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kotykmax [81]
3 years ago
10

Laws that protect businesses involve

Engineering
1 answer:
mr_godi [17]3 years ago
4 0
Laws that protect businesses involve contracts, employment law, intellectual property, real estate, bankruptcy, and many other areas of the law.
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Consider tests of an unswept wing that spans the wind tunnel and whose airfoil section is NACA 23012. Since the wing model spans
Dominik [7]

Answer:

Check the explanation

Explanation:

to know the lift per unit span (N/m) that is expected to be measured when the wing attack angle is 4°

as well as the corresponding section lift coefficient and die moment coefficient .

Kindly check the attached image below to see the step by step explanation to the above question.

3 0
3 years ago
A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage
faltersainse [42]

Answer:

a) 0.489

b) 54.42 kg/s

c) 247.36 kW/s

Explanation:

Note that all the initial enthalpy and entropy values were gotten from the tables.

See the attachment for calculations

4 0
3 years ago
Two previously undeformed rod-shaped specimens of copper are to be plastically deformed by reducing their cross-sectional areas.
mezya [45]
I am not sure I am stuck on this and I have been for 45 min someone please help me and this girl or boy!!
4 0
3 years ago
When do design engineers start on the design improvement step?
ArbitrLikvidat [17]

Answer:

  as soon as there is a design to improve

Explanation:

As a design engineer, I started on the "design improvement" step as soon as I had an initial conceptual design.

__

Then, I started that step again when my boss told me, "make it better."

_____

The more interesting question is, "when do you <em>stop</em> the design improvement step?" (Judging by the constant barrage of software updates, that answer is, "never.")

8 0
3 years ago
Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of subst
Kaylis [27]

Answer:

The specific heat capacity of substance A is 1.16 J/g

Explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

<u>FOR SUBSTANCE A:</u>

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

<u>FOR SUBSTANCE B:</u>

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

<u>C = 1.16 J/g</u>

5 0
3 years ago
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