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 1008C. Air enters the heating section at 108C and 70 percent relative humidity at a rate of 35 m3/min, and it leaves the humidifying section at 208C 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:

B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv

    sending end current : Is = 1.241 ∠ 15.5⁰ KA

    real power = 730.5 Mw

C) percent voltage regulation = 8.7%

D) Transmission line efficiency  = 95.8%

Explanation:

attached is the detailed solution to the problem

Given data:

l = 200 km

z = 0.032 + j0.35 Ω/km

y = j4.2 * 10^-6 S/km

A) find the total series impedance and shunt  admittance

B) sending end voltage : Vs-l-l = 345.8 ∠ 26.14⁰ kv

    sending end current : Is = 1.241 ∠ 15.5⁰ KA

    real power = 730.5 Mw

C) percent voltage regulation = 8.7%

D) Transmission line efficiency  = 95.8%

Answer:

A centrifugal pump is a rotating machine that pumps liquid by forcing it through a paddle wheel or propeller called an impeller. It is a most common type of industrial pump. By the effect of the rotation of the impeller, the pumped fluid is drawn axially into the pump, then accelerated radially, and finally discharged tangentially.

Explanation:

Suppliers generally offer charts in the plan, which present the various things at the nominal operating point using two main arrangements: the inductors and the balancing pants.

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

Answer:

Explanation:

hola friend!!!!!

there is only one crankshaft in V8 engine

Hope this helps

plz mark as brainliest!!!!!!!!

Answer:

Some types of aggregate are susceptible to damage from repeated freezing and thawing due to their porosity. An aggregate being porous allows water molecules to enter in between the rocks.

When freezing occurs, water is known to expand. The expansion of this in the rocks creates a type of pressure which results in the fracture of the rocks. Subsequent freezing and thawing will allow for more fracture between the rock particles which will lead to its disintegration.

Answer:

intrinsic semiconductors

Explanation:

An intrinsic semiconductor is also known as a pure conductor. In such a semiconductor there are no impurities, that is why it is said to be pure.

It has some of these properties:

1. Electrical conductivity is only based on temperature

2. The quantity of electrons is the same as the number of holes in the valence bond

3. Electrical conductivity is not on the high side

4. These materials exist in their pure forms.

Explanation:

a) Given B(n) is the number of the length 'n' bit sequences which have no three consecutive 0s(i.e., they does not contain substring “000”)

Any bit string which has no 000 should have a 1 in at least one of the 1st three positions. Then, the n we will break all the bit strings by avoiding the 000 by when the 1st 1 occurs. i.e., each of the bit of string of the length of n will avoid 000 falls into the exactly any one of these cases:

1 is followed by the bit string of the length of (n-1) avoiding the 000.

01 is followed by some bit string of the length of (n-2) avoiding a 000.

001 which is followed by any bit string of the length(n-3) avoiding the 000.

Therefore, recurrence is

B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(0)=1, B(1)=2, B(2)=4

Or

B(n)=B(n-1)+ B(n-2)+ B(n-3), with B(1)=2, B(2)=4, B(3)=7

b) Let the S(n)={1,2,3,…,n}. We will say that the subset A of S(n) is very good if A does not have any of the two integers that are consecutive.

For any k, let a(k) be a number of any good subsets of a S(k).

There are two types of good subsets of S(n).

Type 1 of  good subsets of S(n) which  contain the element n,

Type 2 of good subsets of S(n)  which do not contain n.

We will first get an expression for a number of Type 1 of good subsets of the S(n) , where n≥2. This a subset does not contain n-1. So any Type 1 of the good subset of the S(n)  is obtainable when adding n to good subset of the S(n-2) . Also, on adding n to a good subset of the S(n-2) , we always get a Type 1 good subset of the S(n). Thus there are exactly as many good Type 1 of subsets of S(n) as there is good subsets of S(n-2) . By the definition there are a(n-2) good subsets of S(n-2) .

A good subset of a S(n)  is either Type 1 or Type 2. Thus the number of the good subsets of S(n)  is a(n-2)+a(n-1)

We have therefore shown that a(n) = a(n-2)+a(n-1)

c). Here firstly see that if the n is not the multiple of a 3, then there will be no chance to tile the rectangle. And also if n is a multiple of a 3, then there may be two ways to tile the 1st three columns:

And the rest of tilling is the tilling of  2x(n-3) rectangle for which there are T(n-3).

So, the recurrence is

T(n )= {2T(n-3), with T(0)=1     if n=0(mod 3)   or 0 else

We could use the base case T(3)=2

So, the following recurrence can be aslo equivalent to T(n)= 2T(n-3), with T(0)=1, T(1)=0,T(2)=0

Or

T(n)= 2T(n-3), with T(1)=0, T(2)=0,T(3)=2

d)A ternary string may be defined as a sequence of some digits, where each digit has either 0,1,or 2.

According to the problem given, we do not have a 2 anywhere after  0, and the dot which represents the binary string of length(n-1) with the property which we cannot use 2 anywhere after we use the 0.

Now for base case,  we should note that any of the ternary string which has a length of 1 satisfies the given required property. Hence the recurrence is

T(1)=3

T(n)=2T(n-1)+2^(n-1)

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:

Test Phase

Explanation:

DevSecOps is an organizational software engineering culture and practice that unifies software development (Dev), security (Sec) and operations (Ops). The main characteristic of DevSecOps is to improve customer outcomes and mission value by automating, monitoring, and applying security at all phases of the software lifecycle.

There are nine phases of the software lifecycle which are: plan, develop, build, test, release, deliver, deploy, operate, and monitor.

The Performance test in the test phase will ensure that applications will perform well under the expected workload. The test focus is on application response time, reliability, resource usage and scalability.

In development phases, database design, development, and testing activities generate database artifacts, which are data models, database schema files, trigger definitions, view definition, test data, test data generation scripts, test scripts, etc. These database artifacts must be under configuration management control. During test phase, database functional test is like application code unit test and functional test to validate the schema, triggers, and data compliance. The non-functional test includes load testing, stress test, and performance test. The security test focuses on vulnerability scan, user authentication and authorization, unauthorized access to data, data encryption, privilege elevation, SQL injection, and denial of service.

In a bid eliminate the vulnerability experienced during the traditional development process, DevSecOps emphasizes reliability, performance and Scaling with the integration of Security phase.

The integration of Security infrastructure into the Development operation(DevOps) process ensures that security challenges experienced by softwares are tackled immediately hence ensuring reliability and reduced vulnerability.

DevSecOps ensures that performance isn't sacrificed for security, hence, softwares are continously checked for security at every phase of the development process during testing.

Therefore, the security phase of the DevSecOps pipeline ensures that satisfactory security and Performance levels are met.

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Answer:

The answer ix below

Explanation:

There are 12 apple pies left.

Given that:

n = number of apple pies left = 12

x = number of food banks = 6

1) For the 12 apple pies to be distributed among 6 food banks. The number of ways in which this can be done is:

C(n + x - 1, x - 1) = C(12 + 6 -1, 6 - 1) = C(17, 5) = [tex]\frac{17!}{(17-5)!5!} =\frac{17!}{12!5!}=6188 \ ways[/tex]

12 apple pies can be distributed among 6 food banks in 6188 ways

2) For the 12 apple pies to be distributed among 6 food banks if each food bank must receive one pie, 6 pies would be remaining. The number of ways in which this can be done is:

C((n - x) + x - 1, n - x) = C(12 - 6 + 6 - 1, 12 - 6) = C(11, 6) = [tex]\frac{11!}{(11-6)!6!} =\frac{11!}{6!5!}=462 \ ways[/tex]

12 apple pies can be distributed among 6 food banks if each food bank must receive one pie in 462 ways

Answer:

Water Hammer is a knocking sound in an water pipe which occurs when the tap is turned off briskly

Explanation:

Answer:

The correct option is A

Explanation:

Thermoplastics are polymers that can be recycled. They can be melted with heat and moulded into any shape which becomes hard when cooled. The heat applied to it in the process of recycling does not affect its chemical composition.

There are two main classes of thermoplastics; semi-crystalline and amorphous.

An amorphous thermoplastic has a disorderly arranged molecular structure which makes it difficult to have a definite melting temperature (Tm). However, a semi-crystalline thermoplastic has a well arranged molecular structure which makes it melt after a certain/definite amount of heat is absolved. Hence, the correct answer is A

Answer:

of 150 pounds per square inch

Explanation:

Note that the unit for measuring water pressure is called pounds per square inch (psi)

In the case of sprinklers and standpipe systems, a pressure of 150 pounds per square inch was used initially.

Answer:

  (523.74 lb)∠-48.59°

Explanation:

a) We can find the sum of the force vectors several ways, but first we need to know the direction of vector BT. The angle PBT is identical to the angle shown as α, which we can find from the lengths BD and AD.

  α = arctan(BD/AD) = arctan((3√3)/6)) ≈ 40.893°

Using the Law of Cosines, we can find the magnitude of PT from ...

  PT^2 = PB^2 +BT^2 -2·PB·BT·cos(α)

  PT^2 = 800^2 +600^2 -2·800·600·2/√7 = 1000000 -1920000/√7

  PT ≈ 523.74 . . . . lb

The direction of PT can be found from the Law of Sines.

  ∠BPT = arcsin(BT/PT·sin(α)) ≈ 1.145597sin(40.893°) ≈ 48.59°

Relative to the +x axis, the resultant force (R) is ...

  R = (523.74 lb)∠-48.59°

__

b) The graphical solution is shown in the attachment. The graphing tool measures the segment lengths and angles. Those measures confirm the above result. (The pound values shown are scaled up by a factor of 100 from the segment lengths on the diagram. They are "captions" for the respective vectors.)

Answer:

C

Explanation:

Answer:

a) x = 5.7791 m ( this a turbulent flow )

b) x = 0.04 m ( this is a Laminar flow )

Explanation:

For Air, take

p = 1.2 kg/m³ and u = 1.8×10⁻⁵ kg/m³

So for A , x = 10 cm

Guess turbulent flow;

(Sturb / x) = (0.16 / (p∪x / u)^1/7)

we substitute

(0.1/x = 0.16) / ( 1.2 × 15 × x / 1.8×10⁻⁵)^1/7

x = 5.7791 m

CHECK

Re = (1.2 × 15 × 5.7791) / 1.8×10⁻⁵ = 5.779×10⁶

Therefore this a turbulent flow

for B, x = 1mm

Guess Laminar flow

(S-laminar / x) = ( 5 / (Reₓ)^1/2)

x = (Stutb)²p∪ / 23u

x = ((0.001)² × 1.2 × 15) / (25 × 1.8×10⁻⁵)

x = 0.04 m

CHECK

Re = (1.2 × 15 × 0.04) / (1.8×10⁻⁵)

= 40000

Therefore  this is a Laminar flow

Answer:

Problem definition

Explanation:

The engineering design is a series of step that the engineer gradually take in other to create a functioning product or process. There are many different models of the engineering design process , but they can be shrunk into four basic steps which are

The students are within the first step, and the problem here is 'an appropriate means for the students to keep their drinks cold, for 3 hours for the ball game.'

Answer:

some parts of your question is missing attached below is the missing part

Answer :  Fa = 4.46 Ib

Explanation:

use the equation

Z = 0.1 sin2∅

next we will differentiate the equation to get the locus of the velocity

z = 0.2 cos2∅∅

differentiate the equation furthermore to get the locus of acceleration in the horizontal axis

z = -0.4sin2∅(∅)^2 + 0.2cos2∅∅

note : express ∅ as 6 rad.[tex]s^{-1}[/tex] for angular velocity and ∅ = 0 for angular acceleration

equation above becomes :

Z = - 0.4 sin 2∅ ( 6)^2 + 0.2 cos 2∅(0)

  = - 14.4 sin 2∅ ( acceleration of the follower in horizontal direction )

next calculate The force at the end of A of the follower

Fa - Kx = mz  

note: m = w / g hence : Fa - Kx = w/g z  ------- (2)

w = weight of the spring-held follower = 0.75 Ib

x = compression of the spring = 0.4

k = spring stiffness = 12 Ib/ft

∅ = 45⁰

g = 32.2 ft/s^2

input these values into equation 2

hence : Fa = 4.46 Ib ( force at the end A of the follower )

Answer:

class helloSarah {

public static void main(String [] args) {

Name = "Hello Sarah";

System.out.print(Name);

}

}

Explanation:

The question seem incomplete; However, since the program is incorrect, I'll assume the question is to correct the given code;

The program segment was written in Java and the correct program is in the answer section

See bold texts for line by line explanation

This line declares the class name

class helloSarah {

This line declares the main method of the program

public static void main(String [] args) {

This line initializes string variable Name

String Name = "Hello Sarah";

This line prints the initialized variable

System.out.print(Name);

}

}

Answer:

class helloSarah {

public static void main(String [] args) {

Name = "Hello Sarah";

System.out.print(Name);

]

Explanation:

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 valies.

Thus;

d_o = 158.15 mm

d_i = 59.31 mm

Answer:

When y = 0.25 ft, velocity gradient = -6 ft/s

When y = 0.5 f, velocity gradient = -1.5 ft/s

Explanation:

Given;

equation for the velocity profile, 3/2 = 4yv

Rearrange this equation, you will get;

[tex]4yv = \frac{3}{2}\\\\v = \frac{3}{2} *\frac{1}{4y} \\\\v = \frac{3}{2}(\frac{1}{4} )(\frac{1}{y} )\\\\v = \frac{3}{8}y^{-1}\\\\ gradient \ of \ velocity \ = \frac{dv}{dy} \\\\\frac{dv}{dy} = -1(\frac{3}{8})y^{-2}[/tex]

When y = 0.25 ft

[tex]\frac{dv}{dy} = -1(\frac{3}{8})(0.25)^{-2}\\\\\frac{dv}{dy} = -6 \ ft/s[/tex]

When y = 0.5 ft

[tex]\frac{dv}{dy} = -1(\frac{3}{8})(0.5)^{-2}\\\\\frac{dv}{dy} = -1.5 \ ft/s[/tex]

Answer:

In order to logically identify alternative solutions to problems

Explanation:

Electricians are specialized in electrical wiring of buildings, transmission lines, stationary machines, and related equipment. They are either employed in the installations of new electrical components, or to maintain an already installed component. The job of an electrician can be mentally tasking, especially in troubleshooting for fault, and methods of fixing of faults. Some problems might require an out-of-norm approach to solve, and the electrician has to be able to logically identify alternative solutions to problems.

Answer:

in order to logically identify alternative solutions to problems

Explanation:

it makes the most sense

Answer:

a) civil engineering.

Explanation:

Civil engineering is a professional engineering program that deals with the construction, design, and maintenance of all the natural and man-made environments including dams, buildings, railways, and roads.

Civil engineering is the branch of engineering that is the practice not restricted because civil engineer is not restricted to academic profession but practice in designing and construction can make someone a professional civil engineer.

Hence, the correct answer is "a)."

The branch of engineering in which the practice is not restricted is; Civil Engineering

             Civil Engineering is a branch of engineering that deals with the design, construction and maintenance of the physical and naturally built environment.

These physically and naturally built environment includes; houses, roads, bridges, airports, railways, canals, dams, sewage systems, pipelines e.t.c

             Now, mechanical engineering involves design, production, operation and maintenance of mechanical systems or machineries.

             In conclusion, we see that the mechanical branch of engineering is restricted to machinery unlike civil engineering that is not restricted to only buildings but also includes pipelines, bridges, roads, railways, dams, sewage systems e.t.c

             Finally, the nuclear engineering branch is also restricted to only nuclear fission and fusion applications.

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Answer:

The answer is the strain gauges.

Explanation:

Inspection systems work or are performed to measure the characteristics of a product, to verify if it meets specified requirements, all using benchmarks and test equipment.

The strain gauges are part of the test equipment used for inspection. These are sensors that measure deformation, pressure and load in resistance tests of materials.

Answer:

the turbine work in MW is   [tex]\mathbf{W_t = 10.47299 \ kW}[/tex]

Explanation:

Given that:

the inlet temperature = 500°C

pressure = 600 kPa

exit temperature = 45°С

mass flow rate = 11 kg/s

The objective is to find the turbine work  in MW given that the mass flow rate of the steam is at 11 kg/s.

From the steam tables, the data obtained for the enthalpies at the inlet temperature of  500°C and a pressure of  600 kPa is:

[tex]\mathtt{h_1 : 3482.7 \ kJ/kg}[/tex]

The turbine expansion process is also the isentropic process, as such:

Inlet entropy [tex]\mathbf{s_1}[/tex] = Exit entropy [tex]\mathbf{s_2}[/tex]

The data obtained from the steam table for [tex]\mathbf{s_1}[/tex] = 8.003 kJ/kg.K

[tex]s_2 = s_{f2}+ x_2 *s_{fg2}[/tex]

The data obtained from the steam tables for this entities are as follows:

[tex]\mathsf{s_{f2} = 0.638 \ kJ/Kg/K }[/tex]

[tex]\mathtt{s_{fg2}=7.528 \ kJ/kg/K}[/tex]

since  [tex]\mathbf{s_1}[/tex] =  [tex]\mathbf{s_2}[/tex] = 8.003 kJ/kg.K

Therefore;

[tex]8.003 = 0.638 + x_2 * 7.528[/tex]

[tex]8.003 -0.638 = 7.528x_2[/tex]

[tex]7.365= 7.528x_2[/tex]

[tex]x_ 2= \dfrac{ 7.365}{7.528}[/tex]

[tex]x_2 = 0.978[/tex]

From the steam tables, the data obtained for the enthalpies at the exit temperature of  45°C and a pressure of  600 kPa is:

[tex]h_{f2}[/tex] = 188.4

[tex]h_{fg2[/tex] =2394.9

Thus;

[tex]h_ 2= 188.4 +0.978*2394.9[/tex]

[tex]h_ 2=2530.61[/tex] kJ/kg

The workdone for the turbine process can be computed as:

[tex]\mathsf{W_t = m(h1-h_2)}[/tex]

[tex]\mathsf{W_t = 11(3482.7-2530.61)}[/tex]

[tex]\mathsf{W_t = 11(952.09)}[/tex]

[tex]\mathsf{W_t = 10472.99}[/tex] kW

To MW, we have

[tex]\mathbf{W_t = 10.47299 \ kW}[/tex]

Answer:

Never

Explanation:

In a reversible adiabatic process, there is not transfer of heat or matter between the system and its environment. An adiabatic reversible process is a  process with constant entropy, i.e ΔQ=0. The internal energy is solely dependent on the work done either due to compression or expansion. So the entropy of the gas will never increase.

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:

1.75 kW

$0.137 kWh

4.61 kW

$3.16 therm

Explanation:

Utilized power input of the burner is

P(ui) = total power input * efficiency

P(ui) = 2400 W * 0.73

P(ui) = 1752 W or 1.75 kW

Unit cost of utilized energy is

C(ui) = Unit cost of electricity/efficiency

C(ui) = $0.1 / 0.73 kWh

C(ui) = $0.137 kWh

Power input to the gas burner is

P(gi) = Utilized power input of the burner / efficiency of the burner

P(gi) = 1.75 / 0.38

P(gi) = 4.61 kW

Unit cost of utilized energy is

C(gi) = Unit cost of gas /efficiency

C(gi) = $1.2 / 0.38 kWh

C(gi) = $3.16 therm

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:

the base-emitter junction is open and the emitter resistor is open

Explanation:

Because here there will be no base current only when the the base emitter junction is kept open and. Also when emitter resistor is kept open or with thus there will be no voltage drop across the Resistor meaning the base voltage Will be equal to that in the voltage divider circuitry