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Sagot :
Sure, let's go through the process of rearranging the scientific formula for density and solving both parts of the problem step-by-step.
### Understanding the Density Formula
The general formula for density [tex]\(D\)[/tex] is expressed as:
[tex]\[ D = \frac{m}{V} \][/tex]
where:
- [tex]\(D\)[/tex] is the density in [tex]\(kg/m^3\)[/tex],
- [tex]\(m\)[/tex] is the mass in kilograms (kg),
- [tex]\(V\)[/tex] is the volume in cubic meters ([tex]\(m^3\)[/tex]).
This formula can be rearranged to solve for either [tex]\(m\)[/tex] (mass) or [tex]\(V\)[/tex] (volume).
### Example 1 - Part (A)
Question: A sinker on a fishing line is made of lead and has a volume of [tex]\(0.000015 m^3\)[/tex]. What is the mass of the sinker?
Given:
- Density of lead ([tex]\(D_{\text{lead}}\)[/tex]) = 11340 [tex]\(kg/m^3\)[/tex]
- Volume of lead ([tex]\(V_{\text{lead}}\)[/tex]) = 0.000015 [tex]\(m^3\)[/tex]
Step-by-step solution:
1. Start with the density formula:
[tex]\[ D = \frac{m}{V} \][/tex]
2. Rearrange the formula to solve for mass ([tex]\(m\)[/tex]):
[tex]\[ m = D \times V \][/tex]
3. Substitute the given values into the formula:
[tex]\[ m = 11340 \, \text{kg/m}^3 \times 0.000015 \, \text{m}^3 \][/tex]
4. Perform the multiplication:
[tex]\[ m = 0.1701 \, \text{kg} \][/tex]
So, the mass of the lead sinker is approximately [tex]\(0.1701\)[/tex] kilograms.
### Example 1 - Part (B)
Question: The design for a life preserver requires 0.3 kilograms of plastic foam to provide proper buoyancy. What is the volume of the plastic foam required?
Given:
- Density of plastic foam ([tex]\(D_{\text{foam}}\)[/tex]) = 75 [tex]\(kg/m^3\)[/tex]
- Mass of plastic foam ([tex]\(m_{\text{foam}}\)[/tex]) = 0.3 [tex]\(kg\)[/tex]
Step-by-step solution:
1. Start with the density formula:
[tex]\[ D = \frac{m}{V} \][/tex]
2. Rearrange the formula to solve for volume ([tex]\(V\)[/tex]):
[tex]\[ V = \frac{m}{D} \][/tex]
3. Substitute the given values into the formula:
[tex]\[ V = \frac{0.3 \, \text{kg}}{75 \, \text{kg/m}^3} \][/tex]
4. Perform the division:
[tex]\[ V = 0.004 \, \text{m}^3 \][/tex]
So, the volume of the plastic foam required is [tex]\(0.004\)[/tex] cubic meters.
### Summary
(A) The mass of the lead sinker is [tex]\(0.1701\)[/tex] kilograms.
(B) The volume of the plastic foam required for the life preserver is [tex]\(0.004\)[/tex] cubic meters.
### Understanding the Density Formula
The general formula for density [tex]\(D\)[/tex] is expressed as:
[tex]\[ D = \frac{m}{V} \][/tex]
where:
- [tex]\(D\)[/tex] is the density in [tex]\(kg/m^3\)[/tex],
- [tex]\(m\)[/tex] is the mass in kilograms (kg),
- [tex]\(V\)[/tex] is the volume in cubic meters ([tex]\(m^3\)[/tex]).
This formula can be rearranged to solve for either [tex]\(m\)[/tex] (mass) or [tex]\(V\)[/tex] (volume).
### Example 1 - Part (A)
Question: A sinker on a fishing line is made of lead and has a volume of [tex]\(0.000015 m^3\)[/tex]. What is the mass of the sinker?
Given:
- Density of lead ([tex]\(D_{\text{lead}}\)[/tex]) = 11340 [tex]\(kg/m^3\)[/tex]
- Volume of lead ([tex]\(V_{\text{lead}}\)[/tex]) = 0.000015 [tex]\(m^3\)[/tex]
Step-by-step solution:
1. Start with the density formula:
[tex]\[ D = \frac{m}{V} \][/tex]
2. Rearrange the formula to solve for mass ([tex]\(m\)[/tex]):
[tex]\[ m = D \times V \][/tex]
3. Substitute the given values into the formula:
[tex]\[ m = 11340 \, \text{kg/m}^3 \times 0.000015 \, \text{m}^3 \][/tex]
4. Perform the multiplication:
[tex]\[ m = 0.1701 \, \text{kg} \][/tex]
So, the mass of the lead sinker is approximately [tex]\(0.1701\)[/tex] kilograms.
### Example 1 - Part (B)
Question: The design for a life preserver requires 0.3 kilograms of plastic foam to provide proper buoyancy. What is the volume of the plastic foam required?
Given:
- Density of plastic foam ([tex]\(D_{\text{foam}}\)[/tex]) = 75 [tex]\(kg/m^3\)[/tex]
- Mass of plastic foam ([tex]\(m_{\text{foam}}\)[/tex]) = 0.3 [tex]\(kg\)[/tex]
Step-by-step solution:
1. Start with the density formula:
[tex]\[ D = \frac{m}{V} \][/tex]
2. Rearrange the formula to solve for volume ([tex]\(V\)[/tex]):
[tex]\[ V = \frac{m}{D} \][/tex]
3. Substitute the given values into the formula:
[tex]\[ V = \frac{0.3 \, \text{kg}}{75 \, \text{kg/m}^3} \][/tex]
4. Perform the division:
[tex]\[ V = 0.004 \, \text{m}^3 \][/tex]
So, the volume of the plastic foam required is [tex]\(0.004\)[/tex] cubic meters.
### Summary
(A) The mass of the lead sinker is [tex]\(0.1701\)[/tex] kilograms.
(B) The volume of the plastic foam required for the life preserver is [tex]\(0.004\)[/tex] cubic meters.
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