Question
×
mnol=50g
xnol=19cm
M=(m+mnol) plus or minus 0.05cm 250, 450, 650,850,1050
loading =25,30,35,40,45,45
unloading =25,30,35,40,45
Mean,pointer (x plus or minus 0.05cm)= 25,30,35,40,45
Extension e=(x-xnol) plus or minus 0.05cm 6cm , 11cm, 16cm, 2cm, 26cm
1) plot a graph of M against e
2) measure the slope and explain its physical meaning. state its unit and estimate the error
3)did your graph pass through the origin? if not, give reasons
4) deduce from your load ( mass) extension, graph the weight W subscription A of the given object, A .
5) estimate the error in W subscrpt A
mnol=50g
xnol=19cm
M=(m+mnol) plus or minus 0.05cm 250, 450, 650,850,1050
loading =25,30,35,40,45,45
unloading =25,30,35,40,45
Mean,pointer (x plus or minus 0.05cm)= 25,30,35,40,45
Extension e=(x-xnol) plus or minus 0.05cm 6cm , 11cm, 16cm, 2cm, 26cm
1) plot a graph of M against e
2) measure the slope and explain its physical meaning. state its unit and estimate the error
3)did your graph pass through the origin? if not, give reasons
4) deduce from your load ( mass) extension, graph the weight W subscription A of the given object, A .
5) estimate the error in W subscrpt A
Answers
GPT-4o mini
To address your questions systematically, let's break each part down step by step:
### 1) Plot a Graph of M Against e
To plot a graph of M (mass plus the mass of the additional load) against e (extension), we start by calculating the values for M by adding the mass \( m \) (in grams) to \( mnol = 50g \) for each loading value.
#### Calculation of M:
Given values of loading are 25g, 30g, 35g, 40g, 45g, and an additional 50g (mass of mnol), we can calculate:
- For loading = 25g: \( M = 50g + 25g = 75g \)
- For loading = 30g: \( M = 50g + 30g = 80g \)
- For loading = 35g: \( M = 50g + 35g = 85g \)
- For loading = 40g: \( M = 50g + 40g = 90g \)
- For loading = 45g: \( M = 50g + 45g = 95g \)
- For loading = 45g (duplicate): \( M = 50g + 45g = 95g \)
Using the extensions provided:
- e values are given as 6cm, 11cm, 16cm, 21cm, and 26cm respectively.
Your final (M, e) pairs will be:
- (75g, 6cm)
- (80g, 11cm)
- (85g, 16cm)
- (90g, 21cm)
- (95g, 26cm)
When you plot these points on a graph with M on the y-axis and e on the x-axis, you will create a scatter plot. If possible, use a graphing tool or software to plot accurately.
### 2) Measure the Slope and Explain Its Physical Meaning
After plotting your graph, you can determine the slope of the line that best fits the data points.
#### Physical Meaning of Slope:
The slope of the graph (Rise/Run) represents the stiffness (spring constant \( k \)) of the material. The relation follows Hooke's Law, \( F = kx \), where:
- \( F \) corresponds to the force due to weight (i.e., \( M \cdot g \)) and
- \( x \) corresponds to the extension \( e \).
#### Unit of Slope:
The unit for the slope will be \( \text{g/cm} \) or \( \text{N/m} \) depending on whether you're calculating the force in grams or converting it to Newtons (where \( 1 g \approx 0.00981 N \)).
#### Estimate the Error:
To estimate the error in the slope, consider the error in the M and e values. If both have an uncertainty of \( \pm 0.05 \) (either mass or extension), you can propagate this uncertainty through your slope calculation.
### 3) Did Your Graph Pass Through the Origin? If Not, Give Reasons
If the graph does not pass through the origin, consider physical reasons:
- There could be a systematic error or offset in measurements, such as the material having an initial extension before any load.
- The spring or material may not start at zero extension due to pre-tension or stress.
If it passes through (0,0), that would suggest the material behaves ideally under Hooke's law with no pre-tension.
### 4) Deduce from Load (Mass) Extension, Graph the Weight \( W_A \) of the Given Object \( A \)
The weight \( W_A \) can be calculated as:
\[ W_A = M \cdot g \]
Where \( g = 9.81 \, \text{m/s}^2 \).
You will consider \( W_A \) in terms of Newtons (after converting from grams). For instance:
- For \( M = 75g \): \( W_A = 0.075 \, kg \times 9.81 \, m/s^2 = 0.73575 \, N \)
Repeat for each mass value to get \( W_A \).
### 5) Estimate the Error in \( W_A \)
Given that the mass has an error of \( \pm 0.05g \):
- The error in weight, using propagation of uncertainties:
\[ \Delta W_A = \Delta M \cdot g \]
Where \( \Delta M = 0.05g = 0.00005 \, kg \)
#### Resulting Error:
Using \( g = 9.81 \):
\[ \Delta W_A = 0.00005 \, kg \cdot 9.81 \, m/s^2 \approx 0.0004905 \, N\]
That's how you can analyze your data with physical principles! Be sure to plot and derive appropriate values based on your calculations.
### 1) Plot a Graph of M Against e
To plot a graph of M (mass plus the mass of the additional load) against e (extension), we start by calculating the values for M by adding the mass \( m \) (in grams) to \( mnol = 50g \) for each loading value.
#### Calculation of M:
Given values of loading are 25g, 30g, 35g, 40g, 45g, and an additional 50g (mass of mnol), we can calculate:
- For loading = 25g: \( M = 50g + 25g = 75g \)
- For loading = 30g: \( M = 50g + 30g = 80g \)
- For loading = 35g: \( M = 50g + 35g = 85g \)
- For loading = 40g: \( M = 50g + 40g = 90g \)
- For loading = 45g: \( M = 50g + 45g = 95g \)
- For loading = 45g (duplicate): \( M = 50g + 45g = 95g \)
Using the extensions provided:
- e values are given as 6cm, 11cm, 16cm, 21cm, and 26cm respectively.
Your final (M, e) pairs will be:
- (75g, 6cm)
- (80g, 11cm)
- (85g, 16cm)
- (90g, 21cm)
- (95g, 26cm)
When you plot these points on a graph with M on the y-axis and e on the x-axis, you will create a scatter plot. If possible, use a graphing tool or software to plot accurately.
### 2) Measure the Slope and Explain Its Physical Meaning
After plotting your graph, you can determine the slope of the line that best fits the data points.
#### Physical Meaning of Slope:
The slope of the graph (Rise/Run) represents the stiffness (spring constant \( k \)) of the material. The relation follows Hooke's Law, \( F = kx \), where:
- \( F \) corresponds to the force due to weight (i.e., \( M \cdot g \)) and
- \( x \) corresponds to the extension \( e \).
#### Unit of Slope:
The unit for the slope will be \( \text{g/cm} \) or \( \text{N/m} \) depending on whether you're calculating the force in grams or converting it to Newtons (where \( 1 g \approx 0.00981 N \)).
#### Estimate the Error:
To estimate the error in the slope, consider the error in the M and e values. If both have an uncertainty of \( \pm 0.05 \) (either mass or extension), you can propagate this uncertainty through your slope calculation.
### 3) Did Your Graph Pass Through the Origin? If Not, Give Reasons
If the graph does not pass through the origin, consider physical reasons:
- There could be a systematic error or offset in measurements, such as the material having an initial extension before any load.
- The spring or material may not start at zero extension due to pre-tension or stress.
If it passes through (0,0), that would suggest the material behaves ideally under Hooke's law with no pre-tension.
### 4) Deduce from Load (Mass) Extension, Graph the Weight \( W_A \) of the Given Object \( A \)
The weight \( W_A \) can be calculated as:
\[ W_A = M \cdot g \]
Where \( g = 9.81 \, \text{m/s}^2 \).
You will consider \( W_A \) in terms of Newtons (after converting from grams). For instance:
- For \( M = 75g \): \( W_A = 0.075 \, kg \times 9.81 \, m/s^2 = 0.73575 \, N \)
Repeat for each mass value to get \( W_A \).
### 5) Estimate the Error in \( W_A \)
Given that the mass has an error of \( \pm 0.05g \):
- The error in weight, using propagation of uncertainties:
\[ \Delta W_A = \Delta M \cdot g \]
Where \( \Delta M = 0.05g = 0.00005 \, kg \)
#### Resulting Error:
Using \( g = 9.81 \):
\[ \Delta W_A = 0.00005 \, kg \cdot 9.81 \, m/s^2 \approx 0.0004905 \, N\]
That's how you can analyze your data with physical principles! Be sure to plot and derive appropriate values based on your calculations.