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antoniya [11.8K]
3 years ago
13

: THE NEXT BIG THING In this lesson, you learned about two revolutionary new ideas that have engineers hard at work. There are m

any other ideas like these out there that will change the future. For this lesson, you will choose one of these new designs, research the science behind it, and learn about the engineering challenges that the idea presents. To complete this project you must: Find an engineering idea that you would like to learn more about. It must be something that is still in the design or prototype phase of development; it can’t be something that is commercially available. If you are having trouble finding an idea, try an online search for “future engineering ideas” or “future of engineering.” Find at least three references on your chosen innovation. These references may be online sources or journal articles. After you have researched the innovation, write a 250-word summary of what the idea is and why it will benefit society. Write a 250-word description of the science behind the innovation. You may use pictures, equations, or charts if they enhance your description. Write a 250-word description of the challenges facing the engineers who are working on the innovation. Submission Guidelines Submit your assignment as one 750-word essay. List all of the references that you used at the end of the essay. Remember to use correct spelling and grammar. Submission Requirements Use standard English and write full phrases or sentences. Do not use texting abbreviations or other shortcuts. Make the information easy to understand.
Engineering
1 answer:
grin007 [14]3 years ago
5 0

Answer:

Okay

Explanation:

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Impact strips may be designed into a bumper cover.<br> True<br> False
d1i1m1o1n [39]

Answer:

true I think

Explanation:

true I think

5 0
3 years ago
A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion
kati45 [8]

Answer:

X_B = 1.8 \times 10^{-3} m = 1.8 mm

Explanation:

Given data:

Diffusion constant for nitrogen is = 1.85\times 10^{-10} m^2/s

Diffusion flux = 1.0\times 10^{-7} kg/m^2-s

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen  concentration is 0.5 kg/m^3   ......?

from fick's first law

J = D \frac{C_A C_B}{X_A X_B}

Take C_A as point  on which nitrogen concentration is 2 kg/m^3

x_B = X_A + D\frac{C_A -C_B}{J}

Assume X_A is zero at the surface

X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}

X_B = 1.8 \times 10^{-3} m = 1.8 mm

4 0
3 years ago
Consider the following relational database that Best Airlines uses to keep track of its mechanics, their skills, and their airpo
Musya8 [376]

Answer:

Explanation:

A)

SELECT MECHNAME,AGE FROM MECHANIC;

B)

SELECT AIRNAME,SIZE FROM AIRPORT WHERE SIZE>=20 AND STATE='CALIFORNIA' AND YEAROPENED >=1935 ORDER BY SIZE ASC;

C)

SELECT AIRNAME,SIZE FROM AIRPORT WHERE (SIZE>=20 OR YEAROPENED >=1935) AND STATE='CALIFORNIA';

D)

SELECT AVG(SIZE) FROM AIRPORT WHERE STATE='CALIFORNIA' AND YEAROPENED >=1935;

E)

SELECT COUNT(AIRNAME) FROM AIRPORT WHERE STATE='CALIFORNIA' AND YEAROPENED >=1935;

F)

SELECT COUNT(AIRNAME),STATE FROM AIRPORT WHERE YEAROPENED>=1935 GROUP BY STATE;

G)

SELECT COUNT(AIRNAME),STATE FROM AIRPORT WHERE YEAR OPENED>=1935 GROUP BY STATE HAVING COUNT(*)>=5;

H)

SELECT MECHNAME FROM MECHANIC A JOIN AIRPORT B

ON A.AIRNAME=B.AIRNAME AND B.STATE='CALIFORNIA';

I)  

SELECT MECHNAME FROM MECHANIC A

JOIN QUALIFICATION B

ON A.MECHNUM=B.MECHNUM

AND B.PROFRATE=4

JOIN SKILL C

ON B.SKILLNUM=C.SKILLNUM

AND SKILLNAME='FAN BLADE RELACEMENT';

J)  SELECT MECHNAME FROM MECHANIC A

JOIN QUALIFICATION B

ON A.MECHNUM=B.MECHNUM

AND B.PROFRATE=4

JOIN SKILL C

ON B.SKILLNUM=C.SKILLNUM

AND SKILL NAME='FAN BLADE REPLACEMENT'

JOIN AIRPORT D

ON A.AIRNAME=D.AIRNAME

AND STATE='CALIFORNIA';

K)   SELECT SUM(SALARY),CITY FROM MECHANIC A

JOIN AIRPORT B

ON A.AIRNAME=B.AIRNAME

AND STATE='CALIFORNIA'

GROUP BY CITY;

L)   SELECT MAX(SIZE) FROM AIRPORT ;

M)  SELECT MAX(SIZE) FROM AIRPORT WHERE STATE='CALIFORNIA';

6 0
3 years ago
Comparison of density values determines whether an item will float or sink in water. For each of the values below, determine the
geniusboy [140]

Answer:

a) the object floats

b) the object floats

c) the object sinks

Explanation:

when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water

a)

volumen for a cube

V=L^3

L=1.53in=0.0388m

V=0.0388 ^3=5.8691x10^-5m^3=58.69ml

density=m/v

density=13.5g/58.69ml=0.23 g/ml

The wooden block floats  because it is less dense than water

b)

m=111mg=0.111g

density=m/v

density=0.111g/0.296ml=0.375g/ml

the metal paperclip floats   because it is less dense than water

c)

V=0.93cups=220.0271ml

m=0.88lb=399.1613g

Density=m/v

density=399.1613/220.027ml=1.8141g/ml

the apple sinks   because it is  denser than water

4 0
3 years ago
ShoppingBay is an online auction service that requires several reports. Data for each auctioned item includes an ID number, item
daser333 [38]

Answer:

START

  READ ID_Number

  READ Item_description

  READ length_of_auction_Days

  READ minimum_required_bid  

  IF minimum_required_bid GREATER THAN 100

      THEN

          DISPLAY

              Item Details are

              Item Id : ID_Number

              Item Description: Item_description

              Length Action days: length_of_auction_Days

              Minimum Required Bid: minimum_required_bid

END

Explanation:

5 0
3 years ago
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