I am just questing to point a ok ok ok this might not be right
Answer: b) 3.47 nj
Explanation:
Given that;
length l = 5m
radius of inner conductor r = 10cm = 0.1m
radius of outer conductor D = 20cm = 0.2m
current I = 100A = 100×10⁻³ = 0.1
medium between conductor in air u₀ = 4π × 10⁻⁷
Energy in a coaxial cable transmission line is
w = u₀ /2π I² en(b/a)
we substitute
L = 4π × 10⁻⁷ /4π ×10⁻²× 5 en (20/10)
L =3.4657 × 10⁻⁹ J
L = 3.4657 nJ ≈ 3.47 nJ
False you have to do the work and perricapate
Answer:
hello your question is incomplete attached below is the missing diagram to the question and the detailed solution
Answer : principal stresses : 0.82 MPa, -33.492 MPa
shear stress = 17.157 MPa
∅ = 9.09 ≈ 10°
Explanation:
The principal stress ( б1 ) = 0.82 MPa
( б2 ) = -33.492 MPa
The shear stress = 17.157 MPa
∅ = 9.09 ≈ 10°
attached below is the detailed solution and the Mohr's circle
Answer:
This is from a website, hope it helped
Explanation:
An Onshape Assembly tab is where you define a hierarchical structure of part and subassembly instances of an Assembly. It is also where you define degrees of freedom and relations. You are able to have more than one Assembly tab in a document. One Assembly can instance another Assembly as a subassembly, and/or instance a part directly. You are able to instance parts from the same document or other documents to which you have permissions (and that are versioned).