compared to the known value (standard) of the frequency f-60 Hz. F=IF-F|X100 P 160-58-41 x 100 = 2.67% 60 5. Suppose the system described in question 4 has a string with a linear density µ-2.95 x 104 kg/m. What is the tension T, in the string for one loops. Show your work. ↑ Answer all questions: 1. Was the experiment accurate in finding the experimental value for the frequency? State clearly the basis for your answer. not accurate, since the experimental value obtained i has 2.67 % error from the actual valve 2. Was the experiment of finding the frequency précised? What would your need to answer this question? Explain your reasoning. the result is precise because the values obtained are cumpled around the Some mean value Page 2
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