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scZoUnD [109]
3 years ago
7

Which best describes the similarities and differences between the Engineering and Technology pathway and the Science and Math pa

thway?
a) Both pathways require workers to use teamwork, leadership, and teaching skills while the Engineering and Technology pathway requires clerical skills.

b) Both pathways require workers to have observational skills while the Engineering and Technology pathway requires CAD knowledge.

c) Both pathways require workers to be creative, while the Science and Math pathway requires specialized knowledge in subject areas like physics or chemistry.

d) Both pathways require workers to have patience and persistence while the Science and Math pathway requires physical stamina, dexterity, and use of technical equipment.
Engineering
1 answer:
melamori03 [73]3 years ago
5 0

Answer:

Both pathways require workers to be creative, while the Science and Math pathway requires specialized knowledge in subject areas like physics or chemistry.

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A particle is moving along a straight line with an initial velocity of 6 m/s when it is subjected to a deceleration of a- (-1.5v
Juli2301 [7.4K]

Answer:

s= 6.53 m

t=3.27 s

Explanation:

velocity = 6 m/s

deceleration = -1.5v^\frac{1}{2}

a=-1.5v^\frac{1}{2}\\v\frac{\mathrm{d} v}{\mathrm{d} s}=-1.5v^\frac{1}{2}\\-1.5ds=v^\frac{1}{2}dv\\\int\ {-1.5} \, ds= \int\ v^\frac{1}{2}dv\\-1.5s=\frac{2}{3}\times v^{\frac{3}{2} }

now inserting value of v=6s we get distance(s)

s= 6.53 m ( distance cannot be negative)

now for time calculation we know that

a=\frac{\mathrm{d}v }{\mathrm{d} t}

-1.5v^\frac{1}{2} =\frac{\mathrm{d}v }{\mathrm{d} t}\\-1.5dt=\frac{dv}{v^\frac{1}{2}} \\\int -1.5 dt=\int v^{-\frac{1}{2}}dt \\1.5t=2v^\frac{1}{2}\\t=\frac{4}{3}v^\frac{1}{2}

putting value of v=6s

t=3.27 s (time cannot be negative)

3 0
3 years ago
Tin atoms are introduced into an FCC copper ,producing an alloy with a lattice parameter of 4.7589×10-8cm and a density of 8.772
slega [8]

Answer:

atomic percentage = 143 %

Explanation:

Let  x be the number of tin atoms and there are 4 atoms / cell in the FCC structure , then 4 -x  be the number of copper atoms . Therefore, the value of x can be determined by using the density equation as shown below:

\mathbf{density (\rho) = \dfrac{(no \ of \ atoms/cell)(atomic \ mass )}{(lattice \ parameter )^3(6.022*10^{23} atoms/ mol)} }

where;

the lattice parameter is given as : 4.7589 × 10⁻⁸ cm

The atomic mass of tin is 118.69 g/mol

The atomic mass of copper is 63.54 g/mol

The density is 8.772 g/cm³

\mathbf{8.772 g/cm^3 = \dfrac{(x)(118.69 \ g/mol) +(4-x)(63.54 \ g/mol)}{(4.7589*10^{-8} cm )^3(6.022*10^{23} atoms/ mol)} }

569.32 = 118.69x + 254.16-63.54x

569.32 - 254.16 = 118.69x - 63.54 x

315.16 = 55.15x

x = 315.16/55.15

x = 5.72 atoms/cell

As there are four atoms per cell in FCC structure for the metal, thus, the atomic percentage of the tin is  calculated as follows :

atomic % = \frac{no \ of \ atoms \ per  \ cell \ in \ tin }{no \ of \ atoms \ per  \ cell \ in \ the \ metal}*100

atomic % = \frac{5.72 \ atoms / cell}{4 \ atoms/ cell} *100

atomic % = 143 %

7 0
3 years ago
How would you expect an increase in the austenite grain size to affect the hardenability of a steel alloy? Why?
seraphim [82]

Answer:

The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.

3 0
3 years ago
An organization sets its standards for quality according to the best product it can produce.
Marianna [84]
I believe it’s True, but please correct me if I’m wrong!
6 0
3 years ago
Read 2 more answers
Q1: The first option should always be to get out safely (RUN)
nekit [7.7K]

Answer:

Q1 true

Q2 true

And other I am confuse

6 0
3 years ago
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