Answer:
The external pressure is p = -21.9 psf or p = -8.85 psf
Explanation:
Given :
Velocity of wind, v = 120 mi / hr
(wind direction factor)
= topographical factor (for flat terrain)
= velocity pressure at height h
But for height h = 30 ft, 
= 0.98 (from table)
= 36.16
Now, 
, so 
(from table)
where, p = external pressure
G = 0.85 = gust factor (for typical rigid building)
(internal pressure co efficient)
Therefore putting the values,
p = -21.9 psf or p = -8.85 psf
Answer:
11.6 mm
Explanation:
With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:
σadm = strength / fos
σadm = 900 / 5 = 180 MPa
The stress is the load divided by the section:
σ = P / A
σ = 4*P / (π*d^2)
Rearranging:
d^2 = 4*P / (π*σ)
Answer:
https://www.origamiway.com/paper-airplane-instructions.shtml
Explanation:
This is a more personal preference, so at the end of the day it is what works best for you :D
Hope this helps!
Answer:
The theoretical density for Niobium is 
.
Explanation:
Formula used :
where,
= density of the unit cell
Z = number of atom in unit cell
M = atomic mass
= Avogadro's number
a = edge length of unit cell
We have :
Z = 2 (BCC)
M = 92.91 g/mol ( Niobium)
Atomic radius for niobium = r = 0.143 nm
Edge length of the unit cell = a
r = 0.866 a (BCC unit cell)
On substituting all the given values , we will get the value of 'a'.
The theoretical density for Niobium is .
