Join IDNLearn.com and become part of a knowledge-sharing community that thrives on curiosity. Whether your question is simple or complex, our community is here to provide detailed and trustworthy answers quickly and effectively.
Sagot :
To find the radius of a small oil drop falling with a terminal velocity of \(4.0 \, \text{m/s}\) through air, given the viscosity of air is \(1.8 \times 10^{-5} \, \text{Ns/m}^2\), the density of the oil is \(900 \, \text{kg/m}^3\), and the acceleration due to gravity \(g\) is \(10 \, \text{m/s}^2\), we proceed as follows:
1. Identify the terminal velocity formula for a falling sphere:
The terminal velocity \(v_t\) of a sphere falling through a fluid can be given by:
[tex]\[ v_t = \frac{2 r^2 (\rho_{\text{oil}} - \rho_{\text{air}}) g}{9 \eta_{\text{air}}} \][/tex]
where:
- \(v_t\) is the terminal velocity,
- \(r\) is the radius of the sphere,
- \(\rho_{\text{oil}}\) is the density of the oil,
- \(\rho_{\text{air}}\) is the density of air,
- \(g\) is the acceleration due to gravity,
- \(\eta_{\text{air}}\) is the viscosity of air.
2. Assumptions and given values:
[tex]\[ v_t = 4.0 \, \text{m/s} \][/tex]
[tex]\[ \eta_{\text{air}} = 1.8 \times 10^{-5} \, \text{Ns/m}^2 \][/tex]
[tex]\[ \rho_{\text{oil}} = 900 \, \text{kg/m}^3 \][/tex]
Since the density of air is negligible compared to the density of oil, we assume \(\rho_{\text{air}} \approx 0\):
[tex]\[ g = 10 \, \text{m/s}^2 \][/tex]
3. Rearrange the formula to solve for \(r\):
[tex]\[ v_t = \frac{2 r^2 \rho_{\text{oil}} g}{9 \eta_{\text{air}}} \][/tex]
Solving for \(r^2\):
[tex]\[ r^2 = \frac{9 \eta_{\text{air}} v_t}{2 g \rho_{\text{oil}}} \][/tex]
4. Substitute the given values:
[tex]\[ r^2 = \frac{9 \times (1.8 \times 10^{-5}) \times 4.0}{2 \times 10 \times 900} \][/tex]
5. Calculate \(r^2\):
[tex]\[ r^2 = \frac{9 \times 1.8 \times 4.0 \times 10^{-5}}{2 \times 10 \times 900} \][/tex]
[tex]\[ r^2 = \frac{64.8 \times 10^{-5}}{18000} \][/tex]
[tex]\[ r^2 = 3.6 \times 10^{-8} \, \text{m}^2 \][/tex]
6. Calculate the radius \(r\):
[tex]\[ r = \sqrt{3.6 \times 10^{-8}} \][/tex]
[tex]\[ r \approx 1.897 \times 10^{-4} \, \text{m} \][/tex]
or
[tex]\[ r \approx 0.0001897 \, \text{m} \][/tex]
Therefore, the radius of the oil drop is approximately [tex]\(0.0001897\)[/tex] meters (or [tex]\(189.7 \, \mu\text{m}\)[/tex]).
1. Identify the terminal velocity formula for a falling sphere:
The terminal velocity \(v_t\) of a sphere falling through a fluid can be given by:
[tex]\[ v_t = \frac{2 r^2 (\rho_{\text{oil}} - \rho_{\text{air}}) g}{9 \eta_{\text{air}}} \][/tex]
where:
- \(v_t\) is the terminal velocity,
- \(r\) is the radius of the sphere,
- \(\rho_{\text{oil}}\) is the density of the oil,
- \(\rho_{\text{air}}\) is the density of air,
- \(g\) is the acceleration due to gravity,
- \(\eta_{\text{air}}\) is the viscosity of air.
2. Assumptions and given values:
[tex]\[ v_t = 4.0 \, \text{m/s} \][/tex]
[tex]\[ \eta_{\text{air}} = 1.8 \times 10^{-5} \, \text{Ns/m}^2 \][/tex]
[tex]\[ \rho_{\text{oil}} = 900 \, \text{kg/m}^3 \][/tex]
Since the density of air is negligible compared to the density of oil, we assume \(\rho_{\text{air}} \approx 0\):
[tex]\[ g = 10 \, \text{m/s}^2 \][/tex]
3. Rearrange the formula to solve for \(r\):
[tex]\[ v_t = \frac{2 r^2 \rho_{\text{oil}} g}{9 \eta_{\text{air}}} \][/tex]
Solving for \(r^2\):
[tex]\[ r^2 = \frac{9 \eta_{\text{air}} v_t}{2 g \rho_{\text{oil}}} \][/tex]
4. Substitute the given values:
[tex]\[ r^2 = \frac{9 \times (1.8 \times 10^{-5}) \times 4.0}{2 \times 10 \times 900} \][/tex]
5. Calculate \(r^2\):
[tex]\[ r^2 = \frac{9 \times 1.8 \times 4.0 \times 10^{-5}}{2 \times 10 \times 900} \][/tex]
[tex]\[ r^2 = \frac{64.8 \times 10^{-5}}{18000} \][/tex]
[tex]\[ r^2 = 3.6 \times 10^{-8} \, \text{m}^2 \][/tex]
6. Calculate the radius \(r\):
[tex]\[ r = \sqrt{3.6 \times 10^{-8}} \][/tex]
[tex]\[ r \approx 1.897 \times 10^{-4} \, \text{m} \][/tex]
or
[tex]\[ r \approx 0.0001897 \, \text{m} \][/tex]
Therefore, the radius of the oil drop is approximately [tex]\(0.0001897\)[/tex] meters (or [tex]\(189.7 \, \mu\text{m}\)[/tex]).
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
