Neglecting the presence of friction, air drag, and other inefficiencies, how much gasoline is consumed when a 1300 kg automobile accelerates from rest to 80 km/h. Assume the density and enthalpy of gasoline are 680 kg/m3 and 45 MJ/kg respectively. Express your answer in the units of mL. Show all work and use SI units for full credit. Box your final answer(s)

Answer:

From the load equation

F=stress*Area

Given stresses are 8 kips and 9 kips.

Hence the minimum weight supported=6.695 lbs.

Answer:

ASD = 306 kips-feet

LRSD = 1387.5 kips-feet

Explanation:

a step by step process to solving this problem.

ΣM at A = 0

where;

RB * 35 - (8+18)15 - (4+9)20 = 0

RB = 18.57k

also E y = 0;

RA + RB = 18 + 8 + 9 +4 = 20.43 k

taking the maximum moment at mid point;

Mc = RA * 35/2 - (8 +18) (35/2 -15)

Mc = 292.525

therefore, MD = RA * 15 = 20.43 * 15 = 306.45 kips-feet

MD = 306.45 kip-feet

ME = 279 kip-feet .IE 18.57 * 15

considering the unsupported  length; 35 - (15*2 = 5ft )

now we have that;

L b = L p = 5ft

where L p = 1.76 r y(√e/f y)

L p = 1.76 r y √29000/50

r y = 1.4 inch

so we have that M r = M p  for L b = L p where

M p = 2 F y ≤ 1.5 s x   F y

Recall from the expression,

RA + RB = (8+4) * 1.2 + (18+9) * 1.6 = 57.6

RA * 35 = 4 * 1.2 * 15 + 9 *1.6 * 15 + 8 * 1.2 * 20 + 18 * 1.6 * 20

RA = 30.17 k

the maximum moment at D = 30.17 * 15 = 452.55 kips-feet

Z required = MD / F y = 452.55 * 12 / 50 = 108.61 inch³

so we have S x = 452.55 * 12 / 1.5 * 50 = 72.4 inch³

also r = 1.41 in

Taking LRFD solution:

where the design strength ∅ M n = 0.9 * Z x * F y

given r = 2.97

Z x = 370 and S x = 81.5, we have

∅ M n = 0.9 * 370 * 50 = 16650 k-inch = 1387.5 kips-feet

this tells us it is safe.

ASD solution:

for L b = L p, and where M n = M p = F c r    S x

we already have value for S x as 81.5 so

F c r = Z x times F y divide by  S x

F c r = 370 * 50 / 81.5 = 227 kips per sq.in

considering the strength;

Strength = M n / Ωb = (0.6 * 81.5 * 50) * (1.5) / 12 = 306 kips-feet  

This justifies that it is safe because is less than 306

Answer:

15.53 m2

Explanation:

Energy needed = 23.8 kW-hr

Solar intensity of required panel = 336 W/m2

Efficiency of panels = 19%

Power needed = 23.8kW-hr = 23800 W-h,

In one day there are 24 hr, therefore power required = 23800/24 = 991.67 W

991.67 = 19% of incident power

991.67 = 0.19x

x = incident power = 991.67/0.19 = 5219.32 W

Surface area required = 5219.32/336

= 15.53 m2

Answer:

#Data section

.data

#Set align

.align 2

#Declare row 1

M1: .word 1, 2, 3

#Declare row 2 elements

M2: .word 4, 5, 6

#Declare row 3 elements

M3: .word 7, 8, 9

#Declare row 4 elements

M4: .word 10, 11, 12

#Declare row 5 elements

M5: .word 13, 14, 15

#Create 5x3 array

M: .word M1, M2, M3, M4, M5

#Set number of rows

nRows: .word 5

#Set number of columns

nCols: .word 3

#Declare string menu

menu: .asciiz "\n The following are the choices:\n"

#Declare string for option 1

op1: .asciiz "1. Replace a value:\n"

#Define string for option 2

op2: .asciiz "2. Calculate the sum of all values\n"

#Define string for option 3

op3: .asciiz "3. Print out the 2D array \n"

#Define string for option 4

op4: .asciiz "4. Exit\n"

#Define string for prompt

urOpt: .asciiz "Enter your option:"

#Define string for row

prompt1: .asciiz "Enter row (1-5):"

#Define string for column input

prompt2: .asciiz "Enter column (1-3):"

#Define string to get the replace value

getVal: .asciiz "Enter the new value:"

#Define strinct to display sum

sumStr: .asciiz "\nThe sum is :"

#Define string

dispStr: .asciiz "The 2D array is:\n"

#Define string to print new line

newLine: .asciiz "\n"

#Define string to put comma

comma: .asciiz ","

#text section

.text

#Main

main:

#Block prints the 2D array

#label

print2DArray:

#Load integer to print string in $v0

li $v0, 4

#Load the address of string to display

la $a0, dispStr

#Display the string

syscall

#Load the base address of the row 1

la $s0, M1

#Load the nRows

lw $t0, nRows

#Initialize the counter for outer loop

li $t7, 0

#Outer loop begins

outLoop1:

#Check condition

beq $t7, $t0, displayMenu

#Load the integer to print string

li $v0, 4

#Load the base address of newLine

la $a0, newLine

#Display newLine

syscall

#Initialize counter for inner loop

li $t2, 0

#Set the limit

lw $t3, nCols

#Decrement value

addi $t3, $t3, -1

#inner loop starts

inLoop1:

#Check condition

beq $t2, $t3, exitInLoop1

#Load the integer to print number

li $v0, 1

#Load the value

lw $a0, ($s0)

#Display the integer

syscall

#Load the integer to print string

li $v0, 4

#Load the base address of the string comma to display

la $a0, comma

#Display comma

syscall

#Increment inner loop counter value

addi $t2, $t2, 1

#Move to next element

addi $s0, $s0, 4

#jump to inner loop

j inLoop1

#Exit from inner loop

exitInLoop1:

#Load integer to print last column value

li $v0, 1

#Load the load column value

lw $a0, ($s0)

#Print value

syscall

#Move to next element

addi $s0, $s0, 4

#Increment outer loop counter

addi $t7, $t7, 1

#Jump to start of the outer loop

j outLoop1

#Prints the menus to the user

displayMenu:

#Load integer value to print string

li $v0, 4

#Load the address of the string "menu"

la $a0, menu

#Print the string

syscall

#Load the integer value to print string

li $v0, 4

#Load the address

la $a0, op1

#Print the string

syscall

#Load the integer value to print string

li $v0, 4

#Load the address

la $a0, op2

#Print string

syscall

#Load integer value to print string

li $v0, 4

#Load address of op3

la $a0, op3

#Print string

syscall

#Load integer value to print string

li $v0, 4

#Load the address

la $a0, op4

#Print string

syscall

#Load integer value to print sting

li $v0, 4

#Load address

la $a0, urOpt

#Print string

syscall

#Load the integer to read int value

li $v0, 5

#Read value

syscall

#Move value

move $t0, $v0

#Initialize values

li $t1, 1

li $t2, 2

li $t3, 3

li $t4, 4

#Check user wishes option 1

beq $t0, $t1, replaceBlock

#Check user wishes option 2

beq $t0, $t2, sumBlock

#Check user wishes option 3

beq $t0, $t3, print2DArray

#Check user wishes to exit

beq $t0, $t4, EndProgram

#Jump to start of the menu

j displayMenu

#Block to replace a value in the 2d array

replaceBlock:

#Load integer

li $v0, 4

#load address

la $a0, prompt1

#Print string

syscall

#Read row value

li $v0, 5

syscall

#Store the row value in $t6

move $t6, $v0

#Get the index

addi $t6, $t6, -1

#Load integer

li $v0, 4

#Load address

la $a0, prompt2

#Print string

syscall

#Read column value

li $v0, 5

syscall

#Store the column value in $t7

move $t7, $v0

#Get the index for the column

addi $t7, $t7, -1

#Load integer

li $v0, 4

#Load address

la $a0, getVal

#Print string

syscall

#Read the new value

li $v0, 5

syscall

#Store the new value in $t5

move $t5, $v0

#Load the base address of M

la $s0, M

#Get M[i]

#two times left shift

sll $t1, $t6, 2

#address of pointer M[i]

add $t1, $t1, $s0

#Get address of M[i]

lw $t3, ($t1)

#Get M[i][j]

#two times left shift

sll $t4, $t7, 2

#Get address of M[i][j]

add $t4, $t3, $t4

Explanation:

See continuation of the code attached

also, See output attached

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

1.443hrs

Explanation:

Please kindly check attachment for the detailed and step by step solution to the problem.

Answer:

c. can destablize the slopes by increasing pore fluid pressure.

Explanation: A reservoir is a term used in Agriculture or geography to describe a location where water is either collected artificially or naturally for later use in farming or to act as dam for the supply of water in large communities. Most countries of the world make use of reservoirs to store water for future use.

Answer:

a) k = 6.4 s/cm²

b) Te = 160 s

Explanation:

a) Given

Solidification time in s: Te = 160 s

L = 50 mm = 5 cm

We use the formula

Te = k*(V/A)ⁿ  

k = Te*(A/V)ⁿ

where

k is the Chvorinov's mold constant

V is casting volume in cm³ = V = L³ = (5 cm)³ = 125 cm³

A is the surface area of the casting in cm² = A = L² = (5 cm)² = 25 cm²

n = 2 (assumed)

⇒  k = 160 s*(25 cm²/125 cm³)²

⇒  k = 6.4 s/cm²

b) Given

D = 25 mm = 2.5 cm  ⇒  R = D/2 = 2.5 cm/2 = 1.25 cm

h = 50 mm = 5 cm

k = 6.4 s/cm²

n = 2

We find the volume as follows

V = π*R²*h ⇒   V = π*(1.25 cm)²*(5 cm) = 24.5436 cm³

and the surface area

A = π*R² = π*(1.25 cm)² = 4.9087 cm²

We apply the equation

Te = k*(V/A)ⁿ  

⇒  Te = (6.4 s/cm²)*(24.5436 cm³/4.9087 cm²)²

⇒  Te = 160 s

Answer:

d/dt[mCp(Ts-Ti)] =  FCp(Ts-Ti) -  FoCp(Ts-Ti) + uA(Ts-Ti)

Explanation:

Differential balance equation on the tank is given as;

Accumulation = d/dt[mcp(Ts-Ti)]

Energy of inlet steam = FCp(Ts-Ti)

Energy of outlet steam =  FoCp(Ts-Ti)

Heat transfer from the steam = uA(Ts-Ti)

Substituting into the formula, we have;

The differential energy balance is [tex]\(\frac{dT}{dt} = \frac{11.5 T_s + 690 - 39.1 T}{1748}\).[/tex]

To write a differential energy balance on the tank contents, we need to consider the energy entering and leaving the system. The system in question is the well-stirred heating tank. Here are the steps to formulate the energy balance:

1. Define the system and parameters:

  - The solvent enters the tank at a flow rate of 12.0 kg/min and at a temperature of 25°C.

  - The solvent has a heat capacity  [tex]\( C_p = 2.30 \text{ kJ/kg°C} \).[/tex]

  - The tank is initially filled with 760 kg of solvent at 25°C.

  - Heat transfer rate from the steam coil to the solvent is given by

[tex]\( UA(T_s - T) \)[/tex]  where  [tex]\( UA = 11.5 \text{ kJ/min·°C} \), \( T_s \)[/tex]   is the steam temperature, and ( T ) is the solvent temperature.

  - The tank is well-stirred, ensuring uniform temperature throughout, and the outlet temperature equals the tank temperature.

2. Energy balance:

  The energy balance for a well-stirred tank in differential form is given by:

[tex]\[ \frac{d(U)}{dt} = \dot{Q}_{\text{in}} + \dot{m} C_p T_{\text{in}} - \dot{m} C_p T_{\text{out}} \][/tex]

  Where:

  - ( U ) is the internal energy of the tank contents.

[tex]- \( \dot{Q}_{\text{in}} \)[/tex]  is the heat transfer rate from the steam coil.

[tex]- \( \dot{m} \)[/tex]  is the mass flow rate of the solvent.

[tex]- \( T_{\text{in}} \)[/tex] is the inlet temperature of the solvent.

[tex]- \( T_{\text{out}} \)[/tex] of the solvent (equal to tank temperature ( T ).

  Since  [tex]\( \dot{m}_{\text{in}} = \dot{m}_{\text{out}} = \dot{m} \) and \( T_{\text{out}} = T \):[/tex]

 [tex]\[ \frac{d(U)}{dt} = UA(T_s - T) + \dot{m} C_p (T_{\text{in}} - T) \][/tex]

3. Internal energy change:

  The internal energy change of the tank contents can be expressed as:

  [tex]\[ \frac{d(U)}{dt} = m C_p \frac{dT}{dt} \][/tex]

  Where ( m ) is the mass of the solvent in the tank (760 kg) and ( T ) is the solvent temperature in the tank.

4. Combine the equations:

  Substituting  [tex]\( \frac{d(U)}{dt} = m C_p \frac{dT}{dt} \)[/tex] into the energy balance:

[tex]\[ m C_p \frac{dT}{dt} = UA(T_s - T) + \dot{m} C_p (T_{\text{in}} - T) \][/tex]

5. Simplify the equation:

  Substitute the given values:

[tex]- \( m = 760 \text{ kg} \)\\ - \( C_p = 2.30 \text{ kJ/kg°C} \)\\ - \( \dot{m} = 12.0 \text{ kg/min} \) - \( UA = 11.5 \text{ kJ/min·°C} \)\\ - \( T_{\text{in}} = 25 \text{ °C} \)[/tex]

[tex]\[ 760 \cdot 2.30 \frac{dT}{dt} = 11.5 (T_s - T) + 12.0 \cdot 2.30 (25 - T) \][/tex]

  Simplify to:

  [tex]\frac{dT}{dt} = 11.5 (T_s - T) + 27.6 (25 - T) \]\[ 1748[/tex]

  Further simplify:

[tex]\[ 1748 \frac{dT}{dt} = 11.5 T_s - 11.5 T + 690 - 27.6 T \][/tex]

  Combine like terms:

 [tex]\[ 1748 \frac{dT}{dt} = 11.5 T_s + 690 - 39.1 T \][/tex]

  Finally:

[tex]\[ \frac{dT}{dt} = \frac{11.5 T_s + 690 - 39.1 T}{1748} \][/tex]

This is the differential energy balance equation for the tank contents.

Answer:

175.5 mm

Explanation:

a hollow shaft of diameter ratio 3/8 is required to transmit 600kw at 110 rpm, the maximum torque being 20% greater than the mean. the shear stress is not to exceed 63 mpa and twist in a length of 3 meters not t exeed 1.4 degrees Calculate the minimum external diameter satisfying these conditions. G = 80 GPa

Let

D = external diameter of shaft

Given that:

d = internal diameter of the shaft = 3/8 × D = 0.375D,

Power (P) = 600 Kw, Speed (N) = 110 rpm, Shear stress (τ) = 63 MPa = 63 × 10⁶ Pa, Angle of twist (θ) = 1.4⁰, length (l) = 3 m, G = 80 GPa = 80 × 10⁹ Pa

The torque (T) is given by the equation:

[tex]T=\frac{60 *P}{2\pi N}\\ Substituting:\\T=\frac{60*600*10^3}{2\pi*110} =52087Nm[/tex]

The maximum torque ([tex]T_{max[/tex]) = 1.2T = 1.2 × 52087 =62504 Nm

Using Torsion equation:

[tex]\frac{T}{J} =\frac{\tau}{R}\\ J=\frac{T.R}{\tau} \\\frac{\pi}{32}[D^4-(0.375D)^4]=\frac{62504*D}{2(63*10^6)} \\D^3(0.9473)=0.00505\\D=0.1727m=172.7mm[/tex]

[tex]\theta=1.4^0=\frac{1.4*\pi}{180}rad[/tex]

From the torsion equation:

[tex]\frac{T}{J} =\frac{G\theta }{l}\\ J=\frac{T.l}{G\theta} \\\frac{\pi}{32}[D^4-(0.375D)^4]=\frac{62504*3}{84*10^9*\frac{1.4*\pi}{180} } \\D=0.1755m=175.5mm[/tex]

The conditions would be satisfied if the external diameter is 175.5 mm

Answer:

Explanation:

r=72.3 is my thought

I hope it is helpful

Answer:

radius = 9.1 × [tex]10^{-3}[/tex] m

Explanation:

given data

applied load = 5560 N

flexural strength = 105 MPa

separation between the support =  45 mm

solution

we apply here minimum radius formula that is

radius = [tex]\sqrt[3]{\frac{FL}{\sigma \pi}}[/tex]      .................1

here F is applied load and  is length

put here value and we get

radius =  [tex]\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}[/tex]  

solve it we get

radius = 9.1 × [tex]10^{-3}[/tex] m

Answer:

See Explaination

Explanation:

#include <stdio.h>

int main(void)

{

const double G = 6.673e-11;

const double M = 5.98e24;

double accelGravity = 0.0;

double distCenter = 0.0;

distCenter = 6.38e6;

//<StudentCode>

accelGravity = (G * M) / (distCenter * distCenter);

printf("accelGravity: %lf\n", accelGravity);

return 0;

}

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = [tex] c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}[/tex]

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

[tex] -0.40= 1.005(ln T_2 - 5.68697)- 0.5968[/tex]

Solving for T2 we have:

[tex] T_2 = 5.8828[/tex]

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

[tex] w_in = c_p(T_2 - T_1)+q_out[/tex]

[tex] w_in = 1.005(358.8 - 295)+120[/tex]

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = [tex] \frac{q_out}{T_1}[/tex]

[tex] \frac{120kJ/kg.k}{295K}[/tex]

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

Consider the expression for the change in entropy of the air.  

[tex]\to \Delta S_{air}=c_p \ \In \frac{T_2}{T_1} - R \In \frac{P_2}{P_1} \\\\[/tex]

Here, change in entropy of air in a compressor is[tex]\Delta S_{air}[/tex], specific heat at constant pressure is [tex]c_p[/tex], the inlet temperature is [tex]T_1[/tex], outlet temperature is [tex]T_2[/tex], the gas constant is R, inlet pressure is [tex]P_1[/tex], and outlet pressure is[tex]P_2[/tex].  

From the ideal gas specific heats of various common gases table, select the specific heat at constant pressure[tex]c_p[/tex] and gas constant (R) at air and temperature:

[tex]\to 22\ C \ \ or \ \ 295\ K\ \ as 1.005 \ \frac{kJ}{kg \cdot K} \ \ and \ \ 0.287\ \frac{kJ}{kg \cdot K}[/tex]

Substituting

Take exponential on both sides of the equation.  

[tex]\to T_2 = exp(5.8828) = 358.8\ K \\\\[/tex]

Hence, the exit temperature of the air is [tex]358.8\ K[/tex]. Apply the energy balance to calculate the work input.  

[tex]\to W_{in}=c_p(T_2-T_1 )+q_{out}[/tex]

Here, work input is [tex]w_{in}[/tex] initial temperature is [tex]T_1[/tex], exit temperature is [tex]T_2[/tex], and heat transfer outlet is [tex]q_{out}[/tex] 

Substituting

Hence, the work input to the compressor is

Express the entropy generated in the process.  

[tex]\to S_{gen} =\Delta S_{air}+\Delta S_{swr}[/tex]

Here, entropy generated is [tex]S_{gen}[/tex], change in entropy of air in a compressor is [tex]\Delta S_{air}[/tex], and change in entropy in the surrounding is [tex]\Delta S_{swr}[/tex].  

Finding the changes into the surrounded entropy.  

[tex]\to \Delta S_{swr} = \frac{q_{out}}{T_{swr}}[/tex]

Here, the heat transfer outlet is [tex]q_{out}[/tex] and the surrounded temperature is [tex]T_{swr}[/tex].  

Substituting

[tex]120\ \frac{kJ}kg } \ for\ q_{out} \ and\ 22\ C \ for\ T_{swr}[/tex]  

[tex]\Delta S_{swr} = \frac{ 120\frac{kJ}{kg}}{(22+273)\ K} =0.4068 \frac{kJ}{kg\cdot K}[/tex]

Finding the entropy generated process.  

[tex]S_{gen} = \Delta S_{air} +\Delta S_{swr}\\[/tex]

Substituting

[tex]-0.40 \frac{kJ}kg\cdot K} \ for \ \Delta S_{air},\ and\ 0.4068 \frac{kJ}{kg\cdot K}\ for\ \Delta S_{swr}\\\\[/tex]

[tex]S_{gen}=-0.40 \frac{kJ}{kg\cdot K} +0.4068 \frac{kJ}{kg\cdot K} = 0.0068 \frac{kJ}{kg\cdot K}[/tex]

Therefore, the entropy generated in the process is [tex]0.0068\ \frac{kJ}{kgK}[/tex].

Learn more:

brainly.com/question/16392696

Answer:

1. Maximum flow = 0.7768 L/s

2. The flow would reduced if other faucets were open. This is due to increase pipe flow and frictional resistance between the water main and the faucets.

Explanation:

See the attached file for the calculation.

Answer:

Explanation:

Mass of the reel is given as,

M = 30kg

Radius of gyration about A is given as,

Ka = 120mm = 0.12m

Mass of the cylinder suspended

Mc = 40kg

Then, weight of the cylinder is

W = mg = Mc × g

W = 40 × 9.81 = 392.4 N

The exert force

P = 300N

Radius of the reel

R = 150mm = 0.15m

r = 75mm = 0.075m

Check attached for diagram of this problem

A. Moment of inertial of the reel?

We will assume the reel is a thin hoop shape, so we will use the thin hoop shape formula

I = M•Ka²

I = 30 × 0.12²

I = 0.432 kgm²

Therefore, Moment of inertia of reel is 0.432 kg.m²

B. Velocity if the cylinder after it moves 2m?

Applying torque equation

Στ = Iα

Where

α is angular acceleration

I is moment of inertia

τ is the torque..

Where τ = F × r

The two force acting on the reel is the weight of the cylinder and it is acting downward and the force applied

Then, taking torque about point A

Στ = Iα.

P × R — W × r = Iα

300 × 0.15 - 392.4 × 0.075 = 0.432 × α

45 — 29.43= 0.432α

15.57 = 0.432α

α = 15.57 / 0.432

α = 36.042 rad/s²

Now, The angle turned by the reel when the cylinder moves 2m above can be determined using

S= rθ

Then, θ = S/r

θ = 2 / 0.075

θ = 26.667 rad

Initially, the reel is at rest, then, the initial angular velocity is 0

ωo = 0 rad/s

Now, apply the kinematic equation and calculate the final angular velocity of the reel,

ω² = ωo² + 2αθ

ω² = 0² + 2 × 36.042 × 26.667

ω² = 0 + 1922.24

ω = √1922.24

ω = 43.84 rad/s

Now, using the relationship between linear velocity and angular velocity

Then, the velocity of cylinder

V = rω

V = 0.075 × 43.84

V = 3.2883 m/s

Therefore, Velocity of Cylinder is 3.288 m/s.

C. Time taken to travel 2m

From equation of circular motion

ω = ωo + αt

43.84 = 0 + 36.042t

43.84 = 36.042t

t = 43.84 / 36.042

t = 1.216 seconds

The time taken to travel 2m is 1.216 seconds

Answer:

Explanation:

Arbitrary means That no restrictions where placed on the number rather still each number is finite and has finite length. For the answer to the question--

Find(A,n,i)

for j =0 to 10000 do

frequency[j]=0

for j=1 to n do

frequency[A[j]]= frequency[A[j]]+1

for j =1 to n do

if i>=A[j] then

if (i-A[j])!=A[j] and frequency[i-A[j]]>0 then

return true

else if (i-A[j])==A[j] and frequency[j-A[j]]>1 then

return true

else

if (A[j]-i)!=A[j] and frequency[A[j]-i]>0 then

return true

else if (A[j]-i)==A[j] and frequency[A[j]-i]>1 then

return true

return false

Answer:

Beta values can be from the equation=change in Vcc/nominal Vcc

Beta=16-3/15=3/15=1/5=0.20

Maximum power=I^2*R=40 W

Answer:

Please see the attached file for the complete answer.

Explanation:

Answer:

(a) 3.455

(b) 21.143

(c) 16.36L/min

Explanation:

In this question, we’d be providing solution to the working process of a refrigerator given the data in the question.

Please check attachment for complete solution and step by step explanation

Answer:

18.6h

Explanation:

To solve this Duck's second law in form of Diffusion will be used.

Also note that since the temperature is constant D (change) will also be constant.

Please go through the attached files for further explanation and how the answer Is gotten.

Answer:

Explanation:

Step by step solution is found in the attachment.

Answer:

See the attached file for the answer.

Explanation:

See the attached file for the explanation

Answer:

a) The final equilibrium temperature is 83.23°F

b) The entropy production within the system is 1.9 Btu/°R

Explanation:

See attached workings

a) Equilibrium temp. ≈ 77.01°F.

b) Total entropy change ≈ 104.58 Btu/°R.

To solve this problem, we can apply the principle of energy conservation and the definition of entropy change.

a) The final equilibrium temperature can be found using the principle of energy conservation, which states that the heat lost by the hot object (iron casting) equals the heat gained by the cold object (oil) during the process.

The equation for energy conservation is:

[tex]\[ m_{\text{iron}} \times C_{\text{iron}} \times (T_{\text{final}} - T_{\text{initial, iron}}) = m_{\text{oil}} \times C_{\text{oil}} \times (T_{\text{final}} - T_{\text{initial, oil}}) \][/tex]

Where:

- [tex]\( m_{\text{iron}} \)[/tex] = mass of iron casting = 50 lbm

- [tex]\( C_{\text{iron}} \)[/tex] = specific heat of iron casting = 0.10 Btu/lbm °R

- [tex]\( T_{\text{initial, iron}} \)[/tex] = initial temperature of iron casting = 700 °F

- [tex]\( m_{\text{oil}} \)[/tex] = mass of oil = 2121 lbm

- [tex]\( C_{\text{oil}} \)[/tex] = specific heat of oil = 0.45 Btu/lbm °R

- [tex]\( T_{\text{initial, oil}} \)[/tex] = initial temperature of oil = 80 °F

- [tex]\( T_{\text{final}} \)[/tex] = final equilibrium temperature (unknown)

Now, let's solve for [tex]\( T_{\text{final}} \)[/tex]:

[tex]\[ 50 \times 0.10 \times (T_{\text{final}} - 700) = 2121 \times 0.45 \times (T_{\text{final}} - 80) \][/tex]

[tex]\[ 5(T_{\text{final}} - 700) = 954.45(T_{\text{final}} - 80) \][/tex]

[tex]\[ 5T_{\text{final}} - 3500 = 954.45T_{\text{final}} - 76356 \][/tex]

[tex]\[ 0 = 949.45T_{\text{final}} - 72856 \][/tex]

[tex]\[ T_{\text{final}} = \frac{72856}{949.45} \][/tex]

[tex]\[ T_{\text{final}} \approx 77.01 \, ^\circ F \][/tex]

So, the final equilibrium temperature is approximately [tex]\( 77.01 \, ^\circ F \).[/tex]

b) The total entropy change for the process can be calculated using the formula:

[tex]\[ \Delta S = \Delta S_{\text{iron}} + \Delta S_{\text{oil}} \][/tex]

Where:

- [tex]\( \Delta S_{\text{iron}} = \frac{Q_{\text{iron}}}{T_{\text{initial, iron}}} \)[/tex]

- [tex]\( \Delta S_{\text{oil}} = \frac{Q_{\text{oil}}}{T_{\text{initial, oil}}} \)[/tex]

- [tex]\( Q_{\text{iron}} \) = heat lost by the iron casting[/tex]

- [tex]\( Q_{\text{oil}} \) = heat gained by the oil[/tex]

Let's calculate:

[tex]\[ Q_{\text{iron}} = m_{\text{iron}} \times C_{\text{iron}} \times (T_{\text{final}} - T_{\text{initial, iron}}) \][/tex]

[tex]\[ Q_{\text{iron}} = 50 \times 0.10 \times (77.01 - 700) \][/tex]

[tex]\[ Q_{\text{iron}} \approx -3175.495 \, \text{Btu} \][/tex]

[tex]\[ Q_{\text{oil}} = m_{\text{oil}} \times C_{\text{oil}} \times (T_{\text{final}} - T_{\text{initial, oil}}) \][/tex]

[tex]\[ Q_{\text{oil}} = 2121 \times 0.45 \times (77.01 - 80) \][/tex]

[tex]\[ Q_{\text{oil}} \approx 8729.535 \, \text{Btu} \][/tex]

Now, calculate entropy changes:

[tex]\[ \Delta S_{\text{iron}} = \frac{-3175.495}{700} \][/tex]

[tex]\[ \Delta S_{\text{iron}} \approx -4.5364 \, \text{Btu/°R} \][/tex]

[tex]\[ \Delta S_{\text{oil}} = \frac{8729.535}{80} \][/tex]

[tex]\[ \Delta S_{\text{oil}} \approx 109.118 \, \text{Btu/°R} \][/tex]

[tex]\[ \Delta S = -4.5364 + 109.118 \][/tex]

[tex]\[ \Delta S \approx 104.5816 \, \text{Btu/°R} \][/tex]

So, the total entropy change for this process is approximately [tex]\( 104.5816 \, \text{Btu/°R} \).[/tex]

Answer:

a) [tex]Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}[/tex], b) [tex]Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}[/tex], c) 1600

Explanation:

a) The Reynolds Number is modelled after the following formula:

[tex]Re = \frac{\rho \cdot v \cdot D}{\mu}[/tex]

Where:

[tex]\rho[/tex] - Fluid density.

[tex]\mu[/tex] - Dynamics viscosity.

[tex]D[/tex] - Diameter of the tube.

[tex]v[/tex] - Fluid speed.

The formula can be expanded as follows:

[tex]Re = \frac{\rho \cdot \frac{4Q}{\pi\cdot D^{2}}\cdot D }{\mu}[/tex]

[tex]Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}[/tex]

b) The Reynolds Number has this alternative form:

[tex]Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}[/tex]

c) Since the diameter is the same than original tube, the Reynolds number is 1600.

Answer:

1913meter per second square.

Explanation:

From the Context the vacuum can be said be the presence of 5e difference below the outside ambient temperature.

For the Venturi.

Please go through the attached file for the rest of the solutions and the answer.

Answer:

(a) Precipitation hardening

(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.

(2) The hardening/strengthening effect is not retained at elevated temperatures for this process.

(4) The strength is developed by a heat treatment.  

(b) Dispersion strengthening

(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.  

(3) The hardening/strengthening effect is retained at elevated temperatures for this process.

(5) The strength is developed without a heat treatment.  

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

[tex]q(x)=\frac{x}{L} q_{0}[/tex]

σ = 120 MPa = 120*10⁶ Pa

[tex]L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\[/tex]

We can see the pic shown in order to understand the question.

We apply

∑MB = 0  (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6   (↑)

Then, we apply

[tex]v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x[/tex]

Then, we can get the maximum bending moment as follows

[tex]M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m[/tex]

then we get  

[tex]M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}[/tex]

We get the inertia as follows

[tex]I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}[/tex]

We use the formula

σ = M*y/I

⇒ M = σ*I/y

where

[tex]y=\frac{h}{2} =\frac{0.3m}{2}=0.15m[/tex]

If M = Mmax, we have

[tex](\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}[/tex]

Answer:

Decrease to typical from utilizing lambda-decrease:  

The given lambda - math terms is, (λf.λx.f(f(fx)))(λy.y×3)2

The of taking the terms is significant in lambda - math,  

For the term, (λy, y×3)2, we can substitute the incentive to the capacity.  

Therefore apply beta-decrease on “(λy, y×3)2,“ will return 2 × 3 = 6  

Presently the tem becomes, (λf λx f(f(fx)))6

The main term, (λf λx f(f(fx))) takes a capacity and a contention and substitute the contention in the capacity.  

Here it is given that it is conceivable to substitute, the subsequent increase in the outcome.  

In this way by applying next level beta - decrease, the term becomes f(f(f(6))), which is in ordinary structure.

Answer:

Diameter will be 27394.76 m

Explanation:

Power P = 0.5 kW = 500 W

Time t required for grinding = 1 hr = 3600 sec

Energy required E = P x t

E = 500 x 3600 = 1800000 J

Flow rate of water Q = 400 ltr/min

We convert to m3/sec

400 ltr/min = 400/(1000 x 60) m3/ses

Q = 0.0067 m3/sec

Energy provided by flow will be

E = pgQd

Where p = density of water = 1000 kg/m3

g = acceleration due to gravity 9.81 m/s2

d = diameter of wheel.

Equating both energy, we have,

1800000 = 1000 x 9.81 x 0.0067 x d

1800000 = 65.73d

d = 1800000/65.73

d = 27394.76 m