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Sagot :
Let's address the task of calculating the volumes of the samples provided in the given table and filling in the missing values. We'll process each material step-by-step.
### Aluminum Bar
Given Data:
- Mass: 39 g
- Dimensions: [tex]\( \text{length} = 9.7 \text{ cm} \)[/tex], [tex]\( \text{width} = 1.6 \text{ cm} \)[/tex], [tex]\( \text{height} = 1.0 \text{ cm} \)[/tex]
- Displacement Volume: [tex]\( 40 \text{ mL} - 25 \text{ mL} = 15 \text{ mL} \)[/tex]
Calculated Volume:
To find the calculated volume of the aluminum bar, we use:
[tex]\[ \text{Volume} = \text{length} \times \text{width} \times \text{height} \][/tex]
[tex]\[ \text{Volume} = 9.7 \text{ cm} \times 1.6 \text{ cm} \times 1.0 \text{ cm} = 15.52 \text{ cm}^3 \][/tex]
### PE (Polyethylene) Rod
Given Data:
- Mass: 21 g
- Dimensions: [tex]\( \text{diameter} = 1.7 \text{ cm} \)[/tex], [tex]\( \text{length} = 10.5 \text{ cm} \)[/tex]
- Displacement Volume: [tex]\( 48 \text{ mL} - 25 \text{ mL} = 23 \text{ mL} \)[/tex]
Calculated Volume:
To find the calculated volume of the PE rod (cylinder), we use the formula for the volume of a cylinder:
[tex]\[ \text{Volume} = \pi \times (\text{radius})^2 \times \text{length} \][/tex]
The radius is half of the diameter:
[tex]\[ \text{radius} = \frac{\text{diameter}}{2} = \frac{1.7 \text{ cm}}{2} = 0.85 \text{ cm} \][/tex]
[tex]\[ \text{Volume} = \pi \times (0.85 \text{ cm})^2 \times 10.5 \text{ cm} \approx 23.832907268295564 \text{ cm}^3 \][/tex]
### Iron Bolt
Given Data:
- Mass: 39 g
- Dimensions: Irregular shape (cannot use length, width, and height)
- Displacement Volume: [tex]\( 31.9 \text{ mL} - 25 \text{ mL} = 6.9 \text{ mL} \)[/tex]
Calculated Volume:
The iron bolt has an irregular shape, so its volume is found using water displacement only. Therefore, we don't calculate the volume using geometric measurements.
The volume of the iron bolt is provided by the displacement method as [tex]\( 6.9 \text{ mL} \)[/tex].
### Summary
Here are the calculated volumes and displacement volumes:
Aluminum Bar:
- Calculated Volume: [tex]\( 15.52 \text{ cm}^3 \)[/tex]
- Displacement Volume: [tex]\( 15 \text{ mL} \)[/tex]
PE Rod:
- Calculated Volume: [tex]\( 23.832907268295564 \text{ cm}^3 \)[/tex]
- Displacement Volume: [tex]\( 23 \text{ mL} \)[/tex]
Iron Bolt:
- Calculated Volume: Not determined directly (irregular shape)
- Displacement Volume: [tex]\( 6.9 \text{ mL} \)[/tex]
### Explanation
We could not use the geometric method (length x width x height) to calculate the volume of the bolt because it has an irregular shape. For such objects, the displacement method is a reliable way to measure volume, as it accounts for the total volume of the object by measuring the amount of water it displaces.
### Aluminum Bar
Given Data:
- Mass: 39 g
- Dimensions: [tex]\( \text{length} = 9.7 \text{ cm} \)[/tex], [tex]\( \text{width} = 1.6 \text{ cm} \)[/tex], [tex]\( \text{height} = 1.0 \text{ cm} \)[/tex]
- Displacement Volume: [tex]\( 40 \text{ mL} - 25 \text{ mL} = 15 \text{ mL} \)[/tex]
Calculated Volume:
To find the calculated volume of the aluminum bar, we use:
[tex]\[ \text{Volume} = \text{length} \times \text{width} \times \text{height} \][/tex]
[tex]\[ \text{Volume} = 9.7 \text{ cm} \times 1.6 \text{ cm} \times 1.0 \text{ cm} = 15.52 \text{ cm}^3 \][/tex]
### PE (Polyethylene) Rod
Given Data:
- Mass: 21 g
- Dimensions: [tex]\( \text{diameter} = 1.7 \text{ cm} \)[/tex], [tex]\( \text{length} = 10.5 \text{ cm} \)[/tex]
- Displacement Volume: [tex]\( 48 \text{ mL} - 25 \text{ mL} = 23 \text{ mL} \)[/tex]
Calculated Volume:
To find the calculated volume of the PE rod (cylinder), we use the formula for the volume of a cylinder:
[tex]\[ \text{Volume} = \pi \times (\text{radius})^2 \times \text{length} \][/tex]
The radius is half of the diameter:
[tex]\[ \text{radius} = \frac{\text{diameter}}{2} = \frac{1.7 \text{ cm}}{2} = 0.85 \text{ cm} \][/tex]
[tex]\[ \text{Volume} = \pi \times (0.85 \text{ cm})^2 \times 10.5 \text{ cm} \approx 23.832907268295564 \text{ cm}^3 \][/tex]
### Iron Bolt
Given Data:
- Mass: 39 g
- Dimensions: Irregular shape (cannot use length, width, and height)
- Displacement Volume: [tex]\( 31.9 \text{ mL} - 25 \text{ mL} = 6.9 \text{ mL} \)[/tex]
Calculated Volume:
The iron bolt has an irregular shape, so its volume is found using water displacement only. Therefore, we don't calculate the volume using geometric measurements.
The volume of the iron bolt is provided by the displacement method as [tex]\( 6.9 \text{ mL} \)[/tex].
### Summary
Here are the calculated volumes and displacement volumes:
Aluminum Bar:
- Calculated Volume: [tex]\( 15.52 \text{ cm}^3 \)[/tex]
- Displacement Volume: [tex]\( 15 \text{ mL} \)[/tex]
PE Rod:
- Calculated Volume: [tex]\( 23.832907268295564 \text{ cm}^3 \)[/tex]
- Displacement Volume: [tex]\( 23 \text{ mL} \)[/tex]
Iron Bolt:
- Calculated Volume: Not determined directly (irregular shape)
- Displacement Volume: [tex]\( 6.9 \text{ mL} \)[/tex]
### Explanation
We could not use the geometric method (length x width x height) to calculate the volume of the bolt because it has an irregular shape. For such objects, the displacement method is a reliable way to measure volume, as it accounts for the total volume of the object by measuring the amount of water it displaces.
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