Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃
taking natural log on both sides
Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)
Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0
Answer:
Relative density = 0.7 or 70%
Explanation:
The following information was provided by this question
Pd = 1.72mg/mg³
Pd max = 1.81 mg/mg³
Pd min = 1.54 mg/mg³
We substitute into the formula. This formula is contained in the attachment.
[(1/1.54)-(1/1.72)]/[1/1.54 - 1/1.81]
= 0.649350 - 0.581395 / 0.649350 - 0.552486
= 0.067955/0.096864
= 0.7015
= 0.7
The relative density is Therefore 0.7 or 70% when converted to percentage
Answer:
Average heat transfer =42.448w/m^2k
Nud = 13.45978
Explanation:
See attachment for step by step guide
Answer:
293 kg
Explanation:
Let's say the tension in each cable is Tb, Tc, and Td.
First, find the length of cable AD:
r = √(2² + 2² + 1²)
r = 3
Using similar triangles:
Tdx = 2/3 Td
Tdy = 2/3 Td
Tdz = 1/3 Td
Sum of the forces in the x direction:
∑F = ma
Tb − 2/3 Td = 0
Td = 3/2 Tb
Sum of the forces in the y direction:
∑F = ma
2/3 Td − Tc = 0
Td = 3/2 Tc
Sum of the forces in the z direction:
∑F = ma
1/3 Td − mg = 0
Td = 3mg
From the first two equations, we know Td is greater than Tb or Tc. So we need to set Td to 8.6 kN, or 8600 N.
8600 N = 3mg
m = 8600 N / (3 × 9.8 m/s²)
m ≈ 292.5 kg
Rounded to three significant figures, the maximum mass of the crate is 293 kg.