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Sagot :
Let's solve the problem step by step.
### Given Data:
- Dimensions for the first configuration: [tex]\(1\, \text{m} \times 1\, \text{m} \times 2\, \text{m}\)[/tex]
- Dimensions for the second configuration: [tex]\(2\, \text{m} \times 1\, \text{m} \times 1\, \text{m}\)[/tex]
- Density of water, [tex]\(\rho = 1000 \, \text{kg/m}^3\)[/tex]
- Gravitational acceleration, [tex]\(g = 9.8 \, \text{m/s}^2\)[/tex]
### (a) Work Done to Fill the Tank in the First Configuration (Height = 2m)
1. Volume Calculation:
The volume of the tank is:
[tex]\[ V = 1 \, \text{m} \times 1 \, \text{m} \times 2 \, \text{m} = 2 \, \text{m}^3 \][/tex]
2. Mass Calculation:
The mass of the water filling the tank is:
[tex]\[ m = \rho \times V = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^3 = 2000 \, \text{kg} \][/tex]
3. Work Calculation:
To pump the water to various heights from 0 to 2 meters, we need to integrate the work done:
[tex]\[ W = \int_0^2 \rho g A h \, dh \][/tex]
Here [tex]\(A = 1 \, \text{m} \times 1 \, \text{m} = 1 \, \text{m}^2\)[/tex] and [tex]\(A\)[/tex] is constant.
Thus,
[tex]\[ W = \rho g A \int_0^2 h \, dh = \rho g A \left[ \frac{h^2}{2} \right]_0^2 = \rho g A \times \frac{2^2}{2} = \rho g \times 2 = 2000 \times 9.8 = 19600 \, \text{J} \][/tex]
### (b) Work Done to Fill the Tank in the Second Configuration (Height = 1m)
1. Volume Calculation:
The volume of the tank remains the same:
[tex]\[ V = 2 \, \text{m} \times 1 \, \text{m} \times 1 \, \text{m} = 2 \, \text{m}^3 \][/tex]
2. Mass Calculation:
The mass of the water filling the tank is:
[tex]\[ m = \rho \times V = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^3 = 2000 \, \text{kg} \][/tex]
3. Work Calculation:
To pump the water to various heights from 0 to 1 meter, we need to integrate the work done:
[tex]\[ W = \int_0^1 \rho g A h \, dh \][/tex]
Here [tex]\(A = 2 \, \text{m} \times 1 \, \text{m} = 2 \, \text{m}^2\)[/tex] and [tex]\(A\)[/tex] is constant.
Thus,
[tex]\[ W = \rho g A \int_0^1 h \, dh = \rho g A \left[ \frac{h^2}{2} \right]_0^1 = \rho g A \times \frac{1^2}{2} = \rho g \times 1 = 2000 \times 9.8 = 19600 \, \text{J} \][/tex]
### (c) Work Done to Fill by Pouring Water into the Top
For both configurations:
1. Mass is the same: [tex]\(2000 \, \text{kg}\)[/tex]
2. Height for the work done:
- First configuration: [tex]\(2 \, \text{m}\)[/tex]
- Second configuration: [tex]\(1 \, \text{m}\)[/tex]
3. Work Calculation when Pouring into the Top:
- For the first configuration:
[tex]\[ W_{\text{top, first}} = m g h = 2000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2 \, \text{m} = 39200 \, \text{J} \][/tex]
- For the second configuration:
[tex]\[ W_{\text{top, second}} = m g h = 2000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2 \, \text{m} = 39200 \, \text{J} \][/tex]
### Comparisons:
1. Difference in Works:
- For the first configuration:
[tex]\[ \Delta W_{\text{first}} = |W_{\text{fill}} - W_{\text{top}}| = |19600 - 39200| = 19600 \, \text{J} \][/tex]
- For the second configuration:
[tex]\[ \Delta W_{\text{second}} = |W_{\text{fill}} - W_{\text{top}}| = |19600 - 39200| = 19600 \, \text{J} \][/tex]
### Explanation:
The work done to fill the tank by pouring directly into the top is larger than the work done by filling it from the bottom because, in the latter method, the water needs to be pumped incrementally from the bottom to the top, meaning each incremental volume of water is lifted to progressively higher heights as the tank fills. In contrast, pouring into the top involves lifting the entire mass of water to the maximum height of 2 meters in an idealized scenario.
Thus, the work required to pour water into the top directly is always more by exactly [tex]\(19600 \, \text{J}\)[/tex] than filling from the bottom for the given dimensions.
### Given Data:
- Dimensions for the first configuration: [tex]\(1\, \text{m} \times 1\, \text{m} \times 2\, \text{m}\)[/tex]
- Dimensions for the second configuration: [tex]\(2\, \text{m} \times 1\, \text{m} \times 1\, \text{m}\)[/tex]
- Density of water, [tex]\(\rho = 1000 \, \text{kg/m}^3\)[/tex]
- Gravitational acceleration, [tex]\(g = 9.8 \, \text{m/s}^2\)[/tex]
### (a) Work Done to Fill the Tank in the First Configuration (Height = 2m)
1. Volume Calculation:
The volume of the tank is:
[tex]\[ V = 1 \, \text{m} \times 1 \, \text{m} \times 2 \, \text{m} = 2 \, \text{m}^3 \][/tex]
2. Mass Calculation:
The mass of the water filling the tank is:
[tex]\[ m = \rho \times V = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^3 = 2000 \, \text{kg} \][/tex]
3. Work Calculation:
To pump the water to various heights from 0 to 2 meters, we need to integrate the work done:
[tex]\[ W = \int_0^2 \rho g A h \, dh \][/tex]
Here [tex]\(A = 1 \, \text{m} \times 1 \, \text{m} = 1 \, \text{m}^2\)[/tex] and [tex]\(A\)[/tex] is constant.
Thus,
[tex]\[ W = \rho g A \int_0^2 h \, dh = \rho g A \left[ \frac{h^2}{2} \right]_0^2 = \rho g A \times \frac{2^2}{2} = \rho g \times 2 = 2000 \times 9.8 = 19600 \, \text{J} \][/tex]
### (b) Work Done to Fill the Tank in the Second Configuration (Height = 1m)
1. Volume Calculation:
The volume of the tank remains the same:
[tex]\[ V = 2 \, \text{m} \times 1 \, \text{m} \times 1 \, \text{m} = 2 \, \text{m}^3 \][/tex]
2. Mass Calculation:
The mass of the water filling the tank is:
[tex]\[ m = \rho \times V = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^3 = 2000 \, \text{kg} \][/tex]
3. Work Calculation:
To pump the water to various heights from 0 to 1 meter, we need to integrate the work done:
[tex]\[ W = \int_0^1 \rho g A h \, dh \][/tex]
Here [tex]\(A = 2 \, \text{m} \times 1 \, \text{m} = 2 \, \text{m}^2\)[/tex] and [tex]\(A\)[/tex] is constant.
Thus,
[tex]\[ W = \rho g A \int_0^1 h \, dh = \rho g A \left[ \frac{h^2}{2} \right]_0^1 = \rho g A \times \frac{1^2}{2} = \rho g \times 1 = 2000 \times 9.8 = 19600 \, \text{J} \][/tex]
### (c) Work Done to Fill by Pouring Water into the Top
For both configurations:
1. Mass is the same: [tex]\(2000 \, \text{kg}\)[/tex]
2. Height for the work done:
- First configuration: [tex]\(2 \, \text{m}\)[/tex]
- Second configuration: [tex]\(1 \, \text{m}\)[/tex]
3. Work Calculation when Pouring into the Top:
- For the first configuration:
[tex]\[ W_{\text{top, first}} = m g h = 2000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2 \, \text{m} = 39200 \, \text{J} \][/tex]
- For the second configuration:
[tex]\[ W_{\text{top, second}} = m g h = 2000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2 \, \text{m} = 39200 \, \text{J} \][/tex]
### Comparisons:
1. Difference in Works:
- For the first configuration:
[tex]\[ \Delta W_{\text{first}} = |W_{\text{fill}} - W_{\text{top}}| = |19600 - 39200| = 19600 \, \text{J} \][/tex]
- For the second configuration:
[tex]\[ \Delta W_{\text{second}} = |W_{\text{fill}} - W_{\text{top}}| = |19600 - 39200| = 19600 \, \text{J} \][/tex]
### Explanation:
The work done to fill the tank by pouring directly into the top is larger than the work done by filling it from the bottom because, in the latter method, the water needs to be pumped incrementally from the bottom to the top, meaning each incremental volume of water is lifted to progressively higher heights as the tank fills. In contrast, pouring into the top involves lifting the entire mass of water to the maximum height of 2 meters in an idealized scenario.
Thus, the work required to pour water into the top directly is always more by exactly [tex]\(19600 \, \text{J}\)[/tex] than filling from the bottom for the given dimensions.
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