Respuesta :
Answer:
t = 224 s
Explanation:
given,
length of container = 15 cm = 0.15 m
diameter of spherical particle = 2.5 μm
mass of particle = 1.9 x 10⁻¹⁴ kg
viscosity of air = μ = 1.18 x 10⁻⁵ kg/m.s
time taken by the particle to stop = ?
radius of particle = 2.5/2 = 1.25 μm
volume of particle = [tex]\dfrac{4}{3}\pi r^3[/tex]
=[tex]\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3[/tex]
Density =[tex]\dfrac{mass}{volume}[/tex]
Density =[tex]\dfrac{1.9 \times 10^{-14}}{\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3}[/tex]
ρ = 2322 kg/m³
terminal velocity
[tex]v_t = \dfrac{2}{9}\ \dfrac{gR^2(\rho - \rho_{air})}{\mu}[/tex]
[tex]v_t = \dfrac{2}{9}\ \dfrac{9.8 \times (1.25 \times 10^{-6})^2(2322 - 1)}{1.18 \times 10^{-5}}[/tex]
v_t = 6693 x 10⁻⁷ m/s
[tex]t = \dfrac{d}{v_t}[/tex]
[tex]t = \dfrac{0.15}{6693 \times 10^{-7}}[/tex]
t = 224 s
The time taken for the spherical particle to settle to the bottom of the container is 0.85 s.
The given parameters;
- length of the container, L = 15 cm = 0.15 m
- diameter of the particle, d = 2.5 μm
- radius of the particle, r = 1.25 μm
- mass of the particle, m = 1.9 x 10⁻¹⁴ kg
The area of the spherical particle is calculated as;
[tex]A = 4\pi r^2\\\\A = 4 \times \pi \times (1.25\times 10^{-6})^2\\\\A = 1.964 \times 10^{-11} \ m^2[/tex]
The terminal velocity of the particle is calculated as;
[tex]V_t = \sqrt{\frac{2mg}{\rho A C_d} }[/tex]
where;
- [tex]\rho[/tex] is density of air = 1.225 kg/m³
- [tex]C_d[/tex] is the drag coefficient of sphere in air = 0.5
- m is mass of the sphere
- g is acceleration due to gravity
[tex]V_t = \sqrt{\frac{2mg}{\rho A C_d} }\\\\V_t = \sqrt{\frac{2\times 1.9 \times 10^{-14}\times 9.8}{1.225 \times 1.964 \times 10^{-11} \times 0.5 } }\\\\V_t = 0.176 \ m/s[/tex]
The time taken for the spherical particle to settle to the bottom of the container is calculated as follows;
[tex]V_t = \frac{d}{t} \\\\t = \frac{d}{V_t} \\\\t = \frac{0.15}{0.176} \\\\t = 0.85 \ s[/tex]
Thus, the time taken for the spherical particle to settle to the bottom of the container is 0.85 s.
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