Answer :
Sure, let's break down the problem step-by-step.
### (a) Determine the average density of the box.
1. Calculate the Volume:
- The tin box has dimensions 5 cm × 5 cm × 4 cm.
- Volume [tex]\( \text{(V)} = \text{length} \times \text{width} \times \text{height} \)[/tex]
- [tex]\( \text{V} = 5 \, \text{cm} \times 5 \, \text{cm} \times 4 \, \text{cm} \)[/tex]
- [tex]\( \text{V} = 100 \, \text{cm}^3 \)[/tex]
2. Calculate the Density:
- Density [tex]\( \text{(d)} = \frac{\text{mass}}{\text{volume}} \)[/tex]
- Given mass [tex]\( \text{(m)} = 80 \, \text{g} \)[/tex]
- [tex]\( \text{d} = \frac{80 \, \text{g}}{100 \, \text{cm}^3} \)[/tex]
- [tex]\( \text{d} = 0.8 \, \text{g/cm}^3 \)[/tex]
The average density of the box is [tex]\( 0.8 \, \text{g/cm}^3 \)[/tex].
### (b) Tin has a density of 7.3 g/cm³. Explain why this box would float on water.
To understand why the box would float on water, let's consider the principle of buoyancy which states that an object will float if its density is less than the density of the fluid it is placed in.
1. Density Comparison:
- The density of water is typically [tex]\( 1 \, \text{g/cm}^3 \)[/tex].
- The average density of the box is [tex]\( 0.8 \, \text{g/cm}^3 \)[/tex], which is less than the density of water.
2. Reasoning:
- Since the density of the box ([tex]\( 0.8 \, \text{g/cm}^3 \)[/tex]) is less than the density of water, it means that the box will displace a mass of water equal to the box's mass before it is completely submerged.
- Therefore, the box will float on water.
### (c) Suppose the box has a hole in each side. Would it continue to float? Give a reason for your answer.
If the box has a hole in each side, it introduces a new factor:
1. Water Entry:
- Water would enter the box through the holes, increasing the internal mass of the box without increasing its volume.
2. Effect on Density:
- As more water enters, the mass of the box increases while the volume remains the same.
- This causes the density of the box to increase.
3. New Density and Buoyancy:
- Eventually, the density of the box will exceed the density of water ([tex]\( 1 \, \text{g/cm}^3 \)[/tex]) because of the added water mass.
- When the density of the box becomes greater than the density of water, the box will no longer be buoyant.
Conclusion:
- The box would not continue to float if it has a hole in each side. The reason is that water entering the box increases its mass, leading to an increase in density beyond that of water, ultimately causing the box to sink.
### (a) Determine the average density of the box.
1. Calculate the Volume:
- The tin box has dimensions 5 cm × 5 cm × 4 cm.
- Volume [tex]\( \text{(V)} = \text{length} \times \text{width} \times \text{height} \)[/tex]
- [tex]\( \text{V} = 5 \, \text{cm} \times 5 \, \text{cm} \times 4 \, \text{cm} \)[/tex]
- [tex]\( \text{V} = 100 \, \text{cm}^3 \)[/tex]
2. Calculate the Density:
- Density [tex]\( \text{(d)} = \frac{\text{mass}}{\text{volume}} \)[/tex]
- Given mass [tex]\( \text{(m)} = 80 \, \text{g} \)[/tex]
- [tex]\( \text{d} = \frac{80 \, \text{g}}{100 \, \text{cm}^3} \)[/tex]
- [tex]\( \text{d} = 0.8 \, \text{g/cm}^3 \)[/tex]
The average density of the box is [tex]\( 0.8 \, \text{g/cm}^3 \)[/tex].
### (b) Tin has a density of 7.3 g/cm³. Explain why this box would float on water.
To understand why the box would float on water, let's consider the principle of buoyancy which states that an object will float if its density is less than the density of the fluid it is placed in.
1. Density Comparison:
- The density of water is typically [tex]\( 1 \, \text{g/cm}^3 \)[/tex].
- The average density of the box is [tex]\( 0.8 \, \text{g/cm}^3 \)[/tex], which is less than the density of water.
2. Reasoning:
- Since the density of the box ([tex]\( 0.8 \, \text{g/cm}^3 \)[/tex]) is less than the density of water, it means that the box will displace a mass of water equal to the box's mass before it is completely submerged.
- Therefore, the box will float on water.
### (c) Suppose the box has a hole in each side. Would it continue to float? Give a reason for your answer.
If the box has a hole in each side, it introduces a new factor:
1. Water Entry:
- Water would enter the box through the holes, increasing the internal mass of the box without increasing its volume.
2. Effect on Density:
- As more water enters, the mass of the box increases while the volume remains the same.
- This causes the density of the box to increase.
3. New Density and Buoyancy:
- Eventually, the density of the box will exceed the density of water ([tex]\( 1 \, \text{g/cm}^3 \)[/tex]) because of the added water mass.
- When the density of the box becomes greater than the density of water, the box will no longer be buoyant.
Conclusion:
- The box would not continue to float if it has a hole in each side. The reason is that water entering the box increases its mass, leading to an increase in density beyond that of water, ultimately causing the box to sink.