A cylindrical specimen of some metal alloy 11.2 mm (0.4409 in.) in diameter is stressed elastically in tension. A force of 15600 N (3507 lbf) produces a reduction in specimen diameter of 5 × 10-3 mm (1.969 × 10-4 in.). Compute Poisson's ratio for this material if its elastic modulus is 100 GPa (14.5 × 106 psi).

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

Answer:

FALSE.

Explanation:

This statement can be considered false, because it is important to know essential aspects of the prospect before starting the sales dialogue.

In addition to obtaining basic information about the client, it is relevant for the company to know the prospect's organization.

We can define the prospect as the one who, for the company, has the ideal customer profile, but who has not yet shown interest in consuming its products and services.

Therefore, knowing information about it, about its basic and complex characteristics will help the organization to develop sales and marketing strategies aimed at attracting the prospect.

This Physics question is directed towards High School students and pertains to the maximum speed and height a 2-kg block achieves when released from a compressed spring, following the removal of a 3-kg block initially resting on it.

The subject of this question is Physics, and it is appropriate for a High School grade level. The question is regarding the oscillatory motion and energy conversion of a block attached to a spring system. Specifically, it involves understanding the concepts of potential energy stored in a spring, conservation of mechanical energy, and kinematics of simple harmonic motion.

To solve part (a) of the problem for the 2-kg block, you'll need to use the conservation of energy principle. Initially, when the 3-kg block is also on the spring, the potential energy stored in the compressed spring is equal to the kinetic energy the 2-kg block will have when it reaches its maximum speed. After the upper block is removed, the spring force will only accelerate the 2-kg block.

To find the maximum speed reached by the 2-kg block, you would calculate the kinetic energy equivalent to the potential energy stored in the spring and set it equal to \\((1/2)mv^2\\). For part (b), you could use the conservation of energy principle again to find the maximum height the 2-kg block reaches by equating the initial kinetic energy to the potential energy at the maximum height (\\(mgh\\)).

Answer:

221.929 hp

98.19%

Explanation:

See attached picture.

Answer:

Hello there, see step by step explanation for answers

Explanation:

A database design for a University Registrar. The following requirements are for designing a Database Schema.

It should include:

1. Information about Students .

2. Information about Departments.

3. Information about Professors.

4. Information about courses.

5 . student Grades.

6. TA's for a course.

7. Department offering different courses.

For Designing Database for Registrar System We need to define various

1. Entity: Student, Course, Instructor, Course offering.

2. Attributes of Entities:

(a) Student entity has Sid, name, program as its attributes

(b)Course has Course_n, title, credits, and syllabus as its attributes

(c) instructor has iid ,name ,dept, title as its attributes

(d) course offering has section_no ,time, room, year ,semester as its attributes

3. Relationship among various entities

(a) enrolls

(b) teaches

© is offered

E-R Diagram for University Registrar

This Diagram shows student entity enrolls various courses which can be having teaches relationship with instructor. Course is being offered by relationship is offered by course offering .

A database design for a University Registrar. The following requirements are for designing a Database Schema.

It should include:

1. Information about Students .

2. Information about Departments.

3. Information about Professors.

4. Information about courses.

5 . student Grades.

6. TA's for a course.

7. Department offering different courses.

For Designing Database for Registrar System We need to define various

1. Entity: Student, Course, Instructor, Course offering.

2. Attributes of Entities:

(a) Student entity has Sid, name, program as its attributes

(b)Course has Course_n, title, credits, and syllabus as its attributes

(c) instructor has iid ,name ,dept, title as its attributes

(d) course offering has section_no ,time, room, year ,semester as its attributes

3. Relationship among various entities

(a) enrolls

(b) teaches

© is offered

E-R Diagram for University Registrar

This Diagram shows student entity enrolls various courses which can be having teaches relationship with instructor. Course is being offered by relationship is offered by course offering .

Answer:

(a) 0.00053

(b) 0.1 mΩ

Explanation:

           New Per-unit reactance = [tex]2* \frac{100}{600} * (\frac{13.8}{345}) ^{2} = 0.00053[/tex]

Given Information:

Zpu_old = 2 pu

Sbase_new = 100 MVA

Sbase_old = 600 MVA

kV_old = 13.8 kV

kV_new = 345 kV

Required Information:

Zpu_new = ?

ZΩ = ?

Answer:

Zpu_new = 0.000533 pu

ZΩ = 0.634 Ω

Explanation:

a) Find the per-unit value of the generator synchronous reactance on the specified bases.

When the base kVA and base kV are changed then we use following relation to update the per unit values.

Zpu_new = Zpu_old*(Sbase_new/Sbase_old)*(kV_old/kV_new)²

Zpu_new = 2*(100x10⁶/600x10⁶)*(13.8x10³/345x10³)²

Zpu_new = 0.000533 pu

b) Find the ohmic value of the synchronous reactance.

ZΩ = Zbase*Zpu_new

Where Zbase is calculated as

Zbase = (kVbase)²/Sbase

Zbase = (345x10³)²/100x10⁶

Zbase = 1190.25 Ω

ZΩ = Zbase*Zpu_new = 1190.25*0.000533 = 0.634 Ω

Answer:

a) 5 V.

b) Energy also become half.

Explanation:

See attached picture.

Answer:

The detailed answer to the question is explained in the attached file.

Explanation:

Answer:

a) [tex]Re_{D} = 111896.745[/tex], b) [tex]Re_{D} = 1.119\times 10^{-7}[/tex]

Explanation:

a) The Reynolds number for the water flowing in a circular tube is:

[tex]Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}[/tex]

Let assume that density and dynamic viscosity at 25 °C are [tex]997\,\frac{kg}{m^{3}}[/tex] [tex]0.891\times 10^{-3}\,\frac{kg}{m\cdot s}[/tex], respectively. Then:

[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]

[tex]Re_{D} = 111896.745[/tex]

b) The result is:

[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]

[tex]Re_{D} = 1.119\times 10^{-7}[/tex]

Answer:

Isothermal  expansion W₁ =-37198.9 J

Polytropic Compression W₂ =-34872.82 J

Isobaric Compression W₃ =  -6974.566 J

The net work for the cycle = -79046.29 J

Explanation:

Mass of air = 0.15 kg = 150 g

Molar mass = 28.9647 g/mol

Number of moles = 150 g /28.9647 g/mol = 5.179 moles of air

PV = nRT therefrore V = nRT/(P) = 5.179*8.314*(350+273.15)/(2×10⁶) = 0.0134167 m³

For isothermal expansion we have

P₁V₁ = P₂V₂ or V₂ = P₁V₁/P₂ = 2×10⁶*0.0134167 / (5×10⁵) = 0.0536668 m³

Therefore work done

W₁ = -nRTln(V₂/V₁) = -26833ln(4) = -37198.9 J

Stage 2

Compression polytropically we have

[tex]\frac{P_2}{P_3} = (\frac{V_3}{V_2} )^n[/tex]  where P₃ = 2 MPa

Therefore V₃ = [tex](\frac{1}{4} )^{\frac{1}{1.2} }*V_2[/tex]  = 1.6904×10⁻² m³

Work = W₂ = [tex]\frac{P_2V_2-P_3V_3}{n-1}[/tex] =  -34872.82 J

[tex]\frac{P_2}{P_3} = (\frac{T_2}{T_3} )^\frac{n}{n-1}[/tex]     or T₃ = [tex]T_2*(\frac{P_3}{P_2})^\frac{n-1}{n}[/tex] = 785.12 K

Isobaric compression we have  thus

Work done W₃ = P(V₁ -V₃) = -6974.566 J

Total work = W₁ + W₂ + W₃ = -37198.9 J + -34872.82 J + -6974.566 J = -79046.29 J

Answer:

I = 14.44A

Explanation:

calculating 80% of the circuit breaker current

I = (80/100)20A = 16A

eight 500W lamps are to be installed on the circuit. assume they are connected in parallel

total power = 500 x 8 = 4000W

power = voltage x current

current = power/voltage = 4000W/277V = 14.4A

The current obtained is less than 80% of the circuit breaker.

20A circuit breaker is large enough to carry the load.

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

Answer:

See explanation

Explanation:

Airflow t

T¥, h D

Ts

w

Heater

h

(q•, k ) L

(a) Under conditions for which a uniform surface temperature Ts is maintained around the circum- ference of the heater and the temperature Too and convection coefficient h of the airflow are known, obtain an expression for the rate of heat transfer per unit length to the air. Evaluate the

heat rate for Ts = 300°C, D = 20 mm, an alu- minum sleeve (ks = 240 W/m · K), w = 40 mm, N = 16, t = 4 mm, L = 20 mm, Too = 50°C, and h = 500 W/m2 · K.

Question:

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.

For a power line that supplies power to 10 000 households, we can conclude that

a) IV < I²R

b) I²R = 0

c) IV = I²R

d) IV > I²R

e) I = V/R

Answer:

d) IV > I²R

Explanation:

In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.

The power delivered to households is given by

P = IV

The losses in the transmission line are given by

Ploss = I²R

Therefore, the relation IV > I²R  holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.

Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.

Answer:

a) 500 rev/min.

b) 50 Hz.

Explanation:

See the attached pictures.

Answer:

[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

[tex]\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}[/tex]

The initial pressure is:

[tex]P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}[/tex]

The speed at outlet is:

[tex]v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}[/tex]

[tex]v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }[/tex]

[tex]v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )[/tex]

The initial pressure is:

[tex]P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}[/tex]

[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]

Answer:

P1 = 42.93 psi

Explanation:

For incompressible fluid, we know that;

A1V1 = A2V2

Making V1 the subject, we obtain;

V1 = A2V2/A1

Now A2V2 is the volumetric flow rate (V') .

Thus; V1 = V'/A1

A1 = πD²/4

Thus, V1 = 4V'/πD²

V' = 250 gal/min

But the diameter is in inches, let's convert to inches³/seconds.

Thus, V' = 250 x 3.85 = 962.5 in³/s

Substituting the relevant values to obtain,

V1 = (4 x 962.5)/(π x 3²) = 136.166 in/s.

Now let's convert to ft/s;

V1 = 136.166 x 0.0833 = 11.34 ft/s

Also for V2;

V2 = (4 x 962.5)/(π x 1.125²) = 968.29 in/s.

Now let's convert to ft/s;

V2 = 968.29 x 0.0833 = 80.66 ft/s

Setting bernoulli equation between the hose and the exit, we obtain;

(p1/γ) + (V1²/2g) = V2²/2g

Where V1 and V2 are intial and final velocities and γ is specific weight of water which is 62.43 lb/ft³ and g i acceleration due to gravity which is 32.2 ft/s²

Making p1 the subject, we obtain;

p1 = (γ/2g)(V2² - V1²)

p1 = (62.43/(2x32.2))(80.66² - 11.34²)

p1 = 6182.35 lb/ft²

So Converting to psi, we have;

p1 = 6182.35/144 = 42.93 psi

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

[tex]y = ax^{2} +bx+c\\dy/dx=2ax+b\\[/tex]

At highest point of the curve [tex]dy/dx=o[/tex]

[tex]2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275[/tex]

[tex]-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\[/tex]

Station PVT

[tex]Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)[/tex]

Answer:

1.10 %

Explanation:

The enrichment of the product can be calculated using the following equation:

[tex] \frac{W}{P} = \frac{x_{p} - x_{f}}{x_{f} - x_{w}} [/tex]   (1)

where W: is waste or tails = 26000, P: is the product = 32000, xp: is the wt. fraction of ²³⁵U in product, xf: is the wt. fraction of ²³⁵U in feed = 0.72% and xw: is the wt. fraction of ²³⁵U in waste or tails = 0.25%.      

By solving equation (1) for xp, we can find the enrichment of the product:

[tex] x_{p} = \frac {W}{P} \cdot (x_{f} - x_{w}) + x_{f} [/tex]        

[tex] x_{p} = \frac {26000}{32000} \cdot (0.72 - 0.25) + 0.72 = 1.10% [/tex]  

Therefore, the enrichment of the product is 1.10 %.

I hope it helps you!

Answer:

The data file with 1000 integers

for merge sort the time efficiency is (1000×㏒1000) = 1000 × 3 = 3000

for insertion sort the time efficiency is (1000 × 2) = 2000

The data file with 1,000,000 integers

for merge sort the time efficiency is (1,000,000×㏒1,000,000) = 1,000,000 × 6 = 6,000,000

for insertion sort the time efficiency is (1,000,000 × 2) = 2,000,000

Explanation:

The execution time or temporal complexity refers to how the execution time of an algorithm increases as the size of input increases

For merge sort, the time efficiency  is given by the formula

For insertion sort the time efficiency  is given by the formula

Where n refers to the size of input

Answer:

639.4psi

Explanation:

Pressure at the closed valve = Air pressure+ (density*gravity*height)

=23psi+(49.27lb/ft^3*32.17ft/s^2*56ft^3)

=23psi+88760.89psft

=23psi+(88760.89/144)psi

=23+616.4

=639.4psi

Answer:

Air void content of core 2 = 5.87%

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Answer:

The solution is given in the attachments.

Answer:

#include<stdio.h>

void main()

{

// using file pointer to print output to txt file

FILE *fptr;

float assessedValue, taxableAmount, taxRate = 1.05, propertyTax;

/* open for writing */

fptr = fopen("output.txt", "w");

if (fptr == NULL)

{

printf("File does not exists \n");

return;

}

// prompting user to enter assessed value and storing it in assessedValue variable

printf("Enter the Assessed Value of property : ");

scanf("%f", &assessedValue);

//writing assessed value to output.txt file using fprintf file i/o function

fprintf(fptr, "AssessedValue : $ %.2f\n", assessedValue);

//calculating taxableAmount based on given condition in the question

taxableAmount = (assessedValue * 0.92);

//writing taxable Amount to output.txt file using fprintf file i/o function

fprintf(fptr, "TaxableAmount: $ %.2f\n", taxableAmount);

//writing tax Rate to output.txt file using fprintf file i/o function

fprintf(fptr, "Tax Rate for each $100.00: $ %.2f\n", taxRate);

//calculating propertyTax based on given condition in the question

propertyTax = ((taxableAmount/100)*taxRate);

//writing property Tax Amount to output.txt file using fprintf file i/o function

fprintf(fptr, "propertyTax: $ %.2f\n", propertyTax);

//closing file using fclose function

fclose(fptr);

}

Explanation :

I used Turbo C compiler to compile and run the C program. The below program compiles and at the run time, automatically, prints output to a file called output.txt.

When you compile the program, remember to check the BIN folder in Turbo c folder of C drive where your turbo c has been installed.

Output:

Assessed value: $200000

Taxable amount: $184000

Tax Rate for each $100.00: $1.05

Answer: For the center plate to remain stationed in one position without rotating, the bottom plate has to move to the left at a speed of 2m/s, so as to cancel the force acting on it from the top.

The center plate will not move when the bottom plate is moving left in a speed of 2m/s to counter the speed of the top plate, because a body will continue to be at rest if all the forces acting towards the body are equal. The center plate will be at rest because we have directed equal force from the top and bottom of the plate.

Answer:

Negative feedback

Explanation:

In Biology, negative feedback refers to the counteraction of an effect by its own influence on the process producing it. For instance, the presence of   a high level of a particular hormone in the blood may inhibit further secretion of that hormone.

In other words, in negative feedback, the result of a certain action may inhibit further performance of that action

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

Answer:

Ai=2300 in² , Ao=Ai*1.44=3312 in²

m=10 lbm/s

Pi=5 psia , Po=5.4 psia , Pa=5.5 psia

Vi=120 m/s , Vo=78 m/s

a) Force =m(Vo-Vi) = -190.5 N = -42.82 lbf (towards the inlet)

b) since force is negative it will slow down the system.

Explanation:

Without specific property data for refrigerant-134a at the given conditions, the final temperature and change in specific internal energy cannot be calculated directly. Typically, such tasks require referencing property tables or using equations of state for refrigerant-134a.

The question concerns the process of compressing a refrigerant-134a in a piston-cylinder device while keeping the pressure constant. Given that 100 kg of refrigerant-134a at 200 kPa is initially contained within a volume of 12.322 m3 and then is compressed to half its volume without changing the pressure, we are tasked with determining the final temperature and the change in specific internal energy of the refrigerant.

Since the pressure remains constant during the compression, this scenario can be classified as an isobaric process. However, without specific data such as the initial temperature or specific internal energy values for refrigerant-134a at the given states, we cannot directly calculate the final temperature and change in specific internal energy. Typically, these calculations would require consulting refrigerant-134a property tables or equations of state appropriate for refrigerant-134a to find values at the specified initial and final volumes while maintaining constant pressure.

To find the final temperature, one would typically use the relationship between pressure, volume, and temperature given by the ideal gas law or the specific equations pertaining to refrigerant-134a. The change in specific internal energy could then be determined using specific heat capacities if assuming an ideal gas behavior, or more accurately, through the refrigerant-134a property tables looking at the specific internal energy values at the initial and final states.

Answer:

mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h

Explanation:

The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h

The rate of evaporation will be "2.871 Kg/hour".

According to the question,

Rate of heat supplied:

= 60% of 3 kW

= 1.8 kW

= 1.8 KJ/s

Vaporization of water, [tex]\Delta H = 2257 \ KJ/Kg[/tex]

Time taken will be:

= [tex]\frac{2257}{1.8}[/tex]

= [tex]1254 \ s[/tex]

= [tex]\frac{1254}{3600} \ hour[/tex]

= [tex]0.3482 \ hour[/tex]

hence,

The rate of evaporation,

= [tex]\frac{Mass \ of \ water}{Time}[/tex]

= [tex]\frac{1}{0.3482}[/tex]

= [tex]2.871 \ Kg/hour[/tex]

Thus the above answer is right.

Find out more information about evaporation here:

https://brainly.com/question/4406110

Answer:

The pumping power per ft of pipe length required to maintain this flow at the specified rate 0.370 Watts

Explanation:

See calculation attached.

- First obtain the properties of water at 60⁰F. Density of water, dynamic viscosity, roughness value of copper tubing.

- Calculate the cross-sectional flow area.

- Calculate the average velocity of water in the copper tubes.

- Calculate the frictional factor for the copper tubing for turbulent flow using Colebrook equation.

- Calculate the pressure drop in the copper tubes.

- Then finally calculate the power required for pumping.