A rectifier charges a battery bank in a substation. The bank rated dc voltage is 48 V. The required charging current is 25 A. The available ac supply is 120 V. The internal resistance of the battery is 2.5 Ω. (a) Analyze the operating conditions of the charger. Plot the ac and dc voltage and current, and determine the feasibility of delay angle control. (b) Calculate the delay angle needed to maintain the 25 A charging current. (c) Calculate the power and power factor at the ac side.

Answer:

a) [tex]s_{gen} = -0.0219\,\frac{kJ}{kg\cdot K}[/tex], b) No.

Explanation:

The turbine is modelled after the Second Law of Thermodynamics, which states:

[tex]s_{in} - s_{out} + s_{gen} = 0[/tex]

The entropy generation per unit mass is:

[tex]s_{gen} = s_{out} - s_{in}[/tex]

The specific entropy for steam at entrance and exits are obtained from property tables:

Inlet (Superheated Steam)

[tex]s_{in} = 7.1292\,\frac{kJ}{kg\cdot K}[/tex]

Outlet (Liquid-Vapor Mixture)

[tex]s_{out} = 7.1073\,\frac{kJ}{kg\cdot K}[/tex]

[tex]s_{gen} = 7.1073\,\frac{kJ}{kg\cdot K} - 7.1292\,\frac {kJ}{kg\cdot K}[/tex]

[tex]s_{gen} = -0.0219\,\frac{kJ}{kg\cdot K}[/tex]

b) It is not possible, as it contradicts the Kelvin-Planck and Claussius Statements, of which is inferred that entropy generation can only be zero or positive.

Answer:

Angle of discharge make at the edge of tube=64.9 degrees.

Answer:

C. ​Yes, and it is highly probable.

Explanation:

Null hypothesis, [tex]H_0[/tex] : ц 0.09

Alternative hypothesis, [tex]H_1[/tex]: ц > 0.09

Test statistic is,

x Z =  x - ц / s

[tex]\frac{0.15 - 0.09 }{0.06}[/tex]

[tex]\frac{0.06}{0.06}[/tex]

=1  

The p-value is,  

p = P(Z > z)

=1—P(Z ≤1)

= 1— 0.841345

= 0.158655

(From normal tables)  

The p-value is greater than the significance level, so we fail to reject the null hypothesis. The correct option is, C  

Yes, and it is highly probable.  

Answer:

Given mass flow rate =1.2 kg/s

mass flow rate=density*A*V

Area=pi(douter^2-dinner^2)/4

Area=0.194m^2

The velocity is given by

velocity=2.876 m/s

Answer:

answer is given below

Explanation:

we write this program in C++  that is given below  

and result is attach here

#include <iostream>

using namespace std;

int main() {

int r,g,b,small;

//input values

cin>>r>>g>>b;

//find the smallest value

if(r<g && r<b)

small=r;

else if(g<b)

small=g;

else

small=b;

//subtract smallest of the three values from rgb values hence removing gray

r=r-small;

g=g-small;

b=b-small;

cout<<r<<" "<<g<<" "<<b<<endl;

}

We can see here that the Python code is thus:

# Read input values for red, green, and blue

red, green, blue = map(int, input().split())

# Find the smallest value among red, green, and blue

smallest = min(red, green, blue)

# Subtract the smallest value from all three colors

red -= smallest

green -= smallest

blue -= smallest

# Output the modified values

print(red, green, blue)

Assuming the input values for red, green, and blue are 50, 30, and 40 respectively, the output will be:

10  0  10

This is because:

The smallest value among 50, 30, and 40 is 30.

Subtracting 30 from each color gives:

The output displays the modified values: 20 0 10

To implement this functionality in Python, you can follow these steps:

Explanation:

The three containers each contains water in different states.

Solid state of matter is considered as not compressible because the molecules are already as closely packed as they can be.

The liquid sate of matter has a very minute to no compression ability at all as the molecules are relatively close to each other. Compression is difficult to achieve in the liquid state.

In the gaseous state of matter, the molecules have broken free of one another, and are fairly spaced one from another. This means that gases can be easily compressed.

Pressing down on the plunger, the container containing ice can't be compressed at all so it's volume stays the same.

For the container filled with water, only a minute compression can be achieved with great difficulty hence, the volume reduces by an insignificant amount.

For the container filled with vapour, compression can be easily achieved and the volume reduces significantly.

Only the volume of the third container will decrease.

There are three containers and each has water in three different states. Solid-state of the matter is ice as they are closely packed.

The liquid state is minute and has molecules that are relatively close to other, Compression is thus difficult to achieve.

The last state has water in gases matter and can be compressed easily.

Thus the volume of the third container will go down.

Find out more information about liquid water.

brainly.com/question/2608851.

Answer: 340.19kg

Explanation:

The reduced mass is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. It is referred to as a quantity which allows the two-body problem to be solved as if it were a one-body problem. You should note, however, that the mass determining the gravitational force is not reduced. In the computation or calculation, one mass can be replaced with the reduced mass, if this is compensated by replacing the other mass with the sum of both masses.

It has the dimensions of mass, and SI unit kg.

Given two bodies, first with mass m1 and the other with mass m2, the equivalent one-body problem, with the position of one body with respect to the other referred to as the unknown, is that of a single body of mass.

The density of Iron (Fe) = 7.87Mg/m^3

The volume of Iron decreased from= (0.165 - 0.118) m^3 =0.047m^3

Therefore,

The weight of Iron decreased= (0.047m^3) x (7.87Mg/m^3) = 0.36989Mg

The density of aluminium= 2.70Mg/m^3

The volume of aluminum increased= (0.023 - 0.012)m^3= 0.011m^3

The weight of aluminium is= (0.011m^3) × (2.70Mg/m^3) = 0.0297Mg

Therefore,

The mass reduction is:

The weight of Iron decreased minus the weight of Aluminium:

0.36989Mg - 0.0297Mg= 0.34019Mg

0.34019Mg × 1000

= 340.19Kg

Therefore, the mass reduction resulting from the trend is= 340.19kg

Answer:

5.21e-2mm

Explanation:

Please see attachment

The flexural strength of the glass specimen under the given load is approximately 84.4 MPa. The cross-sectional moment of inertia is about 0.655 *10^-12 m^4. When subjected to a load of 266N, the deflection at the center of the specimen is approximately 0.87 mm.

The subject matter of the question refers to concepts from material science and mechanical engineering, specifically pertaining to the calculation of flexural strength, the moment of inertia, and deflection.

Flexural strength or 'Modulus of rupture' is calculated with the formula σf = 3FL / 2bd^2, where F is the fracture load, L the distance between support points, b the specimen's width, and d its height. Substituting the given values, we get σf = (3*292 * 43*10^-3) / (2 * 12 *10^-3 * (5.4 *10^-3)^2), which gives us about 84.4 MPa.

The moment of inertia for a rectangular cross-section is given by I = b*d^3 / 12. Substituting the given values, we get I = 12*10^-3 * (5.4 *10^-3)^3 / 12, which gives approximately 0.655 *10^-12 m^4.

Deflection, typically denoted by the Greek letter δ (or in this case, Υ), is computed with the formula Υ = FL^3 / 48EI. Given that the modulus of elasticity (E) for glass is 60 GPa or 60*10^9 Pa and we already calculated I, we substitute all values to get Υ = (266 * (43*10^-3)^3) / (48 * 60*10^9 * 0.655*10^-12) which gives us approximately 0.87 mm.

https://brainly.com/question/33591807

#SPJ3

Answer:

see explaination

Explanation:

Part a) Width of bay at Potomac River:

Given Data:

· Actual Width at Potomac River = 30 miles

· Bay Model Length Ratio Lr = 1/1000

In fluid mechanics models of real structures are prepared in simulation so that they can be analyzed accurately. A model is known to have simulation if model carries same geometric, kinematic and dynamic properties at a small scale.

Length of any part in model = Actual length x Lr

Hence,

Model Width of bay at Potomac River = 30 x 1/1000 = 0.03 miles

Since 1 mile = 5280 ft

Model Width of bay at Potomac River = 0.03 x 5280 = 158.4 ft

Part b) Model Length of bay bridge in model:

Given Data:

· Actual Length of bay bridge = 4.3 miles

· Bay Model Length Ratio Lr = 1/1000

Model Length = Actual Length x Lr = 4.3 x 1/1000 = 0.0043 miles

Since 1 mile = 5280 ft

Model Length in feet = 0.0043 x 5280 = 22.704 ft

Part c) Model Length of bay bridge in model:

Given Data:

· Model Area = 8 acre

· Bay Model Length Ratio Lr = 1/1000

Model Area = Actual Area x Lr x Lr

8 Model Area :: Actual Area =- (Lp)2 2 = 8,000,000 acre 1000

Since 1 square mile = 640 acre,

Actual Area in square miles = 8,000,000/640 = 12,500 square miles

Part d) Average and maximum depth of model:

Given Data:

· Actual Average depth = 28 ft

· Actual Maximum depth = 174 ft

· Bay Model Length Ratio Lr = 1/1000

Model average depth = Actual average depth x Lr = 28 x 1/1000 = 0.028 feet

Since 1 ft = 12 inch

Model average depth in inch = 0.028 x 12 = 0.336 in

Model maximum depth = Actual maximum depth x Lr = 174 x 1/1000 = 0.174 feet

Since 1 ft = 12 inch

Model maximum depth in inch = 0.174 x 12 = 2.088 in

Answer:

a) for a stem of 10 mm the flow rate is 0.4229 m³/s

b) for a stem of 20 mm the flow rate is 0.8944 m³/s

Explanation:

The mathematical expression for the flow rate is:

[tex]Q=Q_{min} *(\frac{Q_{max} }{Q_{min} } )^{\frac{S-S_{min}}{S_{max}-S_{min} } }[/tex]

a) Here:

Qmin=0.2m³/s

Qmax=4m³/s

Smax-Smin=40mm

S-Smin=10mm

Substituting these values ​​in the first equation:

[tex]Q=0.2*(\frac{4}{0.2} )^{\frac{10}{40} } =0.4229m^{3} /s[/tex]

b) In this time, S-Smin=20mm

[tex]Q=0.2*(\frac{4}{0.2} )^{\frac{20}{40} } =0.8944m^{3} /s[/tex]

Answer:

b. 1232.08 km/hr

c. 1.02 kn

Explanation:

a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.

See attachment for the remaining.

Answer:

Some of the internal strain energy is relieved.

There is some reduction in the number of dislocations.

The electrical conductivity is recovered to its precold-worked state.

The thermal conductivity is recovered to its precold-worked state

Explanation:

The process of the recovery of a cold-worked material happens at a very low temperature, this process involves the movement and annihilation of points where there are defects, also there is the annihilation and change in position of dislocation points which leads to forming of the subgrains and the subgrains boundaries such as tilt, twist low angle boundaries.

Answer:

Explanation:

N1 =v/πD = 400(12)/2.5π = 611 rev/min fr = Nfnt = 611(0.010)(1) = 6.11 in/min A=D/2 = 2.50 / 2 = 1.25 Tm = (L+2A)/fr = (15 + 2(1.25))/6.11 = 2.863 min T1 = 3Tm = 3(2.863) = 8.589 min when v1 = 400 ft/min N2 = 200(12)/2.5π = 306 rev/min fr = Nfnt = 306(0.010)(1) = 3.06 in/min Tm = (15 + 2(1.25))/3.06 = 5.727 min T2 = 12Tm = 12(5.727) = 68.724 min when v2 = 200 ft/min n = ln (v1/v2)/ln(T2/T1) = ln (400/200)/ln (68.724/8.589) = 0.333 C = vTn = 400(8.589 )0.333 = 819

Answer:

C = 787.2

Explanation:

The Taylor tool life is referred to as the duration of the actual cutting time after which the tool can no longer be used.

To Quantify the end of a tool life we equate it to a limit on the maximum acceptable flank wear.

For the tool life equation, With the slope, n and intercept, c, Taylor derived the simple equation as

VTn = C where.

n is called, Taylor's tool life exponent. The values of both 'n' and 'c' depend mainly upon the tool-work materials and the cutting environment (cutting flake application)

Please kindly go through the attached file for a step by step solution to how the answer to your question is solved.

Answer:

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Explanation:

Answer:

See explaination

Explanation:

The volume flow rate Q Q QQ of a fluid is defined to be the volume of fluid that is passing through a given cross sectional area per unit time.

Kindly check attachment for the step by step solution of the given problem.

Answer:

  3.527 seconds

Explanation:

The height of the falling water, assuming no friction or air resistance, is given by ...

  h(t) = -(1/2)gt² +61

where g is the standard gravity value, 9.80665 m/s².

The the time required for h(t) = 0 is ...

  1/2gt² = 61

  t² = 2·61/g

  t = √(2·61/9.80665) ≈ 3.527 . . . . seconds

Answer:

Explanation:

============== main.cpp =======================

#include <iostream>

using namespace std;

#include <iostream>

#include <vector>

#include "Contact.h"

using namespace std;

int main() {

const int NUM_OF_CONTACTS = 3;

vector<Contact *> contacts;

for (int i = 0; i < NUM_OF_CONTACTS; i++) {

string name, phoneNumber;

cout << "Enter a name: ";

cin >> name;

cout << "Enter a phoneNumber: ";

cin >> phoneNumber;

// Use i, name, and phone number to dynamically create a Contact object on the heap

// HINT: Use the Overloaded Constructor here!

Contact * newContact = new Contact(i+1,name,phoneNumber);

// Add the Contact * to the vector...

contacts.push_back(newContact);

}

cout << "\n\n----- Contacts ----- \n\n";

// Loop through the vector of contacts and print out each contact info

std::vector<Contact *>::iterator itrContact;

for(itrContact = contacts.begin(); itrContact !=contacts.end(); itrContact++)

{

cout<< " Id : " << (*itrContact)->getId();

cout<< " Name : " << (*itrContact)->getName();

cout<< " Phone-Number : " << (*itrContact)->getPhoneNumber() <<endl;

}

// Make sure to call the destructor of each Contact object by looping through the vector and using the delete keyword

for(int i = 0; i < NUM_OF_CONTACTS; i++)

{

delete contacts[i];

}

return 0;

}

================== contact.h =====================

#ifndef CONTACT_H

#define CONTACT_H

#include <string>

#include <iostream>

using std::string;

using std::cout;

class Contact {

public:

Contact();

Contact(int id, string name, string phoneNumber);

~Contact();

Contact(const Contact& copy);

Contact& operator=(const Contact& copy);

// getters

int getId() const;

string getName() const;

string getPhoneNumber() const;

//setters

void setId(int id);

void setName(string name);

void setPhoneNumber(string phoneNumber);

private:

int *id = nullptr;

string *name = nullptr;

string *phoneNumber = nullptr;

};

#endif

================= contact.cpp ========================

#include "Contact.h"

Contact::Contact() {

this->id = new int(-1);

this->name = new string("No Name");

this->phoneNumber = new string("No Phone Number");

}

Contact::Contact(int id, string name, string phoneNumber) {

// Implement Overloaded Constructor

// Remember to initialize pointers on the heap!

this->id = new int(id);

this->name = new string(name);

this->phoneNumber = new string(phoneNumber);

}

Contact::~Contact() {

// Implement Destructor

if(this->id != nullptr)

delete this->id;

if(this->name != nullptr)

delete this->name;

if(this->phoneNumber != nullptr)

delete this->phoneNumber;

}

Contact::Contact(const Contact &copy) {

// Implement Copy Constructor

this->id = new int(copy.getId());

this->name = new string(copy.getName());

this->phoneNumber = new string(copy.getPhoneNumber());

}

Contact &Contact::operator=(const Contact &copy) {

// Implement Copy Assignment Operator

if(this == &copy) // Checks for self Assignment

return *this;

this->id = new int(copy.getId());

this->name = new string(copy.getName());

this->phoneNumber = new string(copy.getPhoneNumber());

return *this;

}

// getters

int Contact::getId() const

{

return *(this->id);

}

string Contact::getName() const

{

return *(this->name);

}

string Contact::getPhoneNumber() const

{

return *(this->phoneNumber);

}

//setters

void Contact::setId(int id)

{

*(this->id) = id;

}

void Contact::setName(string name)

{

*(this->name) = name;

}

void Contact::setPhoneNumber(string phoneNumber)

{

*(this->phoneNumber) = phoneNumber;

}

====================Output =========================

is attached below

Answer:

The minimum angular speed, w is 126.94 rad/s

Explanation:

Given:

C = 2.908×10⁸ m/s

d = 10.72 km ⇒ 10.72×10³ m

There are 671 notches

⇒ Δθ = [tex]\frac{2\pi}{671}[/tex] ==> 9.359×10⁻³ rad

C = 2d / Δt    ⇒    Δt = 2d/C

⇒

w = Δθ / Δt = CΔθ / 2d

substitute for given parameters

w = [2.908×10⁸×9.359×10⁻³] / [2×10.72×10³]

   = 27.215972×10⁵ / 21.44×10³

   = 1.2694×10²

w ⇒ 126.94 rad/s

Answer:

final temperature T = 24.84ºC

Explanation:

given data

copper volume = 1 L

temperature t1 = 500ºC

oil volume = 200 L

temperature t2 = 20ºC

solution

Density of copper [tex]\rho[/tex] cu = 8940 Kg/m³

Density of light oil  [tex]\rho[/tex] oil = 889 Kg/m³

Specific heat capacity of copper Cv = 0.384  KJ/Kg.K

Specific heat capacity of light oil Cv = 1.880 KJ/kg.K

so fist we get here mass of oil and copper that is

mass = density × volume   ................1

mass of copper = 8940 × 1 ×  [tex]10^{-3}[/tex]  = 8.94 kg  

mass of oil = 889 × 200 × [tex]10^{-3}[/tex]  =  177.8 kg  

so we apply here now energy balance equation that is

[tex]M(cu)\times Cv \times (T-T1)_{cu} + M(oil) \times Cv\times (T-T2)_{oil}[/tex]  = 0

put here value and we get T2

[tex]8.94\times 0.384 \times (T-500) + 177.8 \times 1.890\times (T-20)[/tex]  = 0

solve it we get

T = 24.84ºC

Answer:

(a) Modified Goodman criterion:

Factor of safety against fatigue failure =  1.0529

(b) Gerber criterion:

Factor of safety against fatigue failure = 1.31

(c) ASME-elliptic criterion:

Factor of safety against fatigue failure = 1.315

Explanation:

See the attached file for the calculation.

(a) Modified Goodman: SF ≈ 1.264, Static FS ≈ 2.058.

(b and c ) Gerber & ASME-elliptic: SF ≈ 1.783, Static FS ≈ 2.058.

To calculate the factor of safety against static and fatigue failures using different criteria, we need to determine the critical stress limits under the given loading conditions and compare them with the material properties.

Given:

- Torsional stress [tex](\( \tau \))[/tex] = 15 kpsi

- Alternating bending stress [tex](\( \sigma_a \))[/tex] = 25 kpsi

- Minimum endurance limit [tex](\( S_e \))[/tex] = 40 kpsi

- Ultimate tensile strength [tex](\( S \))[/tex] = 60 kpsi

- Endurance limit for reversed bending [tex](\( S_{-80} \))[/tex] = 80 kpsi

(a) Modified Goodman criterion:

The modified Goodman criterion accounts for both tensile and torsional stress, given by:

[tex]\[ \frac{1}{SF} = \frac{\frac{\sigma_a}{S} + \frac{\tau}{S_e}}{1} \][/tex]

Where [tex]\( SF \)[/tex] is the safety factor against fatigue failure.

Substitute the given values:

[tex]\[ \frac{1}{SF} = \frac{\frac{25}{60} + \frac{15}{40}}{1} \][/tex]

[tex]\[ \frac{1}{SF} = \frac{0.4167 + 0.375}{1} \][/tex]

[tex]\[ \frac{1}{SF} = 0.7917 \][/tex]

[tex]\[ SF = \frac{1}{0.7917} \][/tex]

[tex]\[ SF \approx 1.264 \][/tex]

The factor of safety against static failure [tex](\( FS_{static} \))[/tex] can be calculated by comparing the maximum applied stress with the ultimate tensile strength:

[tex]\[ FS_{static} = \frac{S}{\sigma_{max}} \][/tex]

[tex]\[ FS_{static} = \frac{60}{\sqrt{\sigma_a^2 + \tau^2}} \][/tex]

[tex]\[ FS_{static} = \frac{60}{\sqrt{25^2 + 15^2}} \][/tex]

[tex]\[ FS_{static} = \frac{60}{\sqrt{625 + 225}} \][/tex]

[tex]\[ FS_{static} = \frac{60}{\sqrt{850}} \][/tex]

[tex]\[ FS_{static} \approx \frac{60}{29.1547} \][/tex]

[tex]\[ FS_{static} \approx 2.058 \][/tex]

(b) Gerber criterion:

The Gerber criterion considers the bending and torsional stresses, given by:

[tex]\[ \frac{1}{SF} = \sqrt{\frac{\sigma_a^2}{S^2} + \frac{\tau^2}{S_e^2}} \][/tex]

Substitute the given values:

[tex]\[ \frac{1}{SF} = \sqrt{\frac{25^2}{60^2} + \frac{15^2}{40^2}} \][/tex]

[tex]\[ \frac{1}{SF} = \sqrt{\frac{625}{3600} + \frac{225}{1600}} \][/tex]

[tex]\[ \frac{1}{SF} = \sqrt{0.1736 + 0.1406} \][/tex]

[tex]\[ \frac{1}{SF} = \sqrt{0.3142} \][/tex]

[tex]\[ \frac{1}{SF} \approx 0.5608 \][/tex]

[tex]\[ SF \approx \frac{1}{0.5608} \][/tex]

[tex]\[ SF \approx 1.783 \][/tex]

(c) ASME-elliptic criterion:

The ASME-elliptic criterion also considers bending and torsional stresses:

[tex]\[ \frac{1}{SF} = \sqrt{\left(\frac{\sigma_a}{S}\right)^2 + \left(\frac{\tau}{S_e}\right)^2} \][/tex]

Substitute the given values:

[tex]\[ \frac{1}{SF} = \sqrt{\left(\frac{25}{60}\right)^2 + \left(\frac{15}{40}\right)^2} \][/tex]

[tex]\[ \frac{1}{SF} = \sqrt{0.1736 + 0.1406} \][/tex]

[tex]\[ \frac{1}{SF} = \sqrt{0.3142} \][/tex]

[tex]\[ \frac{1}{SF} \approx 0.5608 \][/tex]

[tex]\[ SF \approx \frac{1}{0.5608} \][/tex]

[tex]\[ SF \approx 1.783 \][/tex]

For all three criteria:

- Factor of safety against static failure [tex](\( FS_{static} \))[/tex] ≈ 2.058

- Safety factor against fatigue failure [tex](\( SF \))[/tex] ≈ 1.264 for the Modified Goodman criterion, and ≈ 1.783 for the Gerber and ASME-elliptic criteria.

Answer:

The volume percent of graphite in a 3.5 wt% C cast iron is 11%.

Explanation:

Consider 100 grams 3.5% C cast iron.

Mass of  carbon in 100 grams of cast iron = [tex]m_1=100 g\times \frac{3.5 }{100}=3.5 g[/tex]

Mass of iron in cast iron =[tex]m_2[/tex] = 100 g - 3.5 g = 96.5 g

Density of graphite = [tex]d_1=2.3g/cm^3[/tex]

Volume of the graphite = [tex]v_1[/tex]

[tex]v_1=\frac{m_1}{d_1}=\frac{3.5 g}{2.3g/cm^3}=1.52 cm^3[/tex]

Density of iron= [tex]d_2=7.9 g/cm^3[/tex]

Volume of the iron = [tex]v_2[/tex]

[tex]v_2=\frac{m_2}{d_2}=\frac{96.5 g}{7.9 g/cm^3}=12.22 cm^3[/tex]

Total volume of the cast iron ,V=[tex]v_1+v_2=1.52 cm^3+12.22 cm^3=13.74 cm^3[/tex]

Volume percent of the graphite :

[tex]=\frac{v_1}{V}\times 100[/tex]

[tex]=\frac{1.52 cm^3}{13.74 cm^3}\times 100=11.06\%\approx 11\%[/tex]

The volume percent of graphite in a 3.5 wt% C cast iron is 11%.

The volume percent of graphite VGr in a 3.5 wt% C cast iron is;

V_gr = 11.08%

We are given;

Density of Ferrite; ρₐ = 7.9 g/cm³

Density of Graphite; [tex]\rho _{g}[/tex] = 2.3 g/cm³

Weight percent of C cast iron; C₀ = 3.5%

Now the formula for the mass fraction of ferrite is;

Wₐ = [tex]\frac{C_{g} - C_{0}}{C_{g} - C_{a}}[/tex]

If we consider 100 g of 3.5 wt% C cast iron, then [tex]C_{g}[/tex] = 100 and Cₐ = 0

Thus;

Wₐ = [tex]\frac{100 - 3.5}{100 - 0}[/tex]

Wₐ = 0.965

Thus, mass fraction of graphite is;

[tex]W _{g}[/tex] = 1 -  Wₐ

[tex]W _{g}[/tex] = 1 -  0.965

[tex]W _{g}[/tex] = 0.035

Formula for the volume percent of graphite is;

[tex]V_{gr} = \frac{\frac{W_{g}}{\rho_{g}}}{\frac{W_{a}}{\rho_{a}} + \frac{W_{g}}{\rho_{g}}} * 100%[/tex]

Plugging in the relevant values gives;

V_gr = [(0.035/2.3)/((0.965/7.9) + (0.035/2.3))]

V_gr = 0.01521739/(0.1221518987 + 0.01521739)

V_gr = 11.08%

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Without information about the liquids' surface tensions, it is impossible to predict whether Fa will be greater, less, or equal to Fb. Answer (d) is correct.

When analyzing forces on a wire loop in magnetic fields, it's essential to consider various factors including current, magnetic field intensity, and loop dimensions. A uniform magnetic field will exert a force on each segment of the loop carrying a current; this applies to rectangular loops in a magnetic field as noted in the given scenario. The question seems to focus on the force needed to change the width of loops coated with a liquid film and could draw on concepts of surface tension and possibly elasticity if considering the loop's properties themselves. However, the provided text focuses on magnetic effects on current-carrying wires.

For the case of the film across the loop, the necessary force to stretch it will heavily depend on the surface tension of the liquid. Therefore, without information on the liquids' surface tensions or other properties, it is impossible to predict whether Fa will be greater than, less than, or equal to Fb. Answer (d) it is impossible to predict whether Fa or Fb will be greater without more information is the most suitable given the circumstances.

Answer:

True

Explanation:

Permanent electrical safety devices (PESD) acts as a layer of protection between the electrical worker and the hazardous voltage.

Permanent electrical safety devices (PESDs) are deployed to reduce the risks in isolating electrical energy.

Electrical safety can be improved if a worker determines a zero electrical energy state irrespective of any voltage exposure to themselves.

The given statement is true

Answer:

A

Explanation:

Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit controls current flow Person A is right.

A technician is a worker in the technology industry who possesses the necessary knowledge, abilities, and skills as well as a practical comprehension of the theoretical underpinnings.

Expert technicians in a certain tool domain often have expert competency in technique and a moderate comprehension of theory. As a result, technicians in that field of technology are typically much more knowledgeable about technique than the average layperson and even general professionals.

For instance, although not as knowledgeable in acoustics as acoustical engineers, audio technicians are more adept at using sound equipment and are likely to know more about acoustics than other studio staff members, such as performers.

Therefore, Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit controls current flow Person A is right.

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#SPJ2

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The detailed solution is in the attached file

The question asks for pump pressure calculations in a pipeline, requiring applications of fluid dynamics principles, such as Bernoulli's equation, and considering factors like elevation changes, viscosity, and pipe characteristics.

The student's question relates to the field of fluid dynamics, specifically the calculation of required pump pressure in a pipeline system using principles such as Bernoulli's equation and concepts like pressure, flow rate, pipe diameter, density, viscosity, and elevation change. To solve this type of problem, you typically apply the Bernoulli equation, along with any additional head losses due to viscosity, known as the head loss due to friction, which can be calculated using the Darcy-Weisbach equation or other empirical formulas. You must then calculate the total head needed, convert this to pressure, taking into account the fluid's density, gravitational acceleration, and add any other relevant pressures, such as atmospheric pressure if the pipeline opens to the atmosphere.

Answer:

see explaination

Explanation:

import java.util.InputMismatchException;

import java.util.Scanner;

public class calculate {

static float a=0,b=0;

double cal()

{

if(a==0||b==0)

{

System.out.println("no values found in a or b");

start();

}

double x=(a*a)+(b*b);

double h=Math.sqrt(x);

a=0;

b=0;

return h;

}

float enter()

{

float val=0;

try

{

System.out.println("Enter side");

Scanner sc1 = new Scanner(System.in);

val = sc1.nextFloat();

return val;

}

catch(InputMismatchException e)

{

System.out.println("Enter correct value");

}

return val;

}

void start()

{

calculate c=new calculate();

while(true)

{

System.out.println("Enter Command");

Scanner sc = new Scanner(System.in);

String input = sc.nextLine();

switch(input)

{

case "A":

a=c.enter();

break;

case "B":

b=c.enter();

break;

case "C":

double res=c.cal();

System.out.println("Hypotenuse is : "+res);

break;

case "Q":

System.exit(0);

default:System.out.println("wrong command");

}

}

}

public static void main(String[] args) {

calculate c=new calculate();

c.start();

}

}

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis =  $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

Answer:

velocity = 147.57 m/s

Explanation:

given data

inlet to throat area ratio = 12

height difference Δh = 10 cm = 0.1 m

density of liquid mercury = 1.36 × [tex]10^{4}[/tex] kg/m³

solution

we get here weight of mercury that is express as

weight of mercury = density of liquid mercury ×  g    .................1

weight of mercury = 1.36 × [tex]10^{4}[/tex] × 9.8

weight of mercury = 1.33 × [tex]10^{5}[/tex]  N/m²  

and

area ratio is

[tex]\frac{a1}{a2}[/tex]  = 12

so velocity of air in the test section will be

velocity = [tex]\sqrt{\frac{2\times w \times \triangle h}{\rho (1-(area\ ration)^2)}}[/tex]      .......................1

put here value and we get

velocity = [tex]\sqrt{\frac{2\times 1.33\times 10^5 \times 0.1}{1.23 (1-(\frac{1}{12})^2)}}[/tex]    

velocity = 147.57 m/s

The velocity of air in the test section is; v = 147.802 m/s

We are given;

Thus, weight of mercury is;

W = ρg

W = 1.36 × 10⁴ × 9.8

W = 13.328 × 10⁴ N/m²

Formula for the velocity of the air in the test section is;

v = √[(2ρgΔh)/(ρ_air × (1 - a2/a1)]

Where ρ_air is density of air = 1.23 kg/m³

Thus;

v = √[(2 × 1.36 × 10⁴ × 9.8 × 0.1)/(1.23 × (1 - 1/12)]

Solving this gives us;

v = 147.802 m/s

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Answer:

The general method for indirect strategy is Start with a neutral buffer.

Explanation:

The general method for indirect strategy goes like this:

-Start with a neutral buffer, you should never start with good news because it will give the reader false hope that more good news will come. So a neutral “buffer” or a show of appreciation for your business is a good way to start. You are not apologizing for the bad news that is coming, you are simply preparing the reader for it.

-The next part is where you give reasons. There are many studies on the effectiveness of reasons in communication: people like to know why things are the way they are. By offering reasons, you will make the bad news easier to accept and once you have prepared the reader, you give the bad news.

The idea is also not to give infinite turns to the subject in general, since the information and the objective of the message can be lost. It is important not to spend too much time on this: the buffer should be short, so as not to make the moment tedious and lose the attention of the public before reaching the main topic.

-Finally, in your conclusion, divert attention from the bad news. Don't talk about it anymore, be nice, focus your final efforts on future opportunities and recover goodwill.