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3 years ago

¿Por qué creen que la Ingeniería Metalúrgica es una carrera estratégica para el desarrollo de nuestro país?

Engineering

1 answer:

galina1969 [7]3 years ago

6 0

Answer:

La ingeniería metalúrgica implica el desarrollo de sistemas de producción industrial a través del desarrollo de distintas aleaciones de metales, mediante los cuales se generan componentes minerales específicos que hacen al desarrollo de nuevos metales de mayor calidad y eficacia en la producción industrial, la construcción, el desarrollo de maquinarias, etc.

Este tipo de ingeniería es una carrera estratégica para el desarrollo económico de las naciones, pues permite la creación de industrias mas eficientes, productos de mejor calidad y un mayor desarrollo del comercio a través de lo producido por las distintas fábricas del país, generando directa e indirectamente ingresos económicos para la nación.

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What kind of abilities are needed to run a business

nordsb [41]

Answer:

Essential business skills:

1. Financial management. Being able to effectively manage your finances is critical. ...

2. Marketing, sales and customer service. ...

3. Communication and negotiation. ...

4. Leadership. ...

5. Project management and planning. ...

6. Delegation and time management. ...

7. Problem solving. ...

8. Networking.

8 0

3 years ago

For high temperature deformation, the bigger the gran sine, the higher the creep rate. a)-True b)- False

Gnom [1K]

Answer:

The given statement is False.

Explanation:

This is because at high temperature the creep rate depends on grain boundary area, increasing with an increase in grain boundary area thus decreasing the grain size. Thus at higher temperatures, grain size has opposite effect , the bigger the grain size the slower the slower is the creep rate, the larger the grain boundary area.

8 0

3 years ago

Read 2 more answers

The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s) in a st

chubhunter [2.5K]

Answer:

$\mathbf{t_f = 1436.96 \ sec }$

Explanation:

Given that :

The strength and stability of tires may be enhanced by heating both sides of the rubber ( 0.14 W/m·K, 6.35 × 10^-8m^2/s)

i.e

k = 0.14 W/mK

∝ = 6.35 × 10⁻⁸ m²/s

L = 0.01 m

$B_1 = \dfrac{hL}{k} \\ \\ B_1 = \dfrac{200*0.01}{0.14} \\ \\ B_1 = 14.2857$

We cannot use the model of Lumped Capacitance; SO Let assume that Fourier Number $F_o > 0.2$

⇒ $\dfrac{T_o - T_ \infty }{T_i - T_ \infty} = C_1 exp (- \zeta_i^2 *F_o)$

From Table 5.1 ; at $B_1$ = 14.2857

$C_1 = 1.265 \\ \\ \zeta_1 = 1.458 \ rad$

$\dfrac{170-200}{35-200} = 1.265 exp [ - (1.458)^2* \dfrac{ \alpha t_f}{L^2}]$

$In ( \dfrac{0.1818}{1.265}) = \dfrac{-1.458^2*6.35*10^{-8}*t_f}{0.01^2}$

$-1.9399=-0.001350 *t_f$

$t_f = \dfrac{-1.9399}{-0.001350}$

$\mathbf{t_f = 1436.96 \ sec }$

8 0

3 years ago

List 10 uses for surveying in areas other than land surveying. Select all that apply. a. topographic surveying b. satellite surv

bazaltina [42]

Answer:

b. satellite surveying

c. aerial surveying

d. optical tooling

e. marketing surveying

f. control surveying

h. statistical surveying

i. telephone surveying

k. alignment surveying

l. mine surveying

m. solar surveying

Explanation:

A Survey is an act of examination of the features of a subject or material under consideration. Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.

Land surveying finds application in construction where a survey is made on all the structures found in a constructed property. Topographic surveying deals with examining the natural and man-made feature of a piece of land. As-built survey as the name implies examines the features and location of a building during or recently after construction. These three are examples of land surveys.

5 0

3 years ago

A confined aquifer with a transmissivity of 300 m2/day and a storativity of 0.0005 and a well radius of 0.3 m. Find the drawdown

Olin [163]

Answer:

8.4627 m

Explanation:

Transmissivity( T ) = 300 m^2/day

Storativity( S ) = 0.0005

well radius ( r ) = 0.3m

Determine the drawdown in well at 100 days

Drawdown at 100 days = ∑ Drawdown at various period

We will use the equation : S = Q / U*π*T [ -0.5772 - In U ] ----- ( 1 )

where : Q = discharge , T = transmissivity

S = drawdown ,

U = r^2*s / 4*T*t --- ( 2 )

r = well radius , S = Storativity, t = time period

i) During 0-20

U1 = r^2*s / u*π*t = 1.875 * 10^-9

Input values into equation 1

S1 = 2.5885

ii) During 20-50

U2 = r^2*s / 4*π*t = 0.3^2 * 30 / u * 300 * 30 = 1.25 * 10^-9

input values into equation 1

S2 = 1.5854 m

iii) During 50 -90

U3 = r^2*s / 4*π*t = 9.375 * 10^-10

input values into equation 1

S3 = 4.2888 m

iv) During 90-100

U4 = 0

s4 = 0

Drawdown at 100 days = ∑ Drawdowns at various period

= s1 + s2 + s3 + s4 = 2.5885 + 1.5854 + 4.2888 + 0

= 8.4627 m

8 0

3 years ago