Answer:
a) V = 0.354
b) G = 25.34 GPA
Explanation:
Solution:
We first determine Modulus of Elasticity and Modulus of rigidity
Elongation of rod ΔL = 1.4 mm
Normal stress, δ = P/A
Where P = Force acting on the cross-section
A = Area of the cross-section
Using Area, A = π/4 · d²
= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²
δ = 50/3.14 × 10⁻⁴ = 159.155 MPA
E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm
Modulus of Elasticity Е = δ/ε
= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA
Also final diameter d(f) = 19.9837 mm
Initial diameter d(i) = 20 mm
Poisson said that V = Е(elasticity)/Е(long)
= - <u>( 19.9837 - 20 /20)</u>
2.33 × 10⁻³
= 0.354,
∴ v = 0.354
Also G = Е/2. (1+V)
= 68.306 × 10⁹/ 2.(1+ 0.354)
= 25.34 GPA
⇒ G = 25.34 GPA
Answer:
The molecular weight will be "28.12 g/mol".
Explanation:
The given values are:
Pressure,
P = 10 atm
= 
= 
Temperature,
T = 298 K
Mass,
m = 11.5 Kg
Volume,
V = 1000 r
= 
R = 8.3145 J/mol K
Now,
By using the ideal gas law, we get
⇒ 
o,
⇒ 
By substituting the values, we get
As we know,
⇒ 
or,
⇒ 
The coil polarity in a waste-spark system is determined by the direction in which the coil is wound (left-hand rule for conventional current flow)and can’t be changed. For example, if a V-8 engine has a firing order of 18436572 and the number 1 cylinder is on compression, which cylinder will be on the exhaust stroke?
Answer:
what are simple machines?
Explanation:
it is 2020 let's be honest all
