For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains prior to necking:
Engineering Stress (MPa) Engineering Strain
315 0.105
340 0.220
On the basis of this information, compute the corresponding true stresses and true strains.

Answer:

If a tube having oval cross section is subjected to pressure its cross section tends to change from oval to circular.

Bourdon tube gauges consist of a circular tube.

One end of the tube is fixed while the other end is free to undergo elastic deformation under the effect of pressure.

Fixed end is open and pressure which is to be measured is applied at the fixed end.

Free end is closed and undergoes deformation under the effect of pressure.

Due to applied pressure the circular tube tends to uncoil and become straight along the dotted line.

Working of Bourdon Gauge

As the pressure is applied at the fixed end free end undergoes deformation.

The free end is attached with sector which further meshes with the pinion on which pointer is mounted.

Deformation of the pointer is transferred to pointer via this mechanism.

As a result point undergoes deflection and shows the pressure reading on calibrated dial.

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Answer:

oooo

Explanation:

oooiiiiijjjii niaje

Answer: Honest

Explanation: In other to establish communication between two or more species, the sender and receiver, It is required of the sender to showcase certain behavior, sound or other demonstrations which are capable of passing a message to the receiver. These demonstrations are often reffered to as signals. Signals could be honest or dishonest. Honest signals are characterized by it's usefulness to the receiver and the conveyance of the actual or true meaning of underlying signal being transmitted. This is the opposite of dishonest signals which are often used to trick the receiving party as the information being transmitted are inaccurate and unreliable.

Answer:

36V, 100A-hr

Explanation:

Since the batteries are connected in series;

i. the output voltage will be the sum of the individual voltages

ii. the current rating (A-hr) will be the same as their individual current rating (A-hr). And this is because the same current flows through the batteries.

From i,

The output voltage, V, is given by the sum of the voltages of the three batteries;

V = 12V + 12V + 12V

V = 36V

From ii,

The A-hr capacity of the connection is the same as that of the individual batteries;

100 A-hr

Therefore, the output voltage and A-hr capacity of this connection is:

36V, 100A-hr

Answer:

External diameter = 158.15 mm mm

Internal diameter = 59.31 mm

Explanation:

We are given;

Diameter ratio; d_i = ⅜d_o

Where d_i is internal diameter and d_o is external diameter

Power;P = 375 KW = 375000 W

Rotational speed;N = 100 rpm

Max torque is 20% greater than mean torque; T_max = 1.2T_avg

Shear stress;τ = 60 N/mm²

Length; L = 4m = 4000 mm

Angle of twist; θ = 2° = 2π/180 radians

Modulus of rigidity;G = 0.85 X 10^(5) N/mm²

Formula for the power transmitted by the shaft is;

P = 2πNT_avg/60

Plugging in the relevant values, we have ;

375000 = 2π × 100T_avg/60

T_avg = (375000 × 60)/(2π × 100) = 35809.862 N.m = 35809862 N.mm

Since T_max = 1.20T_avg

Thus, T_max = 1.20(35809862) = 42971834.4 N.mm

Checking for strength, we'll use;

τ = Tr/J_p

Or since r = d/2

It can be written as;

τ = T(d_o)/2J_p - - - (1)

Where T is T_max

But Polar moment of inertia of hollow shaft is;

J_p = [π(d_o)⁴ - π(d_i)⁴]/32

Now, we are told that d_i = ⅜d_o

Thus;

J_p = [π(d_o)⁴ - π(⅜d_o)⁴]/32

J_p = (π/32) × d_o⁴(1 - 3⁴/8⁴)

J_p = 0.0926 d_o⁴

Plugging this for J_p in eq 1,we have;

τ = T(d_o)/2(0.0926d_o⁴)

Making d_o the subject gives;

d_o³ = T/(2 × 0.0926τ)

Plugging in the relevant values to give;

d_o³ = 42971834.4/(2 × 0.0926 × 60)

d_o³ = 3867155.7235421166

d_o = ∛3867155.7235421166

d_o = 156.96 mm

Thus, d_i = ⅜ × 156.96 = 58.86 mm

Checking for stiffness, we'll use;

T/J_p = Gθ/L

Again T is T_max

Plugging in the relevant values, we have;

42971834.4/0.0926 d_o⁴ = (0.85 × 10^(5) × 2π/180)/4000

464058686.825054/d_o⁴ = 0.7417649321

d_o⁴ = 464058686.825054/0.7417649321

d_o⁴ = 625614216.5028806

d_o = ∜625614216.5028806

d_o = 158.15 mm

d_i = ⅜ × 158.15 = 59.31 mm

So we will pick the highest values.

Thus;

d_o = 158.15 mm

d_i = 59.31 mm

Answer:

16.2 cents

Explanation:

Given that a homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time.

Where Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.

For the first 100 kWh:

16 cent × 100 = 1600 cents = 16 dollars

Since 1 dollar = 100 cents

For the remaining energy:

260 - 100 = 160 kwh

10 cents × 160 = 1600 cents = 16 dollars

The total cost = 10 + 16 + 16 = 42 dollars

Note that the base monthly of 10 dollars is added.

The cost of 260 kWh of energy consumption in July is 42 dollars

To determine the average cost per kWh for the month of July, divide the total cost by the total energy consumed.

That is, 42 / 260 = 0.1615 dollars

Convert it to cents by multiplying the result by 100.

0.1615 × 100 = 16.15 cents

Approximately 16.2 cents

Answer:

Both B and G ( Hexagonal and Tetragonal )

Explanation:

The crystals system listed below has the following relationship for the unit cell edge lengths; a = b ≠ c ( hexagonal and Tetragonal )

hexagonal ; represents  a crystal system  which has three equal axes that have an angle of 60⁰ between them while Tetragonal denotes crystals that have  three axes which have only two of its axes equal in length.

The option B and G that is Hexagonal and the Tetragonal are correct.

These both represent the crystal system that follows the relationships for the units of the cell edge lengths as a = b  not equal to c.

The hexagonal shows the crustal system having equal access of the 60-degree angle in them. While the tetragonal shows the crystal that has three axes having two equal lengths.

Find out more information about the crystal systems.

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Correct question is;

An air-conditioning system operates at a total pressure of 1 atm and consists of a heating section and a humidifier that supplies wet steam (saturated water vapor) at 100°C. Air enters the heating section at 10°C and 70 percent relative humidity at a rate of 35 m3/min, and it leaves the humidifying section at 20°C and 60 percent relative humidity. Determine (a) the temperature and relative humidity of air when it leaves the heating section, (b) the rate of heat transfer in the heating section, and (c) the rate at which water is added to the air in the humidifying section.

Answer:

A) Temperature at state 2: T2 = 19.5°C and Relative Humidity at state 2: Φ2 = 37.8%

B) Q' = 420.01 KJ/min

C) m'w = 0.1472 Kg/min

Explanation:

This question depicts a steady state process and as such the mass flow rate of dry air will remain constant during the entire process.

A) Now, from the psychometric chart attached, at temperatures of T1 = 10°C & T3 = 20°C, Relative humidities; Φ2 = 70% & Φ2 = 60% and at pressure of 1 atm, we have;

Enthalpy at state 1;h1 = 23.5 Kj/kg dry air

Absolute humidity at state 1;ω1 = ω2 = 0.0053 kg of water per kg dry air

Enthalpy at state 3;h3 = 42.3 KJ/Kg dry air

Absolute humidity at state 3; ω3 = 0.0087 kg of water per Kg dry air

Specific volume at state 1;υ1 = 0.809 m³/kg

The formula for energy balance for the humidifying is given as;

h3 = h2 + hg(ω3 - ω2)

Where hg is the enthalpy of wet steam.

From second table attached, hg at 100°C is 2675.6 KJ/kg

Thus;

Making h2 the subject, we have;

h2 = h3 + hg(ω2 - ω3)

Plugging in the relevant values we have;

h2 = 42.3 + 2675.6(0.0053 - 0.0087)

h2 = 33.2 KJ/kg

Still using the psychrometric chart attached at ω2 = 0.0053 and h2 = 33.2 KJ/kg, we have;

Temperature at state 2: T2 = 19.5°C and Relative Humidity at state 2: Φ2 = 37.8%

B) to determine the rate of heat transfer, let's first find the mass flow rate first;

m' = V1'/υ1

Thus, m' = 35/0.809

m' = 43.3 kg/min

Thus, rate of heat transfer is given by;

Q' = m'(h2 - h1)

Plugging in the relevant values, gives;

Q' = 43.3(33.2 - 23.5)

Q' = 420.01 KJ/min

C) the rate at which water is added to the air in the humidifying section is given by;

m'w = m'( ω3 - ω2)

m'w = 43.3(0.0087 - 0.0053)

m'w = 0.1472 Kg/min

Answer:

T₂ = 19.5 °C

∅₂ = 37.8%

Q = 420.01 kJ/min

m[tex]_{w}[/tex] = 0.015 kg/min

Explanation:

Given:

total pressure = p = 1 atm

Temperatures:

T₁ = 10°C

T₃ = 20°C

Relative Humidity:

∅₁ = 70%

∅₃ = 60%

Volume = V₁ = 35 m³/min

Solution:

a) Use psychometric chart to determine the enthalpies by using the given values of temperatures and relative humidity:

h₁ = 23.5 kJ/kg

w₁ = 0.0053

w₁ = w₂

h₃ = 42.3 kJ/kg

w₃ = 0.0087

v₁ = 0.809 m³/kg

When the airs flows through the heating section, the amount of moisture in it remains constant. So

w₁ = w₂

When the airs flows through the humidifying section, the amount of moisture in it increases. So

w₃ > w₂

Compute enthalpy h₂

h₂ = h₃ - ( w₃ - w₂) hs

where hs is the enthalpy of wet steam at 100°C from steam table

hs = 2676 kJ/kg

h₂ = 42.3 - ( 0.0087  - 0.0053 ) 2676

    = 42.3 -  (0.0034) 2676

    = 42.3 - 9.0984

    = 33.2016

h₂ = 33.2 kJ/kg

Compute T₂ and ∅₂

Using h₂ = 33.2 kJ/kg and w₂ = 0.0053 and psychometric chart:

T₂ = 19.5 °C

∅₂ = 37.8%

Compute mass flow rate:

mass flow rate = m = V₁ /v₁

                                = 35/0.809

                                = 43.26 kg/min

     m = 43.3 kg/min

b) Compute heat transfer in the heating section:

Q = m x (h₂  - h₁)

   = 43.3 kg/min ( 33.2 kJ/kg - 23.5 kJ/kg )

   = 43.3  ( 9.7 )

Q = 420.01 kJ/min

c) Compute rate at which water is added to the air in the humidifying section

Let m[tex]_{w}[/tex] be the mass flow rate equation of water in  humidifying section is:

m[tex]_{w}[/tex]  = m(w₃ - w₂)

      = 43.3 ( 0.0087  - 0.0053 )

      =  43.3(0.0034)

      = 0.14722

      = 0.147 kg/min

m[tex]_{w}[/tex] = 0.015 kg/min

Answer:

the minimum volume flow rate needed to tip the block is 2.66 × 10⁻⁴ m/s

Explanation:

Given that;

diameter of the jet d = 10 mm

weight W = 6 N

Now we say

Fₓ Lfₓ - Wlw= 0

horizontal force

Fₓ = W (lw/lfₓ)

Fₓ = 6 ( 0.015/2)

Fₓ = 0.9 N

X-component of momentum

v₁p(-v₁)A₁ = - Fₓ

pA₁v₁² = Fₓ

v₁² = Fₓ / pA₁

v₁ = √( Fₓ / pA₁ )

WE SUBSTITUTE

v₁ = √ ( 0.9 / ((999)(π/4)(0.01))²

v₁ = 3.39 m/s

Now Discharge Q = A₁v₁

Q = π/4 (0.01)² (3.39)

Q = 2.66 × 10⁻⁴ m/s

therefore the minimum volume flow rate needed to tip the block is 2.66 × 10⁻⁴ m/s

Answer:

they could add a play structure, with the stream they can put ducks and fish in it and picnic places

brainliest plz

Explanation:

Answer:

just took the quiz (k12) answer is...

Ask questions to identify a problem, develop a model, and carry out the plan/desgn.

Explanation:

Answer:

Batteries are safe when handled properly.

Explanation:

Just like the battery in your phone, the battery in some variant of an electric car is just as safe. If you puncture/smash just about any common kind of charged battery, it will combust. As long as you don't plan on doing anything extreme with the battery (or messing with high voltage) you should be fine.

The safety risks are the same for technicians who work on hybrid electric vehicles (HEVs) or EVs as those who work on conventional gasoline vehicles: False.

Safety risks can be defined as an assessment of the risks and occupational hazards associated with the use, operation or maintenance of an equipment or automobile vehicle that is capable of leading to the;

Hybrid electric vehicles (HEVs) or EVs are typically designed and developed with parts or components that operates through the use of high voltage electrical systems ranging from 100 Volts to 600 Volts. Also, these type of vehicles have an in-built HEV batteries which are typically encased in sealed shells so as to mitigate potential hazards to a technician.

On the other hand, conventional gasoline vehicles are typically designed and developed with parts or components that operates on hydrocarbon such as fuel and motor engine oil. Also, conventional gasoline vehicles do not require the use of high voltage electrical systems and as such poses less threat to technicians, which is in contrast with hybrid electric vehicles (HEVs) or EVs.

This ultimately implies that, the safety risks for technicians who work on hybrid electric vehicles (HEVs) or EVs are different from those who work on conventional gasoline vehicles due to high voltage electrical systems that are being used in the former.

In conclusion, technicians who work on hybrid electric vehicles (HEVs) or EVs are susceptible (vulnerable) to being electrocuted to death when safety risks are not properly adhered to unlike technicians working on conventional gasoline vehicles.

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Answer:

a) the repeat unit molecular weight of PTFE MW = 100.015 g/mole

b) the number-average molecular weight for a PTFE for which the degree of polymerization is 10,000 щₙ = 1000150 g/mole

Explanation:

Given that;

PTFE which is also called Polytetrafluoroethylene

structure  of repeat unit of Polytetrafluoroethylene

(C2F4)n

a)

To compute the repeat unit molecular weight

we say

MW = 2( atomic weight of C ) + 4( atomic weight of F)

MW = 2 (12.0107) + 4 ( 18.9984)

MW = 100.015 g/mole

therefore the repeat unit molecular weight of PTFE MW = 100.015 g/mole

b)

To compute the number-average molecular weight for  PTFE of which the degree of polymerization is 10,000

we say

DP = щₙ / MW

where щₙ is the number of average molecular weight,

MW is the repeat unit molecular weight give as 100.015 g/mole

DP is degree of polymerization which is 10,000

Now we substitute

10,000 = щₙ / 100.015

щₙ = 10,000 × 100.015

щₙ = 1000150 g/mole

Therefore the number-average molecular weight for a PTFE for which the degree of polymerization is 10,000 щₙ = 1000150 g/mole

Answer: the line-to-line voltage at the high-voltage terminals of the transformer and the sending-end of the feeder is 249.71∠1.8° kV

Explanation:

First we find the phase voltage per phase at the primary side connected in Y, so we say

V₂ = 240K/√3 = 138.56 kV

Now we find the primary current

I₁ = ((400 × 10⁶) / 3(138.56 × 10³)) ∠ -cos⁻¹ (0.8)

I₁  = 962.28∠ -36.87° A

To find the voltage V₁, we say

V₁ = ( 1.2 + j6) I₁ + V₂

we substitute

V₁ =  ( 1.2 + j6) 962.28∠ -36.87° + 138.56 × 10³

V₁  = 143∠1.57° kV

Now we find the phase voltage at the sending end

Vₓ = ( 0.6 + J1.2 )I₁ + V₁  

Vₓ = ( 0.6 + J1.2 ) 962.28∠ -36.87° + 143∠1.57° K

Vₓ = 144.17∠1.8° kV

So to Determine the line to line voltage at the sending end, we say:

Vₓ (line to line) = √3 × 144.17∠1.8° kV

Vₓ (line to line) = 249.71∠1.8° kV

Answer:

Technician A

Explanation:

Technician A is correct. Technician B is incorrect, as the brake booster aids in the drivers ability to depress the brake pedal, but the force applied is the same if the booster was absent.

Answer: b) 3.47 nj

Explanation:

Given that;

length l = 5m

radius of inner conductor r = 10cm = 0.1m

radius of outer conductor D = 20cm = 0.2m

current I = 100A = 100×10⁻³ = 0.1

medium between conductor in air u₀ = 4π × 10⁻⁷

Energy in a coaxial cable transmission line is

w = u₀ /2π I² en(b/a)

we substitute

L = 4π × 10⁻⁷ /4π ×10⁻²× 5 en (20/10)

L =3.4657 × 10⁻⁹ J

L = 3.4657 nJ ≈ 3.47 nJ

Answer:

Both are right

Explanation:

It all depends on the customer. The technicians job is to inform the customer about what can be done, rest depends on the customer that what he wants to be done. If he prefers to get the parts replaced then he should do it, and he he thinks that after repair they will work well then he should go for the repair.

Answer:

b

Explanation: because it says input file.read

the switch should always be placed immediately adjacent to the non-grounded terminal of the power supply.

Answer:

Cite two important additional refinements that resulted from the Wave-mechanical

atomic model.

Answer:

Explanation:

Electric burner

For a 2.4 kW electric burner, the rate of energy consumption is 2.4 kW.

If efficiency is 73%, the cost of utilized energy is ...

  ($0.10 /kWh) / 0.73 ≈ $0.1370 per kWh utilized

__

Gas burner

If the utilized energy provided by the gas burner is the same as the utilized energy of the electric burner, then the rate of energy consumption will be ...

  2.4 kW(0.73)/(0.38) = 4.61 kW

In terms of kWh, the cost of gas is ...

  $1.20/(105,500 kJ)·(1 kJ/(kW·s))·(3600 s/h) = $0.04095 /kWh

If efficiency is 38%, the cost of utilized gas energy is ...

  ($0.04095 /kWh) / 0.38 ≈ $0.1078 per kWh utilized

_____

The cost of gas is about 21% less per utilized joule.

_____

Comment on rates

The "unit rate" will depend on the unit chosen. In order to avoid unnecessary units conversions, we have elected to stick with kWh as the unit of energy for both cases. You may be asked for different units. We trust you or Google can make the necessary conversions.

Answer:

Explanation:

Brittle material are materials that don not undergo plastic deformation, they have very low plasticity that is while cracks can form without plastic deformation

The major/ common examples are glass, ceramics, graphite

In other words brittle materials break instead of bending, they have very low energy absorption as they don not undergo plastic deformation

Answer:

Answer:

"Test Phase " is the correct choice.

Explanation:

The DevSecOps can be described as a software development life cycle which has seen security introduced into the continous development and operations pipeline. Hence, the phase of DevSecOps which emphasizes reliability, performance and scaling is the Security phase

Therefore, ensuring that applications are reliable and performs well without having to sacrifice security in the process.

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Answer:

Ampacity

Explanation:

Ampacity is a word used to expalin ampere capacity defined by National Electrical Codes.

Ampacity is defined as the maximum current, in amperes, that can flow in a conductor continuously under the conditions of use without exceeding the temperature rating of the conductor.

Therefore, in the National Electrical Code, the current carrying abilities of conductors are called the Ampacity.

Answer:

Explanation:

given data :

output voltage ( Van(t) ) = (Vd /2) + (0.85 Vd/2 sin ( w1 t ) )

w1 = 2[tex]\pi[/tex]60 rad/sec

find the value of da(t) by inputting the value of Van (t) into

da = Van(t) / Vd

hence: da(t) = 0.5 + 0.425 sin ((2[tex]\pi[/tex]60)t)

attached below is the plot of the da(t) against time  

Answer: Neutrals

Explanation: System grounding on a power system is a term used to describe the entire processes involved when a neutral is used as the conductor to connect to the solid earth. This ensures that power is generated. This is usually done using either an inductor, an impendance or a resistor. It is very important and necessary to carry out a proper grounding of a power system in order to ensure the safety of the equipment and the personnel etc

Answer:

The total energy of the motor of the electric vehicle is 1.902 × 10⁸ joules.

Explanation:

Power is the rate of change of work in time, since given input is average power, the total energy ([tex]\Delta E[/tex]) of the motor of the electric vehicle, measured in joules, is determined by this formula:

[tex]\Delta E = \dot W \cdot \Delta t[/tex]

Where:

[tex]\dot W[/tex] - Average power, measured in watts.

[tex]\Delta t[/tex] - Time, measured in seconds.

Now, let convert average power and time into watts and seconds, respectively:

Average Power

[tex]\dot W = (50\,hp)\times \frac{746\,W}{1\,hp}[/tex]

[tex]\dot W = 3.730\times 10^{4}\,W[/tex]

Time

[tex]\Delta t = (1\,h)\times \frac{3600\,s}{1\,h} + (25\,min)\times \frac{60\,s}{1\,min}[/tex]

[tex]\Delta t = 5.100\times 10^{3}\,s[/tex]

Then, the total energy is:

[tex]\Delta E = (3.730\times 10^{4}\,W)\cdot (5.100\times 10^{3}\,s)[/tex]

[tex]\Delta E = 1.902\times 10^{8}\,J[/tex]

The total energy of the motor of the electric vehicle is 1.902 × 10⁸ joules.

Answer:

As the impurity concentration in solid solution is increased, the tensile and yield strengths increases.

Explanation:

The addition of impurities in solid solutions shows an improved tensile and yield strength due to the grain refinement and obstacles to the motion of dislocation.

Example, the addition of carbon as impurity into iron, which forms steel shows a significant increase in the tensile and yield strengths of iron.

Blacksmiths also use work hardening to introduce dislocation into solid solutions in order to increase their  tensile and yield strengths.

Therefore, as the impurity concentration in solid solution is increased, the tensile and yield strengths increases.

A solid solution is a mixture of crystalline solids and is soluble over the partial or evenly complete range.

Hence the option Increases is correct.

Learn more about the concentration in solid solution is increased.

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Answer:

Steady flow

Explanation:

Flows in fluids can be categorized into different classes depending on the type of flow and the variations in their characteristics such as velocity, pressure, density, temperature, e.t.c

When these characteristics do not change when measured over time at a single point, then the flow is said to be steady. For a steady flow, the mathematical expression, amidst other conditions, is given as follows;

[tex]\frac{dV}{dt} = 0, \frac{dP}{dt} = 0, \frac{dT}{dt} = 0[/tex]

Where;

V, P and T are the velocity, pressure and temperature of the fluid.

PS:

Other types of flows include:

i. Unsteady flow

ii. Laminar flow

iii. Turbulent flow

iv. Uniform flow

v Non-uniform flow

vi. Rotational flow

vii. Irrotational flow

Explanation:

From the table of conduction shape factor and dimensionless conduction rates for selected systems, for vertical cylinder in a semi-infinite medium

[tex]S=& \frac{2 \pi L}{\ln \left(\frac{4 L}{D}\right)}[/tex]

[tex]S=& \frac{2 \pi L}{\ln \left(\frac{4 L}{D}\right)}[/tex]

[tex]\frac{2 \pi \times 0.1}{\ln \left(\frac{4 \times 0.1}{0.005}\right)}[/tex]

[tex]=0.1433 \mathrm{m}[/tex]

Answer:

d) A mosfet

Explanation:

MOSFET is the most common type of insulated gate Field Effect Transistor (FET),  used in electronic circuits and it stands for Metal Oxide Semiconductor Field Effect Transistor.

To configure MOSFET to act as an amplifier, a small AC signal is applied,   which is superimposed on to DC bias at the gate input, then the MOSFET will act as a linear amplifier.

Therefore, the correct option is (d) A mosfet