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elena-14-01-66 [18.8K]
3 years ago
6

Recommend the types of engineers needed to collaborate on a city project to build a skateboard park near protected wetlands.

Engineering
1 answer:
4vir4ik [10]3 years ago
4 0

Answer:

A civil engineer.

Explanation:

Civil engineering is the science that deals with the design, creation and maintenance of constructions for civil use on the earth's surface. Thus, this specialty seeks to adapt soils to the needs of life in society, creating buildings, bridges, and all other constructions adapted to civil life, while taking care of the correct use of soils and the correct distribution of spaces and resources. to be used for such constructions.

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Chemical milling is used in an aircraft plant to create pockets in wing sections made of an aluminum alloy. The starting thickne
Lelu [443]

Answer:

a) metal removal rate is 1915.37 mm³/min

b) the time required to etch to the specified depth is 500 min or 8.333 hrs

Explanation:

Given the data in the question;

starting thickness of one work part of interest = 20 mm

depth of series of rectangular-shaped pockets = 12 mm

dimension of pocket = 200 mm by 400 mm

radius of corners of each rectangle = 15 mm

penetration rate = 0.024 mm/minute

etch factor = 1.75

a)

To get the metal removal rate MRR;

The initial area will be smaller compare to the given dimensions of 200mm by 400mm and the metal removal rate would increase during the cut as area is increased. so'

A = 200 × 400 - ( 30 × 30 - ( π × 15² ) )

= 80000 - ( 900 - 707 )      

= 80000 - 193

A = 79807 mm²

Hence, metal removal rate MRR = penetration rate × A

MRR = 0.024 mm/minute × 79807 mm²

MRR = 1915.37 mm³/min

Therefore, metal removal rate is 1915.37 mm³/min

b) To get the time required to etch to the specified depth;

Time to machine ( etch ) =  depth of series of rectangular-shaped pockets / penetration rate

we substitute

Time to machine ( etch ) = 12 mm / 0.024 mm/minute

Time to machine ( etch ) = 500 min or 8.333 hrs

Therefore, the time required to etch to the specified depth is 500 min or 8.333 hrs

3 0
3 years ago
An aircraft is in a steady level turn at a flight speed of 200 ft/s and a turn rate about the local vertical of 5 deg/s. Thrust
notka56 [123]

Answer:

L= 50000 lb

D = 5000 lb

Explanation:

To maintain a level flight the lift must equal the weight in magnitude.

We know the weight is of 50000 lb, so the lift must be the same.

L = W = 50000 lb

The L/D ratio is 10 so

10 = L/D

D = L/10

D = 50000/10 = 5000 lb

To maintain steady speed the thrust must equal the drag, so

T = D = 5000 lb

5 0
3 years ago
Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p
oee [108]

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

5 0
3 years ago
A 150 MVA, 24 kV, 123% three-phase synchronous generator supplies a large network. The network voltage is 27 kV. The phase angle
Aleks04 [339]

Answer:

the generator induced voltage is 60.59 kV

Explanation:

Given:

S = 150 MVA

Vline = 24 kV = 24000 V

X_{s} =1.23(\frac{V_{line}^{2}  }{s} )=1.23\frac{24000^{2} }{1500} =4723.2 ohms

the network voltage phase is

V_{phase} =\frac{V_{nline} }{\sqrt{3} } =\frac{27}{\sqrt{3} } =15.58kV

the power transmitted is equal to:

|E|=\frac{P*X_{s} }{3*|V_{phase}|sinO } ;if-O=60\\|E|=\frac{300*4.723}{3*15.58*sin60} =34.98kV

the line induced voltage is

|E_{line} |=\sqrt{3} *|E|=\sqrt{3} *34.98=60.59kV

7 0
3 years ago
Which of the eight diagnostic steps for locating an engine performance problem is performed first?
Kay [80]

Answer:

D. Perform a thorough visual inspection.

4 0
3 years ago
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