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Phantasy [73]
3 years ago
8

What would be the most likely scale factor to use for an n-gauge model train setup? (An n-gauge layout uses locomotives that are

typically about five centimeters long, on a train platform that may be one meter by one meter.)
A. 1:100
B. 1:500
C. 100:1
D. 500:1
Engineering
1 answer:
slega [8]3 years ago
8 0
A or b for sure brother
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Gfci stands for ground fault circuit interference.
dybincka [34]

Answer:

The ground-fault circuit interrupter, or GFCI, is a fast-acting circuit breaker designed to shut off electric power in the event of a ground-fault within as little as 1/40 of a second.

Explanation:

4 0
2 years ago
In the final stages of production, a pharmaceutical is sterilized by heating it from 25 to 75°C as it moves at 0.2 m/s through a
stepan [7]

Answer:

The required heat flux = 12682.268 W/m²

Explanation:

From the given information:

The initial = 25°C

The final = 75°C

The volume of the fluid = 0.2 m/s

The diameter of the steel tube = 12.7 mm = 0.0127 m

The fluid properties for density \rho = 1000 kg/m³

The mass flow rate of the fluid can be calculated as:

m = pAV

m = \rho \dfrac{\pi}{4}D^2V

m = 1000 \times \dfrac{\pi}{4} \times ( 0.0127)^2 \times 0.2

m = 0.0253 \ kg/s

To estimate the amount of the heat by using the expression:

q = mc_p(T_{final}-T_{initial})

q = 0.0253 × 4000(75-25)

q = 101.2 (50)

q = 5060 W

Finally, the required heat of the flux is determined by using the formula:

q" = \dfrac{q}{A_s}

q" = \dfrac{q}{\pi D L}

q" = \dfrac{5060}{\pi \times 0.0127 \times 10}

q" =  12682.268 W/m²

The required heat flux = 12682.268 W/m²

3 0
3 years ago
Whats 47,000 resistance converted to kilo or mega ohms.
Julli [10]
  1. The conversion of 47,000 Ohms to kilo-ohms is equal to 47 kilo-ohms.
  2. The conversion of 47,000 Ohms to mega-ohms is equal to 0.047 kilo-ohms.

<h3>What is resistance?</h3>

Resistance can be defined as an opposition to the flow of current in an electric circuit. Also, the standard unit of measurement of the resistance of an electric component is Ohms, which can be converted to kilo-ohms or mega-ohms.

For Ohms to kilo-ohms, we have:

1 Ohms = 0.001 kilo-ohms

47,000 Ohms = X kilo-ohms

Cross-multiplying, we have:

X = 0.001 × 47000

X = 47 kilo-ohms.

For Ohms to mega-ohms, we have:

1,000,000 ohms = 1 mega-ohms

47,000 Ohms = X mega-ohms

Cross-multiplying, we have:

X1,000,000 = 47,000

X = 47,000/1,000,000

X = 0.047 kilo-ohms.

Read more resistance here: brainly.com/question/19582164

#SPJ1

7 0
1 year ago
How many trips would one rubber-tired Herrywampus have to make to backfill a space with a geometrical volume of 5400 cubic yard?
nikklg [1K]

Answer:

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

Explanation:

a) By volume.

The shrinkage factor is:

\frac{5400cu-yd}{1-0.25} =7200cu-yd

The volume at loose is:

V_{loose} =V_{bank} (1+swell-factor)=7200(1+0.2)=8640cu-yd

If the Herrywampus has a capacity of 30 cubic yard:

\frac{8640cu-yd}{30cu-yd/trip} =288trip

b) By weight

The swell factor in terms of percent swell is equal to:

pounds-per-cubic-yard-loose=\frac{pounds-per-cubic-yard-bank}{\frac{percent-swell}{100}+1 }

pounds-per-cubic-yard-loose=\frac{3000}{\frac{20}{100} +1} =2500lb/cu-yd

The weight of backfill is:

8640cu-yd*2500\frac{lb}{cu-yd} *\frac{1ton}{2000lb} =10800ton

The Herrywampus has a capacity of 40 ton:

\frac{10800}{40ton/trip} =270trip

If analyzed by volume capacity, more trips are needed to fill the space, thus the required trips are 288

8 0
3 years ago
A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejectin
lord [1]

Answer:

hello your question is incomplete attached below is the missing part and also attached is the solution

answer: a) 0.4801

              b) 5.398 kw

              c) 2.14

              d) 12.72

Explanation:

The quality of the refrigerant at the evaporator inlet

h4 = hf4 + x4(hfx4)

Refrigeration load

Ql = m(h1-h4)

COP of the refrigerator

Ql / m(h2-h1) - Qm

Theoretical maximum refrigeration load

( Ql )max = COPr.rev * [m(h2-h1) - Qin]

5 0
3 years ago
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