A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measures 4 m in width and 3 m in height. The inside air is at 27ºC with a convection coefficient of 12 W/m2K, and the outer surface of the glass is at -10ºC. (a) Sketch the thermal circuit. (b) Find the temperature of the glass surface inside the room. (c) Calculate the heat loss through the window. (d) Which, if any, thermal resistances are negligible (less than 1% of the total)?

Answer:

u_e = 9.3 * 10^-8 J / m^3  ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o =  8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

                                        u_e = 0.5*e_o * E^2

- Plug in the values given:

                                        u_e = 0.5*8.854 *10^-12 *145^2

                                        u_e = 9.30777 * 10^-8  J/m^3

Answer and Explanation

Picking room temperature to be 20°C, the viscosity of water (μ) obtained from literature is:

1 centipoise = 0.01 poise = 0.001 Pa.s = 0.001 N.s/m² (This viscosity is the dynamic viscosity of water at 20°C)

Kinematic viscosity of water, η = μ/ρ

At 20°C, μ = 0.001 Pa.s, ρ = 998.23 kg/m³

η = 0.001/998.23 = 1.0 × 10⁻⁶ m²/s

Picking the room temperature to be 25°C, the viscosityof water (μ) is

0.89 centipoise = 0.0089 poise = 0.00089 Pa.s = 8.9 × 10⁻⁴ N.s/m²

Kinematic viscosity of water, η = μ/ρ

At 25°C, μ = 0.00089 Pa.s, ρ = 997 kg/m³

η = 0.00089/997 = 8.9 × 10⁻⁷ m²/s

The viscosity of water at room temperature in

We want to find the viscosity of water at room temperature in N.s/m² and mPa.s is;

0.001 N.s/m² and 1 mPa.s

Room temperature is 20°C.

Now, from online research, the viscosity of water at room temperature is 0.01 poise

We are told that;

1 centipoise = 0.01 poise

Also, that;

1 centipoise = 0.001 N⋅s/m²

Thus; viscosity of water at room temperature in N. s/m² is; 0.001 N.s/m²

Also, also 1 centipoise = 1 mPa.s

Thus;

Viscosity of water at room temperature in mPa.s is; 1 mPa.s

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Answer:

the time, in hours = 4.07hrs

Explanation:

The detailed step by step derivation and appropriate integration is as shown in the attached files.

Answer:

(A) architectural sheet metal roofing

Explanation:

 By the name itself we can judge that the 'Architectural sheet metal roofing' is a kind of metal roofing.

And these type of metal roofing is primarily used for small and big houses, small buildings and as well as in a building that is for commercial use they can be totally flat as well as little bit sloped.  

And the words similarly like batten and standing seam, and flat seam all tells us that these are the types of architectural sheet metal roofing.

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]

To determine how much a wire stretches under a load, one must consider Hooke's Law and use Young's modulus, but the original length of the wire and the type of load distribution must be known.

The student's question relates to the concept of elasticity and specifically the calculation of how much a wire stretches under a given load within the elastic limit.

Using the provided diameter of the wire, which is 0.6 inches, and the fact that an elastic modulus (Est) of 29.0(103) ksi has been given, the stretch can be calculated using Hooke's Law, which relates stress and strain through Young's modulus. However, to calculate the stretch, we would require the initial length of the wire and the format of the distributed load (whether it's uniform or varies along the length of the wire) which have not been provided in the student’s query.

Assuming uniform distributed load and an initial length L, the stretch (delta L) can be found using  delta L =  rac{wL^2}{AE}, where w is the load per unit length, L is the initial length, A is the cross-sectional area, and E is Young's modulus. If the total weight of the load is given instead, then the formula  delta L =  rac{FL}{AE} can be used with F representing the total force or weight affecting the wire.

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.

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:

The Mach number of the throat for supersonic flow = M*  = 1

and the Mach number at exit = 2.44

For

a. Pa/Pt = 0.06, Me = 2.484 Supersonic flow

b. when Pa/Pt = 0.9725 Me = 0.1999 ≅ 0.2 or subsonic flow The mach number at the throat could also be determined given the temperature parameter

Explanation:

To solve the question we note that for a supersonic nozzle, the mach number at the throat = 1

Therefore M* = 1

[tex]\frac{A_{e} }{A^{*} } = 2.5[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(\gamma -1)M_{e} ^{2} }{\gamma +1} )^{\frac{\gamma +1}{2(\gamma -1)} }[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(0.4)M_{e} ^{2} }{2.4} )^{3 }[/tex] = [tex]\frac{1}{M_{e} } ({2+(0.4)M_{e} ^{2} })^{3 } = 34.56[/tex]

34.56Me = (2+(0.4)M²)³ expanding and collecting like terms we have

Possible solutions of  Me = 0.2395, 2.44, 0.90

Since flow is supersonic, Me = 2.44

a)

Solving for [tex]M_{e}[/tex] we have [tex]\frac{P_{a} }{P_{t} } =(1+\frac{\gamma -1}{2} M^{2} _{e} )^{\frac{-\gamma}{\gamma -1} }[/tex]

When Pa/Pt = 0.06 =[tex](1+\frac{1.4 -1}{2} M^{2} _{e} )^{\frac{-1.4}{1.4 -1} }[/tex] = [tex](1+0.2M^{2} _{e} )^{-3.5 }[/tex]

Solving, we get Me = 2.484 Supersonic flow

b)

When Pa/Pt = 0.9725,  Me = 0.1999 ≅ 0.2 or subsonic flow

The Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.

To find the Mach number at the throat and at the exit plane of a supersonic nozzle with an exit area 2.5 times the throat area for a steady, isentropic flow (gamma=1.4) discharging into an atmosphere with pressure Pa, we can use the following steps:

Calculate the critical pressure ratio:

pr_crit = (2 / (gamma + 1)) ** (gamma / (gamma - 1))

pr_crit = 0.5283

Calculate the Mach number at the throat:

Mt = sqrt((1 - pr_crit) / (gamma - 1))

Mt = 1.0859

Calculate the Mach number at the exit plane:

Me = sqrt((1 - (Pa / Pt)) / (gamma - 1))

Part a:

Pa/Pt = 0.06

Me = sqrt((1 - 0.06) / (1.4 - 1))

Me = 1.5329

Part b:

Pa/Pt = 0.9725

Me = sqrt((1 - 0.9725) / (1.4 - 1))

Me = 0.2622

Therefore, the Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.

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Answer:

//Program is written in C++ language

// Comments explains difficult lines

#include<iostream>

using namespace std;

int main (){

//Variable declaration

double price, shipping;

cout<<"Item Price: ";

cin>>price;

//Test price for shipping fee

shipping = 0; //Initialise shipping fee to 0 (free)

if(price <100)

{

shipping = 0.2 * price;

}

//The above if statement tests if item price is less than 100.

//If yes, then the shipping fee is calculated as 20% of the item price

//Else, (if item price is at least 100), shipping fee is 0, which means free

if(shipping != 0)

{

cout<<"Shipping fee is "<<shipping;

}

else

{

cout<<"Item will be shipped for free";

}

return 0;

}

Answer:

attached below

Explanation:

Answer:

Given

Decimal variable: currentSales

Decimal test value: 1000

Boolean variable: newCustomer

Boolean default value: true

The following code segment is written in Java

if(currentSales == 1000 || newCustomer)

{

//Some statements

}

The Program above tests two conditions using one of statement

The first condition is to check if currentSales is 1000

== Sign is s a relational operator used for comparison (it's different from=)

|| represents OR

The second condition is newCustomer, which is a Boolean variable

If one or both of the conditions is true, the statements within the {} will be executed

Meaning that, both conditions doesn't have to be true;

At least 1 condition must be true for the statement within the curly braces to be executed

Answer:

The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.

Xₚբᵣ = 0.632

X꜀ₘբᵣ = 0.5

Xₚբᵣ > X꜀ₘբᵣ

Explanation:

From the reaction rate coefficient, it is evident the reaction is a first order reaction

Performance equation for a CMFR for a first order reaction is

kτ = (X)/(1 - X)

k = reaction rate constant = 0.05 /day

τ = Time constant or holding time = V/F₀

V = volume of reactor = 280 m³

F₀ = Flowrate into the reactor = 14 m³/day

X = conversion

k(V/F₀) = (X)/(1 - X)

0.05 × (280/14) = X/(1 - X)

1 = X/(1 - X)

X = 1 - X

2X = 1

X = 1/2 = 0.5

For the PFR

Performance equation for a first order reaction is given by

kτ = In [1/(1 - X)]

The parameters are the same as above,

0.05 × (280/14) = In (1/(1-X)

1 = In (1/(1-X))

e = 1/(1 - X)

2.718 = 1/(1 - X)

1 - X = 1/2.718

1 - X = 0.3679

X = 1 - 0.3679

X = 0.632

The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.

To determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste stream, we compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation r = kC. In a CMFR, the volume is constant, while in a PFR, the volume varies. Therefore, a PFR may be more efficient depending on the reactor design.

To determine whether a CMFR (Continuous Mixed Flow Reactor) or a PFR (Plug Flow Reactor) is more efficient in removing a reactive compound from the waste stream, we need to compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation: r = kC, where r is the reaction rate, k is the reaction rate coefficient, and C is the concentration of the reactive compound.

In a CMFR, the volume is constant, so the reactor volume (280 m3) is equal to the product of the flow rate (14 m3·day−1) and the residence time (t), which is the time it takes for the fluid to pass through the reactor. Therefore, t = V/Q = 280/14 = 20 days.

In a PFR, the volume varies along the length of the reactor, and the residence time is defined as the integral of the volume divided by the flow rate. Using the equation t = ∫V/Q, we can calculate the residence time for a PFR.

Since the residence time for a CMFR is fixed at 20 days, and the residence time for a PFR can be longer or shorter depending on the reactor design, a PFR may be more efficient in removing the reactive compound from the waste stream under steady-state conditions with a first-order reaction.

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Answer:True

Explanation:

Answer:

True

Explanation:

Input bias current:

It is a small current that flows in parallel with the input terminals of op-amp to bias the input transistors. This current gets converted into voltage and amplified which results in incorrect output results. This bias current Ib+ and Ib- flows in the positive and negative input terminals of the op-amp.  

Ib+ and Ib- create errors of opposite polarity. Therefore, bias current can be minimized by carefully adding a resistor in the positive input terminal.

Input offset current:

Unfortunately, a small error still remains due to the mismatch between input currents Ib+ and Ib-.

This input offset current error can be adjusted by adding a potentiometer and resistor in the negative input terminal.

Answer:

Product segmentation

Explanation:

Product segmentation is an adaptable method for gathering items. Likewise to an objective gathering, an item fragment contains all items that have a specific blend of item qualities.  

You can utilize item sections in a battle to improve key figure arranging.  

The significance of market division is that it permits a business to exactly arrive at a purchaser with explicit needs and needs.

Group of answer choices:

A) differences in willingness to pay

B) horizontal differentiation

C) segmentation

D) vertical differentiation

E) mass customization

Answer:

The correct answer is letter "C": segmentation.

Explanation:

Market segmentation is the classification of companies that make up their customers based on features such as age, gender, income, and profession, just to mention a few. By segmenting the market, firms group consumers with certain characteristics that enable the institution to specialize in the analysis of that particular sector to provide them with a tailored product or service that they are more likely to purchase.

Therefore, Toyota is segmenting its market in economic vehicles (Prius), size of the family (Odyssey), and large SUV (Highlander) to better fit consumer needs and preferences.

Answer:

C_h = 0.166 nF

C_L = 0.153 nF  

Explanation:

Given:

- Ideal frequency f_o = 500 KHz

- Bandwidth of frequency BW = 40 KHz

- The resistance identical to both low and high pass filter = 2 Kohms

Find:

Design a passive band-pass filter to do this by cascading a low and high pass filter.

Solution:

- First determine the cut-off frequencies f_c for each filter:

           f_c,L for High pass filter:

                f_c,L = f_o - BW/2 = 500 - 40/2

                f_c,L = 480 KHz

          f_c,h for Low pass filter:

                f_c,h = f_o + BW/2 = 500 + 40/2

                f_c,h = 520 KHz

- Now use the design formula for R-C circuit for each filter:

           General design formula:

                 f_c = 1 /2*pi*R*C_i

           C,h for High pass filter:

                  C_h = 1 /2*pi*R*f_c,L

                  C_h = 1 /2*pi*2000*480,000

                  C_h = 0.166 nF          

           C,L for Low pass filter:

                  C_L = 1 /2*pi*R*f_c,h

                  C_L = 1 /2*pi*2000*520,000

                  C_L = 0.153 nF          

Answer:

Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L

Explanation:

The lagoon can be modelled as a Mixed flow reactor.

From the value of the decay constant (0.2/day), one can deduce that the decay reaction of the pollutant is a first order reaction.

The performance equation of a Mixed flow reactor is given from the material and component balance thus:

(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)

V = volume of the reactor (The lagoon) = ?

C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³

F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s

C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L

For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³

(-r) = kC (Since we know this decay process is a first order reaction)

This makes the performance equation to be:

(kVC₀/F₀) = (C₀ - C)/C

V = F₀(C₀ - C)/(kC₀C)

k = 0.2/day = 0.2/(24 × 3600s) = 2.31 × 10⁻⁶/s

V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)

V = 86580 m³

Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.

The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)

Hope this helps!

The volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".

The volume for the lagoon relation:

[tex]\to Q_1C_1 = Q_2C_2+KC_2V\\\\[/tex]

[tex]\to Q_1=0.10\\\\\to C_1=30\\\\\to Q_2=0.2\\\\\to C_2=10\\\\\to K=0.10[/tex]

Putting the value into the formula and calculating the volume:

[tex]\to (0.10 \times 30)=(0.2\times 10)+(0.10\times 10\times V \times (\frac{1}{24 \ hrs}) \times (\frac{1 \ hr}{3,600 \ sec})) \\\\\to (3)=(2)+(1\times V \times \frac{1}{86400}) \\\\\to 3-2=(1\times V \times \frac{1}{86400}) \\\\\to 1=(1\times V \times \frac{1}{86400}) \\\\\to 1 \times 86400 = V\\\\\to V= 8.64 \times 10^4 \ m^3\\\\[/tex]

Therefore, the calculated volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".

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Answer: Check the attached

Explanation:

Answer:

[tex]Power=11.52hp[/tex]

Explanation:

Given data

[tex]p_{1}=15psia\\p_{2}=70psia\\V_{ol}=0.8ft^{3}/s[/tex]

As

[tex]m=p*V_{ol}[/tex]

Assuming in-compressible flow p is constant

The total change in system mechanical energy calculated as:

Δe=(p₂-p₁)/p

The Power can be calculated as

[tex]P=W\\P=m(delta)e\\P=p*V_{ol}*(p_{2}-p_{1})/p\\ P=V_{ol}*(p_{2}-p_{1})\\P=44(psia.ft^{3}/s )*[(\frac{1Btu}{5.404psia.ft^{3} } )-(\frac{1hp}{0.7068Btu/s } )]\\P=11.52hp[/tex]    

The power input required to pump 0.8 ft³/s of water is; P = 11.52 hP

We are given;

Initial pressure; P1 = 15 psia

Final pressure; P2 = 70 psia

Volume flow rate; V' = 0.8 ft³/s

The formula for the mass flow rate is;

m' = ρV'

The total change in the mechanical energy of the system is;

△E = (P2 - P1)/ρ

Now, formula for power is;

P = m' × △E

P = ρV' × (P2 - P1)/ρ

P = V'(P2 - P1)

P = 0.8(70 - 15)

P = 44 psia.ft³/s

Converting to Btu/s gives;

P = 8.142 btu/s

Converting Btu/s to HP from conversion tables gives; P = 11.52 hP

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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:

pressure is  825 lb/ft²

density is 1.20 × [tex]10^{-3}[/tex] slug/ft²

Explanation:

given data

p1 = 1800 lb/ft²

T1 = 500°

T2 = 400°

solution

we use here isentropic flow relation that is

[tex]\frac{P2}{P1} = (\frac{T2}{T1})^{\gamma / \gamma - 1 }[/tex]  

put here value we get pressure P2

P2 = 1800 ×  [tex](\frac{400}{500})^{3.5}[/tex]

P2 = 825 lb/ft²

and we know pressure is

pressure = [tex]\rho RT[/tex]

so for pressure 825 we get here  [tex]\rho[/tex]

825 = [tex]\rho[/tex] × 1716 × 400

[tex]\rho[/tex] = 1.20 × [tex]10^{-3}[/tex] slug/ft²

The question involves calculating the direction a swimmer must aim to reach a specified point across a river, accounting for his swimming speed and the river's current. It illustrates a problem of relative motion and vector resolution in mathematics.

The question concerns a man wanting to cross a 40-ft-wide river to a point 30 ft downstream. This scenario involves relative motion in physics, but the calculation primarily uses mathematics to solve. The man can swim at 4 ft/s in still water, and the river flows at 2 ft/s.

To determine the direction the man must swim to reach point B directly across the river, we consider two components of motion: his speed in still water and the river's current. However, the original question indicates he must not swim directly toward point B due to the river's current. Instead, he should aim upstream at a specific angle that compensates for the downstream drift caused by the current.

Without additional details, we can provide a general explanation. The man's effective speed across the river (perpendicular to the current) remains 4 ft/s. To reach point B 30 ft downstream, he must calculate the angle to offset the river's 2 ft/s current. Typically, this involves using trigonometric functions to resolve the swimmer's velocity vector into components parallel and perpendicular to the river flow.

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:

a) -k* dT / dx = q_o

b) T(x) = -280*x + 80

c) T(L) = -4 C

Explanation:

Given:

- large plane wall of thickness L = 0.3 m

- thermal conductivity k = 2.5 W/m · °C

- surface area A =12 m2.

- left side of the wall at net heat flux q_o = 700 W/m2 @ x = 0

- temperature at that surface is measured to be T1 =80°C.

Find:

- Express the differential equation and the boundary conditions for steady one dimensional heat conduction through the wall.

- Obtain a relation for the variation of temperature in the wall by solving the differential equation

- Evaluate the temperature of the right surface of the wall at x = L.

Solution:

- The mathematical formulation of Rate of change of temperature is as follows:

                                    d^2T / dx^2 = 0

- Using energy balance:

                                    E_out = E_in

                                   -k* dT / dx = q_o

- Integrate the ODE with respect to x:

                                     T(x) = - (q_o / k)*x + C

- Use the boundary conditions, T(0) = T_1 = 80C

                                     80 = - (q_o / k)*0 + C

                                      C = 80 C

-Hence the Temperature distribution in the wall along the thickness is:

                                    T(x) = - (q_o / k)*x + 80

                                    T(x) = -(700/2.5)*x + 80

                                    T(x) = -280*x + 80

- Use the above relation and compute T(L):

                                     T(L) = -280*0.3 + 80

                                     T(L) = -84 + 80 = -4 C

Differential equation: [tex]\(\frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0\).[/tex]Temperature variation: [tex]\(T(x) = T_1\).[/tex]Temperature at x = L is [tex]\(T(L) = T_1\)[/tex].

a. The differential equation for steady one-dimensional heat conduction through the wall is given by Fourier's law:

[tex]\[ \frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0 \][/tex]

This equation states that the rate of change of heat flux with respect to distance ( x ) is constant and equal to zero in steady-state conditions.

The boundary conditions are:

1. At  x = 0 : [tex]\( q = q_0 \)[/tex], [tex]\( T = T_1 \)[/tex]

2. At  x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex], as there is no heat flux across the right surface of the wall.

b. To solve the differential equation, integrate it twice:

[tex]\[ k \frac{{dT}}{{dx}} = C_1 \][/tex]

[tex]\[ \frac{{dT}}{{dx}} = \frac{{C_1}}{{k}} \][/tex]

[tex]\[ T = \frac{{C_1}}{{k}} x + C_2 \][/tex]

Apply the boundary conditions:

At  x = 0 : [tex]\( T = T_1 \)[/tex]

[tex]\[ C_2 = T_1 \][/tex]

At  x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex]

[tex]\[ \frac{{C_1}}{{k}} = 0 \][/tex]

[tex]\[ C_1 = 0 \][/tex]

Therefore, the temperature variation in the wall is given by:

[tex]\[ T(x) = T_1 \][/tex]

c. The temperature of the right surface of the wall at  x = L  is equal to [tex]( T(L) = T_1 \)[/tex], as there is no variation in temperature along the wall according to the solution obtained in part b.

Answer:

Explanation:

Given

Take off speed [tex]v=162\ mph\approx 237.6\ ft/s[/tex]

distance traveled in runway [tex]d=5000 ft[/tex]

using motion of equation

[tex]v^2-u^2=2as[/tex]

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

[tex](237.6)^2=2\times a\times 5000[/tex]

[tex]a=5.64\ ft/s^2[/tex]

Acceleration after take off [tex]a_2=3\ ft/s^2[/tex]

time taken to reach [tex]237.6 ft/s[/tex]

[tex]v=u+at[/tex]

[tex]237.6=0+5.64\times t[/tex]

[tex]t=42.127\ s [/tex]

after take off it take [tex]t_2[/tex] time to reach [tex]220 mph\approx 322.67[/tex]

[tex]322.67=237.6+3\times t_2[/tex]

[tex]t_2=28.35\ s[/tex]

total time taken [tex]t_0=t+t_1[/tex]

[tex]t_0=70.48\ s[/tex]

Water is likely to leak out of the hole in the pipe when the average velocity is 5.0 m/s due to high dynamic pressure. With a slower velocity of 0.5 m/s, air might enter the pipe if the static pressure at the hole is less than atmospheric, but this requires additional details to confirm.

The question addresses the behavior of water within a pipe system under different flow conditions, involving principles of fluid dynamics. Specifically, it asks whether water will leak out of a small hole in a vertical pipe or if air will enter into the pipe through the hole given two different average velocities of water flow.

Case 1: Average velocity of 5.0 m/s

With an average velocity of 5.0 m/s, the dynamic pressure of the flowing water is considerable, and thus, water is likely to leak out of the hole due to the higher pressure inside the pipe compared to atmospheric pressure.

Case 2: Average velocity of 0.5 m/s

With a decreased velocity of 0.5 m/s, the dynamic pressure is significantly lower. If the static pressure at the hole's location is less than the atmospheric pressure, air might enter the pipe; however, if it is still higher than atmospheric, water would continue to leak out. The determination requires additional information, such as the height of the water column above the hole and any applied pressures at the water's source.

Generally, the behavior can be predicted using Bernoulli's principle and the continuity equation for incompressible flow, which together relate the velocities, pressures, and cross-sectional areas in different sections of a pipe.

Answer:

A Not Periodic

B Periodic

C Periodic

D Not Periodic

E Periodic

F Periodic

Explanation:

The detailed steps to ascertain the periodicity or the non periodicity of each functions is as shown in the attached file.

Answer:

1) Glass

2) Rock sheet

3) Lexan

4) Masonite

b) k = 8.75 W/m.K

Explanation:

Given:

The thermal conductivity of certain materials as follows:

-Sheet Rock: k = 0.43 W/(m*K)

-Masonite: k = 0.047 W/(m*K)

-Glass: k = 0.72 W/(m*K)

-Lexan: k = 0.19 W/(m*K)

Data Given:

- Q = 100 W

- A = 8 m^2

- dT = 10 C

- L = 7 m

Find:

a) list the materials in order from most conductive to least conductive

b) calculate the thermal conductivity using Fourier's Equation

Solution:

- We know from Fourier's Law the relation between Heat transfer and thermal conductivity as follows:

                                   Q = k*A*dT / L

- From the relation above we can see that rate of heat transfer is directly proportional to thermal conductivity k.

- Hence, the list in order of decreasing conductivity is as follows:

- The list of materials in the decreasing order of thermal conductivity k is:

           1) Glass                 k = 0.72 W/m.K        

           2) Rock sheet      k = 0.43 W/m.K

           3) Lexan               k = 0.19 W/m.K

           4) Masonite          k = 0.047 W/m.K

- Use the relation given above we can compute the thermal conductivity k with the given data:

                                 k = Q*L / (A*dT)

                                 k = (100 W * 7 m) / (8 m^2*10 C)

                                 k = 8.75 W/m.K

Answer:

a. Calories required = 622710 calories

b. Energy = 1000 times much energy

Calories = 622710000 calories

Explanation:

Given:

h = Depth of pot = 20cm = 0.2m

Diameter of pot = 20cm = 0.2m

r = ½ *diameter = ½ * 0.2

r = 0.1m

Density = 1000kg/m³

Water temperature = 1°C

a.

First, we calculate the volume of the water(pot)

V = Volume = πr²h

V = 22/7 * 0.1² * 0.2

V = 0.044/7

V = 0.00629m³

M = Mass of water = Volume * Density

M = 0.00629m³ * 1000kg/m³

M = 6.29kg

M = 6.29 * 1000 grams

M = 6290g

The water is at 1°C, so it needs to gain 99°C to reach boiling point

So, Calories = 99 * 6290

Calories required = 622710 calories

b.

If we consider that D for the pot is 20 cm, approximately how much more energy is needed to heat a hot tub with D = 2 m? How many calories is that?

Depth of pot = 20cm

Depth of pot = 0.2m

Depth of hot tube = 2m

Energy is directly proportional to D³

Since the depth of hot the is 10 times greater than that of the pot

It'll require 10³ much more energy

Energy = 10³

Energy = 1000 times much energy

Calories required = 622710 * 1000

Calories = 622710000 calories

Answer:

1.42 KJ

Explanation:

solution:

power in beginning [tex]p_{0}[/tex]=(1.5 V).(9×[tex]10^{-3}[/tex] A)

                                    = 13.5 mW

after continuous 37 hours it drops to

                                    [tex]p_{37}[/tex]=(1 V).(9×[tex]10^{-3}[/tex] A)

                                         =9 mW

When the voltage will drop energy will not remain the same but the voltage drop will always remain same if the voltage was drop to for example from 5 V to 4.5 V the drop will remain the same.

                  37 hours= 37.60.60

                                 =133200‬ s

                              w=(9×[tex]10^{-3}[/tex] A×133200‬ )+[tex]\frac{1}{2}(13.5.10^{-3}-9.10^{-3})(133200)[/tex]

                                 =1.42 KJ

NOTE:

There maybe a calculation error but the method is correct.