ما جمع كلمة القوة؟help please

If you are a mechanical engineer answer these questions:

Answer:

1. Yes, they are all necessary.

2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.

3 0

3 years ago

What are the different types of documents used to communicate engineering designs?

Ipatiy [6.2K]

Answer:

COMMON ENGINEERING DOCUMENTS

Inspection or trip reports.

Research, laboratory, and field reports.

Specifications.

Proposals.

Progress reports.

ect...

Explanation:

7 0

3 years ago

Generally, final design results are rounded to or fixed to three digits because the given data cannot justify a greater display.

creativ13 [48]

Answer:

(a) 1.90 kpsi

(b) 0.40 kpsi

(c) 0.61 in.

(d) 0.009

(a) 8 MPa

(b) 1.30 cm⁴

(c) 2.04 cm⁴

(d) 62.2 MPa

Explanation:

(a) σ = M/Z, where M = 1770 lbf·in and Z = 0.943 in³.

1770/0.943 = 1876.988 lbf/in² = 1.90 kpsi

(b) σ = F/A, where F = 9440 lbf and A = 23.8 in².

9440 /23.8 = 396.639 lbf/in² = 0.4 kpsi

(c) y = Fl³/(3EI)

F = 270 lbf

l = 31.5 in.

E = 30 Mpsi

I = 0.154 in.⁴

y = 270×31.5³/(3×30×10⁶×0.154) = 0.61 in.

(d) θ = Tl/(GJ), where T = 9740 lbf·in, l = 9.85 in. G = 11.3 Mpsi, and d = 1.00 in.

J = π·d⁴/32 = π/32 in.⁴

∴ θ = 9740 × 9.85 /(11.3 × 10⁶× π/32) = 0.009

(a) σ = F/wt, where F = 1 kN, w = 25 mm, and t = 5 mm

∴ σ = 1000/(0.025 × 0.005) = 8 MPa

(b) I = bh³/12, where b = 10 mm and h = 25 mm.

10×25³/12 = 1.30 cm⁴

(c) I = π·d⁴/64 where d = 25.4 mm.

I = π × 25.4⁴/64 = 2.04 cm⁴

(d) τ = 16×T/(π×d³), where T = 25 N·m, and d = 12.7 mm.

16×25/(π×0.0127³) = 62.2 MPa.

8 0

3 years ago

Bending is defined as? A. the application of a load tending to distort a member in one direction. B. the application of opposing

amm1812

Hi how are you today

4 0

2 years ago

Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d

alexdok [17]

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number ( $Re$ ) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

$Re=\frac{\rho v D}{\mu}$

where $\rho$ is the density of the fluid, $\mu$ is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

$v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s$

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

$D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm$

8 0

3 years ago