Consider the economies of Hermes and Tralfamadore, both of which produce glops of gloop using only tools and workers. Suppose that, during the course of 30 years, the level of physical capital per worker rises by 5 tools per worker in each economy, but the size of each labor force remains the same.

Answer: 9.9KJ

Explanation: Q = U + W + losses

Q is heat transfered to the water

U is the change in energy of the system

W is work done by the system = 2J

Losses = 80J

Heat into system is 10kJ = 10000KJ

Therefore

U = Q - W - losses

U = 10000 - 2 - 80 = 9990J

= 9.9kJ

Answer:

#include<stdio.h>

void main()

{

// using file pointer to print output to txt file

FILE *fptr;

float assessedValue, taxableAmount, taxRate = 1.05, propertyTax;

/* open for writing */

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

if (fptr == NULL)

{

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

return;

}

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

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

scanf("%f", &assessedValue);

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

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

//calculating taxableAmount based on given condition in the question

taxableAmount = (assessedValue * 0.92);

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

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

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

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

//calculating propertyTax based on given condition in the question

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

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

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

//closing file using fclose function

fclose(fptr);

}

Explanation :

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

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

Output:

Assessed value: $200000

Taxable amount: $184000

Tax Rate for each $100.00: $1.05

Answer:

(a) 0.00053

(b) 0.1 mΩ

Explanation:

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

Given Information:

Zpu_old = 2 pu

Sbase_new = 100 MVA

Sbase_old = 600 MVA

kV_old = 13.8 kV

kV_new = 345 kV

Required Information:

Zpu_new = ?

ZΩ = ?

Answer:

Zpu_new = 0.000533 pu

ZΩ = 0.634 Ω

Explanation:

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

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

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

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

Zpu_new = 0.000533 pu

b) Find the ohmic value of the synchronous reactance.

ZΩ = Zbase*Zpu_new

Where Zbase is calculated as

Zbase = (kVbase)²/Sbase

Zbase = (345x10³)²/100x10⁶

Zbase = 1190.25 Ω

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

Answer:

639.4psi

Explanation:

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

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

=23psi+88760.89psft

=23psi+(88760.89/144)psi

=23+616.4

=639.4psi

Answer:

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

Explanation:

Answer:

attached below

Explanation:

Answer:

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

Explanation:

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

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

According to the question,

Rate of heat supplied:

= 60% of 3 kW

= 1.8 kW

= 1.8 KJ/s

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

Time taken will be:

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

= [tex]1254 \ s[/tex]

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

= [tex]0.3482 \ hour[/tex]

hence,

The rate of evaporation,

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

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

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

Thus the above answer is right.

Find out more information about evaporation here:

https://brainly.com/question/4406110

Answer:

Explanation:

AVERAGE: the average value is given as

[tex]\frac{1}{0.1} \int\limits^\frac{1}{10} _0 {25+10sint} \, dt = \frac{1}{0.1} [ 25t- 10cos\ t]_0^{0.1}[/tex]

=[tex]\frac{1}{0.1} ([2.5-10]-10)=-175[/tex]

RMS= [tex]\sqrt{\int\limits^\frac{1}{3} _0 {y(t)^2} \, dt }[/tex]

[tex]y(t)^2 = (25 + 10sin \ t)^2 = 625 +500sin \ t + 10000sin^2 \ t[/tex]

[tex]\frac{1}{\frac{1}{3} } \int\limits^\frac{1}{3} _0 {y(t)^2} \, dt =3[ 625t -500cos \ t + 10000(\frac{t}{2} - \frac{sin2t}{4} )]_0^{\frac{1}{3} }[/tex]

=[tex]3[[\frac{625}{3} - 500 + 10000(\frac{1}{6} - 0.002908)] + 500] = 2845.92\\[/tex]

therefore, RMS = [tex]\sqrt{2845.92} = 53.3[/tex]

Complete Question

Sites like Zillow get input about house prices from a database and provide nice summaries for readers. Write a program with two inputs, current price and last month's price (both integers). Then, output a summary listing the price, the change since last month, and the estimated monthly mortgage computed as (currentPrice * 0.045) / 12. Ex: If the input is 200000 210000, the output is: This house is $200000. The change is $-10000 since last month. The estimated monthly mortgage is $750.

Use Coral Programming Language

Answer:

// Program is written in Coral Programming Language

// Comments are used for explanatory purpose

// Program starts here

// Variable declaration

int currentprice

int prevprice

int change

float mortgage

Put "Enter current price to output" to output

currentprice = Get next input

Put "Enter last month price to output" to output

prevprice = Get next input

// Calculate Change since last month

change = currentprice - prevprice

// Calculate Monthly Mortgage

mortgage = currentprice * 0.045 / 12

// Print Results

Put "This house is $" to output

Put currentprice to output

Put "\n" to output

Put "This change is $" to output

Put change to output

Put "\n" to output

Put "This house is $" to output

Put currentprice to output

Put "since last month\n" to output

Put "This estimated monthly mortgage is $" to output

Put mortgage to output

// End of Program

Answer:

The net work output from the cycle = 520.67 [tex]\frac{KJ}{kg}[/tex]

The efficiency of Brayton cycle = 0.448

Explanation:

Compressor inlet temperature [tex]T_{1}[/tex] = 300 K

Turbine inlet temperature [tex]T_{3}[/tex] = 1700 K

Pressure ratio [tex]r_{p}[/tex] = 8

For the compressor the temperature - pressure relation is given by the formula,

⇒ [tex]\frac{T_{2} }{T_{1} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]

⇒ [tex]\frac{T_{2} }{300} = 8^{\frac{1.4 - 1}{1.4} }[/tex]

⇒ [tex]T_{2}[/tex] = 543.42 K

This is the temperature at compressor outlet.

Same relation for turbine we can write this as,

⇒ [tex]\frac{T_{3} }{T_{4} }[/tex] = [tex]r_{p}^{\frac{\gamma - 1}{\gamma} }[/tex]

⇒[tex]\frac{1700 }{T_{4} }[/tex] = [tex]8^{0.2857}[/tex]

⇒ [tex]T_{4}[/tex] = 938.5 K

This is the temperature at turbine outlet.

Now the work output from the turbine [tex]W_{T}[/tex] = [tex]m C_{p} (T_{3} - T_{4} )[/tex]

Put all the values in above formula we get,

⇒ [tex]W_{T}[/tex] = 1 × 1.005 × ( 1700 - 938.5 )

    [tex]W_{T} = 765.3 \frac{KJ}{kg}[/tex]

This is the work output from the turbine.

Now the work input to the compressor is [tex]W_{C}[/tex] = [tex]m C_{p} (T_{2} - T_{1} )[/tex]

Put all the values in above formula we get,

⇒ [tex]W_{C}[/tex] = 1 × 1.005 × ( 543.42 - 300 )

⇒ [tex]W_{C}[/tex] = 244.63 [tex]\frac{KJ}{kg}[/tex]

This is the work input to the compressor.

Net work output from the cycle [tex]W_{net} = W_{T} - W_{C}[/tex]

⇒ [tex]W_{net}[/tex] = 765.3 - 244.63

   [tex]W_{net} = 520.67\frac{KJ}{kg}[/tex]

This is the net work output from the cycle.

The thermal efficiency is given by

[tex]E_{cycle} =1 - \frac{1}{r_{p}^{\frac{\gamma - 1}{\gamma} } }[/tex]

[tex]E_{cycle} =1 - \frac{1}{8^{\frac{1.4 - 1}{1.4} } }[/tex]

[tex]E_{cycle} = 0.448[/tex]

This is the efficiency of Brayton cycle.

(b). the graph between plot the net work developed per unit mass flowing and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50 is shown in the image below.

Answer:

note:

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

Answer:

COP = 13.31

Explanation:

We have an allowed temperature difference of 2°C, thus, let's make use of temperature of 20°C in the evaporator.

Now, looking at table A-11 i have attached and looking at temperature of 20°C, we will see that the enthalpy(h1) = 261.59 Kj/Kg

While the enthropy(s1) = 0.92234 Kj/KgK

Now, the enthalpy at the second state will be gotten from the given condenser pressure under the condition s2 = s1.

Thus, looking at table A-13 which i have attached, direct 20°C is not there, so when we interpolate between the enthalpy values at 15.71°C and 21.55°C, we get an enthalpy of  273.18 Kj/Kg.

Now, the enthalpy at the third and fourth states is again obtained from interpolation between values at temperatures of 18.73 and 21.55 of the saturated liquid value in table A-12 i have attached.

Thus, h3=h4 = 107.34 Kj/kg

Formula for COP = QL/w = (h1- h4) / (h2 - h1)

COP = (261.59 - 107.34)/( 273.18 - 261.59) = 13.31

Answer:

Negative feedback

Explanation:

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

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

Answer:

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

Explanation:

See calculation attached.

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

- Calculate the cross-sectional flow area.

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

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

- Calculate the pressure drop in the copper tubes.

- Then finally calculate the power required for pumping.

To achieve an effluent concentration of 20 mg/L in a sewage lagoon receiving sewage with 100 mg/L of a contaminant, a precise biodegradation reaction rate coefficient, determined by the mass balance equation under steady-state conditions, must be achieved.

Calculating the Biodegradation Reaction Rate Coefficient

The question involves determining the biodegradation reaction rate coefficient necessary to reduce the concentration of a contaminant in a sewage lagoon, illustrating principles of environmental engineering. Given a lagoon with a surface area of 100,000 m2 and a depth of 1 m, receiving 8,640 m3/d of sewage that contains 100 mg/L of biodegradable contaminant, the goal is to lower the effluent concentration to no more than 20 mg/L.

To find the required biodegradation reaction rate coefficient (d-1), we must apply the mass balance concept in a steady-state condition, assuming the lagoon is well mixed. The mass balance equation for a contaminant undergoing a first-order degradation reaction can be expressed as: Input = Output + Decay. By substituting the given values and solving for the decay rate, we can find the coefficient that ensures the specified effluent concentration.

The calculation involves deriving relationships between the influent and effluent concentrations, the volume of the lagoon, and the decay process characterized by the reaction rate coefficient. For the lagoon described, achieving an effluent concentration of 20 mg/L from an influent concentration of 100 mg/L through biodegradation requires precise control of the treatment process and understanding of the kinetics of contaminant degradation.

Answer:

Explanation:

1) true ( reason is due to polarizability given in statement)

2)false ( nucleophile does not involve in rate determining step)

3)false ( it is an inverse attack , attacks from backside)

4) true

5) false ( it inverts stereo centres which are being attacked)

6) false (not H-)

7) true ( strong bases are good nucleophiles)

8) true ( because of steric hindrance)

9)false ( ther are stabilized by hyper conjugation)

10) false

Answer:

1) True

2) False

3) False

4)

5)

6) True

7)True

8) True

9) False

10) True

Explanation:

1) Generally, polarization increases down the column of the periodic table.

Answer:

Technician B says that a ratchet is used to loosen fasteners that are very tight.

Explanation:

A ratchet is a common wrench device with a fastener component. A ratchet wrench is an essential tool that is used to fasten or loosen nuts and bolts.

Answer:

A ratchet wrench is usually used to loosen and tighten parts like steering linkages, tie rod end clamps and muffler clamps. Basically when nut is used on long thread a ratchet wrench is being used.

hence the technician B (option b) is correct.

Explanation:

Answer:

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

Explanation:

Answer:

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

Explanation:

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

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

The initial pressure is:

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

The speed at outlet is:

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

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

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

The initial pressure is:

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

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

Answer:

P1 = 42.93 psi

Explanation:

For incompressible fluid, we know that;

A1V1 = A2V2

Making V1 the subject, we obtain;

V1 = A2V2/A1

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

Thus; V1 = V'/A1

A1 = πD²/4

Thus, V1 = 4V'/πD²

V' = 250 gal/min

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

Thus, V' = 250 x 3.85 = 962.5 in³/s

Substituting the relevant values to obtain,

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

Now let's convert to ft/s;

V1 = 136.166 x 0.0833 = 11.34 ft/s

Also for V2;

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

Now let's convert to ft/s;

V2 = 968.29 x 0.0833 = 80.66 ft/s

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

(p1/γ) + (V1²/2g) = V2²/2g

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

Making p1 the subject, we obtain;

p1 = (γ/2g)(V2² - V1²)

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

p1 = 6182.35 lb/ft²

So Converting to psi, we have;

p1 = 6182.35/144 = 42.93 psi

Answer:

Explanation:

The bolt is under double shear because it connects 3 plates.

Now the shear force resisted by each shear joint(V):

V=P/(No. of shear area)=185/2=92.5N.

Answer:

The code is as attached here.

Explanation:

The code is as given below

theta = input(' Enter the value of theta?');

y = sin(theta*pi()/180);

z = cos(theta*pi()/180);

if z < 0.01

fprintf('The value of theta is very low')

else

t=round(y/z,2);

disp(['The value of tangent theta is ',num2str(t)]);

end

In MATLAB, evaluate tangent of theta if the magnitude of cosine is

[tex]> = 10[/tex]⁻²; otherwise, display an error message.

Here are the MATLAB statements to evaluate [tex]\( \tan(\theta) \)[/tex] as long as the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], and display an error message if the magnitude of [tex]\( \cos(\theta) \)[/tex] is less than [tex]\( 10^{-2} \)[/tex]:

```matlab

% Define theta in degrees

theta_deg = input('Enter the value of theta in degrees: ');

% Convert theta to radians

theta_rad = deg2rad(theta_deg);

% Calculate cosine of theta

cos_theta = cos(theta_rad);

% Check if the magnitude of cos(theta) is greater than or equal to 10^-2

if abs(cos_theta) >= 1e-2

   % Evaluate tangent of theta

   tan_theta = sin(theta_rad) / cos_theta;

   disp(['tan(theta) = ', num2str(tan_theta)]);

else

   % Display error message

   disp('Error: The magnitude of cos(theta) is too small. Cannot evaluate tan(theta).');

end

```

This script prompts the user to enter the value of [tex]\( \theta \)[/tex] in degrees. It then converts [tex]\( \theta \)[/tex] to radians and calculates [tex]\( \cos(\theta) \)[/tex]. If the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], it evaluates [tex]\( \tan(\theta) \)[/tex] using the given formula and displays the result. Otherwise, it displays an error message indicating that the magnitude of [tex]\( \cos(\theta) \)[/tex] is too small to evaluate [tex]\( \tan(\theta) \).[/tex]

Answer:

a) 500 rev/min.

b) 50 Hz.

Explanation:

See the attached pictures.

Answer:

807.5N

Explanation:

The combined mass (m) on the back muscle is 55kg + 30kg = 85kg

Acceleration due to gravity (g) = 9.8m/s²

Therefore the force FB = ma = 85*9.8

FB= 807.5N

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

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

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

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

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

Station PVT

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

Answer:

Ai=2300 in² , Ao=Ai*1.44=3312 in²

m=10 lbm/s

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

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

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

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

Explanation:

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

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

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

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

Question:

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

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

a) IV < I²R

b) I²R = 0

c) IV = I²R

d) IV > I²R

e) I = V/R

Answer:

d) IV > I²R

Explanation:

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

The power delivered to households is given by

P = IV

The losses in the transmission line are given by

Ploss = I²R

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

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

Answer:

Explanation:

note:

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

To find an expression for the jet boat's steady speed, equate the water jet force (ρQVj) with the drag force (kV^2). Solve for V to get the equation for speed in terms of ρ, Q, k, and Vj. Then use known values for Vj and V to find Q and k and repeat the process for a different Vj to get new V and Q values.

To find an expression for the steady speed V of a freshwater jet boat in terms of the water density ρ, flow rate Q, constant k, and jet speed Vj, we can apply Newton's second law, assuming that the force provided by the water jet equals the drag force on the boat when at steady speed. The water jet force can be expressed as the rate of change of momentum of the water, which is ρQVj (since Q is the mass flow rate ρQ is the momentum flow rate), and the drag force is given as kV^2. Equating these two forces gives us:

ρQVj = kV^2

Solving for V will give us the steady speed expression we are looking for:

V = √(ρQVj / k)

To calculate the new flow rate Q given Vj = 15 m/s and V = 10 m/s, we plug these values into the expression obtained from above:

Q = kV^2 / (ρVj)

With given values, we would have:

Q = k * 10^2 / (ρ * 15)

The constant k can be determined using the known conditions and solving for k.

k = ρQVj / V^2

For (c) and (d), the same equations can be applied with the jet speed Vj changed to 25 m/s to find the new boat speed and flow rate.

There is a part of the question missing and it says;

The velocity field of a flow is given by V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s, where x and y are in feet.

Answer:

A) At (1,5),angle is -11.31°

B) At (5,2),angle is -68.2°

C) At (1,0), angle is -90°

Explanation:

From the question ;

V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s

Let us assume that u and v are the flow velocities in x and y directions respectively.

Thus we have the expression;

u = [20y/(x² + y²)^(1/2)]

and v = - [20x/(x² + y²)^(1/2)]

Thus, V = √(u² + v²)

V = √[20y/(x² + y²)^(1/2)]² + [-20x/(x² + y²)^(1/2)]²

V = √[400y²/(x²+y²)] +[400x²/(x²+y²)

V = √(400x² + 400y²)/(x²+y²)

Now for the angle;

tan θ = opposite/adjacent

And thus, in this question ;

tan θ = v/u

tan θ = [-20x/(x² + y²)^(1/2)]/ [20y/(x² + y²)^(1/2)]

Simplifying this, we have;

tan θ = - 20x/20y = - x/y

so the angle is ;

θ = tan^(-1)(-x/y)

So let's now find the angle at the various coordinates.

At, 1,5

θ = tan^(-1)(-1/5) = tan^(-1)(-0.2)

θ = -11.31°

At, 5,2;

θ = tan^(-1)(-5/2) = tan^(-1)(-2.5)

θ = -68.2°

At, 1,0;

θ = tan^(-1)(-1/0) = tan^(-1)(-∞)

θ = -90°

At,

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Answer:

A) Thermal conductivity of wall = 50 W/m.c

B) value of the convection heat transfer coefficient, h = 5 W/m².C

Explanation:

I've attached all explanations