All categories

2 years ago

What are the success factors for mechanical engineering?

2 answers:

8 0

Problem-solving skills.
creativity.
interpersonal skills.
verbal and written communication skills.
commercial awareness.
teamworking skills.

Eva8 [605]2 years ago

7 0

Answer:

-effective technical skills.

-the ability to work under pressure.

-problem-solving skills.

-creativity.

-interpersonal skills.

-verbal and written communication skills.

-commercial awareness.

-teamworking skills.

Explanation:

is this what ur looking for? if so there ya go lol

You might be interested in

At the coast on a summer day, the land is hotter than the ocean. Warm air over the land rises and is replaced by cooler air, cau

Alona [7]

B) The convection current will reverse direction, reversing the winds.

8 0

2 years ago

Read 2 more answers

The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:

Serggg [28]

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

$y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1$

Here;

$G_{s}$ is specific gravity of soil solids

$y_{w}$ is unit weight of water = 998 kg/m^3

$w$ is the moisture content = 0.17

$e$ is the void ratio

$y$ is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

$w*G_{s} = S*e ..... Eq2$

S is saturation rate

Substitute Eq 2 into Eq 1

$y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w} }{1+e}$

$14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233$

Specific gravity of soil solids

$G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765$

Saturated Unit Weight

$y_{s} = \frac{(G_{s} + e)*y_{w} }{1+e} \\=\frac{(4.878811765 + 1.38233)*998 }{1+1.38233}\\\\= 2622.902571 N/m^3$

7 0

3 years ago

What is the purpose of the graphic language?

solmaris [256]

Answer:

enables the representation, analysis and communication of various aspects of an information system. These aspects correspond to varying and incomplete views of information systems and the processes therein.

5 0

2 years ago

The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to

Sergeeva-Olga [200]

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N

we will calculate the value of by applying this formula

= $\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12$ = 1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6

= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )

= 0.247 N/mm2

5 0

2 years ago

A Class A fire extingisher is for use on general combustibles such as:

Tatiana [17]

Answer:

used for ordinary combustibles, such as wood, paper, some plastics, and textiles. This class of fire requires the heat-absorbing effects of water or the coating effects of certain dry chemicals.

Explanation:

7 0

3 years ago