Answer: making sure that they are up to date
Explanation:
Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / 
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
= 
given that
= 480 mm
we substitute
=
× 
= 501.998 mm
Now we find the elongation;
Elongation = 
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
• it charges banks more interest
• it sells more securities
• it decreases the money supply
In response to high inflation, the Fed charges banks more interests and pays the banks less interests. It also sells not securities.
Answer:
Total wight =640.7927 KN
Explanation:
Given that
do= 61 cm
L =120
t= 0.9 cm
That is why inner diameter of the pipe
di= 61 - 2 x 0.9 cm
di=59.2 cm
Water density ,ρ = 1 kg/L = 1000 kg/m³
Weight of the pipe ,wt = 2500 N/m
wt = 2500 x 120 N = 300,000 N
The wight of the water
wt ' = ρ V g

wt'=340792.47 N
That is why total wight
Total wight = wt + wt'
Total wight =300,000+ 340792.47 N
Total wight =640,792.47 N
Total wight =640.7927 KN
Answer:
The pressure upstream and downstream of a shock wave are related as

where,
= Specific Heat ratio of air
M = Mach number upstream
We know that 
Applying values we get

Similarly the temperature downstream is obtained by the relation
![\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B1%7D%7D%7BT_%7Bo%7D%7D%3D%5Cfrac%7B%5B2%5Cgamma%20M%5E%7B2%7D-%28%5Cgamma%20-1%29%5D%5B%28%5Cgamma%20-1%29M%5E%7B2%7D%2B2%5D%7D%7B%28%5Cgamma%20%2B1%29%5E%7B2%7DM%5E%7B2%7D%7D)
Applying values we get
![\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B1%7D%7D%7B423%7D%3D%5Cfrac%7B%5B2%5Ctimes%201.4%5Ctimes%201.8%5E%7B2%7D-%281.4-1%29%5D%5B%281.4-1%291.8%5E%7B2%7D%2B2%5D%7D%7B%281.4%2B1%29%5E%7B2%7D%5Ctimes%201.8%5E%7B2%7D%7D%5C%5C%5C%5C%5Ctherefore%20%5Cfrac%7BT_%7B1%7D%7D%7B423%7D%3D1.53%5C%5C%5C%5C%5Ctherefore%20T_%7B1%7D%3D647.85K%3D374.85%5E%7Bo%7DC)
The Mach number downstream is obtained by the relation
