The elastic settlement of an isolated single pile under a working load similar to that of piles in the group it represents, is predicted to be 0.25 inches. What is the expected settlement for the pile group given the following information?
1.Group: 16 piles in a 4x4 group
2.Pile Diameter: 12 inches
3.Pile Center to Center Spacing: 3 feet
Answers
Answer:
The expected settlement for the pile group using the given information is 19.92mm or 0.79 inch
Explanation:
In this question, we are asked to calculate the expected settlement for the pole group given some information.
Please check attachment for complete solution and step by step explanation
Related Questions
Amplifiers are extensively used in the baseband portion of a radio receiver system to condition the baseband signal to produce an output signal ready for digital sampling and storage. Some of the key design features of baseband amplifiers include
i. DC gain,
ii. output swing,
iii. power consumption, and
iv. bandwidth.
Answers
Answer:
Please see the attached file for the complete answer.
Explanation:
Alloy parts were cast with a sand mold that took 160 secs for a cube-shaped casting to solidify. The cube was 50 mm on a side. (a) Determine the value of the Chvorinov's mold constant. (b) If the same alloy and mold type were used, find the total solidification time for a cylindrical casting in which the diameter = 25 mm and length = 50 mm.
Answers
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
An air-conditioner with refrigerant-134a as the working fluid is used to keep a room at 23°C by rejecting the waste heat to the outdoor air at 37°C. The room gains heat through the walls and the windows at a rate of 250 kJ/min while the heat generated by the computer, TV, and lights amounts to 900 W. The refrigerant enters the compressor at 400 kPa as a saturated vapor at a rate of 100 L/min and leaves at 1200 kPa and 70°C. Determine (a) the actual COP, (b) the maximum COP, and (c) the minimum volume flow rate of the refrigerant at the compressor inlet for the same compressor inlet and exit conditions.
Answers
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
Given an unsorted array of distinct positive integers A[1..n] in the range between 1 and 10000 and an integer i in the same range. Here n can be arbitrary large. You want to find out whether there are 2 elements of the array that add up to i. Give an algorithm that runs in time O(n).
Answers
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
An air-standard cycle with constant specific heats at room temperature is executed in a closed system with 0.003 kg of air and consists of the following three processes:
1–2 v = Constant heat addition from 95 kPa and 17°C to 380 kPa
2–3 Isentropic expansion to 95 kPa
3–1 P = Constant heat rejection to initial state
The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4
a) Show cycle on P-v and T-s diagrams
b) Calculate net work per cycle in kJ
c) Determine thermal efficiency
Answers
Answer:
A) I attached the diagrams
B)W_net = 0.5434 KJ
C) η_th = 0.262
Explanation:
A) I've attached the P-v and T-s diagrams
B) The temperature at state 2 can be calculated from ideal gas equation at constant specific volume;
So; P2/P1 = T2/T1
Thus, T2 = P2•T1/P1
We are given that;
P2 = 380 KPa
P1 = 95 KPa
T1 = 17 °C = 17 + 273K = 290K
Thus,
T2 = (380 x 290)/95
T2 = 1160 K
While the temperature at state 3 will be gotten from;
T3 = T2 x (P3/P2)^((γ - 1)/γ)
Where γ = cp/cv = 1.005/0.718 = 1.4
Thus;
T3 = 1160 (95/380)^((1.4 - 1)/1.4)
T3 = 780.6 K
Now, net work done is given by the formula;
W_net = Q_in - Q_out
W_net = Q_1-2 - Q_3-1
W_net = m(u2 - u1) - m(h3 - h1)
W_net = m(u2 - u1 - h3 + h1)
From the first table i attached,
At T1 = 290K, u1 = 206.91 KJ/Kg and h1 = 290.16 KJ/Kg
At T2 = 1160K,u2 = 897.91 KJ/Kg
At T3 = 780K, h3 = 800.03 KJ/Kg
We are also given that m = 0.003 kg
Thus;
W_net = 0.003(897.91 - 206.91 - 800.03 + 290.16)
W_net = 0.5434 KJ
C) The thermal efficiency is given by the formula ;
η_th = W_net/Q_in
η_th = 0.5434/(m(u2 - u1))
η_th = 0.5434/(0.003(897.91 - 206.91))
η_th = 0.262
Water containing a solute is flowing through a 1cm diameter tube, which is 50 cm in length, at an average velocity of 1 cm/s. The temperature of water is 37 C. The walls of the tube are coated with an enzyme that makes the solute disappear instantly. The solute enters the tube at a concentration of 0.1 M and the solute has a MW of 300 g/mol. Estimate the fraction of the solute that leaves the tube. Assume that the viscosity and density of water at these conditions are 0.76 cP and 1g/cm^3, respectively.
Answers
Answer:
Explanation:
See the attachment for a step by step solution to the question.
Calculate the diffusion current density for the following carrier distributions. For electrons, use Dn = 35 cm2/s and for holes, use Dp = 10 cm2/s. a. ; Jn = ______________________________________________ b. , at x = 0: Jp (0) = __________________________________ c. , at x = 2.5 µm: Jp (2.5 µm) = _______________________________________ d. , at x = 20 µm: Jp (20 µm) = _______________________________________________ e. , at x = 5 µm: Jn (5 µm) = _______________________________________________ n (x) = (1010 cm−3 ) 5 μm − x 5 μm
Answers
Answer:
(a) Jn = 64.08 μA/m²
(b) Jp = 80.1 μA/m²
(c) Jp = 22.94 μA/m²
(d) Jp = 3.6357 пA/m²
(e) Jn = 22.63 μA/m²
Explanation:
See the attached file for the explanation.
For the first option your program should then ask the user for which row and column in the array to replace, and what value will it be replaced with. For the second option your program should run through all values held in the 2D array and calculate their summation. For the third option your program should print out the contents of the 2D array one row at a time. When given the fourth option your program should simply exit.
Answers
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
A silicon pn junction diode at t has a cross sectional area of cm the length of the p region is and the length of the n region is the doping concentrations are determine approximately the series resistance of the diode and the current through the diode that will produce a drop across this series resistance.
Answers
Answer:
Explanation:
r=72.3 is my thought
I hope it is helpful
A relatively nonvolatile hydrocarbon oil contains 4.0 mol % propane and is being stripped by direct superheated steam in a stripping tray tower to reduce the propane content to 0.2%. The temperature is held constant at 422 K by internal heating in the tower at 2.026 × 105 Pa pressure. A total of 11.42 kg mol of direct steam is used for 300 kg mol of total entering liquid. The vapor–liquid equilibria can be represented by y = 25x, where y is mole fraction propane in the steam and x is mole fraction propane in the oil. Steam can be considered as an inert gas and will not condense. Plot the operating and equilibrium lines and determine the number of theoretical trays needed.
Answers
Answer:
Number of Trays = Six (6)
Explanation:
Given that: y' = 25x' , in terms of molecular ratio, we can write it as
[tex]\frac{Y'}{1 + Y'} =25 \frac{X'}{1 + X'}[/tex] ......... 1
after plotting this we get equilibrium curve as shown in the attached picture.
inlet concentration and outlet concentration of liquid phase is
x₂ = 4% = 0.04 (inlet)
so that can be converted into molar
[tex]X_2 = \frac{x_2}{1-x_2} = \frac{0.04}{1-0.04} = 0.04167[/tex]
and
x₁ = 0.2% = 0.002
[tex]X_1 = \frac{x_1}{1-x_1} = \frac{0.002}{1-0.002} = 2.004*10^{-3}[/tex]
Now we have to use the balance equation a
[tex]\frac{G_s}{L_s} = \frac{X_2-X_1}{Y_2-Y_1}[/tex] .............. a
here amount of solute is comparably lower than
Here we have
L = 300 kmol (total)
[tex]L_s[/tex] = 300(1 - 0.04) = 288 kmol pure oil
G = [tex]G_s[/tex] = 11.42 kmol
[tex]Y_1[/tex] = 0 , solvent free steam
substitute into the equation a
[tex]\frac{11.42}{288} = \frac{0.04167 - 2*10^{-3}}{Y_2 - 0}[/tex]
Y₂ = 1.0003
Now plot the point A(X₁ , Y₁) and B(X₂ , Y₂) and join them to construct operating line AB.
Starting from point B, stretch horizontal line up to equilibrium curve and from there again go down to operating line as shown in the picture attached. This procedure give one count of tray and continue the same procedure up to end of operating.
at last count, the number of stage, gives 6.
∴ Number of trays = 6
The number of theoretical trays needed is 20.
From the given data,
Equilibrium relation
y = 25x
[tex]L_s=L_2(1-x_2)\\L_s=300(1-0.04)= 288Kmol[/tex]
Applying total balance equation[tex]G_sy_1+L_sx_2=G_sy_2+L_sx_1\\G_s(y_1-y_2)=L_s(x_1-x_2)\\x_1=\frac{0.002}{1-0.002}\\x_1=0.002\\y_1=0, y_2=?\\x_2=\frac{0.04}{1-0.04}; x_2=0.0417[/tex]
substituting the values into the equation;
[tex]11.42(0-y_2)=288(0.002-0.0417)\\0-11.42y_2=-11.4048\\y_2=0.998[/tex]
The numbers of trays[tex]N=\frac{In[(\frac{(x_2-y_1)/m}{(x_1-y_1)/m}(1-A)+A }{In(1/A)}\\[/tex]
But [tex]A=\frac{L_s}{mGs}[/tex]
[tex]N=\frac{x_2-x_1}{(x_1-y_1)/m}=\frac{0.0417-0.002}{0.002}\\N=19.85[/tex]
The numbers of tray is approximately 20.
learn more about steam stripping and numbers of tray here
https://brainly.com/question/9349349
The spacing of rafters in a roof is 48-in o.c. Roof dead load = 5 psf. Snow load=30 psf. Roof sheathing is to be a sheathing grade of plywood, and panels are oriented in the strong direction. Deflection limits are L/240 for snow load and L/180 for total load. Find: The minimum span rating, thickness, and edge support requirements for the roof sheathing using ASD procedures.
Answers
Answer:
Explanation:
Step by step solution is found in the attachment.
A 35 ft simply supported beam is loaded with concentrated loads 15 ft in from each support. On one end, the dead load is 8.0 kips and the live load is 18.0 kips. At the other end, the dead load is 4.0 kips and the live load is 9.0 kips. Include the self-weight of the beam in the design. Lateral supports are provided at the supports and the load points. Determine the least-weight W-shape to carry the load. Use A992 steel and Cb
Answers
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
Problem 1: Energy from Flow. Before the advent of the steam-powered engine, most mechanical processes were driven by extracting power from a nearby river using a water wheel. A miller wants to construct a water wheel to grind grain into an hour. He needs a total power output of 0.5 kW to meet the demand of his mill. If a nearby waterfall will flow at a rate of 400 liters/minute onto the top of the wheel, what is the required diameter of the water wheel to achieve the desired power output?
Answers
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
Consider a turbojet mounted on a stationary test stand at sea level. The inlet and exit areas are the same, both equal to 0.45 m^2. The velocity, pressure and temperature of the exhaust gas are 400 m/s, 1.0 ATM and 750K, respectively. Calculate the static thrust of the engine. (Note: Static thrust of a jet engine is the thrust produced when the engine has no forward motion.
Answers
Answer:
The static thrust of the engine is 20,856.44N
Explanation:
In this question, we are asked to calculate the static thrust of the turbojet.
Please check attachment for complete solution and step by step explanation
Water in a household plumbing system originates at the neighborhood water main where the pressure is 480 kPa, the velocity is 5 m/s, and the elevation is 2.44 m. A 19-mm (3/4-in) copper service line supplies water to a two-story residence where the faucet in the master bedroom is 40 m (of pipe) away from the main and at an elevation of 7.62 m. If the sum of the minor-loss coefficients is 3.5, estimate the maximum (open faucet) flow. How would this flow be affected by the operation of other faucets in the house?
Answers
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.
a hollow shaft 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.
Answers
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
A prototype of a part is to be fabricated using stereolithography. The part is shaped like a right triangle whose base = 36 mm, height 48mm, and thickness = 30 mm. In the stereolithography process, the layer thickness = 0.15 mm. Diameter of the laser beam spot = 0.40 mm, and the beam is moved across the surface of the photopolymer at a velocity of 2200 mm/s. Compute the minimum possible time (use units of hours, 3 significant figs) required to build the part, if 25 s are lost each layer to lower the height of the platform that holds the part. Neglect the time for setup and post-processing.
Answers
Answer:
1.443hrs
Explanation:
Please kindly check attachment for the detailed and step by step solution to the problem.
consider a household that uses 23.8 kw-hour of electricity per day on average. (kw-hours is a measure of energy that will be discussed in detail in a later chapter. at this point we want to establish estimations.) most of that electricity is supplied by fossil fuels. to reduce their carbon footprint, the household wants to install solar panels, which receive on average 336 w/m2 from the sun each day. if the solar panels are 19.0% efficient (fraction of solar energy converted into useable electrical energy), what area of solar panels is needed to power the household
Answers
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
The size of an engine is called the engine
A. bore.
B. stroke.
C. displacement.
D. mass.
Answers
Number pattern Write a recursive method called print Pattern() to output the following number pattern. Given a positive integer as input (Ex: 12), subtract another positive integer (Ex: 3) continually until 0 or a negative value is reached, and then continually add the second integer until the first integer is again reached.
Answers
Answer:
See explaination
Explanation:
Code;
import java.util.Scanner;
public class NumberPattern {
public static int x, count;
public static void printNumPattern(int num1, int num2) {
if (num1 > 0 && x == 0) {
System.out.print(num1 + " ");
count++;
printNumPattern(num1 - num2, num2);
} else {
x = 1;
if (count >= 0) {
System.out.print(num1 + " ");
count--;
if (count < 0) {
System.exit(0);
}
printNumPattern(num1 + num2, num2);
}
}
}
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int num1;
int num2;
num1 = scnr.nextInt();
num2 = scnr.nextInt();
printNumPattern(num1, num2);
}
}
See attachment for sample output
A recursive method in Python to output the following number pattern:
def printPattern(n, m):
# Base case: if n is 0 or negative, return
if n <= 0:
return
# Print the current number
print(n)
# Recursively call the function with n subtracted by m
printPattern(n - m, m)
# Recursively call the function with n added by m
printPattern(n + m, m)
# Example usage:
printPattern(12, 3)
Output:
12
9
12
15
12
...
The method works by recursively calling itself twice, once with n subtracted by m and once with n added by m. This process continues until n reaches 0 or a negative value. At that point, the base case is reached and the function returns.
The following is a breakdown of the recursive calls for the example input of 12 and 3:
printPattern(12, 3)
printPattern(9, 3)
printPattern(6, 3)
printPattern(3, 3)
printPattern(0, 3)
# Base case reached, return
printPattern(3, 3)
printPattern(6, 3)
printPattern(9, 3)
printPattern(15, 3)
printPattern(12, 3)
The output of the method is the sequence of numbers that are printed in the recursive calls.
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Consider a venture with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir. The purpose of the venture is to create vacuum in the reservoir when the venture is placed in an airstream. (The vacuum is defined as the pressure difference below the outside ambient pressure.) The venture has a throat to inlet area ratio of 0.85. Calculate the maximum vacuum obtainable in the reservoir when the venturi is placed in an airstream of 90 m/s at standard sea level conditions.
Answers
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.
A 50-lbm iron casting, initially at 700o F, is quenched in a tank filled with 2121 lbm of oil, initially at 80o F. The iron casting and oil can be modeled as incompressible with specific heats 0.10 Btu/lbm o R, and 0.45 Btu/lbm o R, respectively. For the iron casting and oil as the system, determine: a) The final equilibrium temperature (o F) b) The total entropy change for this process (Btu/ o R) (Hint: Total entropy change is the sum of entropy change of iron casting and oil.)
Answers
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]
Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.
(λf.λx.f(f x))(λy.y≠3) 2
Answers
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.
Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg, and the air experiences an entropy decrease of 0.40 kJ/kg·K. Using constant specific heats, determine
(a) the exit temperature of the air,
(b) the work input to the compressor, and
(c) the entropy generation during this process.
Answers
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].
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From the following numbered list of characteristics, decide which pertain to (a) precipitation hardening, and which are displayed by (b) dispersion strengthening. (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 (3) The hardening/strengthening effect is retained at elevated temperatures for this process (4) The strength is developed by a heat treatment (5) The strength is developed without a heat treatment
Answers
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.
Consider a C.T. system in s-plane below. Draw DF1 (Direct Form 1) realization. (by hand) Perform system realization using MATLAB Simulink. (Use same parameters as [Q 11] for "Step Function" and "Delay Block".) Use the "step function", "delay", and "summer" to build the input, X(s). Use a "scope" as the output, Y(s). Include screen shot of the SIMULINK schematic model page, input scope trace, output scope trace in report.Discuss the results in your own words. s +5 H(S) = 52 + 11s + 10
Answers
Answer:
See the attached file for the answer.
Explanation:
See the attached file for the explanation
A voltage regulator is to provide a constant DC voltage Vl=10V to a load Rl from a nominal Vcc=15V supply voltage. The load can vary from 20Ω to 1KΩ. The supply voltage Vcc can vary from 13V to 16V. The op-amp can provide a maximum output current of 20mA. a)Find the βnecessary for the transistor to provide the needed current. b)Find the maximum power the transistor must dissipate.
Answers
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
Consider the adiabatic compressor from a refrigerator that uses refrigerant R-134a as the working fluid, flowing at 0.05 kg/s. The refrigerant enters the compressor as a saturated vapor at a pressure of 0.14 MPa. The pressure of the refrigerant at the exit of the compressor is 0.8 MPa. Please interpolate for this problem where needed.
a. If the compressor is isentropic, what is the exit temperature?
b. What is the minimum amount of power (work) that this compressor might theoretically use?
c. If the isentropic efficiency of the actual, non-ideal, compressor is 0.85, what is the actual amount of power (work) that this compressor requires?
d. What is the entropy generated by this device? Assume that the compressor is adiabatic.
Answers
Answer:
Please see the attached file for the complete answer.
Explanation:
A 2-kg sphere A strikes the frictionless inclined surface of a 6-kg wedge B at a 90 degree angle with a velocity of magnitude 4 m/s. The wedge can roll freely on the ground and is initially at rest. Knowing that the coefficient of restitution between the wedge and the sphere is 0.5 and that the inclined surface of the wedge forms an angle θ=40 degrees with the horizontal, determine the velocities of the sphere and the wedge immediately after impact.
Answers
Answer:
Final velocities are:
Wedge B: v = 2.334 m/s
Sphere A: v = 5.386 m/s
Explanation:
Given:-
- The mass of sphere A, mA = 2-kg
- The mass of the wedge B, mB = 6-kg
- The sphere collides with " normal " to the wedge face.
- The coefficient of restitution , e = 0.5
- The wedge inclination angle, θ=40 degrees with the horizontal.
- The initial speed of sphere A, vA = 4m/s
- The initial speed of wedge B, vB = 0 m/s ... ( rest )
Find:-
- First step would be to sketch a system of sphere A and wedge B as ( FBD ).
- We will add a sketch of "two" coordinate axes on the ( FBD ).
First coordinate system ( normal ( n ) - tangent ( t ) )
Normal axis at 90 degrees directed towards the wedge in direction of sphere motion - denote as ( n ).Tangent axis along ( parallel ) to the wedge surface directed up the wedge - denote as ( t )Second coordinate system ( horizontal ( x ) - vertical ( y ) )
Horizontal axis parallel to ground directed towards the right in assumed direction of wedge motion after impact - denote as ( x ).Vertical axis normal to the ground directed upwards -denote ( y ).Note:- All the above directions of coordinate axes denote the positive direction of vectors.
- Resolve the angle ( α ) between the normal - ( n ) or velocity vector vA and the horizontal - ( x ) axis.
α = 90° - θ = 90° - 40°
α = 50°
- We will denote the final velocity components of sphere A as ( v'An , v'At ):
And, final velocity components of wedge B as ( v'B ).
- Transform the the final velocity of wedge ( vB' ) in the (n-t) coordinate axis using the resolved coordinate transformation angle ( α ). The normal ( n ) and tangential components of the velocity vector ( vB' ) are: ( vB'n , vB't ).
vB'n = vB'*cos ( α ) vB't = vB'*sin ( α° )
vB'n = vB'*cos ( 50° ) vB't = vB'*sin ( 50° )
- Note: There is no y-component of velocity of wedge. This is because the motion of wedge in the vertical direction is restricted by the ground - equilibrium conditions. Hence, v'By = 0.
- Consider the system of sphere A and wedge B to be isolated with no external forces acting on the system. For such conditions the principle of conservation of momentum ( P ) is valid. Which states:
PA,i + PBi = PA,f + PB,f
Where,
PA,i : The initial momentum of the sphere A
PB,i : The initial momentum of wedge B
PA,f : The final momentum of the sphere A
PB,f : The final momentum of wedge B.
- Using the data given and relations computed in the ( n-t ) coordinate system. We will use the principle of conservation of linear momentum in both axis ( n and t ) combined in vector momentum of system.
Conservation of linear momentum:
[tex]m_A*v_A + m_B*v_B = m_A*v_A' + m_B*v_B'[/tex]
- Using vector notations we have, Taking ( i unit vector in tangential direction and j unit vector in the normal direction ):
[tex]m_A*v_A_n j = m_A*( v_A'_t i + v_A'_n j )+ m_B* ( v_B'_n j + v_B'_t i )\\\\2*4 j = 2*( v_A'_t i + v_A'_n j )+ 6* ( v_B'_n j + v_B'_t i )\\\\0 i + 8 j = ( 2*v_A'_t + 6*v_B'*sin ( 50 ) ) i + ( 2*v_A'_n + 6*v_B'* cos (50 ) ) j[/tex]
- Equate all the ( i - j ) vectors on left and right hand side of the equation respectively,
[tex]0 = 2*v_A'_t + 6*v_B'*sin ( 50 )\\\\\\ 8 = 2*v_A'_n + 6*v_B'* cos(50 )[/tex] .... Eq 1
- The coefficient of restitution ( e ) is a squared loss of kinetic energy of the system that can be expressed in terms of velocities of two objects as relative change in velocity of two objects after and before impact.:
[tex]e = \frac{v_B'_n - v_A'_n}{v_A_n - v_B_n}[/tex]
Note: We have only used the velocities normal to the surface of wedge. This is because the kinetic loss is a scalar dimension; hence, the normal direction to the surface of impact is assumed to cater all the loss in kinetic energy.
We have,
[tex]0.5 = \frac{v_B'*cos ( 50 ) - v_A'_n}{ 4 - 0}\\\\2 = v_B'*cos ( 50 ) - v_A'_n \\\\v_B'*cos ( 50 ) = v_A'_n + 2[/tex] ..... Eq 2
- Now substitute equation 2 into equation 1:
[tex]8 = 2*v_A'_n + 6*( v_A'_n + 2 )\\\\-4 = 8*v_A'_n\\\\v_A'_n = -0.5 \frac{m}{s}[/tex]
- Using Equation 2 compute v'B:
[tex]v_B'*cos ( 50 ) = -0.5 + 2\\\\v_B'= \frac{1.5}{cos ( 50 ) } \\\\v_B'= 2.334 \frac{m}{s}[/tex]
- Using Equation 1 compute v'At:
[tex]0 = 2*v_A'_t + 6*2.334*sin ( 50 )\\\\v_A'_t = - \frac{6*2.334*sin ( 50 )}{2} \\\\v_A'_t = - 5.363 \frac{m}{s}[/tex]
- The final velocity of wedge B along the horizontal direction ( x ) is:
vB' = 2.334 m/s .... x-direction
- The velocity of sphere A after impact is given by:
[tex]v_A' = \sqrt{v_A't^2 + v_A'n^2}\\\\v_A' = \sqrt{5.363^2 + 0.5^2}\\\\v_A' = \sqrt{29.011769}\\\\v_A' = 5.386 \frac{m}{s}[/tex]
The reel has a mass of 30 kg and a radius of gyration about A of kA = 120 mm. The suspended cylinder has a mass of 40 kg. Starting from rest, the motor M exerts a constant force of P = 300 N on the cable. Determine the mass moment of inertia of the reel. Determine the velocity of the cylinder after it has traveled upward 2 m. Determine the time that was taken to travel the 2 m distance.
Answers
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