Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW at a 0.707 pf lagging. Determine:
(a) the pf of the second load,
(b) the parallel element required to correct the pf to 0.9 lagging for the two loads.

Answer: Check the attached

Explanation:

Answer:

The pipe is open ended and the length of pipe is 3.4 m.

Explanation:

For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below

Equation for nth frequency for open end pipe is given as

[tex]f_n=\frac{nv}{2L}[/tex]

For (n+1)th value the frequency is

[tex]f_{n+1}=\frac{(n+1)v}{2L}[/tex]

Taking a ratio of both equation and solving for n such that the value of n is a whole number

[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\[/tex]

So n is a whole number this means that the pipe is open ended.

For confirmation the  nth frequency for a closed ended pipe is given as

[tex]f_n=\frac{(2n+1)v}{4L}[/tex]

For (n+1)th value the frequency is

[tex]f_{n+1}=\frac{(2n+3)v}{4L}[/tex]

Taking a ratio of both equation and solving for n such that the value of n is a whole number

[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\[/tex]

As n is not a whole number so this is further confirmed that the pipe is open ended.

Now from the equation of, with n=6, v=340 m/s and f=300 Hz

[tex]f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m[/tex]

The value of length is 3.4m.

For flow over a plate, the variation of velocity with vertical distance y from the plate, the correct relation for the wall shear stress is  [tex]\( \tau = \mu (a - 2by) \)[/tex]. The correct option is B.

The wall shear stress ([tex]\( \tau \)[/tex]) for flow over a plate is given by the following relation, assuming the fluid has constant viscosity ([tex]\( \mu \)[/tex]):

[tex]\[ \tau = \mu \frac{du}{dy} \][/tex]

Where:

[tex]\( \mu \)[/tex] = dynamic viscosity of the fluid,

[tex]\( \dfrac{du}{dy} \)[/tex] = rate of change of velocity with respect to the vertical distance (y) from the plate.

In your case, the velocity profile [tex]\( u(y) = ay - by^2 \)[/tex] is given. To find the rate of change of velocity with respect to y, we differentiate [tex]\( u(y) \)[/tex] with respect to y:

[tex]\[ \frac{du}{dy} = a - 2by \][/tex]

Now, substitute this into the formula for wall shear stress:

[tex]\[ \tau = \mu \left( a - 2by \right) \][/tex]

So, the correct relation for the wall shear stress ([tex]\( \tau \)[/tex]) in terms of a, b, and [tex]\( \mu \)[/tex] is:

[tex]\[ \tau = \mu \left( a - 2by \right) \][/tex]

Thus, the correct option is B.

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Your question seems incomplete, the probable complete question is:

Question:

For flow over a flat plate, the velocity variation with vertical distance y from the plate is given by [tex]\( u(y) = ay - by^2 \)[/tex], where a and b are constants. What is the correct expression for the wall shear stress ([tex]\( \tau \)[/tex]) in terms of a b, and dynamic viscosity ([tex]\( \mu \)[/tex])?

A) [tex]\( \tau = \mu (a - by) \)[/tex]

B) [tex]\( \tau = \mu (a - 2by) \)[/tex]

C) [tex]\( \tau = \mu (a + by) \)[/tex]

D) [tex]\( \tau = \mu (a + 2by) \)[/tex]

Answer:

[tex]F=6200\ \text{N}\\[/tex]

Explanation:

In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:

In this problem the flexural strength is defined with the following formula:

[tex]\sigma=\cfrac{3FL}{2bd^2}[/tex]

where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.

The force is then defined as:

[tex]F=\cfrac{2\sigma bd^2}{3L}=6200\ \text{N}[/tex]

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = [tex]\int\limits^a_b P \, dV[/tex]  = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

Answer: This is EASY!

Explanation:

To make it easy, you would convert those binary numbers and to denary. And this gives:

84 104 105 115 32 105 115 32 69 65 83 89 33

Then, the denary numbers generated can be converted to ASCII code using this ASCII Table in the attachment below. And the result is: This is EASY!

Answer: For non-precision approaches, the maximum acceptable descent rate acceptable should be one that ensures the aircraft reaches the minimum descent altitude at a distance from the threshold that allows landing in the touch down zone. Otherwise, a decent rate greater than 1000fpm is unacceptable.

Explanation: For non-precision approaches, a descent rate should be used that ensures the aircraft reaches the minimum decent altitude at a distance from the threshold that allows landing in the touchdown zone (TDZ) . On many instrument approach procedures, this distance is annotated by a visual descent point (VDP) If no VDP is annotated, calculate a normal descent point to the TDZ. To determine the required rate of descent, subtract the TDZ elevation (TDZE) from the final approach fix (FAF) altitude and divide this by the time inbound. For illustration, if the FAF altitude is 1,000 feet mean sea level (MSL), the TDZE is 200 feet MSL and the time inbound is two minutes, an 400 fpm rate of descent should be applied.

A descent rate greater than approximately 1,000 fpm is unacceptable during the final stages of an approach (below 1,000 feet AGL). Operational experience and research shows that this is largely due to a human perceptual limitation that is independent of the airplane or helicopter type. As a result, operational practices and techniques must ensure that descent rates greater than 1,000 fpm are not permitted in either the instrument or visual portions of an approach and landing operation.

The question is incomplete! Complete question along with answers and explanation is provided below.

Question:

A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz.

a) Draw the circuit

b) Calculate the power drawn by the lamp

c) Calculate the apparent power

d) Calculate the power factor

e) Calculate the reactive power

f) Calculate the size of the capacitor necessary to provide unity power factor correction

Explanation:

a) draw the circuit

Refer to the attached image.

As you can see in the attached drawing, it is a series circuit containing  two resistors and one inductor.

In a series circuit, current remains same throughout the circuit

The circuit is powered by an AC voltage source having voltage of 120 V and frequency 60 Hz.

The current flowing in the circuit can be found by ohm's law

 I = V/Z

where V is the voltage and Z is the total impedance of the circuit

 Z = R + XL

where  XL is the inductive reactance

XL = j2 π f L

XL = j2*π*60*0.9

XL = j339.29Ω

Total resistance is

R =200 + 80 = 280 Ω

Total impedance is

Z = 280 + j339.29 Ω

b) Calculate the power drawn by the lamp

First calculate the current

I = V/Z

I = 120/(280 + j339.29)

I = 0.272<-50.46° A  (complex notation)

P = I²R

P = (0.272)²200

P ≈ 15 W

Power drawn by the circuit

P=V*I*cos(50.46°)

P=20.77 W

c) Calculate the apparent power

A = VI*

A = 120*0.272<50.46°

A = 32.64<50.46° VA

d) Calculate the power factor

PF = cos(50.46)

PF = 0.63

e) Calculate the reactive power

Q = VIsin(50.46)

Q = 120*0.272<-50.46*sin(50.46)

Q = 25.13<-50.46  VAR

f) Calculate the size of the capacitor necessary to provide unity power factor correction

The required reactive compensation power is

Qc = P (tan(old) - tan(new))

Qc = 20.77 (tan(50.46) - tan(0))

Qc = 25.16 VAR

C = Qc/2πfV²

C = 25.16/2*π*60*120²

C = 4.63 uF

Hence adding a capacitor of 4.63 uF parallel to the load will improve the PF from 0.63 to 1.

Answer:

a= - 2.6 m/s².

Explanation:

u = 32 m/s

The speed after 6 s is half of u

[tex]v= \dfrac{32}{2}=16\ m/s[/tex]

t= 6 s

The average acceleration = a

We know v = u +at

v=final velocity

u=initial velocity

Now by putting the values in the above equation

16= 32 + a x 6

[tex]a=\dfrac{16-32}{6}\ m/s^2[/tex]

[tex]a=-2.6\ m/s^2[/tex]

Therefore the acceleration will be - 2.6 m/s².

a= - 2.6 m/s².

Negative indicates that velocity and acceleration is is opposite direction.

Answer:

Yes and no

Explanation:

The thermodynamic equation for the heat transfer in a constant volume process is the following:

[tex]Q=\Delta U=mC_V\Delta T[/tex]

where Q is the required heat, U is the internal energy, m the mass of the gas, C_V the heat capacity assuming consant volume and [tex]\Delta T[/tex] is the change in temperature.

If you assume the heat capacity doesn't change with temperature at which the gas is currently at then the heat transfer depends solely on the change in temperature. With this assumption the transfered heat would be the same in both cases.

In reality the heat capacity does change with respect to temperature. Depending on the type of gas. In reality there would be difference in heat transfered between 295/205 K and 245/255 . Only then you wouldn't use the [tex]\Delta T[/tex] expression since the integral would be different depending on the heat capacity in relation to temperature.

Question:

Current density is given in cylindrical coordinates as J = −10^6z^1.5az A/m² in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0.

(a) Find the total current crossing the surface z = 0.1 m in the az direction.

(b) If the charge velocity is 2 × 10^6 m/s at z = 0.1 m, find ρν there.

(c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.

Answer:

a. -39.8μA

b. -15.81mC/m³

c. 29.05m/s

Explanation:

Given

Density = J = −10^6z^1.5az A/m²

Region: 0 ≤ ρ ≤ 20 µm

ρ ≥ 20 µm

J = 0.

a. Total current is calculated by.

J * ½((ρ1)² - (ρ0)²) * 2 π * φdza.

Where J = Density = -10^6 * z^1.5

ρ1 = Upper bound of ρ = 20

ρ0 = Lower bound of ρ = 0

π = 22/7

φdza = 10^-6

z = 0.1

Total current

= -10^6 * z^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= 10^6 * 0.1^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6

= −39.7543477278310

= -39.8μA

b. Calculating velocity charge density at (ρv)

Density (J) = ρv * V

Where J = Density = -10^6 * z^1.5

V = 2 * 10^6

z = 0.1

Substitute the above values

-10^6 * 0.1 ^1.5 = ρv * 2 * 10^6

ρv = (-10^6 * 0.1^1.5)/(2 * 10^6)

ρv = -0.1^1.5/(2)

ρv = -0.015811388300841

ρv = -0.01581 --------- Approximated

ρv = -15.81mC/m³

c. Calculating Velocity

Velocity = J/V

Where Velocity Charge Density = -2000 C/m3

Where J = -10^6 * z^1.5

z = 0.15

J = -10^6 * 0.15^1.5

J = -58094.75019311125

Velocity = -58094.75019311125/-2000

Velocity = 29.047375096555625m/s

Velocity = 29.05m/s

A) The total current crossing the surface z = 0.1 m in the z^ direction is; I_tot ≈ -39.8μA

B) If the charge velocity is 2 × 10⁶ m/s, then ρv is; -15.81 mC/m³

C) If the volume charge density at z = 0.15 m is −2000 C/m³, the Charge velocity is; 29.05m/s

We are given;

Current Density; J = −10⁶z^(1.5) (z^) A/m²

Region: 0 ≤ ρ ≤ 20 µm

At ρ ≥ 20µm,  J = 0.

I_tot = J × ½((ρ1)² - (ρ0)²) × 2π × φdza.

Where;

J is current Density = −10⁶z^(1.5) A/m²

ρ1 is Upper bound of ρ = 20 µm

ρ0 is Lower bound of ρ = 0 µm

 φdza = 10⁻⁶

z = 0.1

Thus plugging in the relevant values, we have;

Total current;

I_tot = -10⁶ × 0.1^(1.5) * ½(20² - 0²) × 2 × π × 10⁻⁶

I_tot ≈ -39.8μA

ρv = J/V

where;

J is current density = −10⁶z^(1.5) A/m²

 V is charge velocity = 2 × 10⁶ m/s

z = 0.1

Thus;

 ρv = ( −10⁶ × 0.1^(1.5))/(2 × 10^6)

ρv = -¹/₂(0.1^(1.5))

ρv ≈  -0.01581 C/m³

Thus;

ρv = -15.81 mC/m³

Charge Velocity = J/V

Where;

J is current density = −10⁶z^(1.5) A/m²

V is volume charge density = -2000 C/m³

At z = 0.15;

Charge velocity = (−10⁶ × 0.15^(1.5))/(-2000)

Charge velocity ≈ 29.05m/s

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To determine the minimum cross-sectional area of the steel cables, calculate the allowable stress and use it along with the provided load. The result is a minimum area of 50 mm² for each cable.

The question involves calculating the minimum cross-sectional area of each steel cable designed to support a load with a safety factor, given the yield stress of the material. First, determine the allowable stress by dividing the yield stress by the factor of safety. In this case, the allowable stress is 200 MPa (400 MPa / 2.0). To find the minimum cross-sectional area (A), use the formula A = P / σ, where P = 10 kN = 10,000 N (the load) and σ (sigma) is the allowable stress in N/m². Convert 200 MPa to N/m² to get 200,000 N/m². Therefore, the minimum cross-sectional area required for each cable is 50 mm² (10,000 N / 200,000 N/m²).

Answer:

As a heating engineer and considering a house as a dynamic system , and that without a heater, the average temperature in the house would vary over a 24-h period.

What might you consider for inputs, outputs, and state variables for a simple dynamic model?

State variables: according to the weather conditions of the area where the house was built:

State variable # 1: minimum temperture during a day in an specific season (*4);

State variable # 2: maximum temperature during the day, in an specific season (*4) as well;

State variable # 3: average temperature during the day in an specific season (*4).

That makes 16 state variables all of them in Centigrade degrees.

Input variables:

# 1: one degree over each of the state variables given.

# 2: one degree below each of the state variables, all of them in Centigrade           degrees.

Output variables:

# 1 are the temperatures reached after adding one degree to each of the input variables.

# 2 are the temperatures reached after decreasing one degree, all of them in Centigrade degrees.

How would you expand your model so that it would predict temperatures in several rooms of the house?

I would add output variables in a "Y" system to predict temperatures in several rooms of the house.

How does the installation of a thermostatically controlled heater change your model?

It would change on the "Y" variables as they will get a control system  designed for sensors to produce from some input variables to make the system respond.

Explanation:

State-determined system models using well defined physical systems is of highly interest to engineers.

Answer:

G = 0.424

Explanation:

Ds = ( 0.278tr * V ) + (0.278 * V²)/ ( 19.6* ( f ± G))

Where Ds = stopping sight distance = 415miles = 126.5m

G = absolute grade road

V = velocity of vehicle = 52miles/hr

f = friction = 0 because the road is wet

tr = standard perception / reaction time = 2.5s

So therefore:

Substituting to get G

We have

2479.4G = 705.6G + 751.72

1773.8G = 751.72

G = 751.72/1773.8

G = 0.424

Based on the calculations, the grade of the road is equal to 2.43 %.

Given the following data:

Assumed data:

Mathematically, the stopping distance for an inclined surface with a coefficient of friction is given by this formula:

[tex]SD = 0.278Vt + \frac{V^2}{254} (f \pm 0.01G)[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]135 = 0.278 \times 83.6859 \times 2.5 + [\frac{83.69^2}{254} \times (\frac{3.5}{9.8} + 0.01G)]\\\\135 = 58.14 + [27.58 \times (0.36 + 0.01G)]\\\\135 = 58.14 +9.93+0.2758G\\\\0.2758G=135 - 58.14 -9.93\\\\0.2757G=66.9923\\\\G=\frac{66.93}{0.2757}[/tex]

G = 242.76 ≈ 243

G = 2.43 %

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The work done by the gas is -940 kJ

Explanation:

In this process, we are told that the product of pressure and volume remains constant:

[tex]pV=const.[/tex]

so we can write

[tex]p_1 V_1 = p_2 V_2[/tex]

where

[tex]p_1 = 1.4 bar[/tex] is the initial pressure

[tex]p_2 = 6.8 bar[/tex] is the final pressure

[tex]V_1=4.25 m^2[/tex] is the initial volume

Solving for [tex]V_2[/tex], we find the final volume:

[tex]V_2=\frac{p_1V_1}{p_2}=\frac{(1.4)(4.25)}{6.8}=0.875 m^3[/tex]

Now by looking at the equation of state of an ideal gas:

[tex]pV=nRT[/tex] (1)

we notice that since [tex]pV=const.[/tex], this means that also the absolute temperature of the gas T remains constant (because the number of moles n does not change). Therefore this is an isothermal process: the work done in an isothermal process is given by

[tex]W=nRTln(\frac{V_2}{V_1})[/tex]

And by looking again at (1), we  can substitute (nRT) with (pV), so we get

[tex]W=p_1 V_1 ln (\frac{V_2}{V_1})[/tex]

Converting the pressure into SI units,

[tex]p_1 = 1.4 bar = 1.4\cdot 10^5 Pa[/tex]

So the work done is

[tex]W=(1.4\cdot 10^5)(4.25)ln(\frac{0.875}{4.25})=-9.4\cdot 10^5 J[/tex]

Which means -940 kJ. This value is negative since the work is done by the surroundings on the gas (because the gas is compressed).

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Answer:Name resolution traffic goes to the single zone

Administrative burden

Submit

Centralized Management

Explanation:DNS(Domain name system) is a term used in the internet which Describes the conversion of alphabetical names into Numerical representations,he a large Organisation as stated which spans through different Geographical areas continue with a single Domain name system it will lead to the following.

Name resolution traffic will increase which might delay the execution of tasks

Administrative burden will be Increased as it is carrying out a wide range of activities.

Centralized management which may affect the flow of work.

Answer:

Explanation:

Code used will be like

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

namespace PaintingWall

{

class Room

{

public int length, width, height,Area,Gallons;

public Room(int l,int w,int h)

{

length = l;

width = w;

height = h;  

}

private int getLength()

{

return length;

}

private int getWidth()

{

return width;

}

private int getHeight()

{

return height;

}

public void WallAreaAndNumberGallons()

{

Area = getLength() * getHeight() * getWidth();

if (Area < 350)

{

Gallons = 1;

}

else if (Area > 350)

{

Gallons = 2;

}    

Console.WriteLine ("The area of the Room is " + Area);

Console.WriteLine("The number of gallons paint needed to paint the Room is " + Gallons);

}

}

class PaintingDemo

{

static void Main(string[] args)

{

int l, w, h;

Room[] r = new Room[8];

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room "+(i+1));

Console.Write("Enter Length : ");

l = Convert.ToInt32(Console.ReadLine() );

Console.Write("Enter Width : ");

w = Convert.ToInt32(Console.ReadLine());

Console.Write("Enter Height : ");

h= Convert.ToInt32(Console.ReadLine());

r[i] = new Room(l,w,h);

Console.WriteLine();

}

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room " + (i + 1));

r[i].WallAreaAndNumberGallons();

}

Console.ReadKey();  

}

}

}

Answer:

Your stratified squamous epithelium is difficult to penetrate!

Explanation:

The epithelium is the tissue formed by one or several layers of cells attached to each other that cover all the free surfaces of the organism, and constitute the internal lining of the cavities, organs, hollows, ducts of the body and skin and also form the Mucous and glands. The epithelia also forms the parenchyma of many organs, such as the liver.

Flat or squamous epithelia: Formed by flat cells, with much less height than width and a flattened nucleus. It is one of the most external being part of the epidermis and generating some inconveniences when penetrating with needles to perform blood extractions when you have certain characteristics of hardness.

Answer:

Negotiated speed should be lower. Perception/reaction time is too less than design values.

Explanation:

Given:

- The claimed safe speed V_1 = 65 mi/h

- Sight distance D = 510 ft

- The practical deceleration a = 11.2 ft/s   ... according to standards

Find:

Assuming practical stopping distance, comment on the student whether the claim is correct or not

Solution:

- Calculate the practical stopping distance:

                      d = V_1^2 / 2*a

                      d = ( 65 * 1.46 )^2 / 2*11.2 = 402.054 ft

- Solve for reaction distance d_r is as follows:

                     d_r = D - d = 510 - 402.054 = 107.945 ft

- The perception/time reaction is:

                   t_r = d_r/V_1 = 107.945 / 94.9

                  t_r = 1.17 sec

Answer: The perception/reaction time t_r = 1.17 s is well below the t = 2.3 s.

Hence, the safe speed should be lower.

Answer:

P= 541.5 kW.

Explanation:

Given that

velocity of water after leaving the nozzle ,v= 95 m/s

The mass flow rate of the water , m= 120 kg/s

The power generated P is given as

[tex]P=\dfrac{1}{2}mv^2[/tex]

Now by putting the values in the above equation we get

[tex]P=\dfrac{1}{2}\times 120\times 95^2\ W[/tex]

P=541500  W

The  power in kW will be 541.5 kW.

Therefore the answer will be 541.5 kW

P= 541.5 kW.

The power generation potential of the water jet is 542.25 kW.

To determine the power generation potential of the water jet, we need to calculate the kinetic energy of the water jet and then convert it to power. The kinetic energy of the water jet can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass flow rate of the water and v is the velocity of the water jet. Given that the flow rate is 120 kg/s and the velocity is 95 m/s, we can calculate the kinetic energy to be KE = 0.5 * 120 * 95^2 = 0.5 * 120 * 9025 = 542,250 J/s.

To convert the kinetic energy to power, we divide by the time taken to deliver the energy. Since the flow rate is given in kg/s, we can assume the time taken is 1 second. Therefore, the power generation potential of the water jet is 542,250 J/s, or 542.25 kW.

The expression for the bonding energy E₀ in terms of the equilibrium interionic separation r₀ and the given constants is;

E₀ = rD[(exp(-r₀/ρ) + exp(-r/ρ)] - EN/r₀

We are given the expression;

EN = -C/r + Dexp(-r/ρ)

where;

EN is net potential energy

r is the interionic separation

C, D, and rho(ρ) are constants whose values depend on the specific material.

The formula for the bonding energy is usually;

E₀ = -C/r₀ + D exp(-r₀/ρ)

Step 1; We are to differentiate EN with respect to r. Thus, we have;

dEN/dr = C/r² - D exp(-r/p)

Step 2; We are to solve for C in terms of D, rho(ρ) and r₀. Thus;

E₀ + (C/r₀) = -D*exp(-r₀/ρ)

⇒ C/r₀ = -Dexp(-r₀/ρ) - E₀

Thus, multiplying both sides by r₀ gives;

C = -r₀(Dexp(-r₀/ρ) + E₀)

Step 3; We are to determine the expression for E₀ by substitution for C in the equation given. This gives us;

EN = -r₀(Dexp(-r₀/ρ) + E₀)/r + Dexp(-r/ρ)

EN = r₀*D*exp(-r₀/ρ) - (r₀E₀/r) + D*exp(-r/ρ)

EN + (r₀E₀/r) = r₀*D*exp(-r₀/ρ) + D*exp(-r/ρ)

r₀E₀/r = D[(exp(-r₀/ρ) + exp(-r/ρ)] - EN

Thus;  E₀ = rD[(exp(-r₀/ρ) + exp(-r/ρ)] - EN/r₀

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Answer:

The question continues ; Determine the expressions for the velocity and acceleration of a point P as a function of time t, and determine the maximum velocity of point P during one cycle. Use the values r = 75mm and w = pie-rads/s

Explanation:

The diagram and the detailed step by step explanation is as shown in the attachment

In the Scotch-yoke mechanism, if the displacement is given by x = r sin(t), the velocity v = r cos(t) and acceleration a = -r sin(t). The negative sign in acceleration indicates that it is in the opposite direction to displacement.

The Scotch-yoke mechanism which converts rotary motion into reciprocating motion can be analyzed using principles of kinematics. If we have the displacement given by x = r sin(t), the velocity and acceleration can be derived from this displacement equation.

The velocity (v) is the time derivative of the displacement function, i.e., v = dx/dt = r cos(t).

The acceleration (a) is the time derivative of the velocity function, so a = dv/dt = -r sin(t).

The negative sign signifies that acceleration is in the opposite direction to displacement.

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Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

                           mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

                           BE / BC = 45 / 60 = 0.75

Hence,                E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

                           mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

                           vec(EF) = < -15 , -110 , 30 >

                           unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension            T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

                           mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

                           vec(AD) = < 48 , -12 , 36 >

                           unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

                           unit (AD) = <0.7845 , -0.19612 , 0.58835 >

Next:

                           M_AD = unit(AD) . ( E x T_EF)

                           [tex]M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right][/tex]

                            M_AD = 1835.73 + 116.49528 - 593.0568

                            M_AD = 1359.17 lb-in

The moment of the force about the line joining points A and D is; 617.949 lb.in

We are given;

Force exerted by cable EF at E; T_EF = 46 lb.

From the diagram of the guy wire, we can draw a triangle and we will have the following coordinates;

A(0, 0, 0)

D(48, -12, 36)

E(E_x, 96, E_z)

Also, we can get that;

BC² = 48² + 36²

BC = √(48² + 36²)

BC = 60 in

Also, from similar triangles, we will have the coordinate of E as;

E(36, 96, 27)

Position of Vector of EF is;

EF = {(21 - 36)i + (-14 - 96)j + (57 - 27)k} in

EF = {-15i - 110j + 30k} in

Magnitude of EF from online calculation = 115 in

Force along cable EF is;

F_EF = 46{(-15i - 110j + 30k)/115}

F_EF = {-6i - 44j + 12k} lb

Position vector of AE is {36i + 96j + 27k} in

Position vector of AD is {48i - 12j + 36k} in

Magnitude of AD = 61.188 N

Unit vector of AD; λ_ad = {48i - 12j + 36k}/61.188

λ_ad = 0.7845i - 0.1961j + 0.5883k

M_ad = λ_ad × r_ea × T_EF

M_ad = [tex]\left[\begin{array}{ccc}0.7845&-0.1961&0.5883\\36&96&27\\6&-44&12\end{array}\right][/tex]

Solving this gives;

M_ad = 617.949 lb.in

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Answer:

attached below

Explanation:

Answer:

(a) 1.125 ft, Section factor = 22.78

(b) 42.75 ft

Explanation:

Hydraulic radius is given by [tex]R_{H} = \frac{A}{P}[/tex] Where

A = Cross sectional area of flow and

P = Perimeter  h

Since the cross section is a circle  then at depth 4 of 4.5 the perimeter

[tex]=2 \pi r-\frac{\theta }{360} *2 \pi r[/tex]  where r = 2.25 and θ = 102.1 °

perimeter = 10.1 ft and the  area = [tex]=\pi r^2-\frac{\theta }{360} * \pi r^2[/tex] =  11.39 ft²

Therefore [tex]R_{H} = \frac{11.39}{10.1} = 1.125 ft[/tex]

Section factor is given by for critical flow = Z = A×√D

= 11.39 ft² × √(4 ft) = 22.78

for normal flow Z =[tex]Z_{} ^{2} = \frac{A^{3}}{T}[/tex] = 22.78

(b) The alternate depth of flow is given by

for a given flow rate, we have from chart for flow in circular pipes

Alternative depth = 0.9×45 = 42.75 ft

Answer:

a) Power = Av⁴

b) Energy = Av⁴/2 or (A²v⁶)/2m

Explanation:

Given Y = Av³

A = a constant, v = Velocity, Y = system variable.

a) If Y = Force, F, Find Power in the element.

Power is the dot product of Force and velocity and it's a scalar quantity.

i.e. P = F.v = F.v (cos θ) where θ is the angle between the force and velocity vector.

But in this case, average power is simply given by Fv.

P(avg) = Fv = Yv = (Av³) × v = Av⁴

b) If Y = linear momentum, p, Find the energy stored in the element.

Energy is related to linear momentum by the relationship between kinetic energy and linear momentum.

p = mv and E = mv²/2 = (mv)(v)/2, so,

E = pv/2

For this question, p = Y = Av³

E = Yv/2 = (Av³)v/2 = Av⁴/2

Kinetic energy is often related to momentum through this expression too,

p = mv; p² = m²v²

E = mv²/2; E = (mv²/2) × (m/m) = m²v²/2m = p²/2m

Therefore, E = Y²/2m = (Av³)²/2m = (A²v⁶)/2m

Hope this helps!

Answer:

[tex]W=-52 800\ \text{J}=-52.8\ \text{kJ}[/tex]

Explanation:

First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.

To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:

[tex]W=\int_{0.42}^{0.12}-1200V+500dV=-52.8\ \text{kJ}[/tex]

Answer:

The fluid level difference in the manometer arm = 22.56 ft.

Explanation:

Assumption: The fluid in the manometer is incompressible, that is, its density is constant.

The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

And P(gage) = ρgh

ρ = density of the manometer fluid = 60 lbm/ft³

g = acceleration due to gravity = 32.2 ft/s²

ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

h = fluid level difference between the two arms of the manometer = ?

P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²

1353.6 = ρg × h = 60 lbf/ft³ × h

h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!

Answer:

59.9%

Explanation:

R^2 =(-0.774)^2  = 0.599

59.9% of fuel is accounted for

Answer:

6.30 miles/hour

Explanation:

Newton's second law applies here. In simple terms:

[tex]F = ma[/tex]

where F = Force (Thrust) in N

           a = acceleration (m/s²)

The acceleration can be given by rearranging the  formula to give:

[tex]a = \frac{F}{N}[/tex]

  = [tex]\frac{(686.68*10^{6} )}{24811505120150.2656} \\= 0.0000277 m/s\\= 6.03 miles/hr[/tex]