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RSB [31]
2 years ago
8

Mary is interested in developing new machines that would increase the productivity of small scale farmers. After high school, wh

ich profession should Mary select that would suit her interests and goals?
-farm equipment mechanic
-machine operator
-fuel cell engineering
-agriculture equipment technician
-agricultural engineering

(plato)
Engineering
1 answer:
vovangra [49]2 years ago
8 0

Answer:

Agricultural engineering

Explanation:

Hello,

First, this area of engineering deals with design, construction and upgrade of farm equipment and machines for purpose of production and processing.

This course will equip Mary with the skills and knowledge that can be applied is solving challenges in the field of Agriculture especially in areas of power supplies, machinery efficiency, agricultural products storage and processing, environmental issues, pollution in the ecosystem and structures and facilities that improve agriculture.

The vital skills that Mary should develop while pursing this course are analytical skills to solve complicated systems, technical skills to do proper math, critical thinking skills to be able to think outside the box for solutions and communication skills to proper work with others.

Mary will learn about using software to design systems, machine operation and farm equipment.Her studies will cover ways of modifying the environment for sustainable agricultural practices.She will be able to know ways of testing farm equipment to ensure proper functionality and high output.This is a multi-disciplinary career, hence she will interact with civil engineers, contractors, environmentalists and other stakeholders to ensure the goals of projects are attained.

Hope this Helps Engineer!

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A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
morpeh [17]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.                                          

<u>Explanation</u>:        

<u>Given</u>:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

<u>To calculate</u>:

The magnitude of applied stress in the direction of [101] and [011].

<u>Formula</u>:

zcr=σ cosФ cosλ

<u>Solution</u>:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ  = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.                                                                                                                                                                      

                                                                                   

                                                                                                                                                   

                                                                                                                                                             

5 0
2 years ago
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m
kodGreya [7K]

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

8 0
2 years ago
Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b
grandymaker [24]

Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

3 0
2 years ago
A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses
yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature T_1 = 25° C

T{\infty} =600^0C

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}

Bi = \dfrac{2.5}{231}

Bi = 0.0108

The time constant value \tau_t is :

\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}

\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}

\tau_t = \dfrac{2702* 0.025*1033}{100}

\tau_t = 697.79

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]

where;

Q = -\Delta E _{st} which correlates with the change in the internal energy of the solid.

So;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}

The maximum value for the change in the internal energy of the solid  is :

(pVc)\theta_1 = -\Delta E _{st}max

By equating the two previous equation together ; we have:

\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{  (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

0.75=  [1-e^{\dfrac {-t}{ \tau_1}}]}

So;

0.75=  [1-e^{\dfrac {-t}{ 697.79}}]}

1-0.75=  [e^{\dfrac {-t}{ 697.79}}]}

0.25 =  e^{\dfrac {-t}{ 697.79}}

In(0.25) =  {\dfrac {-t}{ 697.79}}

-1.386294361= \dfrac{-t}{697.79}

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}

\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}

\dfrac{T - 600}{25-600}= 0.25

\dfrac{T - 600}{-575}= 0.25

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0
2 years ago
Instructions given by traffic police or construction flaggers _____. A. Are sometimes important to follow B. Are usually not imp
Anestetic [448]

Answer:

D. Overrule any other laws and traffic control devices.

Explanation:

Laws and traffic control devices are undoubtedly compulsory to be followed at every point in time to control traffic and other related situations. However, there are cases when certain instructions overrule these laws and traffic control devices. For example, when a traffic police is giving instructions, and though the traffic control devices too (such as traffic lights) are displaying their own preset lights to control some traffic, the instructions from the traffic police take more priority. This is because at that point in time, the instructions from the traffic control devices might not be just applicable or sufficient.

Also, in the case of instructions given by construction flaggers, these instructions have priority over those from controlling devices. This is because during construction traffic controls are redirected from the norms. Therefore, the flaggers such be given more importance.

6 0
2 years ago
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