Answer:
The ABC Corporation
a) Total Expected Revenue (in dollars) for 20XX:
Revenue from T1 = 60,000 x $165 = $26,400,000
Revenue from T2 = 40,000 x $250 = $10,000,000
Total Revenue from T1 and T2 = $36,400,000
b) Production Level (in units) for T1 and T2
T1 T2
Total Units sold 160,000 40,000
Add Closing Inventory 25,000 9,000
Units Available for sale 185,000 49,000
less opening inventory 20,000 8,000
Production Level 165,000 units 41,000 units
c) Total Direct Material Purchases (in dollars):
Cost of direct materials used T1 T2
A: (165,000 x 4 x $12) $7,920,000 $2,460,000 (41,000 x 5 x $12)
B: (165,000 x 2 x $5) 1,650,000 615,000 (41,000 x 3 x $5)
C: 0 123,000 (41,000 x 1 x$3)
Total cost $9,570,000 $3,198,000 Total = $12,768,000
Cost of direct per unit = $58 ($9,570,000/165,000) for T1 and $78 ($3,198,000/41,000) for T2
Cost of direct materials used for production $12,768,000
Cost of closing direct materials:
A (36,000 x $12) $432,000
B (32,000 x $5) 160,000
C (7,000 x $3) 21,000 $613,000
Cost of direct materials available for prodn $13,381,000
Less cost of beginning direct materials:
A (32,000 x $12) $384,000
B (29,000 x $5) 145,000
C (6,000 x $3) 18,000 $547,000
Cost of direct materials purchases $12,834,000
d) The Total Direct Manufacturing Labor Cost (in dollars):
T1 T2
Direct labor per unit 2 hours 3 hours
Direct labor rate per hour $12 $16
Direct labor cost per unit $24 $48
Production level 165,000 units 41,000 units
Labor Cost ($) $3,960,000 $1,968,000
Total labor cost $5,928,000 ($3,960,000 + $1,968,000)
e) Total Overhead cost (in dollars):
Overhead rate = $20 per labor hour
Overhead cost per unit: T1 = $40 ($20 x 2) and T2 = $60 ($20 x 3)
T1 overhead = $20 x 2 x 165,000) = $6,600,000
T2 overhead = $20 x 3 x 41,000) = $2,460,000
Total Overhead cost = $9,060,000
Cost of goods produced:
Cost of opening inventory of materials = $547,000
Purchases of directials materials 12,834,000
less closing inventory of materials = $613,000
Cost of materials used for production 12,768,000
add Labor cost 5,928,000
add Overhead cost 9,060,000
Total production cost $27,756,000
f) Total cost of goods sold (in dollars):
Cost of opening inventory = $3,928,000
Total Production cost = $27,756,000
Cost of goods available for sale $31,684,000
Less cost of closing inventory $4,724,000
Total cost of goods sold $26,960,000
g) Total expected operating income (in dollars)
Sales Revenue: T1 and T2 $36,400,000
Cost of goods sold 26,960,000
Gross profit $9,440,000
less marketing & distribution 400,000
Total Expected Operating Income = $9,040,000
Explanation:
a) Cost of beginning inventory of finished goods:
T1, (Direct materials + Labor + Overhead) X inventory units =
T1 = 20,000 x ($58 + 24 + 40) = $2,440,000
T2 = 8,000 ($78 + 48 + 60) = $1,488,000
Total cost of beginning inventory = $3,928,000
b) Cost of closing Inventory of finished goods:
T1 = 25,000 x ($58 + 24 + 40) = $3,050,000
T2 = 9,000 ($78 + 48 + 60) = $1,674,000
Total cost of closing inventory = $4,724,000
The Rappahannock River near Warrenton, VA, has a flow rate of 3.00 m3/s. Tin Pot Run (a pristine stream) discharges into the Rappahannock at a flow rate of 0.05 m3/s. To study mixing of the stream and river, a conserva- tive tracer is to be added to Tin Pot Run. If the instruments that can mea- sure the tracer can detect a concentration of 1.0 mg/L, what minimum concentration must be achieved in Tin Pot Run so that 1.0 mg/L of tracer can be measured after the river and stream mix? Assume that the 1.0 mg/L of tracer is to be measured after complete mixing of the stream and Rappa- hannock has been achieved and that no tracer is in Tin Pot Run or the Rap- pahannock above the point where the two streams mix. What mass rate (kg/d) of tracer must be added to Tin Pot Run?
Find the given attachments for complete explanation
Consider a classroom for 56 students and one instructor, each generating heat at a rate of 100 W. Lighting is provided by 18 fluorescent lightbulbs, 40 W each, and the ballasts consume an additional 10 percent. Determine the rate of internal heat generation in this classroom when it is fully occupied. The rate of internal heat generation in this classroom when it is fully occupied is W.
Answer:
What is the probability of selecting the 4 of spade or black diamond from a deck of 52 playing cards?
a) 2/52
b) 4/52
c) 3/52
d) 1/5
Explanation:
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the central plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)
Answer:
1) The normal reactions at the front wheel is 9909.375 N
The normal reactions at the rear wheel is 8090.625 N
2) The least coefficient of friction required between the tyres and the road is 0.625
Explanation:
1) The parameters given are as follows;
Speed, u = 90 km/h = 25 m/s
Distance, s it takes to come to rest = 50 m
Mass, m = 1.8 tonnes = 1,800 kg
From the equation of motion, we have;
v² - u² = 2·a·s
Where:
v = Final velocity = 0 m/s
a = acceleration
∴ 0² - 25² = 2 × a × 50
a = -6.25 m/s²
Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N
The coefficient of friction, μ, is given as follows;
[tex]\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625[/tex]
Weight transfer is given as follows;
[tex]W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N[/tex]
Therefore, we have for the car at rest;
Taking moment about the Center of Gravity CG;
[tex]F_R[/tex] × 1.3 = 1.7 × [tex]F_F[/tex]
[tex]F_R[/tex] + [tex]F_F[/tex] = 18000
[tex]F_R + \dfrac{1.3 }{1.7} \times F_R = 18000[/tex]
[tex]F_R[/tex] = 18000*17/30 = 10200 N
[tex]F_F[/tex] = 18000 N - 10200 N = 7800 N
Hence with the weight transfer, we have;
The normal reactions at the rear wheel [tex]F_R[/tex] = 10200 N - 2109.375 N = 8090.625 N
The normal reactions at the front wheel [tex]F_F[/tex] = 7800 N + 2109.375 N = 9909.375 N
2) The least coefficient of friction, μ, is given as follows;
[tex]\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625[/tex]
The least coefficient of friction, μ = 0.625.
The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 8.7 kN. Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.
Answer:
[tex] T_A_C = 6.296 kN [/tex]
[tex] R = 10.06 kN [/tex]
Explanation:
Given:
[tex] T_A_B = 8.7 kN[/tex]
Required:
Find the tension TAC and magnitude R of this downward force.
First calculate [tex] \alpha, \beta, \gamma [/tex]
[tex] \alpha = tan^-^1 =\frac{40}{50} = 38. 36 [/tex]
[tex] \beta = tan^-^1 =\frac{50}{30} = 59.04 [/tex]
[tex] \gamma = 180 - 38.36 - 59.04 = 82.6 [/tex]
To Find tension in AC and magnitude R, use sine rule.
[tex] \frac{sin a}{T_A_C} = \frac{sin b}{T_A_B} = \frac{sin c}{R} [/tex]
Substitute values:
[tex]\frac{sin 38.36}{T_A_C} = \frac{sin 59.04}{8.7} = \frac{82.6}{R}[/tex]
Solve for T_A_C:
[tex] T_A_C = 8.7 * \frac{sin 38.36}{sin 59.04} = [/tex]
[tex] T_A_C = 8.7 * 0.724 = 6.296 kN [/tex]
Solve for R.
[tex] R = 8.7 * \frac{sin 82.6}{sin 59.04} = [/tex]
[tex] R = 8.7 * 1.156 [/tex]
R = 10.06 kN
Tension AC = 6.296kN
Magnitude,R = 10.06 kN
In contouring, it is necessary to measure position and not velocity for feedback.
a. True
b. False
In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.
a. True
b. False
Job shop is another term for process layout.
a. True
b. False
Airplanes are normally produced using group technology or cellular layout.
a. True
b. False
In manufacturing, value-creating time is greater than takt time.
a. True
b. False
Answer:
(1). False, (2). True, (3). False, (4). False, (5). True.
Explanation:
The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.
(1). In contouring, it is necessary to measure position and not velocity for feedback.
ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.
(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.
ANSWER: a=> True.
(3). Job shop is another term for process layout.
ANSWER: b => False
JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.
(4). Airplanes are normally produced using group technology or cellular layout.
ANSWER: b => False.
(5). In manufacturing, value-creating time is greater than takt time.
ANSWER: a => True.
The lower half of a 7-m-high cylindrical container is filled with water (rho = 1000 kg/m3) and the upper half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and the bottom of the cylinder. (Round the final answer to one decimal place.)
Answer:
Pressure difference (ΔP) = 63,519.75 kpa
Explanation:
Given:
ρ = 1,000 kg/m³
Height of cylindrical container used (h) = 7m / 2 = 3.5m
Specific gravity (sg) = 0.85
Find:
Pressure difference (ΔP).
Computation:
⇒ Pressure difference (ΔP) = h g [ ρ(sg) + ρ] ∵ [ g = 9.81]
⇒ Pressure difference (ΔP) = (3.5)(9.81) [ 1,000(0.85) + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [8,50 + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [1,850]
⇒ Pressure difference (ΔP) = 63,519.75 kpa
Find the largest number. The process of finding the maximum value (i.e., the largest of a group of values) is used frequently in computer applications. For example, an app that determines the winner of a sales contest would input the number of units sold by each salesperson. The sales person who sells the most units wins the contest. Write pseudocode, then a C# app that inputs a series of 10 integers, then determines and displays the largest integer. Your app should use at least the following three variables:
Counter: Acounter to count to 10 (i.e., to keep track of how many nimbers have been input and to determine when all 10 numbers have been processed).
Number: The integer most recently input by the user.
Largest: The largest number found so far.
Answer:
See Explanation
Explanation:
Required
- Pseudocode to determine the largest of 10 numbers
- C# program to determine the largest of 10 numbers
The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;
The largest of the 10 numbers is then saved in variable Largest and printed afterwards.
Pseudocode (Number lines are used for indentation to illustrate the program flow)
1. Start:
2. Declare Number as 1 dimensional array of 10 integers
3. Initialize: counter = 0
4. Do:
4.1 Display “Enter Number ”+(counter + 1)
4.2 Accept input for Number[counter]
4.3 While counter < 10
5. Initialize: Largest = Number[0]
6. Loop: i = 0 to 10
6.1 if Largest < Number[i] Then
6.2 Largest = Number[i]
6.3 End Loop:
7. Display “The largest input is “+Largest
8. Stop
C# Program (Console)
Comments are used for explanatory purpose
using System;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
int[] Number = new int[10]; // Declare array of 10 elements
//Accept Input
int counter = 0;
while(counter<10)
{
Console.WriteLine("Enter Number " + (counter + 1)+": ");
string var = Console.ReadLine();
Number[counter] = Convert.ToInt32(var);
counter++;
}
//Initialize largest to first element of the array
int Largest = Number[0];
//Determine Largest
for(int i=0;i<10;i++)
{
if(Largest < Number[i])
{
Largest = Number[i];
}
}
//Print Largest
Console.WriteLine("The largest input is "+ Largest);
Console.ReadLine();
}
}
}
Use a delta-star conversion to simplify the delta BCD (40 , 16 , and 8 ) in the
bridge network in Fig. f and find the equivalent resistance that replaces the network
between terminals A and B, and hence find the current I if the source voltage is 52 V.
Answer:
Current, I = 4A
Explanation:
Since the connection is in delta, let's convert to star.
Simplify BCD:
[tex] R1 = \frac{40 * 8}{40 + 16 + 8} = \frac{320}{64} = 5 ohms [/tex]
[tex] R2 = \frac{16 * 8}{40 + 16 + 8} = \frac{128}{64} = 2 ohms [/tex]
[tex] R3 = \frac{40 * 16}{40 + 16 + 8} = \frac{640}{64} = 10 ohms [/tex]
From figure B, it can be seen that 6 ohms and 6 ohms are connected in parallel.
Simplify:
[tex] \frac{6 * 6}{6 + 6} = \frac{36}{12} = 3 \ohms [/tex]
Req = 10 ohms + 3 ohms
Req = 13 ohms
To find the current, use ohms law.
V = IR
Where, V = 52volts and I = 13 ohms
Solve for I,
[tex] I = \frac{V}{R} = \frac{52}{13} = 4A[/tex]
Current, I = 4 A
cubical tank 1 meter on each edge is filled with water at 20 degrees C. A cubical pure copper block 0.46 meters on each edge with an initial temperature of 100 degrees C is quickly submerged in the water, causing an amount of water equal to the volume of the smaller cube to spill from the tank. An insulated cover is placed on the tank. The tank is adiabatic. Estimate the equilibrium temperature of the system (block + water). Be sure to state all applicable assumptions.
Answer:
final temperature = 26.5°
Explanation:
Initial volume of water is 1 x 1 x 1 = 1 [tex]m^{3}[/tex]
Initial temperature of water = 20° C
Density of water = 1000 kg/[tex]m^{3}[/tex]
volume of copper block = 0.46 x 0.46 x 0.46 = 0.097 [tex]m^{3}[/tex]
Initial temperature of copper block = 100° C
Density of copper = 8960 kg/[tex]m^{3}[/tex]
Final volume of water = 1 - 0.097 = 0.903 [tex]m^{3}[/tex]
Assumptions:
since tank is adiabatic, there's no heat gain or loss through the wallsthe tank is perfectly full, leaving no room for cooling airtotal heat energy within the tank will be the summation of the heat energy of the copper and the water remaining in the tank.mass of water remaining in the tank will be density x volume = 1000 x 0.903 = 903 kg
specific heat capacity of water c = 4186 J/K-kg
heat content of water left Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules
mass of copper will be density x volume = 8960 x 0.097 = 869.12 kg
specific heat capacity of copper is 385 J/K-kg
heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules
total heat in the system = 75.59 + 33.46 = 109.05 Mega-joules
this heat will be distributed in the entire system
heat energy of water within the system = mcT
where T is the final temperature
= 903 x 4186 x T = 3779958T
for copper, heat will be
mcT = 869.12 x 385 = 334611.2T
these component heats will sum up to the final heat of the system, i.e
3779958T + 334611.2T = 109.05 x [tex]10^{6}[/tex]
4114569.2T = 109.05 x [tex]10^{6}[/tex]
final temperature T = (109.05 x [tex]10^{6}[/tex])/4114569.2 = 26.5°
A float valve, regulating the flow of water into a reservoir, is shown in the figure. The spherical float (half of the sphere is submerged) is 0.1553 m in diameter. AOB is the weightless link carrying the float at one end, and a valve at the other end which closes the pipe through which flows into the reservoir. The link is mounted on a frictionless hinge at O, and the angle AOB is 135o. The length of OA is 253 mm and the distance between the center of the float and the hinge is 553 mm. When the flow is stopped AO will be vertical. The valve is to be pressed on to the seat with a force of 10,53 N to be completely stop the flow in the reservoir. It was observed that the flow of water is stopped, when the free surface of water in the reservoir is 353 mm below the hinge at O. Determine the weight of the float sphere.
Answer:
9.29 N . . . . weight of 0.948 kg sphere
Explanation:
The sum of torques on the link BOA is zero, so we have ...
(right force at A)(OA) = (up force at B)(OB·sin(135°))
Solving for the force at B, we have ...
up force at B = (10.53 N)(253 mm)/((553 mm)/√2) ≈ 6.81301 N
This force is due to the difference between the buoyant force on the float sphere and the weight of the float sphere. Dividing by the acceleration due to gravity, it translates to the difference in mass between the water displaced and the mass of the sphere.
∆mass = (6.81301 N)/(9.8 m/s^2) = 0.695205 kg
__
The center of the sphere of diameter 0.1553 m is below the waterline by ...
(553 mm)cos(45°) -(353 mm) = 38.0300 mm
The volume of the spherical cap of radius 155.3/2 = 77.65 mm and height 77.65+38.0300 = 115.680 mm can be found from the formula ...
V = (π/3)h^3(3r -h) = (π/3)(115.680^2)(3·77.65 -115.68) mm^3 ≈ 1.64336 L
So the mass of water contributing to the buoyant force is 1.64336 kg. For the net upward force to correspond to a mass of 0.695305 kg, the mass of the float sphere must be ...
1.64336 kg -0.695205 kg ≈ 0.948 kg
The weight of the float sphere is then (9.8 m/s^2)·(0.948 kg) = 9.29 N
The weight of the 0.948 kg float sphere is about 9.29 N.
what is called periodic function give example? Plot the output which is started with zero degree for one coil rotating in the uniform magnetic field and name it. How can you represent this output as the periodic function?
Answer:
A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt
Explanation:
A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt
attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field
B ( wave flux density ) = Bm sinwt and w = 2[tex]\pi[/tex]f = [tex]\frac{2\pi }{T}[/tex] rad/sec
A solid square rod is cantilevered at one end. The rod is 0.6 m long and supports a completely reversing transverse load at the other end of 62 kN. The material is AISI 1080 hot-rolled steel. If the rod must support this load for 104 cycles with a design factor of 1.5, what dimension should the square cross section have
Answer:
The dimension of the square cross section is = 30mm * 30mm
Explanation:
Before proceeding with the calculations convert MPa to Kpsi
Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi
the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut
at 10^6 cycles" for 111.65 Kpsi = f ( fatigue strength factor) = 0.83
To calculate the endurance limit use Se' = 0.5 Sut since Sut < 1400 MPa
Se'( endurance limit ) = 0.5 * 770 = 385 Mpa
The surface condition modification factor
Ka = 57.7 ( Sut )^-0.718
Ka = 57.7 ( 770 ) ^-0.718
Ka = 0.488
Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one
The endurance limit at the critical location of a machine part can be expressed as :
Se = Ka*Kb*Se'
Se = 0.488 * 0.85 * 385 = 160 MPa
Next:
Calculating the constants to find the number of cycles
α = [tex]\frac{(fSut)^2}{Se}[/tex]
α =[tex]\frac{(0.83*770)^2}{160}[/tex] = 2553 MPa
b = [tex]-\frac{1}{3} log(\frac{fSut}{Se} )[/tex]
b = [tex]-\frac{1}{3} log (\frac{0.83*770}{160} )[/tex] = -0.2005
Next :
calculating the fatigue strength using the relation
Sf = αN^b
N = number of cycles
Sf = 2553 ( 10^4) ^ -0.2005
Sf = 403 MPa
Calculate the maximum moment of the beam
M = 2000 * 0.6 = 1200 N-m
calculating the maximum stress in the beam
∝ = ∝max = [tex]\frac{Mc}{I}[/tex]
Where c = b/2 and I = b(b^3) / 12
hence ∝max = [tex]\frac{6M}{b^3}[/tex] = 6(1200) / b^3 = 7200/ b^3 Pa
note: b is in (meters)
The expression for the factor of safety is written as
n = [tex]\frac{Sf}{\alpha max }[/tex]
Sf = 403, n = 1.5 and ∝max = 7200 / b^3
= 1.5 = [tex]\frac{403(10^6 Pa/Mpa)}{7200 / B^3}[/tex] making b subject of the formula in other to get the value of b
b = 0.0299 m * 10^3 mm/m
b = 29.9 mm therefore b ≈ 30 mm
since the size factor assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2
the dimension = 30 mm by 30 mm
An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 4 kg of an ideal gas at 750 kPa and 48°C, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank. (Round the final answers to the nearest whole number.)
Answer:
The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].
Explanation:
An ideal gas is represented by the following model:
[tex]P\cdot V = \frac{m}{M}\cdot R_{u} \cdot T[/tex]
Where:
[tex]P[/tex] - Pressure, measured in kilopascals.
[tex]V[/tex] - Volume, measured in cubic meters.
[tex]m[/tex] - Mass of the ideal gas, measured in kilograms.
[tex]M[/tex] - Molar mass, measured in kilograms per kilomole.
[tex]T[/tex] - Temperature, measured in Kelvin.
[tex]R_{u}[/tex] - Universal constant of ideal gases, equal to [tex]8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex]
As tank is rigid and insulated, it means that no volume deformations in tank, heat and mass interactions with surroundings occur during expansion process. Hence, final pressure is less that initial one, volume is doubled (due to equal partitioning) and temperature remains constant. Hence, the following relationship can be derived from model for ideal gases:
[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex]
Now, final pressure is cleared:
[tex]P_{2} = P_{1}\cdot \frac{T_{2}}{T_{1}}\cdot \frac{V_{1}}{V_{2}}[/tex]
[tex]P_{2} = (750\,kPa)\cdot 1 \cdot \frac{1}{2}[/tex]
[tex]P_{2} = 375\,kPa[/tex]
The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].
Liquid benzene and liquid n-hexane are blended to form a stream flowing at a rate of 1700 lbm/h. An on-line densitometer (an instrument used to determine density) indicates that the stream has a density of 0.810 g/mL. Using specific tractors from Table B.1, estimate the mass and volumetric feed rates of the two hydrocarbons to the mixing vessel (in U.S. customary units). State at least two assumptions required to obtain the estimate from the recommended date.
Air, at a free-stream temperature of 27.0°C and a pressure of 1.00 atm, flows over the top surface of a flat plate in parallel flow with a velocity of 12.5 m/sec. The plate has a length of 2.70 m (in the direction of the fluid flow), a width of 0.65 m, and is maintained at a constant temperature of 127.0°C. Determine the heat transfer rate from the top of the plate due to forced convection.
Answer:
Explanation:
Given that:
V = 12.5m/s
L= 2.70m
b= 0.65m
[tex]T_{ \infty} = 27^0C= 273+27 = 300K[/tex]
[tex]T_s= 127^0C = (127+273)= 400K[/tex]
P = 1atm
Film temperature
[tex]T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K[/tex]
dynamic viscosity =
[tex]\mu =20.9096\times 10^{-6} m^2/sec[/tex]
density = 0.9946kg/m³
Pr = 0.708564
K= 229.7984 * 10⁻³w/mk
Reynolds number,
[tex]Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}[/tex]
[tex]=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043[/tex]
we have,
[tex]Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k[/tex]
we have,
heat transfer rate from top plate
[tex]\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w[/tex]
Consider a series RC circuit at the left where C = 6 µ F, R = 2 MΩ, and ε = 20 V. You close the switch at t = 0. Find (a) the time constant for the circuit, (b) the half-life of the circuit, (c) the current at t = 0, (d) the voltage across the capacitor at t = 0, and (e) the voltage across the resistor after a very long time.
Answer:
(a) 12 seconds (b) t = 8.31 seconds (c) 10µ A (d) V = 20 V (e) V =0
Explanation:
Solution
Given that:
C = 6 µ which is = 6 * 10^ ⁻6
R = 2 MΩ, which is = 2 * 10^ 6
ε = 20 V
(a) When it is at the time constant we have the following:
λ = CR
= 6 * 10^ ⁻6 * 2 * 10^ 6
λ =12 seconds
(b) We solve for the half life of the circuit which is given below:
d₀ = d₀ [ 1- e ^ ⁺t/CR
d = decay mode]
d₀/2 = d₀ 1- e ^ ⁺t/12
2^⁻1 = e ^ ⁺t/12
Thus
t/12 ln 2
t = 12 * ln 2
t = 12 * 0.693
t = 8.31 seconds
(c) We find the current at t = 0
So,
I = d₀/dt
I = d₀/dt e ^ ⁺t/CR
= CE/CR e ^ ⁺t/CR
E/R e ^ ⁺t/CR
Thus,
at t = 0
I E/R = 20/ 2 * 10^ 6
= 10µ A
(d) We find the voltage across the capacitor at t = 0 which is shown below:
V = IR
= 10 * 10^ ⁻6 * 2 * 10^ 6
V = 20 V
(e) We solve for he voltage across the resistor.
At t = 0
I = 0
V =0
Participating in extracurricular activities in high school helps:
Answer:
Develop social skills
Explanation:
Answer:
strengthen your college applications
Explanation:
You are standing at the edge of the roof of the engineering building, which is H meters high. You see Professor Murthy, who is h meters tall, jogging towards the building at a speed of v meters/second. You are holding an egg and want to release it so that it hits Prof Murthy squarely on top of his head. What formulas describes the distance from the building that Prof Murthy must be when you release the egg?
Answer:
s = v√[2(H - h)/g]
This formula describes the distance from the building that Prof Murthy must be when you release the egg
Explanation:
First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:
Vf = Vi + gt
where,
Vf = Velocity of egg at the time of hitting head of the Professor
Vi = initial velocity of egg = 0 m/s (Since, egg is initially at rest)
g = acceleration due to gravity
t = time taken by egg to come down
Therefore,
Vf = 0 + gt
t = Vf/g ----- equation (1)
Now, we use 3rd euation of motion for Vf:
2gs = Vf² - Vi²
where,
s = height dropped = H - h
Therefore,
2g(H - h) = Vf²
Vf = √[2g(H - h)]
Therefore, equation (1) becomes:
t = √[2g(H - h)]/g
t = √[2(H - h)/g]
Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:
s = vt
s = v√[2(H - h)/g]
This formula describes the distance from the building that Prof Murthy must be when you release the egg
two opposite poles repel each other
Answer:
South Pole and South Pole or North Pole and North Pole.
Given in the following v(t) signal.
a. Find the first 7 harmonics of the Fourier series function in cosine form.
b. Plot one side spectrum
c. Find the first 7 harmonics of the Fourier series function in exponential form.
d. Plot two side spectrum Given in the following v(t) signal.
Answer:
Check the v(t) signal referred to in the question and the solution to each part in the files attached
Explanation:
The detailed solutions of parts a to d are clearly expressed in the second file attached.
A cylinder of metal that is originally 450 mm tall and 50 mm in diameter is to be open-die upset forged to a final height of 100 mm. The strength coefficient is 230 MPa and the work hardening exponent is 0.15 while the coefficient of friction of the metal against the tool is 0.1. If the maximum force that the forging hammer can deliver is 3 MN, can the forging be completed
Answer:
Yes, the forging can be completed
Explanation:
Given h = 100 mm, ε = ㏑(450/100) = 1.504
[tex]Y_f = 230 \times 1.504^{0.15} = 244.52[/tex]
V = π·D²·L/4 = π × 50²×450/4 = 883,572.93 mm³
At h = 100 mm, A = V/h = 883,572.93 /100 = 8835.73 mm²
D = √(4·A/π) = 106.07 mm
[tex]K_f[/tex] = 1 + 0.4 × 0.1 × 106.07/100 = 1.042
F = 1.042 × 244.52 × 8835.73 = 2252199.386 N =2.25 MN
Hence the required force = 2.25 MN is less than the available force = 3 MN therefore, the forging can be completed.
Under the normal sign convention, the distributed load on a beam is equal to the:_______A. The rate of change of the bending moment with respect to the shear force. B. The second derivative of the bending moment with respect to the length of the beam. C. The rate of change of the bending moment with respect to the length of the beam. D. Negative of the rate of change of the shear force with respect to the length of the beam.
Answer:
Under the normal sign convention, the distributed load on a beam is equal to the: O The second derivative of the bending moment with respect to the length of the beam O Negative of the rate of change of the shear force with respect to the length of the beam.
Sorry if the answer is wrong
Increase the sampling time by a factor of 10 (to 0.1 seconds), keeping the frequency of the square wave the same, and observe the delay. Discuss relationship between sampling time and delay from one board to another.
Answer:
Time delay increases
Explanation:
Time delay is the delay between occurance of signal. If sampling time that is time between two samples is increased, the delay in the occurance of regenerated samples is also increased.
: Explain why testing can only detect the presence of errors, not their absence?
Answer:
The goal of the software is to observe the software behavior to meet its requirement expectation. In software engineering, validating software might be harder since client's expectation may be vague or unclear.
Explanation:
An aluminium bar 600mm long with a diameter 40mm has a hole drilled in the centre of which 30mm in diameter and 100mm long if the modulus of elasticity is 85GN/M2 calculate the total contraction oon the bar due to comprehensive load of 160KN.
Answer:
Total contraction on the bar = 1.238 mm
Explanation:
Modulus of Elasticity, E = 85 GN/m²
Diameter of the aluminium bar, [tex]d_{Al} = 40 mm = 0.04 m[/tex]
Load, P = 160 kN
Cross sectional area of the aluminium bar without hole:
[tex]A_1 = \frac{\pi d_{Al}^2 }{4} \\A_1 = \frac{\pi 0.04^2 }{4}\\A_1 = 0.00126 m^2[/tex]
Diameter of hole, [tex]d_h = 30 mm = 0.03 m[/tex]
Cross sectional area of the aluminium bar with hole:
[tex]A_2 = \frac{\pi( d_{Al}^2 - d_{h}^2)}{4} \\A_2 = \frac{\pi (0.04^2 - 0.03^2) }{4}\\A_2 = 0.00055 m^2[/tex]
Length of the aluminium bar, [tex]L_{Al} = 600 mm = 0.6 m[/tex]
Length of the hole, [tex]L_h = 100mm = 0.1 m[/tex]
Contraction in aluminium bar without hole [tex]= \frac{P * L_{Al}}{A_1 E}[/tex]
Contraction in aluminium bar without hole [tex]= \frac{160*10^3 * 0.6}{0.00126 * 85 * 10^9 }[/tex]
Contraction in aluminium bar without hole = 96000/107100000
Contraction in aluminium bar without hole = 0.000896
Contraction in aluminium bar with hole [tex]= \frac{P * L_{h}}{A_2 E}[/tex]
Contraction in aluminium bar without hole [tex]= \frac{160*10^3 * 0.1}{0.00055 * 85 * 10^9 }[/tex]
Contraction in aluminium bar without hole = 16000/46750000
Contraction in aluminium bar without hole = 0.000342
Total contraction = 0.000896 + 0.000342
Total contraction = 0.001238 m = 1.238 mm
2) Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.
S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2
S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3
S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2
Answer:
Explanation:
Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.
S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2
S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3
S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2
Strict schedule:
A schedule is strict if it satisfies the following conditions:
Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.
Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.
S3 is not strict because In a strict schedule T3 must read X after C1 but here T3 reads X (r3(X)) before Then T1 has written to X (w1(X)) and T3 commits after T1.
S4 is not strict because In a strict schedule T3 must read X after C1, but here T3 reads X (r3(X)) before T1 has written to X (w1(X)) and T3 commits after T1.
S5 is not strict because T3 reads X (r3(X)) before T1 has written to X (w1(X))
but T3 commits after T1. In a strict schedule T3 must read X after C1.
Cascadeless schedule:
Cascadeless schedule follows the below condition:
Tj reads X only? after Ti has written to X and terminated means aborted or committed.
S3 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S4 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S5 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits or T2 reads Y (r2(Y)) before T3 commits.
But while come to the definition of cascadeless schedules S3, S4, and S4 are not cascadeless, and T3 is not affected if T1 is rolled back in any of the schedules, that is,
T3 does not have to roll back if T1 is rolled back. The problem occurs because these
schedules are not serializable.
Recoverable schedule:
Schedule that follows the below condition:
-----Tj commits after Ti if Tj has?read any data item written by Ti.
Ci > Cj means that Ci happens before Cj. Ai denotes abort Ti. To test if a schedule is
recoverable one has to include abort operations. Thus in testing the recoverability abort
operations will have to used in place of commit one at a time. Also the strictest condition is
------where a transaction neither reads nor writes to a data item, which was written to by a transaction that has not committed yet.
If A1?>C3>C2, then schedule S3 is recoverable because rolling back of T1 does not affect T2 and
T3. If C1>A3>C2. schedule S3 is not recoverable because T2 read the value of Y (r2(Y)) after T3 wrote X (w3(Y)) and T2 committed but T3 rolled back. Thus, T2 used non- existent value of Y. If C1>C3>A3, then S3 is recoverable because roll back of T2 does not affect T1 and T3.
Strictest condition of schedule S3 is C3>C2.
If A1?>C2>C3, then schedule S4 is recoverable because roll back of T1 does not affect T2 and T3. If C1>A2>C3, then schedule S4 is recoverable because the roll back of T2 will restore the value of Y that was read and written to by T3 (w3(Y)). It will not affect T1. If C1>C2>A3, then schedule S4 is not recoverable because T3 will restore the value of Y which was not read by T2.
Find the minimum diameter of an alloy, tensile strength 75 MPa, needed to support a 30 kN load.
Answer:
The minimum diameter to withstand such tensile strength is 22.568 mm.
Explanation:
The allow is experimenting an axial load, so that stress formula for cylidrical sample is:
[tex]\sigma = \frac{P}{A_{c}}[/tex]
[tex]\sigma = \frac{4\cdot P}{\pi \cdot D^{2}}[/tex]
Where:
[tex]\sigma[/tex] - Normal stress, measured in kilopascals.
[tex]P[/tex] - Axial load, measured in kilonewtons.
[tex]A_{c}[/tex] - Cross section area, measured in square meters.
[tex]D[/tex] - Diameter, measured in meters.
Given that [tex]\sigma = 75\times 10^{3}\,kPa[/tex] and [tex]P = 30\,kN[/tex], diameter is now cleared and computed at last:
[tex]D^{2} = \frac{4\cdot P}{\pi \cdot \sigma}[/tex]
[tex]D = 2\sqrt{\frac{P}{\pi \cdot \sigma} }[/tex]
[tex]D = 2 \sqrt{\frac{30\,kN}{\pi \cdot (75\times 10^{3}\,kPa)} }[/tex]
[tex]D = 0.0225\,m[/tex]
[tex]D = 22.568\,mm[/tex]
The minimum diameter to withstand such tensile strength is 22.568 mm.
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2. The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2. (Hint: The police will not go against the law.) a) Find the total time required for the police car to overtake the automobile. (12 marks) b) Find the total distance travelled by the police car while overtaking the automobile. (2 marks) c) Find the speed of the police car at the time it overtakes the automobile
Answer:
A.) Time = 17.13 seconds
B.) Distance = 31.9 m
C.) V = 11.18 m/s
D.) V = 7.1 m/s
Explanation:
The initial velocity U of the automobile is 15.65 m/s.
At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.
For the automobile, let us use first equation of motion
V = U - at.
Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And
V = 0.
Substitute U and a into the formula
0 = 15.65 - 3.05t
15.65 = 3.05t
t = 15.65/3.05
t = 5.13 seconds
But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².
The total time required for the police car to overtake the automobile will be
12 + 5.13 = 17.13 seconds.
b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²
V^2 = U^2 + 2aS
Where S = distance travelled.
Substitute V and a into the formula
11.18^2 = 0 + 2 × 1.96 ×S
124.99 = 3.92S
S = 124.99/3.92
S = 31.88 m
c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s
d.) That will be the final velocity V of the automobile car.
We will use third equation of motion to solve that.
V^2 = U^2 + 2as
V^2 = 15.65^2 - 2 × 3.05 × 31.88
V^2 = 244.9225 - 194.468
V = sqrt( 50.4545)
V = 7.1 m/s
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa. After being compressed, the air is at 450 °C. Determine
(a) the final pressure in [MPa],
(b) the increase in total internal energy in [kJ], and
(c) the total work required in [kJ].
Note that for air R-287 J/kg.K and c.-716.5 J/kg.K, and ?-
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
Scheduling can best be defined as the process used to determine:
Answer:
Overall project duration
Explanation:
Scheduling can best be defined as the process used to determine a overall project duration.