Technician A says that reinforcements may be made of plastic,
Technician B says that reinforcements may be made of nylon.
Who is right?
O A only
B only
Both A and B
of Contents
Neither Anor B

Answer:

- Vernier thrust is 5.37 kN

- mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Explanation:  

Given that;

For individual thrust chambers (vacuum);

Fc = 73.14 kN ,  Cc = 3279 m/sec

For Overall engine with Vernier (vacuum);

Foa = 297.93 kN = ,  Coa = 3197 m/sec.

- determine the Vernier thrust

Vernier thrust Fv =  Foa - ( 4 × Fc )

Vernier thrust Fv  = 297.93 - ( 4 × 73.14)

Vernier thrust Fv  = 297.93 - 292.56

Vernier thrust Fv  = 5.37 kN

Therefore, Vernier thrust is 5.37 kN

-

Vernier mass flow rate;

we know that

[tex]Co_{a}[/tex] = Fc + Fv  / mc + mv

mv = Foa/Coa - Fc/Cc

we convert kilonewton to kilograms

1 kn = 102 kg

Fc = 73.14 kN = 73.14 × 102 = 7460.28 kg

Foa = 297.93 kN = 297.93 × 102 = 30388.86 kg

we substitute

mv = (30388.86 / 3197) - (( 4 × 7460.28) / 3279)

mv = 9.5054 - 9.1006

mv = 0.4048 kg/s

Therefore, mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Answer:

electrical

computer

mechanical

and manufacturing .... I think

The answer is electrical, manufacture, computer, and chemical

Answer:

Hello your question is incomplete attached below is the missing part of the question

answer ;

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

Explanation:

Determine voltage drop from top terminal to bottom terminal ( Vab ) in the right hand branch and Vcd in left hand branch

40v and 50mA are in series hence;  Ix = 50mA

also Vcd = 30V

CD is parallel to AB hence; Vcd = Vab = 30 V

Vab = ∝*Ix + 60 v

 30v  = ∝ ( 50mA ) + 60

therefore ∝ = -600

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

Answer:

They have to make sure that the building is safe, appropriate for city conditions, they are very creative, they can build really well, and can be very focused.

Answer:

The rate of heat supply is 8.901 horse-power.

Explanation:

From Thermodynamics energy efficiency of the electric car ([tex]\eta[/tex]), no unit, is the ratio of translational mechanical power ([tex]\dot E_{out}[/tex]), measured in horse power, to electric power ([tex]\dot E_{in}[/tex]), measured in horse-power. The rate of heat supply ([tex]\dot E_{l}[/tex]), measured in horse-power, by the motor to the engine compartment at full load is difference between electric energy and translational mechanical energy. That is:

[tex]\eta = \frac{\dot E_{out}}{\dot E_{in}}[/tex] (1)

[tex]\dot E_{l} = \dot E_{in}-\dot E_{out}[/tex] (2)

[tex]\dot E_{l} = \left(\frac{1}{\eta}-1\right)\cdot \dot E_{out}[/tex] (3)

If we know that [tex]\eta = 0.91[/tex] and [tex]\dot E_{out} = 90\,hp[/tex], then rate of heat supply is:

[tex]\dot E_{l} = \left(\frac{1}{0.91}-1 \right)\cdot (90\,hp)[/tex]

[tex]\dot E_{l} = 8.901\,hp[/tex]

The rate of heat supply is 8.901 horse-power.

Answer:

what does that mean welp I have no idea sorry for answering

Answer:

Tech A is correct.          

Explanation:

A front-wheel-drive pulling to the left can result from several factors. One of them is definitely a faulty break.

A correct diagnosis linking the problem to the brakes is when there is an internal restriction and the pull is constant to one side and gets worse when the brakes are applied.

To confirm this, one would need to lift the vehicle and rotate each wheel by hand to check for excessive friction.

So the restriction may be caused by:

Cheers

Answer:

False

Explanation:

I got it wrong picking true

Answer:

a) the friction angle of the sand is 36.87°

b) the deviator stress at failure is 450 kN/m³

Explanation:

Given the data in the question;

For a consolidated drained test

The effective major principle stresses

σ₃ = σ₃' = 80 kN/m²

and

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 80 kN/m² + 240 kN/m²  = 320 kN/m²

now

a) friction angle of the sand

σ₁' = σ₃'tan²( 45° + Ф/2' ) + 2c' tan( 45° + Ф/2 )

for sand; c' = 0

so

σ₁' = σ₃'tan²( 45° + Ф/2' )

we substitute

320 = 80 tan²( 45° + Ф/2' )

Ф' = 2 × [ tan⁻¹ (√[tex]\frac{320}{80}[/tex]) - 45° ]

Ф' = 2 × [ 63.4349° - 45° ]

Ф' = 2 × 18.4349

Ф' = 36.87°

Therefore,  the friction angle of the sand is 36.87°

b)  deviator stress

σ₁' = σ₃'tan²( 45° + Ф/2' )

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )

σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )   3.597

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 150tan²( 45° + 36.87°/2 )

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600 - 150

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex]  = 450 kN/m³

Therefore, the deviator stress at failure is 450 kN/m³

Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×[tex]\frac{100}{w}[/tex]

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = [tex]\frac{D60}{D10}[/tex]

⇒ Cu = [tex]\frac{0.4}{0.12} = 3.33[/tex]

d.)

Coefficient of Graduation = Cc = [tex]\frac{D30^{2}}{D10 . D60}[/tex]

⇒ Cc = [tex]\frac{0.22^{2}}{(0.4) . (0.12)}[/tex] = 1

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

Hey there ..

I think the answer is as they put brake ..

I am not sure .. just check the image also provided above .. ..

If u think this helped u ..plz mark me as brainliest ..

And follow me

Answer:

During starts and stops

Explanation:

"Many of the energy inefficiencies in vehicles occur during starts and stops."

found it in the text.

Answer:

Construction, the techniques and industry involved in the assembly and ... Early building materials were perishable, such as leaves, branches, and animal hides. ... The well-developed masonry technology of Mesopotamia was used to build large ... although its precise description is unknown; the concealed faces of stones

Explanation:

Answer:

Technician A Is right

Explanation:

Reinforcements, as the name suggests, are used to enhance the mechanical properties of a plastic. Finely divided silica, carbon black, talc, mica, and calcium carbonate, as well as short fibres of a variety of materials, can be incorporated as particulate fillers.

Complete Question:

Throughout your job search, you’ll find that _____(the economy/education/salary) is closely related to the career of your choosing. It’s important to take the time to find out what to expect now, so you can start developing the ____ (task/communication/skill) that you need to excel in that career.

Answer:

Education; skills.

Explanation:

Throughout your job search, you'll find that education is closely related to the career of your choosing. It's important to take the time to find out what to expect now, so you can start developing the skills you need to excel in that career.

This ultimately implies that, all job openings that are being advertised by various organizations usually have a minimum requirement such as academic experience or level i.e the prospective candidate must have attained a level of education. Also, it is very important for undergraduates and potential employees to ensure that they are being proactive, as well as developing the requisite skills needed to excel in their career.

Answer:

both are c

Explanation:

Answer:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

Explanation:

Given

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,[/tex]

[tex]0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,[/tex]

[tex]0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,[/tex]

[tex]0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.[/tex]

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.[/tex]

[tex]0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,[/tex]

[tex]0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.[/tex]

[tex]0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.[/tex]

[tex]0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.[/tex]

[tex]0.78.[/tex]

The 0.3's is will be plotted as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

The 0.4's is as follows:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}[/tex]

The 0.5's is as follows:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}[/tex]

The 0.6's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

Lastly, the 0.7's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

The combined steam and leaf plot is:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

[tex]\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)[/tex]

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

[tex]\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}[/tex]

where;

[tex]Q_g[/tex] = heat generated per unit volume

[tex]\alpha[/tex] = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

[tex]\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0[/tex]

where;

The radial directional term = [tex]\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})[/tex] and the axial directional term is [tex]\dfrac{\partial^2 T}{\partial z^2 }[/tex]

d) The heat equation for a wire going through a furnace is:

[tex]\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ][/tex]

since;

the steady-state is zero, Then:

[tex]\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ][/tex]'

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

[tex]\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}[/tex]

Answer:

no it has to be removed

Explanation:

It is completely inappropriate to mention that the drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been removed. Therefore, the statement given above is false.

Drum brakes can be referred to or considered as the types of brakes that are useful in application of brakes to an object, such as wheels, in motion. To understand better, it can be stated that the system of braking under drum brakes is completely in contrast to that of disc brakes.

Drum brakes have a hydraulic pressure, which means that if the condition of lining is to be checked, removal of drums becomes essential. If the drums are not removed, correction or alignment of wheels cannot be performed.

Therefore, the significance regarding drum brakes has been aforementioned, and the statement given above with respect to their removal also holds false.

Learn more about drum brakes here:

https://brainly.com/question/14937026

#SPJ2

Answer:

A) attached below

B) Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

Explanation:

attached below is a detailed solution

A) attached below

B) Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A

Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

Answer:

Explanation:

The given equation is :

[tex]\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f[/tex]

Answer:

The fifth deviation is +1

Explanation:

Given

[tex]Deviations = -5, +2, +4,-2[/tex]

Required

Determine the fifth deviation

Represent the fifth deviation with x.

As a theorem, the sum of deviations from mean is always 0.

So, we have:

[tex]-5 +2 +4-2+x = 0[/tex]

[tex]-1+x = 0[/tex]

Add 1 to both sides

[tex]1 -1+x = 0 +1[/tex]

[tex]x = 0 +1[/tex]

[tex]x= +\ 1[/tex]

Hence, the fifth deviation is +1

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

Determine the maximum speed for safe vehicle operation

firstly calculate the stopping sight distance

m = R ( 1 - cos [tex]\frac{28.655*S}{R}[/tex] )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos [tex]\frac{28.65 *s }{678}[/tex]  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + [tex]\frac{u^2}{30(\frac{a}{3.2} )-G1}[/tex]  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + [tex]\frac{u^2}{30[(11.2/32.2)-0 ]}[/tex]

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

Answer:

This band gap also allows semiconductors to convert light into electricity in photovoltaic cells and to emit light as LEDs when made into certain types of diodes. Both these processes rely on the energy absorbed or released by electrons moving between the conduction and valence bands.

Explanation:

On the internet

a) Question Completion:

INPUT HOURS OF LABOR

Country   Cookies      Milk

Atlantis       2 hours   1 hour

Neverland  4 hours   1 hour

Answer:

1. Atlantis has the absolute advantage in the production of cookies.

2. No country has the absolute advantage in the production of milk.

Explanation:

Absolute advantage refers to superior production capability.  It is determined when a country, for example, has the ability to produce a particular good or service at lower cost or more efficiently (i.e. with less resources) than the other country.  In the scenario above, Atlantis has an absolute advantage in the production of cookies because it can produce the same quantity of cookies using 2 labor hours that Neverland can produce using 4 labor hours.  But for the production of milk, Atlantis and Neverland share the same comparative advantage less they can use less labor hours to produce milk than they can produce cookies.

Answer:

50421.6 m³

Explanation:

The river has an average rate of water flow of 59.6 m³/s.

Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:

Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s

The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken

time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds

The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³

Answer:

a. molecular interactions.

Explanation:

Conduction is thermal energy transfer by molecular interactions. Therefore, conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Some examples of conductors include metal, steel, aluminum, copper, graphite, etc.

Hence, conduction is thermal energy transfer as a result of the movement of electrons and collision between the molecules of an object.