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zepelin [54]
3 years ago
5

Develop a simple Business plan as an entrepreneur​

Engineering
1 answer:
Evgesh-ka [11]3 years ago
3 0

Answer:

Executive summary. Briefly tell your reader what your company is and why it will be successful. ...

Company description. ...

Market analysis. ...

Organization and management. ...

Service or product line. ...

Marketing and sales. ...

Funding request. ...

Financial projections.

Explanation:

i hope it helps u

pls make me as brainlist

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True I don’t really know.
7 0
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If you are a mechanical engineer answer these questions:
Natasha_Volkova [10]

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
You are to design two CONCEPTUALLY different synchronous state machines (Mealy and Moore) that perform the task described below.
allochka39001 [22]
Answer:








Explanation:









I hope this helps!
3 0
3 years ago
A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter =
oee [108]

Answer:

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

k = E * L0 / A

And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%

5 0
3 years ago
A cylinder of aluminum-magnesium alloy 0.5 m long is subjected to an elastic tensile stress of 10.2 MPa. The measured elastic el
elena55 [62]

Answer:

E= 15 GPa.

Explanation:

Given that

Length ,L = 0.5 m

Tensile stress ,σ = 10.2 MPa

Elongation ,ΔL = 0.34 mm

lets take young modulus  = E

We know that strain ε given as

\varepsilon =\dfrac{\Delta L}{L}

\varepsilon =\dfrac{0.34}{0.5\times 1000}

\varepsilon =0.00068

We know that

\sigma = \varepsilon  E\\\\E=\dfrac{10.2}{0.00068}\\E= 15000\ MPa\\E=15\ GPa

Therefore the young's modulus will be 15 GPa.

8 0
3 years ago
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