Answer: downward velocity = 6.9×10^-4 cm/s
Explanation: Given that the
Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m
Where radius r = 2.5 × 10^-5 m
Density = 1200 kg/m^3
Area of a sphere = 4πr^2
A = 4 × π× (2.5 × 10^-5)^2
A = 7.8 × 10^-9 m^2
Volume V = 4/3πr^3
V = 4/3 × π × (2.5 × 10^-5)^3
V = 6.5 × 10^-14 m^3
Since density = mass/ volume
Make mass the subject of formula
Mass = density × volume
Mass = 1200 × 6.5 × 10^-14
Mass M = 7.9 × 10^-11 kg
Using the formula
V = sqrt( 2Mg/ pCA)
Where
g = 9.81 m/s^2
M = mass = 7.9 × 10^-11 kg
p = density = 1200 kg/m3
C = drag coefficient = 24
A = area = 7.8 × 10^-9m^2
V = terminal velocity
Substitute all the parameters into the formula
V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]
V = sqrt[ 1.54 × 10^-9/2.25×10-4]
V = 6.9×10^-6 m/s
V = 6.9 × 10^-4 cm/s
Explanation:
McLeod gauge:
It is used to measure very low pressure of gas.It measure gas pressure by the help of mercury(Hg).it measure absolute pressure of gas.
McLeod gauge on the principle of Boyle's law.Boyle's law state that ,at constant temperature the pressure of gas is directly proportional to its volume.From Boyle's law
A sample of gas is taken from vacuum then its pressure is measured by help Hg.
Answer:
Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of 
by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.
Have an amazing day and I hope this can somewhat help :)
Answer:
Rate of heat loss per unit length of pipe, q' = 767.01 W/m
Explanation:
Let q' be the Rate of heat loss per unit length
Let q be the Rate of heat loss
q' = q/L
Where L is the length of the pipe
Diameter, D= 0.6m
The rate of heat loss q is given by the formula: q = Sk(T₂ - T₁)
Where k is the thermal conductivity of the concrete at 300 K
k = 1.4 Wb/m-K (at 300K)
And S is the shape factor given by the formula:
S = 2πL/ ln(1.08w/D)
S = (2π*L) / ln(1.08*1.75/0.6))
S = (2π*L) / 1.147
S = 5.48 L
q = 5.48L*1.4(400-300)
q = 767.01 L
q' = q/L
q' = 767.01L/L
q' = 767.01 W/m
