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3 years ago

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technician a says symbols are used to represent components in electrical wiring diagrams. technician b says components often loo

k different from one manufacturer to another. who is correct?

1 answer:

MArishka [77]3 years ago

3 0

Technician a. Components should always look the same so they are known universally

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What happens when force is placed on a square/rectangle?

mart [117]

im not sure i need to see a photo and also is this science

7 0

3 years ago

Read 2 more answers

Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe

lawyer [7]

A protective equipment which protects workers who are passing by from stray sparks or metal while another worker is welding is: E. Welding Screens.

A wielder refers to an individual who is saddled with responsibility of joining two or more metals together by wielding.

During the process of wielding, sparks and minute metallic objects are produced, which are usually hazardous to both the wielder and other workers within the vicinity.

Hence, the following protective equipment are meant to be worn or used directly by a wielder (worker) who is wielding:

Visors.

Goggles.

Protective clothing.

Dark walls.

However, a protective equipment which protects other workers who are passing by from stray sparks or metallic objects while wielder (worker) is welding is referred to as welding screens.

Find more information: brainly.com/question/15442363

4 0

3 years ago

A cold air standard gas turbine engine with a thermal efficiency of 56.9 % has a minimum pressure of 100 kP

Aleks04 [339]

Answer:

a) 5.2 kPa

b) 49.3%

Explanation:

Given data:

Thermal efficiency ( л ) = 56.9% = 0.569

minimum pressure ( P1 ) = 100 kpa

<u>a) Determine the pressure at inlet to expansion process</u>

P2 = ?

r = 1.4

efficiency = 1 - [ 1 / (rp) $\frac{r-1}{r}$ ]

0.569 = 1 - [ 1 / (rp)^0.4/1.4

1 - 0.569 = 1 / (rp)^0.285

∴ (rp)^0.285 = 0.431

rp = 0.0522

note : rp = P2 / P1

therefore P2 = rp * P1 = 0.0522 * 100 kpa

= 5.2 kPa

b) Thermal efficiency

Л = 1 - [ 1 / ( 10.9 )^0.285 ]

= 0.493 = 49.3%

4 0

3 years ago

Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The at

andriy [413]

Answer: the theoretical density is 4.1109 g/cm³

Explanation:

first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ

θ + ∅ + 90° = 180°

θ = 90° - ∅

θ = 90° - ( 109.5° / 2 )

θ = 35.25°

next we calculate the value of x from the geometry

given that; distance angle d = 0.234

x = dsinθ

= 0.234 × sin35.25°)

= 0.135 nm = 0.135 × 10⁻⁷ cm

next we calculate the length of the unit cell

a = 4x

a = 4(0.135)

a = 0.54 nm = 0.54 × 10⁻⁷ cm

next we calculate number of formula units

n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)

n' = 8 × 1/8) + ( 6 × 1/2)

= 1 + 3

= 4

next we calculate the theoretical density using this equation

P = [n'∑(Ac + AA)] / [Vc.NA]

= [n'∑(Ac + AA)] / [(a)³NA]

where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)

∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)

Na is the Avogadro’s number( 6.023 × 10²³ units/mole)

so we substitute

P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]

= 389.88 / 94.84

= 4.1109 g/cm³

therefore the theoretical density is 4.1109 g/cm³

5 0

3 years ago

Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t

almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

$h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}$

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

$n = \frac{h_1 - h_2*}{h_1 - h_2}$

$n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}$

solving for h2, we have

$h_2 = 1197.77 Btu/Ibm$

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

$(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}$

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

3 0

3 years ago