Between 1975 and 1985, the volume of all iron and steel in a given automobile model decreased from 0.165 m3m3 to 0.118 m3m3 . In the same time frame, the volume of all aluminum alloys increased from 0.012 m3m3 to 0.023 m3m3 . Using the densities of pure FeFe and AlAl, estimate the mass reduction resulting from this trend in materials substitution.

Answer:

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Explanation:

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:

  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:

The voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

Explanation:

Given;

Resistance, R₁ = 50Ω

Resistance, R₂ = 75Ω

Total resistance, R = (R₁R₂)/(R₁ + R₂)

Total resistance, R = (50 x 75)/(125)

Total resistance, R = 30 Ω

According to ohms law, sum of current in a parallel circuit is given as

I = I₁ + I₂

[tex]I = \frac{V}{R_1} + \frac{V}{R_2}[/tex]

Voltage across each resistor is the same

[tex]1.6 = \frac{V}{R_2}[/tex]  

V = 1.6 x R₂

V = 1.6 x 75

V = 120 V

Therefore, the voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.

This voltage is also the same for 50 ohms resistance but the current will be 2.4 A.

Answer:

120 volts

Explanation:

Since the two resistances are connected in parallel across the voltage source, the effective resistance of the circuit can be obtained by using the formula.

[tex]\frac{1}{R_{eq}}=\frac{1}{R_{1}}+ \frac{1}{R_{2}}[/tex]

given that [tex]R_{1} and R_{2}[/tex]  are 50 and 75 ohms respectively, we have the equivalent resistance as:

[tex]\frac{1}{R_{eq}}=\frac{1}{50}+ \frac{1}{75}=\frac{1}{30}[/tex]

hence,

[tex]R_{eq}= 30\Omega[/tex]

From Ohm's law, voltage = current X resistance.

given that the current through the 75 ohm resistor is 1.6 A

[tex]V= I\times R[/tex]

[tex]V= 1.6 \times 75\Omega[/tex]

voltage = 120 Volts.

Because the resistors are connected in parallel, it means that they are connected to the same voltage source.

Hence, the voltage source for the 75 Ohm resistance = 120 volts. This is same for the 50 Ohm resistor.

Answer:

0.34232

Explanation:

See attachment

Answer:

The annual flow at the outlet of Broad River is 13688480 m³

The runoff coefficient for the watershed is 0.342212

Explanation:

Here, we have

The annual storage in  the watershed = 0 cm/y

Annual precipitation = 100 cm/y

Annual evapotranspiration = 50 cm/y

infiltration rate = 5 × 10-7 cm/s

Watershed area = 40 km²

From the question, annual infiltration rate is given by

Annual infiltration rate = 5 × 10-7 cm/s × ‪31557600‬ s/year = 15.7788 cm/y

Therefore, annual flow =

Annual precipitation - Annual evapotranspiration - Annual infiltration rate - Annual storage in  the watershed

Annual flow = 100 cm/y - 50 cm/y  - 15.7788 cm/y - 0 cm/y = 34.2212 cm/y

The area of the watershed = 40 km²

Therefore, the volume of the annual flow is given by;

Height of annual flow × area of the watershed

Volume of annual flow = 34.2212 cm × 40 km²  =  1368.848 cm·km²

= 13688480 m³

The runoff coefficient = [tex]\frac{Amount \hspace{0.1cm} of \hspace{0.1cm}runoff}{Amount \hspace{0.1cm} of \hspace{0.1cm} precipitation \hspace{0.1cm} received}[/tex] =

The amount of precipitation received is given by

100 cm × 40 km² = 4000 cm·km² = 40000000 m³

Runoff coefficient = [tex]\frac{13688480 m^3}{40000000 m^3}[/tex] = 0.342212.

Answer:

A The heat transfer to the river would be 900 MW

B. The actual heat transfer would be lower than the theoretical heat transfer.

Explanation:

Part A

For a steam power plant, the process is continuous and the law of thermodynamic is conserved. The steam power plant consists of the cold and hot reservoir and the rate of heat transfer in the steam power plant can be gotten with the expression below.

W = [tex]Q_{H}[/tex] - [tex]Q_{c}[/tex] ......................1

where W is the work output of the steam power plant

[tex]Q_{H}[/tex]  is the heat transfer at the hot reservoir

[tex]Q_{c}[/tex] is the heat transfer at the cold reservoir( heat transfer to the nearby river)

The efficiency of the steam power plant would be used to obtain the heat transfer at the hot reservoir ([tex]Q_{H}[/tex]). The efficiency can be obtained with equation 2;

e = [tex]\frac{W}{Q_{H} }[/tex]

e is the thermal efficiency of the steam power plant = 40 % = 0.4

W is the work output of the steam power plant = 600 MW

[tex]Q_{H}[/tex]  is the heat transfer at the hot reservoir

Rearranging equation 2 to make  [tex]Q_{H}[/tex]  the subject formula we have;

[tex]Q_{H}[/tex]  = W/e

substituting the values we have;

[tex]Q_{H}[/tex] = 600 MW/ 0.4

[tex]Q_{H}[/tex] = 1500 MW

The heat transfer at the hot reservoir is 1500 MW and putting it into equation 1 to find the heat transfer to the river;

[tex]Q_{c}[/tex] = [tex]Q_{H}[/tex] - W

[tex]Q_{c}[/tex] = 1500 MW - 600 MW

[tex]Q_{c}[/tex] = 900 MW

Therefore the heat transfer to the river would be 900 MW

Part B

The actual heat transfer would be lower than the theoretical heat transfer. Losses dues to friction, head losses and heat loss by evaporation to the surroundings account for the lower value.

The actual heat transfer rate is higher than the power delivered by the steam power plant.

The heat transfer rate released to the river ([tex]\dot Q_{L}[/tex]), in megawatts, is derived from the definition of energy efficiency ([tex]\eta[/tex]), no unit, that is to say:

[tex]\eta = \frac{\dot W}{\dot Q_{L}+\dot {W}}[/tex]

[tex]\eta \cdot \dot Q_{L} + \eta \cdot \dot W = \dot W[/tex]

[tex](1-\eta)\cdot \dot W = \eta \cdot \dot Q_{L}[/tex]

[tex]\dot Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot \dot W[/tex] (1)

Where [tex]\dot W[/tex] is the power delivered by the steam power plant, in megawatts.

If we know that [tex]\eta = 0.4[/tex] and [tex]\dot W = 600\,MW[/tex], then the heat transfer rate released to the river is:

[tex]\dot Q_{L} = \left(\frac{1}{0.4}-1 \right)\cdot (600\,MW)[/tex]

[tex]\dot Q_{L} = 900\,MW[/tex]

Thus, the actual heat transfer rate is higher than the power delivered by the steam power plant.

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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.

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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:

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:

Determine the added thrust required during water scooping, as a function of aircraft speed, for a reasonable range of speeds.= 132.26∪

Explanation:

check attached files for explanation

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:

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:

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:

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:Coding is required in other to compare the time efficiency between two different search approaches.

Explanation

Assuming a linear search approach is to be conducted to find the highest entry in a sequential or descending order, there is need to write a programing code for the two approaches, then compare the time taken for each task to be completed . After, one can now tell the approach that is most time efficient.

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.

Answer:

See explaination

Explanation:

A constant voltage source is called a DC Voltage with a voltage that varies periodically with time is called an AC voltage. Voltage is measured in volts, with one volt being defined as the electrical pressure required to force an electrical current of one ampere through a resistance of one Ohm.

Please check the attached file.

Answer:

Explanation:

The answers and step by step to the solution Can be found in the attached files, please kindly go through them.

Answer:

a. 147lb/dt^3

b. 148lb/dt^3

c. 7318.52lb

Explanation:

Please see attachments

Answer:

Maximum work under this condition (∆G) = Maximum work under Standard Condition (∆G°) + Activities defining this condition

Explanation:

In this equation, the term DGo provides us with a value for the maximal work we could obtain from the reaction starting with all reactants and products in their standard states, and going to equilibrium. The term DG' provides us with a value for the maximal work we could obtain under the conditions defined by the activities in the logarithmic term. The logarithmic term can be seen as modifying the value under standard conditions to account for the actual conditions. In describing the work available in metabolic processes, we are concerned with the actual conditions in the reaction medium (whether that is a test-tube, or the cell cytoplasm); the important term is therefore DG'. If we measure the actual activities (in practice, we make do with concentrations), and look up a value for DGo in a reference book, we can calculate DG' from the above equation.

Values for DGo provide a useful indication through which we can compare the relative work potential from different processes, because they refer to a standard set of conditions.

Therefore both phrases describe the Biochemical and Chemical Standard State

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.

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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:

10 mph

Explanation:

20 / x = 12 / (x - 4)

Cross multiply

12x = 20 (x - 4)

12x = 20x - 80

80 = 20x - 12x

80 = 8x

divide through by '8'

10 = x

x = 10 mph

Answer:

16

Explanation:

Answer:

A student wants to determine experimentally, without disconnecting any wires in the circuit, the DC current moving through a copper wire. Which of the following items of laboratory equipment would be sufficient to make the necessary measurements for this determination?

(E) Voltmeter and current sensor

Explanation:

DC (direct current) is the unidirectional stream or development of electric charge transporters (which are generally electrons). The power of the current can shift with time, yet the general direction of development remains the equivalent consistently. As a descriptor, the term DC is utilized in reference to voltage whose extremity never turns around. In a DC circuit, electrons rise up out of the negative, or less, shaft and move towards the positive, or additionally, post. All things considered, physicists characterize DC as making a trip from in addition to short.

A voltmeter is an instrument utilized for estimating electrical potential contrast between two focuses in an electric circuit. Simple voltmeters move a pointer over a scale in relation to the voltage of the circuit; advanced voltmeters give a numerical presentation of voltage by utilization of a simple to computerized converter. A voltmeter in a circuit outline is spoken to by the letter V around. Voltmeters are made in a wide scope of styles. Instruments for all time mounted in a board are utilized to screen generators or other fixed mechanical assembly. Versatile instruments, typically prepared to likewise gauge flow and obstruction as a multimeter, are standard test instruments utilized in electrical and hardware work. Any estimation that can be changed over to a voltage can be shown on a meter that is reasonably adjusted; for instance, pressure, temperature, stream or level in a compound procedure plant.

A current sensor is a gadget that identifies electric current in a wire and creates a sign relative to that current. The produced sign could be simple voltage or current or even a computerized yield. The produced sign can be then used to show the deliberate current in an ammeter, or can be put away for additional investigation in an information securing framework, or can be utilized with the end goal of control. Current sensors are either open-or shut circle. Open-circle current sensors measure air conditioning and DC currents and give electrical confinement between the circuit being estimated and the yield of the sensor (the essential current is estimated without electrical contact with the essential circuit, giving galvanic detachment). More affordable than their shut circle cousins, open-circle current sensors are commonly favored in battery-controlled circuits given their low-working force prerequisites and little impression highlights.

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:

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:

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:

Check the explanation

Explanation:

First of all the initial or primary and final masses can be calculated with the use of the ideal gas relations.

The net It transfer is determined from the energy balance. The initial and final internal energies and the enthalpy of the air in the supply line are obtained from A-I] for the given temperatures.  

kindly check the attached image below to see working.

Answer:

it would take approximately 232 to 258 D cell batteries to power a human for 1 day.

Explanation:

Estimate the recommended daily food intake (in Food Calories).

An average adult man requires between 2000 to 3000 calories per day.

Estimate the daily energy (in joule) a human needs.

As we know 1 food calorie is equal to 4.184 Joules of energy

2000*4.184 to 3000*4.184

8368 to 12552 Joules

But for engineering calculations 1 food calorie is equal to 1000 engineering calories, so

8368*1000 to 12552*1000

8368000 to 12552000 Joules

Estimate the voltage (in volt) of one D cell battery.

The voltage of a D cell battery is around 1.5 Volts

Estimate the charge (in amp-hours) of one D cell battery.

The amp-hours of a D cell battery varies with the manufacturing company, the typical amp-hours are in the range of 6 to 10 amp-hours.

Estimate the energy of one D cell battery (in watt-hour).

Energy in watt-hour is given by

voltage*amp-hour

1.5*6 to 1.5*10

9 to 15 watt-hour

Estimate the number of D-cell batteries it takes to power a human for 1 day.

First let us calculate the energy in a D cell battery,

1 watt-hour is equal to 3600 Joules

9*3600 to 15*3600

32400 to 54000 Joules

The number of D cell batteries required is found by dividing the energy need of a human by the energy stored in a D cell battery.

12552000/54000 to 8368000/32400

232 to 258 batteries

Therefore, it would take approximately 232 to 258 D cell batteries to power a human for 1 day.