Which option identifies the tool best to use in the following scenario?

Engineering is a broad category that includes a variety of occupations and attempts to solve problems using math and

nadezda [96]

Answer:

Engineering is a broad category that includes a variety of occupations and attempts to solve problems using math and systematic and scientific processes.

Explanation:

Engineering is the scientific discipline that deals with the implementation of science on materials, constructions, machines, systems and processes to realize a specific goal. It is an area of technical activity that includes a number of specialized areas and disciplines aimed at the practical application of scientific, economic, social and practical knowledge in order to turn natural resources for the benefit of humans.

The goals of engineering activities are the invention, development, creation, implementation, repair, maintenance and / or improvement of equipment, materials or processes. Engineering is closely intertwined with science, relying on the postulates of fundamental science and the results of applied research. In this sense, it is a branch of scientific and technical activity.

4 0

3 years ago

can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and a

anyanavicka [17]

Answer:

branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
branch 2: i = 21.466∠63.435°; pf = 0.447 leading
total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

$X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915\times10^{-3}\approx j4.99984\,\Omega$

The capacitive reactance is ...

$X_C=\dfrac{1}{j\omega C}=\dfrac{-j}{100\pi\cdot 318.31\times10^{-6}F}\approx -j10.00000\,\Omega$

Branch 1

The impedance of branch 1 is ...

Z1 = 8 +j4.99984 Ω

so the current is ...

I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

Branch 2

The impedance of branch 2 is ...

Z2 = 5 -j10 Ω

so the current is ...

I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

Total current

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

__

The phasor diagram of the currents is attached.

_____

Additional comment

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

3 0

2 years ago

Which one of the following faults cause the coffee in a brewer to keep boiling after the brewing cycle is finished?

MrRa [10]

Answer:

C. Welded contacts on the thermostat

Explanation:

Any fault that keeps the heating element heating when it should not is a fault that will cause the symptom described. The details depend on the design of the brewer (not given).

"A short at the terminals" depends on what terminals are being referenced. The device on-off switch terminals are normally connected together when the brewer is turned on, so a short there may not be observable.

"Welded contacts on the thermostat" will have the observed effect if the thermostat is the primary means of ending the brewing cycle. If the thermostat of interest is an overheat protective device not normally involved in ending the brewing cycle, then that fault may not cause the observed symptom.

__

If the heating element is open-circuit, no heating will occur. A gasket leak may cause a puddle, but may have nothing to do with the end of the brewing cycle. (Loss of water can be expected to end boiling, rather than prolong it.)

8 0

3 years ago

Precast concrete curtain wall panels: Group of answer choices are typically manufactured on site and then hoisted into place. of

amid [387]

Answer:

c) can be made with a variety of surface finishes.

Explanation:

The missing options are;

When it comes to concrete work in construction, the concrete can be cast either in-situ or in form of pre-cast concrete.

Now in-situ concrete means concrete done on the construction site being built while pre cast concrete simply means concrete cast outside in a factory or yard and brought to site to mount.

These pre cast concrete could have different surface finishes as required as this is one of it's advantages over in situ because there is a lot of space and room to have the desired concrete finish.

a) are typically manufactured on site and then hoisted into place.

b) cannot be fiber-reinforced.

c) can be made with a variety of surface finishes.

d) never include insulation.

e) often are unreinforced.

5 0

2 years ago

Concerned with the number of maintenance visits the rocket can undergo before being out of service, you have been informed that

Ainat [17]

Answer:

(a) Mn = M₁ + (n-1) (M₂ -M₁) = 1 + (n- 1) 1 = n (b) n > 10 (exceed 10) or n =11 (c) n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

Explanation:

Solution

(a) Let Mn denotes the number of maintenance visits after the nth journey

Then M₁ = 1 , M₂ = 1 +M₁ = 2, M₃ = 1 +M₂ = 3

We therefore, notice that M follows an arithmetic sequence

So,

Mn = M₁ + (n-1) (M₂ -M₁)

= 1 + (n- 1) 1 = n

or Mn =n

(b) For what value of n we will get fro Mn > 10

Thus,

n > 10 (exceed 10) or n =11

(c)Similarly of Mn is greater than 50 or Mn>50, the rocket will not be used or reused

So,

n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

7 0

3 years ago