Homework Answers
In the above graph between average mass and acceleration, we have to use automatic curve fitting and apply the inverse relationship => (y= a/x+c)
C is constant term is a term in the above algebraic expression that has a value that is constant or cannot change, because it does not contain any modifiable variables.
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Using the second set of data, plot a graph of Total Mass on the x-axis and...
Table 4.2: Force and acceleration. kg Total mass of the system Mass on the hanger (kg)...
Table 4.2: Force and acceleration. kg Total mass of the system Mass on the hanger (kg) Net force on the system (N) Acceleration Trial 1 (m/s) Acceleration Trial 2 (m/s) Acceleration Trial 3 (m/s) Average acceleration (m/s) 0.685 0.612 0.655 0.005 0.883 0.857 0.009 0.861 0.948 0.928 0.927 0.013 1.34 1.42 1.43 0.023 1.67 1.77 1.75 0.043 Correlation: Intercept: Slope: MEASUREMENTS 1. Record the total mass of the hanging objects in the first column of Table 4.2. Calculate the weight...
1). A cart of mass M = 500 g on 20.0 degree incline plane is connected...
1). A cart of mass M = 500 g on 20.0 degree incline plane is connected by a light string passing over massless pulley to the mass m=200 g hanging on the string. What is the acceleration of the cart (say, in +x downhill direction) acart = m/s', of the hanging mass amass m = m/s”, and how much is the tension in the thread T = 2) Two masses m=1.00 kg and M= 2.00 kg are hanging on a...
data included 3. In which situation did the rotating arm spin faster, with the weights closer...
data included
3. In which situation did the rotating arm spin faster, with the weights closer to or further from the axis of rotation? Explain why. We were unable to transcribe this imagePart 2 - Varying Torque by Changing Pulley Diameter Table 3: Varying torque by changing pulley diameter Distance of masses from axis = Hanging mass value=' Pulley Diameter Average Angular (cm) Acceleration (rad/s) Data Run 1 Large ए.पीड Data Run 2 Medeon 3.7 parts Data Run 3 SMG!...
1. A graph of the acceleration a, against the difference in mass 1 and mass 2...
1. A graph of the acceleration a, against the difference in mass 1 and mass 2 (delta m), with the sum of masses (Mtotal) held constant for the Atwood machine (like in question 2) gives: a) slope = g; Intercept on the y-axis = 0 b) slope = g/Mtotal; Intercept on the y-axis = 1 c) slope = g; Intercept on the y-axis = 1 d) slope = g/Mtotal; Intercept on the y-axis = 0 2. What is the acceleration...
The mass of the cart is 0.365 kg. The rest of the data can be obtained from the graph above. All answers below must be...
The mass of the cart is 0.365 kg. The rest of the data can be obtained from the graph above. All answers below must be correct to 3 significant figures. Units are required if not stated. What the final speed of the cart predicted by the work done by the rubber band, assuming that friction negligible? Vpred What the percent difference between between the predicted and measured final speed, expressed as a percent of the measured speed? Percent difference between...
DATA TABLE Number of books Height of books, h (m) Length of incline, x (m) Acceleration...
DATA TABLE Number of books Height of books, h (m) Length of incline, x (m) Acceleration Trial 1 Trial 2 Trial 3 (m/s) (m/s) (m/s) Average acceleration (m/s) sin(0) 3 4 5 10,005 m 0.092 0.0082m 0.092 6.ollm 0.092 0.0135m20 0.0154m/0.092 0.05 0.785 0.780 0.775 1.78) 10.08/1.02/1.17 1.08 1.09 11.2 1.41 1.32 1.72 1.48 10.14 1.60 1.681.68 1.65 0.15 1.81 1.77 1.80 1679 ANALYSIS 1. Using trigonometry and your value Bongonometry and your values of x and h in the...
looking for the rest of the answers to pages 5-2, 5-6. (graphs are included at the...
looking for the rest of the answers to pages 5-2, 5-6. (graphs
are included at the end)
DATA 722.1 Mass of cart with plastic box mounted on it- Mass constant, resultant force varied 0:78 Total moving mass =_7799-m: Acceleration (in m/s Resultant Force 0.1728 0.196 20 gm wt = nt G-311 gm wt = 0.39Z 40gm wt = 0-$584 Resultant force constant, mass varied Resultant forceo gm wt-9.49 nt Reciprocal of total moving mass (Can be obtained when use Excel...
Question 7 is related to the force vs mass graph that is provided and the first...
Question 7 is related to the force vs mass graph that is
provided and the first section of the excel sheet. Question 3 has
to do with the force bs acceleration graph and second section of
the excel sheet. The first two files are showing the equations that
are supposed to be used to find these answers. Any help would be
greatly appreciated.
I
mainly need assistance on number 1 and 2 now. The question with the
free body diagram...
Please do not answer unless you are sure it is correct. thank you. " A mass...
Please do not answer unless you are sure it is correct. thank
you.
" A mass hanging from a vertical spring is somewhat more complicated than a mass attached to a horizontal spring because the gravitational force acts along the direction of motion. Therefore, the restoring force of the oscillations is not provided by the spring force alone, but by the net force resulting from both the spring force and the gravitational force. Ultimately, however, the physical quantities of motion...
I need help on everything. Measured Force Constant, k (N/m): 5.29 (20 points) Mass, m (kg)_0.100...
I need help on everything.
Measured Force Constant, k (N/m): 5.29 (20 points) Mass, m (kg)_0.100 kg tittettststtst Time (s) 208 2.94 3.844.12 5.60 6.48 1.36 8.249.14 Period (s) .86.9 -88 -88.88.88.88.9 Calculate the experimental values for the period of oscillation between each consecutive time value, then use those values to calculate the experimental average period value. (10 points) Average Period, Tavg (s): -88 Physical Quantity Crest (m) Trough (m) Amplitude (m) Position Wave 501 .348 0.79 Velocity Wave I-0.57...
Table 4.2: Force and acceleration. kg Total mass of the system Mass on the hanger (kg) Net force on the system (N) Acceleration Trial 1 (m/s) Acceleration Trial 2 (m/s) Acceleration Trial 3 (m/s) Average acceleration (m/s) 0.685 0.612 0.655 0.005 0.883 0.857 0.009 0.861 0.948 0.928 0.927 0.013 1.34 1.42 1.43 0.023 1.67 1.77 1.75 0.043 Correlation: Intercept: Slope: MEASUREMENTS 1. Record the total mass of the hanging objects in the first column of Table 4.2. Calculate the weight...
1). A cart of mass M = 500 g on 20.0 degree incline plane is connected by a light string passing over massless pulley to the mass m=200 g hanging on the string. What is the acceleration of the cart (say, in +x downhill direction) acart = m/s', of the hanging mass amass m = m/s”, and how much is the tension in the thread T = 2) Two masses m=1.00 kg and M= 2.00 kg are hanging on a...
data included
3. In which situation did the rotating arm spin faster, with the weights closer to or further from the axis of rotation? Explain why. We were unable to transcribe this imagePart 2 - Varying Torque by Changing Pulley Diameter Table 3: Varying torque by changing pulley diameter Distance of masses from axis = Hanging mass value=' Pulley Diameter Average Angular (cm) Acceleration (rad/s) Data Run 1 Large ए.पीड Data Run 2 Medeon 3.7 parts Data Run 3 SMG!...
The mass of the cart is 0.365 kg. The rest of the data can be obtained from the graph above. All answers below must be correct to 3 significant figures. Units are required if not stated. What the final speed of the cart predicted by the work done by the rubber band, assuming that friction negligible? Vpred What the percent difference between between the predicted and measured final speed, expressed as a percent of the measured speed? Percent difference between...
DATA TABLE Number of books Height of books, h (m) Length of incline, x (m) Acceleration Trial 1 Trial 2 Trial 3 (m/s) (m/s) (m/s) Average acceleration (m/s) sin(0) 3 4 5 10,005 m 0.092 0.0082m 0.092 6.ollm 0.092 0.0135m20 0.0154m/0.092 0.05 0.785 0.780 0.775 1.78) 10.08/1.02/1.17 1.08 1.09 11.2 1.41 1.32 1.72 1.48 10.14 1.60 1.681.68 1.65 0.15 1.81 1.77 1.80 1679 ANALYSIS 1. Using trigonometry and your value Bongonometry and your values of x and h in the...
looking for the rest of the answers to pages 5-2, 5-6. (graphs
are included at the end)
DATA 722.1 Mass of cart with plastic box mounted on it- Mass constant, resultant force varied 0:78 Total moving mass =_7799-m: Acceleration (in m/s Resultant Force 0.1728 0.196 20 gm wt = nt G-311 gm wt = 0.39Z 40gm wt = 0-$584 Resultant force constant, mass varied Resultant forceo gm wt-9.49 nt Reciprocal of total moving mass (Can be obtained when use Excel...
Question 7 is related to the force vs mass graph that is
provided and the first section of the excel sheet. Question 3 has
to do with the force bs acceleration graph and second section of
the excel sheet. The first two files are showing the equations that
are supposed to be used to find these answers. Any help would be
greatly appreciated.
I
mainly need assistance on number 1 and 2 now. The question with the
free body diagram...
Please do not answer unless you are sure it is correct. thank
you.
" A mass hanging from a vertical spring is somewhat more complicated than a mass attached to a horizontal spring because the gravitational force acts along the direction of motion. Therefore, the restoring force of the oscillations is not provided by the spring force alone, but by the net force resulting from both the spring force and the gravitational force. Ultimately, however, the physical quantities of motion...
I need help on everything.
Measured Force Constant, k (N/m): 5.29 (20 points) Mass, m (kg)_0.100 kg tittettststtst Time (s) 208 2.94 3.844.12 5.60 6.48 1.36 8.249.14 Period (s) .86.9 -88 -88.88.88.88.9 Calculate the experimental values for the period of oscillation between each consecutive time value, then use those values to calculate the experimental average period value. (10 points) Average Period, Tavg (s): -88 Physical Quantity Crest (m) Trough (m) Amplitude (m) Position Wave 501 .348 0.79 Velocity Wave I-0.57...
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