Respuesta :
Answer:
t=39.76 min
Explanation:
Given that
D= 44 μm
R= 22 μm
h=310 m
[tex]density = 2700\ kg/m^3[/tex]
[tex]viscosity = 2.0\times 10^{-5} Ns/m^2[/tex]
We know that drag force
F = 6πμRV
V=Terminal velocity
R=Radius
μ=Dynamic velocity
At the terminal condition
m g = 6πμRV -----------1
m=ρ x Volume
[tex]Volume= \dfrac{4}{3}\pi R^3[/tex]
[tex]Volume= \dfrac{4}{3}\times \pi \times (22\times 10^{-6})^3\ m^3[/tex]
[tex]Volume=4.4\times 10^{-14}\ m^3[/tex]
[tex]m=4.4\times 10^{-14}\times 2700\ kg[/tex]
[tex]m=1.1\times 10^{-10}\ kg[/tex]
Now by putting the value in equation 1
[tex]1.1\times 10^{-10}\times 9.81=6\times \pi \times 2.0\times 10^{-5}\times 22\times 10^{-6}\times V[/tex]
[tex]V=\dfrac{1.1\times 10^{-10}\times 9.81}{6\times \pi \times 2.0\times 10^{-5}\times 22\times 10^{-6}}\ m/s [/tex]
V= 0.13 m/s
So time taken by particle
t= h/V
t = 310/0.13
t=2384.6 sec
t=39.76 min
Answer:
Time=185.459 s in seconds
Time=3.091 min in minutes
Explanation:
Given
density=2700 kg/[tex]m^{3}[/tex]
Diameter=44 μm
radius=diameter/2
radius=44/2=22 μm
Volume=(4/3)*π*[tex]r^{3}[/tex]
Volume=[tex](4/3)*(3.14)*(22*10^{-6} )^{3}[/tex]
Volume=[tex]4.46*10^{-14}[/tex] [tex]m^{3}[/tex]
mass=Volume*density
mass=2700*4.46*[tex]10^{-14}[/tex]
mass=1.2042*[tex]10^{10}[/tex]
density of air=1.1839 kg/[tex]m^{3}[/tex]
Projected area=[tex]\pi *r^2[/tex]
Projected area=[tex]3.14*(22*10^{-6})^2[/tex]
Projected area=1.5197*[tex]10^{-9}[/tex]
[tex]C_{d}[/tex] for sphere=0.47
Terminal velocity=v=[tex]\sqrt{(2*m*g)/A*C_{d} *density }[/tex]
where density is the air density.
Substitute values we will get
v=1.6715 m/s
Time=distance/velocity
height=distance=310 m
Time=310/1.6715
Time=185.459 s in seconds
Time=3.091 min in minutes
