what are the conditions for sheet generator to build up its voltage?

Answer:

It is difficult to introduce products modification

Answer:

1. Slope = 53.3 x 10⁻⁶

2. Deflection = -0.00016m

Explanation:

given:

let L = 4 m (span of cantilever beam)

let w = 300 N/m (distributed load)

let EI =60 MNm² (flexural stiffness)

                 dy      w * L³        300 x 4³

1. slope = ------- = --------- =  ------------------- =  53.3 x 10⁻⁶

                  dx        6EI           6 x 60x10⁶

                                  wL⁴               300 x 4⁴

2. Deflection = y = - ----------- =  - ------------------ =   -0.00016m

                                    8EI                8 x 60x10⁶

therefore the deflection is 0.16mm downwards.

Answer:

A) 5 MPa , 55 MPa

B) maximum stress = 55 MPa,  maximum shear stress = 25 MPa

Explanation:

using the given Data

free surface of a solid body

α[tex]_{x}[/tex] = 50 MPa,    α[tex]_{y}[/tex] = 10 MPa , t[tex]_{xy}[/tex] = -15 MPa

attached below is the detailed solution to the question

Answer:

I think its A) 5 MPa , 55 MPa

B) ms = 55 MPa,  mss= 25 MPa

α = 50 MPa,    α = 10 MPa , t = -15 MPa

Answer: 0.002174

Explanation:

Given that the

Yield strength rho = 450 MPa

Length = 2 m

Elastic modulus E= 207 GPa

According to Hook's law, if the elastic limits is not reached, the elastic modulus is the ratio of elastic strength to the elastic strain ə

E = rho/ə

Make ə the subject of formula

ə = rho/ E

ə = (450 × 10^6) / (207 × 10^9)

ə = 2.174 × 10^-3

Therefore, the engineering strain which depends on engineering stress and elastic modulus is 2.174 × 10^-3

Elastic Strain has no S.I Units.

Given:

We have given two statements.

Statement 1: Proper footwear may include both leather and steel-toed shoes.

Statement 2:  Leather-soled shoes provide slip resistance.

Find:

Which statement is true.

Solution:

A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.

Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.

Leather-soled shoes don't provide slop resistance.

Therefore, both the Technicians are wrong.

From the statements made by both technician A and Technician B, we can say that; both technicians are wrong.

We are given the statements made by both technicians;

Technician A: Proper footwear may include both leather and steel-toed shoes.

Technician B: Leather-soled shoes provide slip resistance.

Now, they are talking about safety shoes to be worn in workshops.

A shoe that is Slip resistant will have rubber soles and tread patterns that can help to have better grip of wet or greasy floors.

This is the type of shoe that should be worn by technicians in the workshop.

Thus, Technician A is wrong because proper footwear does not include leather shoes.

Similarly, technician B is also wrong because leather shoes are not safety shoes.

Read more about slip resistant shoes at; https://brainly.com/question/17411739

Answer:

Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing.

There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.

Technical standards exist to provide glossaries of abbreviations, acronyms, and symbols that may be found on engineering drawings. Many corporations have such standards, which define some terms and symbols specific to them; on the national and international level, like BS8110 or Eurocode 2 as an example.

Explanation:

Answer:

An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration

Explanation:

Answer:

hello your question is incomplete here is the complete question

. “ABC construction Inc.” company becomes the lowest in the bed process to get a $21 million construction project for “Northern Inc.”. Now “ABC construction Inc.” planning to make a formal contract agreement with the “Northern Inc.”. What are the main elements of this agreement to consider it as a legal contract?

Answer : elements of the agreement

Explanation:

Offer : an offer is the beginning element for any valid agreement to be started or reached between two or more bodies. ABC construction would have to make an offer first for the agreement to be valid

Acceptance: This is part where by the company "Northern Inc" after receiving the offer from ABC construction Inc would have to consent to the approval of the offer made.

capacity : This the element of the agreement that helps to ensure that both parties have the legal and financial backings to embark on the contract agreement .

certainty : This element ensures that both parties understands the terms and conditions attached to the agreement and this to ensure that there are no bogus conditions

Consideration : This is a very vital element because the both parties have to give something in return while going into a valid agreement

Intention to create legal relation : Legal relations are applied to contract agreements whereby both parties want the contract agreement to b legally enforced and this is important in order to prevent contract breach by any party involved in the agreement

Answer:

We can prevent it by:

a) By wearing GOOGLES.

b) By wearing our Lab coat.

c) Fire extinguisher should always be present in the lab.

d) Hand Gloves must be worn.

e) No playing in the lab.

f) No touching of things/equipment's e.g bottles, in the lab.

g) No eating/snacking in the lab.

h) Always pay attention, no gisting.

i) Adult/qualified person must be present in the lab with pupils/students.

Explanation:

Hope it helps.

Answer:

partner itself

Explanation:

Answer:

limited partner

hope this answer correct :)

Answer:

The discharge temperature is 259.82 K

Explanation:

In this question, we are concerned with calculating the discharge temperature

Please check attachment for complete solution

Answer:

max shear = R = V = 15 kN

Explanation:

given:

load = 10 kn/m

span = 3m

max shear = R = V = wL / 2

max shear = R = V = (10 * 3) / 2

max shear = R = V = 15 kN

Explanation:

1. increase This due to increase in the pore volume.

2.increase . This is due to the fact that more liquid phase will be present at the firing.

3. Increase. This increase is because of the fact that clay on cooling forms glass.Thus, gaining more strength as the liquid phase formed fills in pore volume.

4. Increase, Rate of corrosion depends upon the surface area exposed.Since, upon vitrification surface area would increase, therefore corrosion increases.

5. Increase , glass has higher thermal conductivity than the pores it fills.

Answer:

true

Explanation:

true

Answer:

I believe this is true

Explanation:

If your looking at something and you look at something else everything is still in perfect view and clear, in focus.

hope this helps :)

Answer:

P=11 kW

Explanation:

Given that

Number of poles= 8

I.E.C. 180L motor frame

From data book , for 8 poles motor at 50 Hz

Speed = 730 rpm

Power factor = 0.75

Efficiency at 100 % load= 89.3 %

Efficiency at 50 % load= 89.1 %

Output power = 11 kW

Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.

P=11 kW

Answer:

At temperature is and relative humidity is 86% therefore,  the humidity ratio is 0.0223 and the specific volume is 14.289

At temperature is and Relative humidity is 40% therefore, the humidity ratio is  0.0066 and the specific volume is 13.535.

To calculate the mass of air can be calculated as follows:

Now , we going to calculate the volume,

The time which is required to fill the cistern can be calculated as follows:

Now, putting the value in above formula we get,

Therefore, the hours required to fill the cistern is 4.65 hours.

Explanation:

Answer:

Following are the answer to this question:

Explanation:

1)

Following the four operations in the airlines:

Landside operations:

In Airlines, the airports are divided into areas on the countryside and on the airside, in which landside region is available to the public, although strictly controlled access to the airside zone. Its area covers all areas of the airport across the aircraft, including parts of the buildings which can only be reached by customers and employees.

Airside Operations:

It's also committed to ensuring which air operations military exercises Ballarat airfields are safe and secure. It includes the provision of parking and flight escort services to itinerant and automates. Organizing operational response to incidents, accidents, or emergencies at the airport.

Billing and invoicing Operations:

This requires several steps, each of which must be performed with absolute accuracy to ensure that perhaps the airport operator is adequately paid for supplying passengers with all the services and infrastructure. After this, the receipts want to be produced and sent to customers on the airline.

Information management:

In this, it collects all the data about the customers and employees and it also helps in finding new routes and after collecting the data it processes on them.  

2)

It involves many activities in the airline, including dispatch, flight preparation, flight watch, weather information source, activities control, ground-to-air communications, and staff coordination, scheduling, and maintenance planning. Computing and expert programs are constantly being used to handle unpredictable activities.

Answer:

Net heat transfers:  during summer = 52.253 W/m^2 during winter = 119.375 w/m^2

Explanation:

Given data :

Room temperature throughout the year = 20⁰c = 293 k

Room temperature during summer = 27⁰c = 300 k

Room temperature during winter = 14⁰c = 287 k

surface temperature of a person throughout the year = 32⁰c = 305 k

coefficient of heat transfer by natural convection (h)= 2 w /m^2 k

emissivity = 0.9

An explanation to the condition of feeling chilled during the winter and comfortable during summer  can be explained with the calculation below

The heat transfer from the surface body of a person the the room is carried out by convection and this can be calculated as

q = hΔt = 2 * ( 305  -  293) = 2 * 12 = 24 w /m^2

also calculate heat transfer through radiation using this formula

[tex]q_{rad} =[/tex] εσ [ (Temperature of body)^4- (temperature of room at each season)^4 ]

ε = 0.90,(emissivity)    σ = 5.67 *10^-8 w/^2 . k ( Boltzmann's constant)

during summer :

[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8)* ( 305^4 - 300^4 ) = 28.253 W/m^2

therefore the net heat transfer during the summer = 24 w/m^3 + 28.253 W/m^3 = 52.253 W/m^2

During winter :

[tex]q_{rad}[/tex] = (0.90)*(5.67*10^-8) * ( 305^4 - 287^4 ) = 95.375 W/m^3

therefore the net heat transfer during the winter

= 24 w/m^3 + 95.375w/m^3 = 119.375 w/m^2

Answer:

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.

Explanation:

Heat generated Q = 200 kW

power input W = 75 kW

Temperature of heated region [tex]T_{h}[/tex] = 293 K

Temperature of heat source [tex]T_{c}[/tex] = 273 K

For this engine,

coefficient of performance COP = Q/W  = 200/75 = 2.67

The maximum theoretical COP obtainable for a heat pump is given as

COP = [tex]\frac{T_{h} }{T_{h} - T_{c} }[/tex] =  [tex]\frac{293 }{293 - 273 }[/tex] = 14.65

From the calculation, we can see that the invention's COP of 2.67 does not exceed the maximum theoretical COP of 14.65. Hence his claim is valid and could be possible.

Answer:

[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]

Explanation:

Given that:

The height of a  triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the  triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = [tex]1.2075 \times 10^{-5} \ ft^2/s[/tex]

the density of water ρ = 62.36 lb /ft³

[tex]Re_{max} = \dfrac{Ux}{v}[/tex]

[tex]Re_{max} = \dfrac{2.5 \ ft/s \times 2 \ ft }{1.2075 \times 10 ^{-5} \ ft^2/s}[/tex]

[tex]Re_{max} = 414078.6749[/tex]

[tex]Re_{max} = 4.14 \times 10^5[/tex] which is less than < 5.0 × 10⁵

Now; For laminar flow;  the drag on  the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

[tex]dF_D = (\dfrac{0.664 \times \rho \times U^2 (2-x) dy}{\sqrt{Re_x}})^2[/tex]

where;

[tex](2-x) dy[/tex] = strip area

[tex]Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}[/tex]

Therefore;

[tex]dF_D = (\dfrac{0.664 \times 62.36 \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})[/tex]

[tex]dF_D = 1.136 \times(2-x)^{1/2} \ dy[/tex]

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy[/tex]

[tex]\int\limits \ dF_D = \int\limits^1_0 \ 1.136 \times(2-2y)^{1/2} \ dy[/tex]

Let U = (2-2y)

-2dy = du

dy = -du/2

[tex]F_D = \int\limits^0_2 \ 1.136 \times(U)^{1/2} \ \dfrac{du}{-2}[/tex]

[tex]F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du[/tex]

[tex]F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2[/tex]

[tex]F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2[/tex]

[tex]F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ][/tex]

[tex]F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ][/tex]

[tex]F_D = 1.071031071 \ lbf[/tex]

[tex]\mathbf{F_D \approx 1.071 \ lbf}[/tex]

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3,  c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient  = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

[tex]\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}[/tex]

= 0.4167 = [tex]1.0396e^{-0.4888*Fo}[/tex]

therefore Fo =  3.8264

Now determine how long the plate should be left in the furnace

Fo = [tex](\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )[/tex]

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

Answer:

(a). -1.76i + 0.76j ft/s^2.

(b). -10.42i + 5.7j ft/s^2.

Explanation:

So, we are given the following data or parameters in the question/problem above;

=> "The fire truck is moving forward at a speed = 35 mi /hr and is decelerating at the rate = 10 ft/sec2."

=> "At the instant considered the angle = 30° and is increasing at the constant rate = 10 deg /sec. Also at this instant the extension b of the ladder = 5 ft, with b ˙ = 2 ft/sec and b ¨ =−1 ft/sec2"

STEP ONE: Convert the angular speed to rad/s and speed to ft/s.

Angular speed => π/18 rad/s

Speed => 35 × 1.46667 = 52.33 ft/s

STEP TWO: determine the coordinate form of the acceleration of the truck.

= - 10(cos 30°)i - sin(30°)j [π/18 × K] × 25i[π/18 × K] × 2i × [π/9 × K] - 1i.

= -10.42i + 5.7j ft/s^2.

PLEASE NOTE: The step TWO above is the solution to the second part (b) of this question which is the with respect to the ground.

STEP THREE:

This, from step two above;

( -10.42i + 5.7j ) - ( acceleration of the truck).

= - 10.42i + 5.7j - - 10(cos 30°)i - sin(30°)j = -1.76i + 0.76j ft/s^2.

Answer:

the required stress level  at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa

Explanation:

From the given information; the objective is to compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.

The Critical Stress for a maximum internal crack can be expressed by the formula:

[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]

[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]

where;

[tex]\sigma_c[/tex] = critical stress required for initiating crack propagation

[tex]K_{lc}[/tex] = plain stress fracture toughness = 26 Mpa

Y = dimensionless parameter

a = length of the internal crack

given that ;

the maximum internal crack length is 8.6 mm

half length of the internal  crack will be 8.6 mm/2 = 4.3mm

half length of the internal  crack a = 4.3 × 10⁻³ m

From :

[tex]Y= \dfrac{K_{lc}}{\sigma_c \sqrt{\pi a}}[/tex]

[tex]Y= \dfrac{26}{112 \times \sqrt{\pi \times 4.3 \times 10 ^{-3}}}[/tex]

[tex]Y= \dfrac{26}{112 \times0.1162275716}[/tex]

[tex]Y= \dfrac{26}{13.01748802}[/tex]

[tex]Y=1.99731315[/tex]

[tex]Y \approx 1.997[/tex]

For this same component and alloy, we are to also compute the stress level at which fracture will occur for a critical internal crack length of 3.0 mm.

when the length of the internal crack a = 3mm

half  length of the internal  crack will be 3.0 mm / 2 = 1.5 mm

half length of the internal  crack a =1.5 × 10⁻³ m

From;

[tex]\sigma_c = \dfrac{K_{lc}}{Y \sqrt{\pi a}}[/tex]

[tex]\sigma_c = \dfrac{26}{1.997 \sqrt{\pi \times 1.5 \times 10^{-3}}}[/tex]

[tex]\sigma_c = \dfrac{26}{0.1370877444}[/tex]

[tex]\sigma_c =189.6595506[/tex]

[tex]\sigma_c =[/tex] 189.66 MPa

Thus; the required stress level  at which fracture will occur for a critical internal crack length of 3.0 mm is 189.66 MPa

Answer:

Both technician A and technician B are correct.

Explanation: Vehicle manufacturers always specify the size of the tires required for a given vehiclefor optimal efficiency,this will ensure that the speedometer is accurate and the level of the vehicle is good enough to ensure the vehicle works efficiently.

It is also a known fact that an increase in a vehicle's rpm(revolution per minute) will eventually lead to increased fuel consumption which means the fuel economy of the vehicle will be reduced making the vehicle less efficient in its fuel consumption.

Answer:

The material will not fail

Explanation:

A rod subjected under cyclic stress will fail if the cyclic stress it is subjected to is a constant maximum value that is above the fatigue limit of the rod. but in this problem the Rod is subjected to a cyclic stress that ranges from  200 MPa(maximum stress) and 20 MPa ( minimum stress). this simply means that at all times the Rod will not experience maximum stress of 200 MPa and its Fatigue limit is also set at ~100 MPa

attached is the diagram showing the cyclic stress the rod is subjected to

Answer:

μb = 0.096

μc  = 0.073

Explanation:

member AB:

-800( 4/3 ) + Nb (2) = 0

Nb (2) = 3200/3

Nb = 533.3N

Post BC:

summation of force along the y axis=0

Nc + Nb + 150(3/5 ) -50(9.81)=0

Nc + 533.3 + 150(3/5 ) -50(9.81)=0

Nc = 933.83 N

Also (-4/5)(150)(3) + Fb(0.7)= 0

Fb = (4/5)(150)(3)/0.7 = 51.429 N

Likewise alog the x axis,

4/5(150) - Fc -Fb = 0

4/5(150) - Fc -51.429 = 0

Fc = 4/5(150)  -51.429 =68.571 N

μb = Fb/Nb = 51.429/533.3  = 0.096

μc = Fc/Nc = 68.571 / 933.83 = 0.073

Answer:

A) slip(s1) = (1800 - 1750) / 1800 = 0.0277

B) 1800 rpm, 1350 rpm,  900 rpm, 450 rpm

Explanation:

Given data

frequency = 60 Hz,  

Line - line rms = 440 - V

3 phase induction-motor drive

number of poles = 4

Full-load rated speed = 1750

rated torque = 40 Nm

A) The plot of torque-speed characteristics for the following values of the frequency f : 60 Hz, 45 Hz, 30 Hz, and 15 Hz is attached below

first we calculate the rated speed:

Ns = [tex]\frac{120f}{p}[/tex]

f = 60 Hz .  p (number of poles) = 4

Ns[tex]_{1}[/tex] = [tex]\frac{120(60)}{4}[/tex] = 1800

full loaded rated value = 1750.

slip(s1) = (1800 - 1750) / 1800 = 0.0277

considering a linearly condition the slip is low

[tex]\frac{T1}{T2} = \frac{S1}{S2} * \frac{F2}{F1}[/tex]

S1 = 0.0277

f1 = 60 Hz

hence s2 = 0.018 therefore Ns2 = 1500

B) The speeds of operation at : 60 Hz, 45 Hz, 30 Hz, 15 Hz

for 60 Hz :

Ns = [tex]\frac{120f}{p}[/tex] = (120*60) / 4 = 1800 rpm

for 45 Hz:

Ns = 120(f) / p = (120*45) / 4 = 5400 /4 = 1350 rpm

for 30 Hz:

Ns = 120(f) / p = (120*30) / 4 = 3600 / 4 = 900 rpm

for 15 Hz:

Ns = 120(f) / p = (120*15) / 4 = 1800 / 4 = 450 rpm

Answer:

a) 42 mm

b) 144.4 MPa

Explanation:

Load P = 200 kN = 200 x 10^3 N

Torque T = 1.5 kN-m = 1.5 x 10^3 N-m

maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa

diameter of shaft d = ?

From T = τ * [tex]\frac{\pi }{16}[/tex] * [tex]d^{3}[/tex]

substituting values, we have

1.5 x 10^3 = 100 x 10^6 x [tex]\frac{3.142 }{16}[/tex] x [tex]d^{3}[/tex]

[tex]d^{3}[/tex] = 7.638 x 10^-5

d = [tex]\sqrt[3]{7.638 * 10^-5}[/tex] = 0.042 m = 42 mm

b) Normal stress = P/A

where A is the area

A = [tex]\frac{\pi d^{2} }{4}[/tex] = [tex]\frac{3.142*0.042^{2} }{4}[/tex] = 1.385 x 10^-3

Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = 144.4 MPa

Answer:

from the below explanation...  we can say that, in the series circuit, flowing current remains the same at each part of the circuit. While in parallel circuits, the voltage across two endpoints of the branches is the same as the supplied voltage.

Explanation:

1.

The components in a series circuit are arranged in a single path from one end of supply to another end. However, the multiple components in a parallel circuit are arranged in multiple paths wrt the two end terminals of the battery.

2.

In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit.

3.

In the series circuit, different voltage exists across each component in the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.

4.

A fault in one of the components of the series circuit causes hindrance in the operation of a complete circuit. As against fault in a single component in a parallel network do not hinder the functioning of another part of the circuit.

5.

The detection of a fault in case of a series circuit is difficult, but it is quite easy in parallel circuits.

6.

The equivalent resistance in case of a series circuit is always more than the highest value of resistance in the series connection. While the equivalent resistance in the parallel circuit is always less than any of the individual resistances in parallel combination.

Answer:

A)  222.58 kJ / kg

B)  0.8897 M^3/ kg

c)  0.7737 m^3/kg

D)  746.542 k

E)  536.017 kj/kg

efficiency = 58% ( approximately )

Explanation:

Given Data :

Gas constant (R) =  0.287 kJ/ kg.K

T1 = 310 k

P1 ( Kpa ) = 100

r = 11.5 ( compression ratio )

rp = 1.95 ( pressure ratio )

A ) specific internal energy at state 1

 = Cv*T1 =  0.718 * 310 = 222.58 kJ / kg

B) Relative specific volume at state 1

= P1*V1 = R*T1 ( ideal gas equation )

V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3

V1 = 88.97 / 100 = 0.8897 M^3/ kg

C ) relative specific volume at state 2

Applying  r ( compression ratio) = V1 / V2

11.5 = 0.8897 / V2

V2 = 0.8897 / 11.5 = 0.7737 m^3/kg

D) The temperature (k) at state 2

since the process is an Isentropic process we will apply the p-v-t relation

[tex]\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }[/tex]

hence T2 = [tex]9^{1.4-1} * 310[/tex] = 2.4082 * 310 = 746.542 k

e) specific internal energy at state 2

= Cv*T2 = 0.718  * 746.542 = 536.017 kj/kg

efficiency = output /input = 390.3511 / 667.5448 ≈ 58%

attached is a free hand diagram of an Otto cycle is attached below