Answer:
None of these
Explanation:
There are different types of amplifiers, and each has different characteristics.
Voltage amplifier needs high input and low output resistance.Current amplifier needs Low Input and High Output resistance.Trans-conductance amplifier Low Input and High Output resistance.Trans-Resistance amplifier requires High Input and Low output resistance.Therefore, the correct answer is "None of these "
Talc and graphite are two of the lowest minerals on the hardness scale. They are also described by terms like greasy or soapy. Both have a crystal structure characterized by sheet-structures at the atomic level, yet they don't behave like micas. What accounts for their unusual physical properties
Answer:
The reason for their unusual properties of the greasy feel and low hardness is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets.
Explanation:
Talc is a monoclinic mineral with a sheet structure similar to the micas and also has perfect cleavage that follows planes between the weakly bonded sheets.
Now, these sheets are held together only by van der Waals bonds and this allows them to slip past each other easily. Thus, this unique characteristic is responsible for talc's extreme softness, its greasy, soapy feel, and its value as a high-temperature lubricant.
While for graphite, it's carbon atoms are linked in a hexagonal network which forms sheets that are one atom thick. It's sheets are poorly connected and easily cleave or slide over one another when subjected to a small amount of force. Thus, gives graphite its very low hardness, its perfect cleavage, and its slippery feel.
So, we can conclude that the reason for their unusual properties is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets; hence, the greasy feel and low hardness.
Caulking is recommended around the edges of partitions between apartments to... Group of answer choices reduce the need for trim. reduce sound transmission. reduce heat loss. increase the fire rating of the partition
Answer:
Reduce sound transmission.
Explanation:
A caulking is a flexible material used to seal joints, cracks or gaps formed between building materials and pipes against leakage.
Caulking is recommended around the edges of partitions between apartments to reduce sound transmission.
Hence, in the event that an individual notices that air or sound is gaining entrance into their apartment, a caulking can be used to mitigate this noise or unwanted sound.
The caulking when applied to the gap or edges of partitions between apartments would create a tight seal and block the flow or entry of air, thereby reducing sound transmission.
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();
}
}
}
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
At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15 mm. The bar is 270-mm long. At the 22-kN load, the stress in the polymer bar is less than its proportional limit. Determine Poisson’s ratio.
Answer:
The Poisson's Ratio of the bar is 0.247
Explanation:
The Poisson's ratio is got by using the formula
Lateral strain / longitudinal strain
Lateral strain = elongation / original width (since we are given the change in width as a result of compession)
Lateral strain = 0.15mm / 40 mm =0.00375
Please note that strain is a dimensionless quantity, hence it has no unit.
The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.
Longitudinal strain = 4.1 mm / 270 mm = 0.015185
Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247
The Poisson's Ratio of the bar is 0.247
Please note also that this quantity also does not have a dimension
: 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:
A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 4200 rpm. When the power is turned on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assume uniformly accelerated motion. determine the number of revolutions that the motor executes
(a) in reaching its rated speed,
(b) in coating to rest.
Answer:
a) [tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex], b) [tex]n = 2450\,rev[/tex]
Explanation:
a) The acceleration experimented by the grinding wheel is:
[tex]\ddot n = \frac{4200\,\frac{rev}{min} - 0 \,\frac{rev}{min} }{\frac{5}{60}\,min }[/tex]
[tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex]
Now, the number of revolutions done by the grinding wheel in that period of time is:
[tex]n = \frac{(4200\,\frac{rev}{min} )^{2}-(0\,\frac{rev}{min} )^{2}}{2\cdot \left(50400\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n = 175\,rev[/tex]
b) The acceleration experimented by the grinding wheel is:
[tex]\ddot n = \frac{0 \,\frac{rev}{min} - 4200\,\frac{rev}{min} }{\frac{70}{60}\,min }[/tex]
[tex]\ddot n = -3600\,\frac{rev}{min^{2}}[/tex]
Now, the number of revolutions done by the grinding wheel in that period of time is:
[tex]n = \frac{(0\,\frac{rev}{min} )^{2} - (4200\,\frac{rev}{min} )^{2}}{2\cdot \left(-3600\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n = 2450\,rev[/tex]
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.11 m2 and whose thickness is 4 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 11 m2 and 0.20 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window
Answer:
Explanation:
Given that,
The area of glass [tex]A_g[/tex] = [tex]0.11m^2[/tex]
The thickness of the glass [tex]t_g=4mm=4\times10^-^3m[/tex]
The area of the styrofoam [tex]A_s=11m^2[/tex]
The thickness of the styrofoam [tex]t_s=0.20m[/tex]
The thermal conductivity of the glass [tex]k_g=0.80J(s.m.C^o)[/tex]
The thermal conductivity of the styrofoam [tex]k_s=0.010J(s.m.C^o)[/tex]
Inside and outside temperature difference is ΔT
The heat loss due to conduction in the window is
[tex]Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j[/tex]
The heat loss due to conduction in the wall is
[tex]Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j[/tex]
The net heat loss of the wall and the window is
[tex]Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j[/tex]
The percentage of heat lost by the window is
[tex]=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%[/tex]
You are tasked with designing a thin-walled vessel to contain a pressurized gas. You are given the parameters that the inner diameter of the tank will be 60 inches and the tank wall thickness will be 5/8" (0.625 inches). The allowable circumferential (hoop) stress and longitudinal stresses cannot exceed 30 ksi.
(1) What is the maximum pressure that can be applied within the tank before failure? = psi(2) If you had the opportunity to construct a spherical tank having an inside diameter of 60 inches and a wall thickness of 5/8" (instead of the thin-walled cylindrical tank as described above), what is the maximum pressure that can be applied to the spherical tank? = psi
Answer:
Explanation:
For cylinder
Diameter d = 60 inches
thickness t = 0.625 inches
circumferential (hoop) stress = 30 ksi
[tex]hoop \ \ stress =\sigma_1=\frac{P_1d}{2t}\\\\\sigma_1=30ksi\\\\30000=\frac{P_1\times 60}{2\times0.625}\\\\P_1=624psi[/tex]
[tex]longitudinal \ \ stress =\sigma_2=\frac{P_2d}{2t}\\\\\sigma_2=30ksi\\\\30000=\frac{P_2\times 60}{4\times0.625}\\\\30000=\frac{P_2\times 60}{2.5}\\\\75000=P_2\times60\\\\P_2=\frac{75000}{60} \\\\P_1=1250psi[/tex]
Therefore maximum pressure without failure is P₁ = 625 psi
ii) For Sphere
[tex]\sigma_1=\sigma_2=\frac{Pd}{4t} \\\\P=\frac{30000\times 4 \times 0.625}{60} \\\\=\frac{75000}{60}\\\\=1250\ \ psi[/tex]
The force of T = 20 N is applied to the cord of negligible mass. Determine the angular velocity of the 20-kg wheel when it has rotated 4 revolutions starting from rest. The wheel has a radius of gyration of kO = 0.3 m.
Image of wheel is missing, so i attached it.
Answer:
ω = 14.95 rad/s
Explanation:
We are given;
Mass of wheel; m = 20kg
T = 20 N
k_o = 0.3 m
Since the wheel starts from rest, T1 = 0.
The mass moment of inertia of the wheel about point O is;
I_o = m(k_o)²
I_o = 20 * (0.3)²
I_o = 1.8 kg.m²
So, T2 = ½•I_o•ω²
T2 = ½ × 1.8 × ω²
T2 = 0.9ω²
Looking at the image of the wheel, it's clear that only T does the work.
Thus, distance is;
s_t = θr
Since 4 revolutions,
s_t = 4(2π) × 0.4
s_t = 3.2π
So, Energy expended = Force x Distance
Wt = T x s_t = 20 × 3.2π = 64π J
Using principle of work-energy, we have;
T1 + W = T2
Plugging in the relevant values, we have;
0 + 64π = 0.9ω²
0.9ω² = 64π
ω² = 64π/0.9
ω = √64π/0.9
ω = 14.95 rad/s
The drum has a mass of 50 kg and a radius of gyration about the pin at O of 0.23 o k m = . If the 15kg block is moving downward at 3 / m s , and a force of P N =100 is applied to the brake arm, determine how far the block descends from the instant the brake is applied until it stops. Neglect the thickness of the handle. The coefficient of kinetic friction at the brake pad is 0.5 k = .
Note: The diagram referred to in this question is attached as a file below.
Answer:
The block descended a distance of 9.75m from the instant the brake is applied until it stops.
Explanation:
For clarity and easiness of expression, the calculations and the Free Body Diagram are contained in the attached file. Check the attached file below.
The block descended a distance of 9.75 m
The uniform sign has a weight of 1500 lb and is supported by the pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in. If the face of the sign is subjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at points C and D. Show the results on a differential volume element located at each of these points. Neglect the thickness of the sign, and assume that it is supported along the outside edge of the pipe.The uniform sign has a weight of 1500 lb and is supported bythe pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in.. If the face of the sign issubjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at pointsC and D. Show the results on a differentialvolume element located at each of these points. Neglect the thickness of the sign, and assume that it issupported along the outside edge of the pipe.
Answer:
See explanation
Explanation:
See the document for the complete FBD and the introductory part of the solution.
Static Balance ( Sum of Forces = 0 ) in all three directions
∑[tex]F_G_X = W - G_x = 0[/tex]
[tex]G_X = W = 1500 lb[/tex]
∑[tex]F_G_Y = P - G_Y = 0[/tex]
[tex]G_Y = P = -10,800 lb[/tex]
∑[tex]F_G_Z = - G_Z = 0[/tex]
Where, ( [tex]G_X, G_Y, G_Z[/tex] ) are internal forces at section ( G ) along the defined coordinate axes.
Static Balance ( Sum of Moments about G = 0 ) in all three directions
[tex]M_G = r_O_G x F_O[/tex]
Where,
r_OG: The vector from point O to point G
F_OG: The force vector at point O
- The vector ( r_OG ) and ( F_OG ) can be written as follows:
[tex]r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\[/tex]
[tex]F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb[/tex]
- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):
[tex]( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k[/tex]
- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:
[tex]T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft[/tex]
[tex]V = P = -10,800 lb[/tex]
- For the state of stress at point "C" we need to determine the the normal stress along x direction ( σ_x ) and planar stress ( τ_xy ) as follows:-
σ_x = [tex]-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}[/tex]
Where,
A: The area of pipe cross section
[tex]A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2[/tex]
z*: The distance of point "C" along z-direction from central axis ( x )
[tex]z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft[/tex]
I_YY: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
y*: The distance of point "C" along y-direction from central axis ( x )
[tex]y* = 0[/tex]
- The normal stress ( σ_x ) becomes:
σ_x = [tex][-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi[/tex]
- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):
τ_xy = [tex]- \frac{T.c}{J} + \frac{V.Q}{I.t}[/tex]
Where,
c: The radial distance from central axis ( x ) and point "C".
[tex]c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft[/tex]
J: The polar moment of inertia of the annular cross section of pipe:
[tex]J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4[/tex]
Q: The first moment of area for point "C" = semi-circle
[tex]Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3[/tex]
I: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
t: The effective thickness of thin walled pipe:
[tex]t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft[/tex]
- The planar stress is ( τ_xy ) becomes:
τ_xy = [tex][ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi[/tex]
- The principal stresses at point "C" can be determined from the following formula:-
σ_x = 15.55 ksi, σ_y = 0 ksi , τ_xy = 52.4 ksi
σ_1 =[tex]\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_2 = [tex]\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_1 = [tex]\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi[/tex]
σ_2 =[tex]-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi[/tex]
- The angle of maximum plane stress ( θ ):
θ = [tex]0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg[/tex]
Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at Brainly, i'm unable to post here.
The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa
Complete Question:
Grain diameter 1 (mm) = 4.4E-02
Yield stress 1 (MPa) = 131
Grain diameter 2 (mm) = 7.7E-03
Yield Stress 2 (MPa) = 268
The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa
Answer:
d = 1.3 * 10⁻² m
Explanation:
According to the Hall Petch equation:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
At [tex]d_{1} = 4.4 * 10^{-2} mm[/tex], [tex]\sigma_{y1} = 131 MPa = 131 N/ mm^2[/tex]
[tex]131 = \sigma_0 + k/\sqrt{4.4 * 10^{-2}} \\k = 27.45 - 0.2096 \sigma_0[/tex]
At [tex]d_{2} = 7.7 * 10^{-3} mm[/tex], [tex]\sigma_{y2} = 131 MPa = 268 N/ mm^2[/tex]
[tex]268 = \sigma_0 + (27.45 - 0.2096 \sigma_0)/\sqrt{7.7 * 10^{-3}} \\23.5036 = 27.47 - 0.1219 \sigma_0\\ \sigma_0 = 32.45 N/mm^2[/tex]
k = 27.45 - 0.2096(32.45)
k = 20.64
At [tex]\sigma_y = 217 MPa[/tex], reapplying Hall Petch law:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
[tex]217 =32.45 + 20.64/\sqrt{d} \\217 - 32.45 = 20.64/\sqrt{d}\\184.55 = 20.64/ \sqrt{d} \\\sqrt{d} = 20.64/184.55\\\sqrt{d} = 0.11184\\d = 0.013 mm[/tex]
d = 1.3 * 10⁻² m
Technician A says that one planetary gear set can provide gear reduction, overdrive, and reverse. Technician B says that most transmissions today use compound (multiple) planetary gear sets. Which technician is correct?
Answer:
Both technician A and technician B are correct
Explanation:
A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.
One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.
So, both technician A and technician B are correct.
An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,500 kg/hr. The feed is known to contain174 kg of aluminum and 2,326 kg of reject. After operating for 1 hour, a total of 256 kg ofmaterials is collected in the product stream. On close inspection, it is found that 140 kg ofproduct is aluminum. Estimate the % recovery of aluminum product and the % purity of thealuminum produc
Answer:
the % recovery of aluminum product is 80.5%
the % purity of the aluminum product is 54.7%
Explanation:
feed rate to separator = 2500 kg/hr
in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine
of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.
After the separation, 256 kg is collected in the product stream.
of this 256 kg, 140 kg is aluminium.
% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material
% recovery of aluminium = 140kg/174kg x 100% = 80.5%
% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream
% purity of the aluminium product = 140kg/256kg
x 100% = 54.7%
Choose two consumer services careers and research online to determine what kinds of professional organizations exist for these professions. Write a paragraph describing the purpose of the organization, the requirements for joining, and the benefits of membership.
Bank loan facilitator, and hospital emergency care specialist are the two consumer or customer services careers.
Bank loan facilitator is a consumer service facilitator who ask and provide people loan in emergency, for the purpose of education, treatment, family events, and for other reasons. For bank loan facilitator the professional organizations should be banking and finance sector. The purpose of these organizations is to help people in financial matter seeking benefit by getting interest from customers. The requirements for joining of the employee must include strong convincing power for the employee, time management, strong and tactful communication skills. Benefits of membership of the customers can help them to seek loans on need basis on lower interest. Hospital emergency care specialist provides help to the staff and the customers in medical emergency. These professionals are necessary for the hospital, clinics, and rehabilitation centers. Purpose of the organization is to provide medical care to the patients. The requirements for joining of the employee includes ability to give information to patients and staff during emergency conditions, facilitating ambulance to rescue patients from their homes, and from other areas, providing medicine, medical equipment, and other facilities to the patients and other medical staff necessary for treatment. Benefits of membership in clinical or hospital settings can help the patient in frequent visits for treatment, concession in laboratory tests, and medication.Learn more about customer:
https://brainly.com/question/13735743
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P
Answer:
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
Explanation:
The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"
Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:
[tex]n = \frac{S_{uts}}{\tau_{max}}[/tex]
Where:
[tex]n[/tex] - Safety factor, dimensionless.
[tex]S_{uts}[/tex] - Ultimate shear strength, measured in pascals.
[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.
The maximum allowable shear stress is consequently cleared and computed: ([tex]n = 4.2[/tex], [tex]S_{uts} = 320\times 10^{6}\,Pa[/tex])
[tex]\tau_{max} = \frac{S_{uts}}{n}[/tex]
[tex]\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}[/tex]
[tex]\tau_{max} = 76.190\times 10^{6}\,Pa[/tex]
Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:
[tex]\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}[/tex]
Where:
[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.
[tex]V[/tex] - Shear force, measured in kilonewtons.
[tex]A[/tex] - Cross section area, measured in square meters.
As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:
[tex]V = \frac{P}{5}[/tex]
[tex]V = \frac{450\,kN}{5}[/tex]
[tex]V = 90\,kN[/tex]
The minimum allowable cross section area is cleared in the shearing stress equation:
[tex]A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}[/tex]
If [tex]V = 90\,kN[/tex] and [tex]\tau_{max} = 76.190\times 10^{3}\,kPa[/tex], the minimum allowable cross section area is:
[tex]A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}[/tex]
[tex]A = 1.640\times 10^{-3}\,m^{2}[/tex]
The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:
[tex]A = \frac{\pi}{4}\cdot D^{2}[/tex]
The diameter is now cleared and computed:
[tex]D = \sqrt{\frac{4}{\pi}\cdot A}[/tex]
[tex]D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})[/tex]
[tex]D = 0.0457\,m[/tex]
[tex]D = 45.7\,mm[/tex]
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
We have that the minimum allowable bolt diameter is mathematically given as
d = 26.65mmFrom the question we are told
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of Assuming P to be P = 425 kN.DiameterGenerally the equation for the stress is mathematically given as
[tex]\mu= 320/4.2 \\\\\mu= 76.190 N/mm^2[/tex]
Therefore
Force = Stress * area
Force = P/2
F= 425,000 N / 2 = 212,500 N
Hence area of each bolt is given as
212,500 = 76.190*( 5* area of each bolt)
area of each bolt = 557.815
Since
area of each bolt=\pi*d^2/4
\pi*d^2/4 = 557.815
d = 26.65mmFor more information on diameter visit
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While having a discussion about O-rings at the bottom of filters, Technician A says that the Automotive Filter Manufacturers Council recommends that the filter O-ring be lubricated with oil after installing the filter. Technician B says that the filter O-ring should be lubricated before installation. Who is correct
Answer:
Technician B is correct
Explanation:
O- rings are used with oil transmission filters to avoid transmission failures but some people use lip seals as well. either of them is inserted onto the outer part of the transmission system i.e it is inserted/found in-between Transmission filters and the transmission systems and it main purpose is to avoid leaks and transmission failure in the short and long term.
0-rings should be lubricated before installation this is because the o-rings are usually super tight when installing and would require lubrication to ease the installation process else the rubber might get ruptured and this would lead to instant transmission failure.
Which statements describe how the Fed responds to high inflation? Check all that apply.
It charges banks more interest.
It pays banks less interest.
It sells more securities.
It decreases the money supply.
It increases the money supply.
Answer:
Answer:
• it charges banks more interest
• it sells more securities
• it decreases the money supply
Explanation:
hope this help edge 21
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
Effluents from metal-finishing plants have the potential of discharging undesirable quantities of metals, such as cadmium, nickel, lead, manganese, and chromium, in forms that are detrimental to water and air quality. A local metal-finishing plant has identified a wastewater stream that contains 5.15 wt% chromium (Cr) and devised the following approach to lowering risk and recovering the valuable metal. The wastewater stream is fed to a treatment unit that removes 95% of the chromium in the feed and recycles it to the plant. The residual liquid stream leaving the treatment unit is sent to a waste lagoon. The treatment unit has a maximum capacity of 4500 kg wastewater/h. If wastewater leaves the finishing plant at a rate higher than the capacity of the treatment unit, the excess (anything above 4500 kg/h) bypasses the unit and combines with the residual liquid leaving the unit, and the combined stream goes to the waste lagoon.
(a) Without assuming a basis of calculation, draw and label a flowchart of the process. (b) Waste water leaves the finishing plant at a rate m_ 1 ? 6000 kg/h. Calculate the flow rate of liquid to
the waste lagoon, m_ 6?kg/h?, and the mass fraction of Cr in this liquid, x6(kg Cr/kg). (c) Calculate the flow rate of the liquid to the waste lagoon and the mass fraction of Cr in this liquid for m_1 varying from 1000 kg/h to 10,000 kg/h in 1000 kg/h increments. Generate a plot of x6 versus m_ 1 .
(Suggestion: Use a spreadsheet for these calculations.) (d) The company has hired you as a consultant to help them determine whether or not to add capacity to the treatment unit to increase the recovery of chromium. What would you need to know to make this determination? (e) What concerns might need to be addressed regarding the waste lagoon?
Answer:
Explanation:
The solution of all the four parts is provided in the attached figures
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.
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.
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.
4. In its natural state, a soil weighs 2800 lb/cy, while in the loose and compacted states, it weighs 2500 lb/cy and 3300 lb/cy, respectively. a. Find the load and shrinkage factors for this soil. b. How many trucks loads with a capacity of 5 lcy/truck would be required to haul 750,000 ccy of this soil to a project
Answer:
a. load factor = 0.893
shrinkage factor = 0.848
b. Number of Trucks loads = 113,585 Trucks loads
Explanation:
Here, we start by identifying the factors as given in the question.
γn = 2800 lb/cy
γloose = 2500 lb/cy
and γcompacted = 3300 lb/cy
a. Mathematically,
Load factor = γloose/γn = 2500/2800 = 0.893
Shrinkage factor = γn/γcompacted = 2800/3300 = 0.848
b. To find the number of trucks loads with a capacity of 5 lcy/truck, we use the mathematical formula as follows;
ρlcy = 5
Load factor × Shrinkage factor = ρloose/γn × γn/γcompacted = ρlcy/ρccy
0.893 × 0.848 = 5/ρccy
ρccy =5/(0.893 × 0.848) = 6.603
The number of truck loads = 750,000/6.603 = 113,584.7 which is approximately 113,585 trucks loads
Suppose you used the pipette to make 10 additions to a flask, and suppose the pipette had a 10% random error in the amount delivered with each delivery. Use equation 1 on page 25 to calculate the percent error in the total volume delivered to the flask using the number of clicks you were permitted to make. Report that total percentage below.
Here is the equation: random error of average= error in one measurement/n^1/2
Answer:
The total percentage is 3.16237%
Explanation:
Solution
Now,
We have to know what a random error is.
A random error is an error in measured caused by factors or elements which varies from one measurement to another.
The random error is shown as follows:
The average random error is = the error found in one measurement/n^1/2
Where
n =Number ( how many times the experiment was done)
Now that we added 10 times we have that,
n → 10
Thus,
The error in one measurement = 10%
So,
The average random error = 10 %/(10)^1/2
= (10)^1/2 %
√10%
The total percentage is = 3.16237%
Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5.5 m/s. The refrigerant gains heat as it flows and leaves the pipe at 180 kPa and 40°C. The specific volumes of R-134a at the inlet and exit are 0.1142 m3/kg and 0.1374 m3/kg. Determine (a) the volume flow rate of the refrigerant at the inlet, (b) the mass flow rate of the refrigerant, and (c) the velocity and volume flow rate at the exit.
Answer:
(a) The volume flow rate of the refrigerant at the inlet is 0.3078 m3/s
(b) The mass flow rate of the refrigerant is 2.695 kg/s
(c) The velocity and volume flow rate at the exit is 6.017 m/s
Explanation:
According to the given data we have the following:
diameter of the pipe=d=28 cm=0.28 m
inlet pressure P1=200 kPa
inlet temperature T1=20°C
inlet velocity V1=5.5 m/s
Exit pressure P2=180 kPa
Exit Temperature T2=40°C
a. To calculate the volume flow rate of the refrigerant at the inlet we would have to use the following formula:
V1=AV1
=π/4(0.28∧2)5
V1=0.3078 m3/s
b. To calculate the mass flow rate of the refrigerant we would have to use the following formula:
m=p1 V1
m=V1/v1
=0.3078/0.11418
=2.695 kg/s
c. To calculate the velocity and volume flow rate at the exit we would have to use the following formula:
m=m1=m2
V1/v1=V2/v2
V2=(v2/v1)v1
=(0.13741/0.11418)5
=6.017 m/s
g A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. The overall length of the bar is L = 570 mm, and the elastic modulus of the material is 77 GPa. If the average normal stress in the central portion of the bar is 200 MPa, calculate the overall elongation δ of the bar.
Answer:
the overall elongation δ of the bar is 1.2337 mm
Explanation:
From the information given :
According to the principle of superposition being applied to the axial load P of the system; we have:
[tex]\delta = \delta_{AB} +\delta_{BC} + \delta_{CD}[/tex]
where;
δ = overall elongation
[tex]\delta _{AB}[/tex] = elongation of bar AB
[tex]\delta _{BC}[/tex] = elongation of bar BC
[tex]\delta _{CD} =[/tex] elongation of bar CD]
If we replace; [tex]\dfrac{PL}{AE}[/tex] for δ and bt for area;
we have:
[tex]\delta = \dfrac{P_{AB}L_{AB}}{(b_{AB}t)E} +\dfrac{P_{BC}L_{BC}}{(b_{BC}t)E}+\dfrac{P_{CD}L_{CD}}{(b_{CD}t)E}[/tex]
where ;
P = load
L = length of the bar
A = area of the cross-section
E = young modulus of elasticity
Let once again replace:
P for [tex]P_{AB}, P_{BC} , P_{CD}[/tex] (since load in all member of AB, BC and CD will remain the same )
[tex]\dfrac{L}{4}[/tex] for [tex]L_{AB}[/tex],
[tex]\dfrac{L}{2}[/tex] for [tex]L_{BC}[/tex] and
[tex]\dfrac{L}{4}[/tex] for [tex]L_{CD}[/tex]
[tex]2\dfrac{b}{3}[/tex] for [tex]b_{BC}[/tex]
b for [tex]b_{CD}[/tex]
[tex]\delta = \dfrac{P (\dfrac{L}{4})}{btE}+ \dfrac{P (\dfrac{L}{2})}{2 \dfrac{b}{3}tE}+\dfrac{P (\dfrac{L}{4})}{btE}[/tex]
[tex]\delta = \dfrac{PL}{btE}[\dfrac{1}{4}+ \dfrac{1}{2}*\dfrac{3}{2}+ \dfrac{1}{4}][/tex]
[tex]\delta = \dfrac{5}{4}\dfrac{PL}{btE} --- \ (1)[/tex]
The stress in the central portion can be calculated as:
[tex]\sigma = \dfrac{P}{A}[/tex]
[tex]\sigma = \dfrac{P}{\dfrac{2}{3}bt}[/tex]
[tex]\sigma = \dfrac{3P}{2bt}[/tex]
So; Now:
[tex]\delta = \dfrac{5}{4}* \dfrac{2 * \sigma}{3}*\dfrac{L}{E}[/tex]
[tex]\delta= \dfrac{5}{4}* \dfrac{2 * 200}{3}*\dfrac{570}{77*10^3 \ MPa}[/tex]
δ = 1.2337 mm
Therefore, the overall elongation δ of the bar is 1.2337 mm
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.
A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter in a steel alloy when a load of 500 kg was used. Calculate the Brinell hardness (in HB) of this material. Enter your answer in accordance to the question statement HB
Answer:
HB = 3.22
Explanation:
The formula to calculate the Brinell Hardness is given as follows:
[tex]HB = \frac{2P}{\pi D\sqrt{D^{2}- d^{2} } }[/tex]
where,
HB = Brinell Hardness = ?
P = Applied Load in kg = 500 kg
D = Diameter of Indenter in mm = 10 mm
d = Diameter of the indentation in mm = 1.55 mm
Therefore, using these values, we get:
[tex]HB = \frac{(2)(500)}{\pi (10)\sqrt{10^{2}- 1.55^{2} } }[/tex]
HB = 3.22
