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leonid [27]
2 years ago
11

Select three ways an engineer can create a view of a design.

Engineering
1 answer:
Angelina_Jolie [31]2 years ago
5 0

Sketching projection

  • Sketching will give the brief planning of the engineering work to be done so it's important

Isometric projection

  • Yes doing a work in 3D factor out the mistakes which can be optimised

Third angle projection:-

  • Viewing angles matter a lot
  • So it will help the engineer to make it perfect from all angles
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True or false? Engineering degree programs and engineering technology degree programs have a different requirements
sergey [27]
That’s true brother
4 0
4 years ago
a metal rod 24mm diameter and 2m long is subjected to an axial pull of 40 kN. If the rod is 0.5mm, then find the stress-induced
ozzi

Answer:

i dont know but i will take the points tho hahah

Explanation:

8 0
3 years ago
Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and
Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in  

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that   PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒  P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0
4 years ago
The statement that is NOT true about the concept of a boundary layer on an object is: a. the Reynolds number is greater than uni
Ilya [14]

Answer:

Option E

Explanation:

All the given statements are true except the velocity gradients normal to the flow direction are small since these are not normally small. It's true that viscous effects are present only inside the boundary layer and the fluid velocity equals the free stream velocity at the edge of the boundary layer. Moreover, Reynolds number is greater than unity and the fluid velocity is zero at the surface of the object.

5 0
3 years ago
The y component of velocity in a steady, incompressible flow field in the xy plane is v = -Bxy3, where B = 0.4 m-3 · s-1, and x
love history [14]

Answer:

attached below

Explanation:

3 0
3 years ago
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