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3 years ago

Which branch of engineering studies the physical behavior of metallic elements?

Engineering

2 answers:

olga55 [171]3 years ago

6 0

Material engineering studies the physical behavior of metallic elements.

Answer: Option C

<u>Explanation: </u>

Material Engineering is the creation and learning about the materials at an atomic level. An engineers from this branch focus on material and model its characteristics using the computer.

Also, they combine the knowledge of solid-states, metallurgy, chemistry and ceramics to the application level. It also has a great role in building the future with the advancing study in nanotechnology, biotechnology, etc. Simply, these are meant to have vivid applications in future life.

steposvetlana [31]3 years ago

3 0

Answer:

your answer should be c.

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A 3-phase induction motor with 4 poles is being driven at 45 Hz and is running in its normal operating range. When connected to

Dennis_Churaev [7]

Answer:

a) The slip for the given conditions is 0.2074 or 20.74% and the developed torque is 7.14 Nm.

b) The new slip after reducing the torque to 4 Nm is 0.1162 or 11.62% and the new motor speed is 1193.13 rpm.

Explanation:

In order to find the slip we can use the definition as the relative difference between Synchronous speed and the rotor speed and that the developed torque is the ratio between output power and rotor angular velocity.

Slip.

We can find the synchronous speed u sing the following formula

$N_s =\cfrac{120f}p$

Where f stands for the frequency and p the number of poles, so we have

$N_s = \cfrac{120(45)}{4}\\N_s =1350 \,rpm$

Replacing on the slip formula

$S = \cfrac{N_s-N_r}{N_s}$

we get

$S = \cfrac{1350-1070}{1350}\\ S = 0.2074$

Thus the slip for the given conditions is 0.2074 or 20.74%.

Developed torque.

We can find the angular velocity in radians per second

$\omega_r =1070 \cfrac{rev}{min} \times \cfrac{1 \, min}{60 \, s}\times \cfrac{2\pi rad}{1 \, rev}\\\omega_r =112.05 \, \cfrac{rad}{s}$

Thus we can replace on the torque formula

$\tau = \cfrac{P}{\omega_r}\\\tau=\cfrac{800 W}{112.05 \, \cfrac{rad}{s}}$

We get

$\tau = 7.14\, N m$

The developed torque is 7.14 Nm.

Slip for the reduced torque.

The torque is proportional to the slip, so we can write

$\cfrac{\tau_1}{\tau_2}= \cfrac{S_1}{S_2}$

Thus solving for the new slip $S_2$ we have:

$S_2 = S_1 \cfrac{\tau_2}{\tau_1}$

Replacing the values obtained on the previous part we have

$S_2 = 0.2074 \cfrac{4 Nm}{7.14 Nm}\\ S_2=0.1162$

So the new slip after reducing the torque to 4 Nm is 0.1162 or 11.62%

Motor speed.

We can use the slip definition

$S_2 =\cfrac{N_s-N_{r_2}}{N_s}$

Solving for the motor speed we have

$S_2N_s =N_s-N_{r_2}$

$N_{r_2}=N_s-S_2N_s \\N_{r_2}=N_s(1-S_2)$

Replacing values we have

$N_{r_2}=1350 \, rpm(1-0.1162)\\N_{r_2}=1193.13 rpm$

The new motor speed is 1193.13 rpm.

5 0

3 years ago

What must engineers keep in mind so that their solutions will be appropriate? O abstract knowledge O context O scientists persev

Margaret [11]

Answer: Context

Explanation: It is always very important for an engineer to keep the context of his/her expirament in mind.

3 0

3 years ago

Jen is developing the positioning statement for a new line of sunglasses. In a meeting, the marketing team tells Jen that she ha

Allushta [10]

Answer:

Unique Selling Proposition

Explanation:

Unique Selling Proposition (USP) is the distinguishing feature that makes one product or business better than its competitor in the market. Jen has developed a competitive advantage of the new sunglasses brand making the company to sell their products. This is an example of USP.

8 0

3 years ago

A heat engine receives heat from a heat source at 1453 C and has a thermal efficiency of 43 percent. The heat engine does maximu

xxMikexx [17]

Answer:

a) 1253 kJ

b) 714 kJ

c) 946 C

Explanation:

The thermal efficiency is given by this equation

η = L/Q1

Where

η: thermal efficiency

L: useful work

Q1: heat taken from the heat source

Rearranging:

Q1 = L/η

Replacing

Q1 = 539 / 0.43 = 1253 kJ

The first law of thermodynamics states that:

Q = L + ΔU

For a machine working in cycles ΔU is zero between homologous parts of the cycle.

Also we must remember that we count heat entering the system as positiv and heat leaving as negative.

We split the heat on the part that enters and the part that leaves.

Q1 + Q2 = L + 0

Q2 = L - Q1

Q2 = 539 - 1253 = -714 kJ

TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:

η = 1 - T2/T1

T2/T1 = 1 - η

T2 = (1 - η) * T1

The temperatures must be given in absolute scale (1453 C = 1180 K)

T2 = (1 - 0.43) * 1180 = 673 K

673 K = 946 C

8 0

3 years ago

A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground

7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

T ≈ 17199 lbf

4 0

3 years ago