Answer :
Certainly! Let's go through each part step by step:
### Part A: Factors determining fluid pressure
Fluid pressure is influenced by several factors:
1. Depth (or height):
- Example: The pressure at the bottom of a swimming pool is greater than at the surface. This is because the pressure in a fluid increases with depth due to the weight of the fluid above.
- Explanation: The deeper you go into a fluid, the greater the weight of the fluid above that point, hence higher pressure.
2. Density of the fluid:
- Example: Pressure exerted by seawater at a particular depth is higher than the pressure by freshwater at the same depth. This is because seawater is denser than freshwater.
- Explanation: A denser fluid means more mass per unit volume, thereby exerting more pressure at a given depth.
3. Gravitational acceleration:
- Example: The fluid pressure at a certain depth on Earth is greater than that on the Moon at the same depth, considering both have the same fluid. This is because gravitational acceleration on the Moon is less than on the Earth.
- Explanation: Fluid pressure is directly proportional to gravitational acceleration. Higher gravitational force results in higher pressure.
### Part B: Balancing the see-saw
Given:
- Weight of the mother [tex]\( W_m = 600 \, \text{N} \)[/tex]
- Distance of the mother from the pivot [tex]\( d_m = 3 \, \text{m} \)[/tex]
- Weight of the son [tex]\( W_s = 400 \, \text{N} \)[/tex]
- Distance of the son from the pivot [tex]\( d_s = 4 \, \text{m} \)[/tex]
- Weight of the sister [tex]\( W_{\text{sis}} = 100 \, \text{N} \)[/tex]
The see-saw is balanced when the moments on both sides of the pivot are equal.
### Moment calculation
- Moment due to mother = [tex]\( W_m \times d_m \)[/tex]
- Moment due to son = [tex]\( W_s \times d_s \)[/tex]
To balance:
[tex]\( \text{Moment due to mother} + \text{Moment due to sister} = \text{Moment due to son} \)[/tex]
Substituting the given values:
[tex]\[ 600 \times 3 + 100 \times d_{\text{sis}} = 400 \times 4 \][/tex]
Solving for [tex]\( d_{\text{sis}} \)[/tex]:
[tex]\[ 1800 + 100 \times d_{\text{sis}} = 1600 \][/tex]
[tex]\[ 100 \times d_{\text{sis}} = 1600 - 1800 \][/tex]
[tex]\[ 100 \times d_{\text{sis}} = -200 \][/tex]
[tex]\[ d_{\text{sis}} = \frac{-200}{100} = -2 \, \text{m} \][/tex]
Thus, the sister should sit at a distance of [tex]\(-2\)[/tex] meters on the side where her mother is seated.
### Part C: Effort required by the lifting machine
Given:
- Efficiency of the lifting machine [tex]\( \eta = 0.80 \)[/tex]
- Distance effort moves [tex]\( d_e = 6 \, \text{m} \)[/tex]
- Distance load moves [tex]\( d_l = 0.060 \, \text{m} \)[/tex] (60 mm converted to meters)
- Load [tex]\( L = 300 \, \text{N} \)[/tex]
### Calculation
Using the principle of work and efficiency, the work done by the effort equals the work done by the load divided by efficiency.
Work output (load work):
[tex]\[ \text{Work}_{\text{output}} = L \times d_l \][/tex]
Work input (effort work):
[tex]\[ \text{Work}_{\text{input}} = \frac{\text{Work}_{\text{output}}}{\eta} \][/tex]
Then the effort required can be found by:
[tex]\[ \text{Effort} = \frac{\text{Work}_{\text{input}}}{d_e} \][/tex]
Substituting the given values:
[tex]\[ \text{Work}_{\text{output}} = 300 \times 0.060 = 18 \, \text{N·m} \][/tex]
[tex]\[ \text{Work}_{\text{input}} = \frac{18}{0.80} = 22.5 \, \text{N·m} \][/tex]
[tex]\[ \text{Effort} = \frac{22.5}{6} = 3.75 \, \text{N} \][/tex]
Thus, the effort required to lift the load of 300 N is [tex]\(3.75 \, \text{N}\)[/tex].
### Part A: Factors determining fluid pressure
Fluid pressure is influenced by several factors:
1. Depth (or height):
- Example: The pressure at the bottom of a swimming pool is greater than at the surface. This is because the pressure in a fluid increases with depth due to the weight of the fluid above.
- Explanation: The deeper you go into a fluid, the greater the weight of the fluid above that point, hence higher pressure.
2. Density of the fluid:
- Example: Pressure exerted by seawater at a particular depth is higher than the pressure by freshwater at the same depth. This is because seawater is denser than freshwater.
- Explanation: A denser fluid means more mass per unit volume, thereby exerting more pressure at a given depth.
3. Gravitational acceleration:
- Example: The fluid pressure at a certain depth on Earth is greater than that on the Moon at the same depth, considering both have the same fluid. This is because gravitational acceleration on the Moon is less than on the Earth.
- Explanation: Fluid pressure is directly proportional to gravitational acceleration. Higher gravitational force results in higher pressure.
### Part B: Balancing the see-saw
Given:
- Weight of the mother [tex]\( W_m = 600 \, \text{N} \)[/tex]
- Distance of the mother from the pivot [tex]\( d_m = 3 \, \text{m} \)[/tex]
- Weight of the son [tex]\( W_s = 400 \, \text{N} \)[/tex]
- Distance of the son from the pivot [tex]\( d_s = 4 \, \text{m} \)[/tex]
- Weight of the sister [tex]\( W_{\text{sis}} = 100 \, \text{N} \)[/tex]
The see-saw is balanced when the moments on both sides of the pivot are equal.
### Moment calculation
- Moment due to mother = [tex]\( W_m \times d_m \)[/tex]
- Moment due to son = [tex]\( W_s \times d_s \)[/tex]
To balance:
[tex]\( \text{Moment due to mother} + \text{Moment due to sister} = \text{Moment due to son} \)[/tex]
Substituting the given values:
[tex]\[ 600 \times 3 + 100 \times d_{\text{sis}} = 400 \times 4 \][/tex]
Solving for [tex]\( d_{\text{sis}} \)[/tex]:
[tex]\[ 1800 + 100 \times d_{\text{sis}} = 1600 \][/tex]
[tex]\[ 100 \times d_{\text{sis}} = 1600 - 1800 \][/tex]
[tex]\[ 100 \times d_{\text{sis}} = -200 \][/tex]
[tex]\[ d_{\text{sis}} = \frac{-200}{100} = -2 \, \text{m} \][/tex]
Thus, the sister should sit at a distance of [tex]\(-2\)[/tex] meters on the side where her mother is seated.
### Part C: Effort required by the lifting machine
Given:
- Efficiency of the lifting machine [tex]\( \eta = 0.80 \)[/tex]
- Distance effort moves [tex]\( d_e = 6 \, \text{m} \)[/tex]
- Distance load moves [tex]\( d_l = 0.060 \, \text{m} \)[/tex] (60 mm converted to meters)
- Load [tex]\( L = 300 \, \text{N} \)[/tex]
### Calculation
Using the principle of work and efficiency, the work done by the effort equals the work done by the load divided by efficiency.
Work output (load work):
[tex]\[ \text{Work}_{\text{output}} = L \times d_l \][/tex]
Work input (effort work):
[tex]\[ \text{Work}_{\text{input}} = \frac{\text{Work}_{\text{output}}}{\eta} \][/tex]
Then the effort required can be found by:
[tex]\[ \text{Effort} = \frac{\text{Work}_{\text{input}}}{d_e} \][/tex]
Substituting the given values:
[tex]\[ \text{Work}_{\text{output}} = 300 \times 0.060 = 18 \, \text{N·m} \][/tex]
[tex]\[ \text{Work}_{\text{input}} = \frac{18}{0.80} = 22.5 \, \text{N·m} \][/tex]
[tex]\[ \text{Effort} = \frac{22.5}{6} = 3.75 \, \text{N} \][/tex]
Thus, the effort required to lift the load of 300 N is [tex]\(3.75 \, \text{N}\)[/tex].
