A thick-walled tube of stainless steel having a k = 21.63 W/m∙K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with 0.0254 m thick layer of an insulation (k = 0.2423 W/m∙K). The inside-wall temperature of the pipe is 811 K and the outside surface of the insulation is 310.8 K. For a 0.305 m length of pipe, calculate the heat loss and the temperature at the interface between the metal and the insulation.

Answer:

D. Any of the above

Explanation:

All of the following strategies maximize insulation levels:

Answer:

Mean.m

fid = fopen('input.txt');

Array = fscanf(fid, '%g');

Amean = AM(Array);

Gmean = GM(Array);

Rmean = RMS(Array);

display('Amlan Das Statistical Package')

display(['arithmetic mean = ', sprintf('%.5f',Amean)]);

disp(['geometric mean = ', sprintf('%.5f',Gmean)]);

disp(['RMS average = ', sprintf('%.5f',Rmean)]);

AM.m

function [ AMean ] = AM( Array )

n = length(Array);

AMean = 0;

for i = 1:n

AMean = AMean + Array(i);

end

AMean = AMean/n;

end

GM.m

function [ GMean ] = GM(Array)

n = length(Array);

GMean = 1;

for i = 1:n

GMean = GMean*Array(i);

end

GMean = GMean^(1/n);

end

RMS.m

function [ RMean ] = RMS( Array)

n = length(Array);

RMean = 0;

for i = 1:n

RMean = RMean + Array(i)^2;

end

RMean = (RMean/n)^(0.5);

end

Answer:

False, True , True.

Explanation:

Buckling is defined as  large lateral deflection of member of under compression with slight increase in load beyond a critical load.  

1) Buckling occurs in axially loaded member in tension. is false as buckling occurs in tension.

2) Buckling is caused by lateral deflection of the member and not the axial  deflection. hence true.

3) Buckling is an instability phenomenon that can lead to failure. Hence True.

Answer:

magnitude of thrust uis  11061.65 lb/ft

location is 5 ft from bottom

Explanation:

Given data:

Height of vertical wall is 15 ft

OCR  is 1.5

[tex]\phi = 33^o[/tex]

saturated uit weight[tex] \gamma_{sat} = 115.0 lb/ft^3[/tex]

coeeficent of earth pressure [tex]K_o[/tex]

[tex]K_o = 1 -sin \phi[/tex]

        = 1 - sin 33 = 0.455

for over consolidate

[tex]K_{con} = K_o \times OCR[/tex]

            [tex] = 0.455 \times 1.5 = 0.683[/tex]

Pressure at bottom of wall is

[tex]P =K_{con} \times (\gamma_{sat} - \gamma_{w}) + \gamma_w \times H[/tex]

   [tex]= 0.683 \times (115 - 62.4) \times 15 + 62.4 \times 15[/tex]

P = 1474.88 lb/ft^3

Magnitude pf thrust is

[tex]F= \frac{1}{2} PH[/tex]

   [tex]=\frac{1}{2} 1474.88\times 15 = 11061.65 lb/ft[/tex]

the location must H/3 from bottom so

[tex]x = \frac{15}{3} = 5 ft[/tex]

Answer:

Explanation:

1. The problems in the above pseudocode include (but not limited to) the following

1. The end year is not properly represented.

2. Though, the factor of 5 is declared, it's not implemented as increment in the pseudocode

3. Incorrect use of control structures. (While ...... And .......Endif)

2.

Start

Start_Year = 2017

Kount = 1

While Kount <= 30

Display Start_Year

Start_Year = Start_Year + 5

Kount = Kount + 1

End While

Stop

3. Yes, by doing the following

* Apply increment of 5 to start year within the whole loop, where necessary

* Use matching control structures

While statement ends with End While (not end if, as it is in the question)

4. Using VBA

Dim Start_Year: Start_Year = 2017 ' Declare and initialise year to 2017

Dim Counter: Counter = 1 ' Declare and Initialise Counter to 1

While Count <= 30 ' Test Value of Counter; to check if the desired number of year has gotten to 30. While the condition remains value, the following code will be executed.

msgbox Start_Year ' Display the value of start year

Start_Year = Start_Year + 5 'Increase value of start year by a factor of 5

Counter = Counter + 1 'Increment Counter

Wend

5. Yes

Answer:

h=1.122652m

Explanation:

Assuming density of air = 1.2kg/m³

the differential pressure is given by:

[tex]h^{i} =h(\frac{density of manometer}{density of flowing air}-1)\\h^{i} =h(\frac{1000}{1.2}-1)\\ h^{i}=832.33h...(1)\\\\but\\ h^{i} =\frac{change in pressure}{air density*g} \\\\h^{i} =\frac{11*10^3}{1.2*9.81}\\\\h^{i}= 934.42...(2)\\\\equating, \\\\934.42=832.33h\\\\h=1.122652m[/tex]

The distance  h  between the two fluid levels of the water-filled manometer is approximately 1.121 meters.

To determine the distance  h  between the two fluid levels of the water-filled manometer, we need to use the pressure readings from both the pressure gauge and the manometer. Here's how we can approach the problem:

Given:

- Reading on the pressure gauge: [tex]\( P_{gauge} = 11 \, \text{kPa} \)[/tex]- Atmospheric pressure: [tex]\( P_{atm} = 101 \, \text{kPa} \)[/tex]

- Density of water: [tex]\( \rho = 1000 \, \text{kg/m}^3 \)[/tex]

- Acceleration due to gravity: [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]

Step-by-Step Solution:

1. Convert the gauge pressure to absolute pressure:

[tex]\[ P_{absolute} = P_{gauge} + P_{atm} \][/tex]

[tex]\[ P_{absolute} = 11 \, \text{kPa} + 101 \, \text{kPa} = 112 \, \text{kPa} \][/tex]

2. Determine the pressure difference between the air tank and atmospheric pressure:

  The pressure difference [tex](\( \Delta P \))[/tex] that the manometer measures is equal to the gauge pressure:

[tex]\[ \Delta P = P_{absolute} - P_{atm} = 112 \, \text{kPa} - 101 \, \text{kPa} = 11 \, \text{kPa} \][/tex]

3. Use the pressure difference to find the height ( h ):

  The pressure difference is related to the height ( h ) of the water column by the hydrostatic equation:

 [tex]\[ \Delta P = \rho g h \][/tex]

  Solving for ( h ):

[tex]\[ h = \frac{\Delta P}{\rho g} \][/tex]

  Convert [tex]\(\Delta P\)[/tex] to Pascals [tex](since \(1 \, \text{kPa} = 1000 \, \text{Pa}\))[/tex]:

 [tex]\[ \Delta P = 11 \, \text{kPa} = 11000 \, \text{Pa} \][/tex]

  Now, calculate ( h ):

[tex]\[ h = \frac{11000 \, \text{Pa}}{1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2} \][/tex]

[tex]\[ h \approx \frac{11000}{9810} \, \text{m} \][/tex]

[tex]\[ h \approx 1.121 \, \text{m} \][/tex]

Answer:

a. 2.08, b. 1110 kJ/min

Explanation:

The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. Assume that the air conditioner operates steadily.

a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is

COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat

COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08

The COP of this air conditioner is 2.08.

b. The rate of heat discharged to the outside air is determined from the energy balance.

Q(H) = Q(L) + W(net in)

Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min

The rate of heat transfer to the outside air is 1110 kJ for every minute.

a) The COP of the air conditioner is 2.083.

b) Rate of heat transfer to the outside air is 18.5 kW.

Step 1

Given Data:

- Heat removal rate from the house, [tex]\( \dot{Q}_{in} = 750 \, \text{kJ/min} \)[/tex]

- Electric power input, [tex]\( \dot{W}_{in} = 6 \, \text{kW} \)[/tex]

Converting Units:

[tex]\[ \dot{Q}_{in} = 750 \, \text{kJ/min} = \frac{750 \, \text{kJ}}{60 \, \text{s}} = 12.5 \, \text{kW} \][/tex]

Calculations:

(a) Coefficient of Performance (COP):

The COP of an air conditioner is defined as the ratio of the heat removed from the house to the work input (electric power):

[tex]\[ \text{COP} = \frac{\dot{Q}_{in}}{\dot{W}_{in}} \][/tex]

Step 2

Substituting the given values:

[tex]\[ \text{COP} = \frac{12.5 \, \text{kW}}{6 \, \text{kW}} = 2.083 \][/tex]

(b) Rate of Heat Transfer to the Outside Air:

The rate of heat transfer to the outside air, [tex]\( \dot{Q}_{out} \)[/tex], can be determined using the first law of thermodynamics for a steady-state system:

[tex]\[ \dot{Q}_{out} = \dot{Q}_{in} + \dot{W}_{in} \][/tex]

Substituting the given values:

[tex]\[ \dot{Q}_{out} = 12.5 \, \text{kW} + 6 \, \text{kW} = 18.5 \, \text{kW} \][/tex]

Answer: 33.35 minutes

Explanation:

A(t) = A(o) *(.5)^[t/(t1/2)]....equ1

Where

A(t) = geiger count after time t = 100

A(o) = initial geiger count = 400

(t1/2) = the half life of decay

t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

.25= (.5)^[66.7/(t1/2)]

Take log of both sides

Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

(t1/2) = (66.7/2 ) = 33.35 minutes

Answer:

Alternatively we produce a complex (hash) value for key which mean that "to obtain a numerical value for every single string" that we are looking for.  Then compare that values along the range of list, that turns out be numerical values so that comparison becomes faster.

Explanation:

Every individual key is a big character  so to compare every keys, it is required to conduct a quite time consuming string reference procedure at every node. Alternatively we produce a complex (hash) value for key which mean that "to obtain a numerical value for every single string" that we are looking for.  Then compare that values along the range of list, that turns out be numerical values so that comparison becomes faster.

Answer:

Q = 2.532 x 10³ W

Explanation:

The properties of air at the film temperature of

T(f) = (T(s) + T()∞)/2

T(f) = (120 + 20)/2 = 70 °C

From the table of properties of air at T = 20 °C we know that ρ = 1.028 kg/m³ , k = 0.02881W/mK, Pr = 0.7177 , v = 1.995 x 10⁻⁵ m²/s

Calculate the Reynold’s Number

Re(l) = V x L/ν, where V is the speed of air flowing, L is the length of the surface in meters and ν is the kinematic viscosity

Re(l) = 62 x 0.45/(1.995 x 10⁻⁵) = 1.398 x 10⁶

Re(l) is greater than the critical Reynold Number. Thus, we have combined laminar and turbulent flow and the average Nusselt number for the entire plate is determined by

Nu = (0.037 x (Re(l)^0.8) - 871) x Pr^1/3, where Pr is the Prandtl’s Number

Nu = (0.037 x (1.398 x 10⁶)^0.8) – 871) x (0.7177)^1/3

Nu = 1952.85

We also know that

Nu = h x L/k, where h is the heat transfer coefficient, L is the length of the surface and  k is the thermal conductivity

Rearranging to solve for h

h = (Nu x k)/l

h = 1952.85 x 0.02881/0.45 = 125.03 W/m²C

The heat transfer rate Q, may be determined by

Q = h x A x (T(s) - T(∞))

Q = 125.03 x 0.45 x 0.45 x (120 - 20)

Q = 2.532 x 10³ W

The heat transfer rate is 2.532 x 10³ W to the air.

Answer:

Stories that discuss how he dealt with a crisis

Stories that illustrate how he went above and beyond expectations

Stories that discuss how he handled a tough interpersonal situation

Explanation:

For effective story-telling during an interview, Jamar needs to highlight stories by discussing how he dealt with certain crisis which relates to the job and tell the panel why he opted for a particular solution. The solution should provide a long-term solution to the crisis. He also needs to illustrate stories on how he went above and beyond expectations  to prove that he's creative. Moreover, it's important that he highlights how he handled a tough interpersonal situation to show how he deals with situations at a personal level.

Answer:

import java.util.HashMap;

import java.util.Map;

import java.util.Scanner;

public class PhoneBook {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       Map<String, String> map = new HashMap<>();

       String name, number, choice;

       do {

           System.out.print("Enter name: ");

           name = in.next();

           System.out.print("Enter number: ");

           number = in.next();

           map.put(name, number);

           System.out.print("Do you want to try again(y or n): ");

           choice = in.next();

       } while (!choice.equalsIgnoreCase("n"));

       System.out.print("Enter name to search for: ");

       name = in.next();

       if (map.containsKey(name)) {

           System.out.println(map.get(name));

       } else {

           System.out.println(name + " is not in the phone book");

       }

   }

}

Answer:

a) [tex]h_L=-3.331ft[/tex]

b) The flow would be going from section (b) to section (a)

Explanation:

1) Notation

[tex]p_a =31.1psi=4478.4\frac{lb}{ft^2}[/tex]

[tex]p_b =27.3psi=3931.2\frac{lb}{ft^2}[/tex]

For above conversions we use the conversion factor [tex]1psi=144\frac{lb}{ft^2}[/tex]

[tex]z_a =56.7ft[/tex]

[tex]z_a =68.8ft[/tex]

[tex]h_L =?[/tex] head loss from section

2) Formulas and definitions

For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.

The formula is given by:

[tex]\frac{p_a}{\gamma}+\frac{V_a^2}{2g}+z_a =\frac{p_b}{\gamma}+\frac{V_b^2}{2g}+z_b +h_L[/tex]

Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:

[tex]\frac{p_a}{\gamma}+z_a =\frac{p_b}{\gamma}+z_b +h_L[/tex]

3)Part a

And on this case we have all the values in order to replace and solve for [tex]h_L[/tex]

[tex]\frac{4478.4\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+56.7ft=\frac{3931.2\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+68.8ft +h_L[/tex]

[tex]h_L=(71.769+56.7-63-68.8)ft=-3.331ft[/tex]

4)Part b

Analyzing the value obtained for [tex]\h_L[/tex] is a negative value, so on this case this means that the flow would be going from section (b) to section (a).

Answer:

magnitude of the maximum stress is 3263 MPa

Explanation:

given data

radius of curvature = 5.5 × [tex]10^{-4}[/tex] mm

crack length = 5 × [tex]10^{-2}[/tex] mm

tensile stress = 220 MPa

to find out

magnitude of the maximum stress

solution

we know that magnitude of the maximum stress is express as

magnitude of the maximum stress = [tex]2\sigma_o ( \frac{\alpha }{2 \rho} )^{0.5}[/tex]      ..........................1

here σo is tensile stress and α is crack length and ρ is radius of curvature

so put all value in equation 1 we get

magnitude of the maximum stress = [tex]2*220 ( \frac{5.5*10^{-2}}{2 *5*10^{-4}} )^{0.5}[/tex]

solve it we get

magnitude of the maximum stress = 3263 MPa

so magnitude of the maximum stress is 3263 MPa

Answer: Temperature T = 218.399K.

Explanation:

Q= 129W/m2

σ = 5.67 EXP -8W/m2K4

Q= σ x t^4

t = (Q/σ)^0.25

t= 218.399k

Answer:

a. 149.74 KJ/KG

b. 97.9%

c. 0.81 kJ/kg K

Explanation:

Answer:The awnser is 5

Explanation:Just divide all of it

Answer:

q= 1.77 W

Explanation:

Given data, Dimensions = 20mm x 20mm, Unheated starting length (ε) = 20 mm, Air flow temperature = 25 C, Velocity of flow = 25m/s, Surface Temperature = 75 C

To find maximum allowed power dissipated use the formula Q = hA(ΔT), where Q = Maximum allowed power dissipated from surface , h = heat transfer coefficiant, A = Area of Surface, ΔT = Temperature difference between surface and surrounding

we need to find h, which is given by (Nu x K)/x, where Nu is the Nusselt's Number, k is the thermal conductivity at film temperature and x is the length of the substrate.

For Nu use the Churchill and Ozoe relation used for parallel flow over the flat plate , Nu = (Nux (ε=0))/[1-(ε/x)^3/4]^1/3

Nux (ε=0) = 0.453 x Re^0.5 x Pr^1/3, where Re is the Reynolds number calculated by [Rex = Vx/v, where V is the velocity and v is the kinematic velocity, x being the length of the substrate] and Pr being the Prandtl Number

The constants, namel Pr, k and v are temperature dependent, so we need to find the film temperature

Tfilm = (Tsubstrate + Tmax)/2 = (25 + 75)/2 = 50 C

At 50 C, Pr = 0.7228, k = 0.02735w/m.k , v = 1.798 x 10^-5 m2/s

First find Rex and keep using the value in the subsequent formulas until we reach Q

Rex = Vx/v = 25 x (0.02 + 0.02)/ 1.798 x 10^-5  = 55617.35 < 5 x 10^5, thus flow is laminar, (x = L + ε)

Nux = (Nux (ε=0))/[1-(ε/x)^3/4]^1/3 = 0.453 x Re^0.5 x Pr^1/3/[1-(ε/x)^3/4]^1/3

= 0.453 x 55617.35 ^0.5 x 0.7228^1/3/ [1 - (0.02/0.04)^3/4]^1/3 = 129.54

h = (Nu x K)/x = (129.54 x 0.02735)/0.04 = 88.75W/m2.k

Q = 88.75 x (0.02)^2 x (75-25) = 1.77 W

Answer:

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes

Explanation:

In order to find the actual heat transfer rate is lower or higher than its value we will first find the rate of heat transfer to power plant:

[tex]Efficiency=\frac{work}{heat transfer to power plant}[/tex]

[tex]Heat transfer=\frac{work}{Efficiency\\} \\\\Heat transfer=\frac{800}{0.40}\\\\Heat transfer=2000MW[/tex]

From First law of thermodynamics:

Rate of heat transfer to river=heat transfer to power plant-work done

Rate of heat transfer to river=2000-800

Rate of heat transfer to river=1200MW

So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes.

Answer:

\epsilon = 0.028*0.3 = 0.0084

Explanation:

\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2

where P_1 = P_2 = 0

V1 AND V2  =0

Z1 =0

h_P = \frac{w_p}{\rho Q}

=\frac{40}{9.8*10^3*0.2} = 20.4 m

20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10

we know thaTV  =\frac{Q}{A}

V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec

20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10

f  = 0.0560

Re =\frac{\rho v D}{\mu}

Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5

fro Re = 7.53*10^5 and f = 0.0560

\frac{\epsilon}{D] = 0.028

\epsilon = 0.028*0.3 = 0.0084

Answer:

Explanation:

given data

loading rate = 8.00 m/h

filtration rate = 7.70 m/h

dimensions = 10 m × 8 m

filter cycle duration = 52 h

time = 20 min

to find out

flow rate  and  volume of water is used for back washing plus rinsing the filter  

solution  

we consider here production efficiency is 96%

so here flow rate will be  

flow rate = area × rate of filtration  

flow rate = 10 × 8 × 7.7  

flow rate = 616 m³/h

and  

we know back washing generally 3 to 5 % of total volume of water per cycle so  

volume of water is = 616 × 52

volume of water is  32032 m³

and  

volume of water of back washing is = 4% of 32032  

volume of water of back washing is 1281.2 m³

Answer:

a) [tex]-7.4\frac{lb}{ft^3}[/tex]

b) [tex]-69.8\frac{lb}{ft^3}[/tex]

c) [tex]55 \frac{lb}{ft^3}[/tex]

Explanation:

1) Notation

[tex]\tau=1.85\frac{lb}{ft^2}[/tex] represent the shear stress  defined as "the external force acting on an object or surface parallel to the slope or plane in which it lies"

R represent the radial distance

L the longitude

[tex]\theta=0\degree[/tex] since at the begin we have a horizontal pipe, but for parts b and c the angle would change.

D represent the diameter for the pipe

[tex]\gamma=62.4\frac{lb}{ft^3}[/tex] is the specific weight for the water  

2) Part a

For this case we can use the shear stress and the radial distance to find the pressure difference per unit of lenght, with the following formula

[tex]\frac{2\tau}{r}=\frac{\Delta p -\gamma Lsin\theta}{L}[/tex]

[tex]\frac{2\tau}{r}=\frac{\Delta p}{L}-\gamma sin\theta[/tex]

If we convert the difference's into differentials we have this:

[tex]-\frac{dp}{dx}=\frac{2\tau}{r}+\gamma sin\theta[/tex]

We can replace [tex]r=\frac{D}{2}[/tex] and we have this:

[tex]\frac{dp}{dx}=-[\frac{4\tau}{D}+\gamma sin\theta][/tex]

Replacing the values given we have:

[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin0]=-7.4\frac{lb}{ft^3}[/tex]

3) Part b

When the pipe is on vertical upward position the new angle would be [tex]\theta=\pi/2[/tex], and replacing into the formula we got this:

[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin90]=-69.8\frac{lb}{ft^3}[/tex]

4) Part c

When the pipe is on vertical downward position the new angle would be [tex]\theta=-\pi/2[/tex], and replacing into the formula we got this:

[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin(-90)]=55 \frac{lb}{ft^3}[/tex]

Answer:

[tex]\dfrac{Q_B}{Q_A}=\sqrt{5}[/tex]

Explanation:

Lets take

Length of pipe B = L

Length of pipe A = 5 L

Discharge in pipe A = Q₁

Discharge in pipe B = Q₂

We know that head loss in the pipe given as

[tex]h_f=\dfrac{FLQ^2}{12.1d^5}[/tex]

F=Friction factor, Q=Discharge,L=length

d=Diameter of pipe

here all only Q and L is varying and all other quantity is constant

So we can say that

LQ²= Constant

L₁Q₁²=L₂Q²₂

By putting the values

5LQ₁²=LQ²₂

[tex]\dfrac{Q_2}{Q_1}=\sqrt{5}[/tex]

Therefore

[tex]\dfrac{Q_B}{Q_A}=\sqrt{5}[/tex]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is    [tex]= 1.751 \times 10^{-5} \% per hr[/tex]

Explanation:

we know Arrhenius expression is given as

[tex]\dot \epsilon =Ce^{\frac{-Q}{RT}[/tex]

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is[tex] \dot \epsilon = 5.5\times 10^{-2} [/tex]% per hr

At 800 degree C  creep rate is[tex] \dot \epsilon = 1 [/tex]% per hr

activation energy for creep is [tex]\frac{\epsilon_{800}}{\epsilon_{700}}[/tex] = [tex]= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}[/tex]

[tex]\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}[/tex]

[tex]\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}][/tex]

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

[tex]C = \epsilon e^{\frac{Q}{RT}}[/tex]

    [tex]=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr[/tex]

[tex]\epsilon_{500} = C e^{\frac{Q}{RT}}[/tex]

                         [tex]= 1.804 \times 10^{12}  e{\frac{251758}{8.314(500+273}}[/tex]

                         [tex]= 1.751 \times 10^{-5} \% per hr[/tex]

The question pertains to the use of functions and pointers in programming. The task involves writing a function call that manipulates the tens and ones digit of an integer by passing the first two arguments as pointers. An example solution involves creating a function named ExtractDigits to perform this task and using pointers to directly modify specified variables.

The question refers to the concept of functions and pointers in programming, which is an advanced topic typically covered at the college level in computer science or engineering courses. The task is to write a function call that uses pointers for the first two arguments, tensPlace and onesPlace, and a regular argument for userInt. This function could be crafted to manipulate or assess the tens and ones place of a given integer, userInt.

Suppose we have a function named ExtractDigits designed to extract the tens and ones place of an integer. This function might look like:

void ExtractDigits(int* tensPlace, int* onesPlace, int userInt){
 *tensPlace = (userInt / 10) % 10;
 *onesPlace = userInt % 10;
}

To call this function with the variables tensPlace, onesPlace, and a specific integer (for example, 41), you can use the following code:

int main() {
 int tens, ones;
 ExtractDigits(&tens, &ones, 41);
 printf("tensPlace = %d, onesPlace = %d", tens, ones);
 return 0;
}

This function call passes the addresses of tens and ones to the function so that their values can be modified directly, which is a fundamental use of pointers in C and C++ programming.

Answer:

L=0.0074 mm

Explanation:

Given that

d= 6.5 micron meter

d= 6.5 x 10⁻³ mm

Yield strength ,Sy = 73 MPa

Bond strength ,σ = 32 MPa

The fiber length given as

[tex]L=\dfrac{S_y \times d}{2 \times \sigma }[/tex]

By putting the values

[tex]L=\dfrac{S_y \times d}{2 \times \sigma }[/tex]

[tex]L=\dfrac{73 \times 6.5\times 10^{-3}}{2\times 32 }\ mm[/tex]

L=0.0074 mm

Therefore the minimum fiber length is 0.0074 mm.

Answer:

Please see explanation below .

Explanation:

Since you have not provided the db schema, I am providing the queries by guessing the table and column names. Feel free to reach out in case of any concerns.

1. SELECT ISBN, TITLE, PUBLICATION_YEAR FROM BOOKS WHERE AUTHOR_ID = (SELECT AUTHOR_ID FROM AUTHOR WHERE AUTHOR_NAME = 'C.J. Cherryh');

2. SELECT COUNT(*) FROM BOOKS;

3. SELECT COUNT(*) FROM ORDERS WHERE CUSTOMER_ID = 11 AND order_filled = 'No';

Answer:

6.4 m/s

Explanation:

From the equation of continuity

A1V1=A2V2 where A1 and V1 are area and velocity of inlet respectively while A2 and V2 are the area and velocity of outlet respectively

[tex]A1=\pi (r1)^{2}[/tex]

[tex]A2=\pi (r2)^{2}[/tex]

where r1 and r2 are radius of inlet and outlet respectively

v1 is given as 1.6 m/s

Therefore

[tex]\pi (0.01)^{2}\times 1.6 = \pi (0.005)^{2}v2[/tex]

[tex]V2=\frac {\pi (0.01)^{2}\times 1.6}{\pi (0.005)^{2}}=6.4 m/s[/tex]