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2 years ago

What is the key to being a good engineer?

Engineering

2 answers:

wariber [46]2 years ago

4 0

Answer:

Approaching a problem and finding a solution within given guidelines.

Explanation:

To be a good engineer, you simply need to know how to create a solution to a problem with a given set of restraints or guidelines.

For instance, let's say your boss wants you to build a machine that can automate some trivial task so he can use the worker elsewhere. He wants the machine to be low maintenance and under $40,000.

As an engineer, your first thought should be, let's see what this "trivial task" is, and then after your observation, you should begin to pull upon your experience and resources to build a solution that can be low maintenance and under $40,000.

Put simply, when you approach a problem, consider all aspects of the problem and then build a solution that satisfies all requirements.

Cheers.

padilas [110]2 years ago

3 0

Answer: ‍♀️

Explanation:

Many engineers focus their professional development efforts on improving their technical skills. Don't underestimate the importance of soft skills, however, such as communication, teamworking, and presentation skills. You should also take time to acquire and improve essential skills like these.

You might be interested in

Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

$\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%$

Where:

$V_{gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{fe}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

$\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%$

Where:

$m_{fe}$ , $m_{gr}$ - Masses of the ferrite and graphite phases, measured in grams.

$\rho_{fe}, \rho_{gr}$ - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

$\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%$

$\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%$

If $\rho_{gr} = 2.3\,\frac{g}{cm^{3}}$ , $\rho_{fe} = 7.9\,\frac{g}{cm^{3}}$ , $m_{gr} = 3.2\,g$ and $m_{fe} = 96.8\,g$ , the volume percentage of graphite is:

$\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%$

$\%V_{gr} = 10.197\,\%V$

The volume percentage of graphite is 10.197 per cent.

5 0

3 years ago

An existing building is suffering from cracks in the exterior walls. The investigating engineer wants to ensure that the foundat

jek_recluse [69]

Answer:

$18 ft^{2}$

Explanation:

Soil bearing pressure= $\frac {Load}{Area}$

Since we're given pressure of 2500 psf and load of 45000 pounds

The area= $\frac {45000}{2500}=18$

Therefore, the smallest area of safe footings should not be less than $18 ft^{2}$

6 0

2 years ago

A car is about to start but it blows up. what is the problem with the car<br> ?

ratelena [41]

Answer:

because there is a bomb

6 0

3 years ago

Read 2 more answers

The combustion chamber has different shapes depending on the make and model of the engine. True or false

Alisiya [41]

Answer:

svdsdfdfsdfssdf

Explanation:

fsdsdfsdffsdsfdsdf

5 0

2 years ago

A house is losing heat at a rate of 1700 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp

Furkat [3]

Answer:

1700kJ/h.K

944.4kJ/h.R

944.4kJ/h.°F

Explanation:

Conversions for different temperature units are below:

1K = 1°C + 273K

1R = T(K) * 1.8

= (1°C + 273) * 1.8

1°F = (1°C * 1.8) + 32

Q/delta T = 1700kJ/h.°C

T (K) = 1700kJ/h.°C

= 1700kJ/K

T (R) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8R

= 944.4kJ/h.R

T (°F) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8°F

= 944.4kJ/h.°F

Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C

= 2°C

(273+5) - (273+3)

= 2 K

5 0

3 years ago