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In the image seen here, a fish (10N) swims in an aquarium (20N) filled with water (40N) upon a table (50N). Identify the action...Question
In the image seen here, a fish (10N) swims in an aquarium (20N) filled with water (40N) upon a table (50N). Identify the action/reaction force pairs seen here and place them properly in the image. Assume that down is negative in all cases.
A. +120N
B. -120N
C. +10N
D. -10N
E. +70N
F. -70N
You have to use all the numbers
A. +120N
B. -120N
C. +10N
D. -10N
E. +70N
F. -70N
You have to use all the numbers
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Answered by
GPT-4o mini
In the scenario described, we have several action/reaction force pairs based on Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Here’s how the forces would correspond to the situation:
1. **Fish (10N)**:
- Action: The fish exerts a downward force on the water (10N).
- Reaction: The water exerts an equal and opposite force upward on the fish (10N).
So we can assign:
- **C. +10N** (upward force on the fish)
- **D. -10N** (downward force exerted by the fish)
2. **Water (40N)**:
- Action: The water exerts a downward force on the aquarium (40N).
- Reaction: The aquarium exerts an equal and opposite force upward on the water (40N).
We can assign:
- **E. +70N** (the aquarium plus the fish, since 40N + 10N that the fish adds)
- **F. -70N** (the total downward force on the table)
3. **Aquarium (20N)**:
- Action: The aquarium exerts a downward force on the table (20N).
- Reaction: The table exerts an equal and opposite force upward on the aquarium (20N).
So we can assign:
- **A. +120N** (the total upward force on the table, sum of the forces: 50N from the table + 70N down from water)
- **B. -120N** (the total downward force acting on the table)
Putting it all together, we have the following action/reaction force pairs:
- Fish on Water:
- C. +10N (water force on fish)
- D. -10N (fish force on water)
- Water on Aquarium:
- E. +70N (upward force on table)
- F. -70N (downward force on aquarium)
- Aquarium on Table:
- A. +120N (reaction force from table)
- B. -120N (action force on table)
Thus, the forces are arranged as follows:
- **C. +10N** (upward force on the fish)
- **D. -10N** (downward force by the fish)
- **E. +70N** (from the aquarium)
- **F. -70N** (downward on the table)
- **A. +120N** (upward from table forces)
- **B. -120N** (total forces acting downward on the table)
1. **Fish (10N)**:
- Action: The fish exerts a downward force on the water (10N).
- Reaction: The water exerts an equal and opposite force upward on the fish (10N).
So we can assign:
- **C. +10N** (upward force on the fish)
- **D. -10N** (downward force exerted by the fish)
2. **Water (40N)**:
- Action: The water exerts a downward force on the aquarium (40N).
- Reaction: The aquarium exerts an equal and opposite force upward on the water (40N).
We can assign:
- **E. +70N** (the aquarium plus the fish, since 40N + 10N that the fish adds)
- **F. -70N** (the total downward force on the table)
3. **Aquarium (20N)**:
- Action: The aquarium exerts a downward force on the table (20N).
- Reaction: The table exerts an equal and opposite force upward on the aquarium (20N).
So we can assign:
- **A. +120N** (the total upward force on the table, sum of the forces: 50N from the table + 70N down from water)
- **B. -120N** (the total downward force acting on the table)
Putting it all together, we have the following action/reaction force pairs:
- Fish on Water:
- C. +10N (water force on fish)
- D. -10N (fish force on water)
- Water on Aquarium:
- E. +70N (upward force on table)
- F. -70N (downward force on aquarium)
- Aquarium on Table:
- A. +120N (reaction force from table)
- B. -120N (action force on table)
Thus, the forces are arranged as follows:
- **C. +10N** (upward force on the fish)
- **D. -10N** (downward force by the fish)
- **E. +70N** (from the aquarium)
- **F. -70N** (downward on the table)
- **A. +120N** (upward from table forces)
- **B. -120N** (total forces acting downward on the table)
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