54 volts
Ohms law. E= I x R
Answer:
If the turbulent velocity profile in a pipe of diameter 0.6 m may be approximated by u/U=(y/R)^(1/7), where u is in m/s and y is in m and 0.15 m from the pipe.
Explanation:
hope it helps
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
Answer:
Explanation:
The stress experimented by the circular bar is:
![\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cleft%5B%5Cfrac%7B2000%5C%2C%20lbf%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20%280.5%5C%2Cin%29%5E%7B2%7D%7D%5Cright%5D%5Ccdot%20%5Cleft%28%5Cfrac%7B1%5C%2Ckpsi%7D%7B1000%5C%2Cpsi%7D%20%5Cright%29)
The safety factor is:
