Answer:
In Java:
import java.util.Scanner;
import java.lang.Math;
public class MyClass {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
float x, y, z;
System.out.print("Enter three numbers: ");
x = input.nextFloat(); y = input.nextFloat(); z = input.nextFloat();
System.out.print("Square root of x: ");
System.out.printf("%.2f",Math.sqrt(x));
System.out.println();
System.out.print("Absolute of y - z: ");
System.out.printf("%.2f",Math.abs(y - z));
System.out.println();
int fact = 1;
for(int i = 1;i<=(int)Math.ceil(z);i++){
fact*=i;
}
System.out.print("Factorial of z: "+fact);
}
}
Explanation:
This line declares x, y and z as float
float x, y, z;
This line prompts user for three numbers
System.out.print("Enter three numbers: ");
The next line get input for x, y and z
x = input.nextFloat(); y = input.nextFloat(); z = input.nextFloat();
The italicized prints the square root of x rounded to 2 decimal places
System.out.print("Square root of x: ");
System.out.printf("%.2f",Math.sqrt(x));
System.out.println();
The italicized prints the absolute of y - z to 2 decimal places
System.out.print("Absolute of y - z: ");
System.out.printf("%.2f",Math.abs(y - z));
System.out.println();
This initializes fact to 1
int fact = 1;
The following iteration calculates the factorial of ceil z
for(int i = 1;i<=(int)Math.ceil(z);i++){
fact*=i;
}
This prints the calculated factorial
System.out.print("Factorial of z: "+fact);
}
}
Write the heat equation for each of the following cases:
a. A wall, steady state, stationary, one-dimensional, incompressible and no energy generation.
b. A wall, transient, stationary, one-dimensional, incompressible, constant k with energy generation.
c. A cylinder, steady state, stationary, two-dimensional (radial and axial), constant k, incompressible, with no energy generation.
d. A wire moving through a furnace with constant velocity, steady state, one-dimensional (axial), incompressible, constant k and no energy generation.
e. A sphere, transient, stationary, one-dimensional (radial), incompressible, constant k with energy generation.
Answer:
Explanation:
a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:
[tex]\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)[/tex]
b) For a transient, 1-D, constant with energy generation
suppose T = f(x)
Then; the equation can be expressed as:
[tex]\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}[/tex]
where;
[tex]Q_g[/tex] = heat generated per unit volume
[tex]\alpha[/tex] = Thermal diffusivity
c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:
[tex]\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0[/tex]
where;
The radial directional term = [tex]\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})[/tex] and the axial directional term is [tex]\dfrac{\partial^2 T}{\partial z^2 }[/tex]
d) The heat equation for a wire going through a furnace is:
[tex]\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ][/tex]
since;
the steady-state is zero, Then:
[tex]\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ][/tex]'
e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:
[tex]\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}[/tex]
A consolidated-drained triaxial test is carried out on a sand specimen that is subjected to 80 kN/m2 confining pressure. The vertical deviator stress was increased slowly such that there is no built-up of pore water pressure within the specimen. The specimen failed when the addition axial stress reached 240 kN/m2. Find the friction angle of the sand. If another identical sand specimen was subjected to 150 kN/m2 confining pressure, what would be the deviator stress at failure.
Answer:
a) the friction angle of the sand is 36.87°
b) the deviator stress at failure is 450 kN/m³
Explanation:
Given the data in the question;
For a consolidated drained test
The effective major principle stresses
σ₃ = σ₃' = 80 kN/m²
and
σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 80 kN/m² + 240 kN/m² = 320 kN/m²
now
a) friction angle of the sand
σ₁' = σ₃'tan²( 45° + Ф/2' ) + 2c' tan( 45° + Ф/2 )
for sand; c' = 0
so
σ₁' = σ₃'tan²( 45° + Ф/2' )
we substitute
320 = 80 tan²( 45° + Ф/2' )
Ф' = 2 × [ tan⁻¹ (√[tex]\frac{320}{80}[/tex]) - 45° ]
Ф' = 2 × [ 63.4349° - 45° ]
Ф' = 2 × 18.4349
Ф' = 36.87°
Therefore, the friction angle of the sand is 36.87°
b) deviator stress
σ₁' = σ₃'tan²( 45° + Ф/2' )
σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )
σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' ) 3.597
150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 150tan²( 45° + 36.87°/2 )
150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600
(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600 - 150
(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 450 kN/m³
Therefore, the deviator stress at failure is 450 kN/m³
Select the correct answer from each drop-down menu. Choose the correct words.To complete the statements about career planning. Throughout your job search, you'll find that is closely related to the career of your choosing. It's important to take the time to find out what to expect now, so you can start developing the you need to excel in that career. Reset Next
Complete Question:
Throughout your job search, you’ll find that _____(the economy/education/salary) is closely related to the career of your choosing. It’s important to take the time to find out what to expect now, so you can start developing the ____ (task/communication/skill) that you need to excel in that career.
Answer:
Education; skills.
Explanation:
Throughout your job search, you'll find that education is closely related to the career of your choosing. It's important to take the time to find out what to expect now, so you can start developing the skills you need to excel in that career.
This ultimately implies that, all job openings that are being advertised by various organizations usually have a minimum requirement such as academic experience or level i.e the prospective candidate must have attained a level of education. Also, it is very important for undergraduates and potential employees to ensure that they are being proactive, as well as developing the requisite skills needed to excel in their career.
Answer:
both are c
Explanation:
A Russian rocket engine (RD-110 with LOX-kerosene) consists of four thrust chambers supplied by a single turbopump. The exhaust from the turbine of the turbopump then is ducted to four vernier nozzles (which can be rotated to provide some control of the flight path). Using the information below, determine the thrust and mass flow rate of the four vernier gas nozzles. For individual thrust chambers (vacuum): F= 73.14 kN, c = 3279 m/sec Overall engine with verniers (vacuum): F= 297.93 kN, c = 3197 m/sec.
Answer:
- Vernier thrust is 5.37 kN
- mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s
Explanation:
Given that;
For individual thrust chambers (vacuum);
Fc = 73.14 kN , Cc = 3279 m/sec
For Overall engine with Vernier (vacuum);
Foa = 297.93 kN = , Coa = 3197 m/sec.
- determine the Vernier thrust
Vernier thrust Fv = Foa - ( 4 × Fc )
Vernier thrust Fv = 297.93 - ( 4 × 73.14)
Vernier thrust Fv = 297.93 - 292.56
Vernier thrust Fv = 5.37 kN
Therefore, Vernier thrust is 5.37 kN
-
Vernier mass flow rate;
we know that
[tex]Co_{a}[/tex] = Fc + Fv / mc + mv
mv = Foa/Coa - Fc/Cc
we convert kilonewton to kilograms
1 kn = 102 kg
Fc = 73.14 kN = 73.14 × 102 = 7460.28 kg
Foa = 297.93 kN = 297.93 × 102 = 30388.86 kg
we substitute
mv = (30388.86 / 3197) - (( 4 × 7460.28) / 3279)
mv = 9.5054 - 9.1006
mv = 0.4048 kg/s
Therefore, mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s
A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. An old building (sight obstruction) is located 30 feet from the edge of the innermost lane. The road is level and the superelevation is 0.06. Please determine the maximum speed for safe vehicle operation on this horizontal curve.
Answer:
maximum speed for safe vehicle operation = 55mph
Explanation:
Given data :
radius ( R ) = 678 ft
old building located ( m )= 30 ft
super elevation = 0.06
Determine the maximum speed for safe vehicle operation
firstly calculate the stopping sight distance
m = R ( 1 - cos [tex]\frac{28.655*S}{R}[/tex] ) ---- ( 1 )
R = 678
m ( horizontal sightline ) = 30 ft
back to equation 1
30 = 678 ( 1 - cos (28.655 *s / 678 ) )
( 1 - cos (28.655 *s / 678 ) ) = 30 / 678 = 0.044
cos [tex]\frac{28.65 *s }{678}[/tex] = 1.044
hence ; 28.65 * s = 678 * 0.2956
s = 6.99 ≈ 7 ft
next we will calculate the design speed ( u ) using the formula below
S = 1.47 ut + [tex]\frac{u^2}{30(\frac{a}{3.2} )-G1}[/tex] ---- ( 2 )
t = reaction time, a = vehicle acceleration, G1 = grade percentage
assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0
back to equation 2
6.99 = 1.47 * u * 2.5 + [tex]\frac{u^2}{30[(11.2/32.2)-0 ]}[/tex]
3.675 u + 0.0958 u^2 - 6.99 = 0
u ( 3.675 + 0.0958 u ) = 6.99
why you so mean to me? leave my questions please. answer them
Answer: Why is even here then.
Explanation:
A freezer compartment consists of a cubical cavity that is 2 m on a side. Assume the bottom to be perfectly Problems 49 CH001.qxd 2/24/11 12:03 PM Page 49 insulated. What is the minimum thickness of styrofoam insulation (k 0.030 W/m K) that must be applied to the top and side walls to ensure a heat load of less than 500 W, when the inner and outer surfaces are 10 and 35 C
Answer:
30 mm is the minimum thickness that must be applied.
Explanation:
Given the data in the question;
Using Fourier's equation. the heat rate is
q = kA(ΔT/Δx)
where
A is the surface area, we must consider all surfaces through which the heat can dissipate through
i.e 2×2 for one wall gives you 4m²,
there are 5 walls, so we will have 20m² for surface area.
k is thermal conductivity of the styrofoam ( 0.030 W/m K)
q is the heat loss (500 W )
ΔT is the Temperature difference ( 35 - 10) = 25°C
Δx = ?
So we substitute
500 = (0.030)(20)(25/Δx)
500 = 0.6 (25/Δx)
500 = 15 / Δx
Δx = 15 / 500
Δx = 0.03 m = 30 mm
Therefore, 30 mm is the minimum thickness that must be applied.
The accompanying specific gravity values describe various wood types used in construction. 0.320.350.360.360.370.380.400.400.40 0.410.410.420.420.420.420.420.430.44 0.450.460.460.470.480.480.490.510.54 0.540.550.580.630.660.660.670.680.78 Construct a stem-and-leaf display using repeated stems. (Enter numbers from smallest to largest separated by spaces. Enter NONE for stems with no values.)
Answer:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]
Explanation:
Given
[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,[/tex]
[tex]0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,[/tex]
[tex]0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,[/tex]
[tex]0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.[/tex]
Required
Plot a steam and leaf display for the given data
Start by categorizing the data by their tenth values:
[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.[/tex]
[tex]0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,[/tex]
[tex]0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.[/tex]
[tex]0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.[/tex]
[tex]0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.[/tex]
[tex]0.78.[/tex]
The 0.3's is will be plotted as thus:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]
The 0.4's is as follows:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}[/tex]
The 0.5's is as follows:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}[/tex]
The 0.6's is as thus:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]
Lastly, the 0.7's is as thus:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]
The combined steam and leaf plot is:
[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]
Select four types of engineers who might be involved in the development of a product such as an iPhone.
(select four)
-hydraulic engineers
-geodetic engineers
-chemical engineers
-electrical engineers
-computer engineers
-manufacturing engineers
-mechanical engineers
-civil engineers
Answer:
electrical
computer
mechanical
and manufacturing .... I think
The answer is electrical, manufacture, computer, and chemical
PWM input and output signals are often converted to analog voltage signals using low-pass filters. Design and simulate the following: a PWM signal source with 1 kHz base frequency and adjustable pulse-width modulation(PWM source), an analog filter with time constant of 0.01 s. Simulate the input and output of the low pass filter for a PWM duty cycle of 25, 50, and 100%
Answer:
Attached below
Explanation:
PWM signal source has 1 KHz base frequency
Analog filter : with time constant = 0.01 s
low pass transfer function = [tex]\frac{1}{0.01s + 1 }[/tex]
PWM duty cycle is a constant block
Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively
A site is underlain by two layers of normally consolidated clayey sand. The unit weights of the top layer are 19 and 21 kN/m2 and the layer is 6 meters thick. The unit weights of the bottom layer are 20 and 22 kN/m, and the layer is 8 meters thick. Below the bottom layer lies bedrock. The water table is located 2 meters below the ground surface. The top layer soil has a friction angle of 38 degrees, a cohesion intercept of 20 kPa, and an undrained strength of 160 kPa. The bottom layer soil has a friction angle of 33 degrees, a cohesion intercept of 10 kPa, and an undrained strength of 120 kPa. Point A is located 9 meters underground. All layers have a lateral stress ratio of 0.5.
Required:
a. Draw the profile neatly, add dimensions, and draw a tree on the ground surface.
b. Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A.
c. Draw the Mohr Circle to determine the effective stress that acts at point A on a plane inclined 10 degrees counter-clockwise from the horizontal plane. Make sure the Mohr Circle is a well-drawn circle
Answer:
A) attached below
B) Geostatic vertical effective stress ( бv )
= 119.33 KN/m^2
Geostatic horizontal effective stress ( бn )
= 59.66 KN/m^2
C) attached below
Explanation:
attached below is a detailed solution
A) attached below
B) Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A
Geostatic vertical effective stress ( бv )
= 119.33 KN/m^2
Geostatic horizontal effective stress ( бn )
= 59.66 KN/m^2
C) attached below
A driver counts 21 other vehicles using 3 EB lanes on one section of I-80 between her rented car and an overpass ahead. It turned out to be 0.78 mile to the overpass from the point at which she started her vehicle count. The vehicles were moving at approximately 52 mph at the time. Traffic was distributed evenly over the three EB lanes.
Required:
a. Calculate the vehicle density per lane (excluding the driver's rental vehicle) on that stretch of EB I-80.
b. What was the flow rate for that section of I-80?
Answer:
vehicle density = 28.205 veh/mile
flow rate = 0.909 veh/hr
Explanation:
given data
count n = 21
distance = 0.78 miles
speed = 52 mph
solution
we get here vehicle density that is express as
vehicle density = n ÷ distance ...............1
vehicle density = ( 21 + 1 ) ÷ 0.78
vehicle density k = 28.205 veh/mile
and
now we get here flow rate that is express as
flow rate = k × vs .................2
flow rate = 28.205 × ( 52 × 0.00062 ÷ 1m )
flow rate = 0.909 veh/hr
There are four people who want to cross a rickety bridge; they all begin on the same side. You have 17 minutes to get them all across to the other side. It is night, and they have one flashlight. A maximum of two people can cross the bridge at one time. Any party that crosses, either one or two people, must have the flashlight with them. The flashlight must be walked back and forth; it cannot be thrown, for example. Person 1 takes 1 minute to cross the bridge, person 2 takes 2 minutes, person 3 takes 5 minutes, and person 4 takes 10 minutes. A pair must walk together at the rate of the slower person’s pace.
Required:
What is the shortest time needed for all four of them to cross the bridge?
Answer:
The shortest time needed for all four of them to cross the bridge = 17 minutes
Explanation:
As given , to cross the bridge
Person 1 takes 1 minute,
Person 2 takes 2 minutes,
Person 3 takes 5 minutes, and
Person 4 takes 10 minutes.
For the first time ,
Person 1 and Person 2 will go to cross the bridge
As Person 2 takes 2 minutes
So, total time taken by Person 1 and person 2 together will be 2 minutes
Now,
Person 1 will come back with flashlight.
He takes 1 minutes.
So, Total time becomes 2 + 1 = 3 minutes
Then,
Person 3 and Person 4 will cross the bridge
As Person 4 takes 10 minutes
So, total time taken by Person 3 and person 4 together will be 10 minutes
So, Total time becomes 3 + 10 = 13 minutes
Now,
Person 2 will come back with flashlight.
He takes 2 minutes.
So, Total time becomes 13 + 2 = 15 minutes
Then,
Person 1 and Person 2 will cross the bridge
As Person 2 takes 2 minutes
So, total time taken by Person 1 and person 2 together will be 2 minutes
So, Total time becomes 15 + 2 = 17 minutes
∴ we get
The shortest time needed for all four of them to cross the bridge = 17 minutes
Ethylene glycol, the ingredient in antifreeze, does not cause health problems because it is a clear liquid
Answer:
False
Explanation:
I got it wrong picking true
3/4 + 1/2
Ashskfnrjcisj
The values of four out of a sample of five deviations are: -5, +2, +4, -2. What is The value of the fifth deviation?
Answer:
The fifth deviation is +1
Explanation:
Given
[tex]Deviations = -5, +2, +4,-2[/tex]
Required
Determine the fifth deviation
Represent the fifth deviation with x.
As a theorem, the sum of deviations from mean is always 0.
So, we have:
[tex]-5 +2 +4-2+x = 0[/tex]
[tex]-1+x = 0[/tex]
Add 1 to both sides
[tex]1 -1+x = 0 +1[/tex]
[tex]x = 0 +1[/tex]
[tex]x= +\ 1[/tex]
Hence, the fifth deviation is +1
A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength is found to be 460 MPa. If the cross-sectional diameter at fracture is 10.7 mm, determine (max. pts. 8): a. The ductility in terms of percent reduction in area b. The true stress at fracture
Answer:
a) The ductility = -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) the true stress at fracture is 658.26 Mpa
Explanation:
Given that;
Original diameter [tex]d_{o}[/tex] = 12.8 mm
Final diameter [tex]d_{f}[/tex] = 10.7
Engineering stress [tex]\alpha _{E}[/tex] = 460 Mpa
a) determine The ductility in terms of percent reduction in area;
Ai = π/4([tex]d_{o}[/tex] )² ; Ag = π/4([tex]d_{f}[/tex] )²
% = π/4 [ ( ([tex]d_{f}[/tex] )² - ([tex]d_{o}[/tex] )²) / ( π/4 ([tex]d_{o}[/tex] )²) ]
= ( ([tex]d_{f}[/tex] )² - ([tex]d_{o}[/tex] )²) / ([tex]d_{o}[/tex] )² × 100
we substitute
= [( (10.7)² - (12.8)²) / (12.8)² ] × 100
= [(114.49 - 163.84) / 163.84 ] × 100
= - 0.3012 × 100
= -30.12%
the negative sign means reduction
Therefore, there is 30.12% reduction
b) The true stress at fracture;
True stress [tex]\alpha _{T}[/tex] = [tex]\alpha _{E}[/tex] ( 1 + [tex]E_{E}[/tex] )
[tex]E_{E}[/tex] is engineering strain
[tex]E_{E}[/tex] = dL / Lo
= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49
= 49.35 / 114.49
[tex]E_{E}[/tex] = 0.431
so we substitute the value of [tex]E_{E}[/tex] into our initial equation;
True stress [tex]\alpha _{T}[/tex] = 460 ( 1 + 0.431)
True stress [tex]\alpha _{T}[/tex] = 460 (1.431)
True stress [tex]\alpha _{T}[/tex] = 658.26 Mpa
Therefore, the true stress at fracture is 658.26 Mpa
Engineer drawing:
How can i draw this? Any simple way?
The input and output signals of a system is related by the following equation: fraction numerator d squared y over denominator d t squared end fraction plus sin (3 y )fraction numerator d y over denominator d t end fraction plus y equals t fraction numerator d f over denominator d t end fraction plus f. Then, the system is:
Answer:
Explanation:
The given equation is :
[tex]\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f[/tex]
Drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been
removed.
Answer:
no it has to be removed
Explanation:
It is completely inappropriate to mention that the drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been removed. Therefore, the statement given above is false.
What is the significance of drum brakes?Drum brakes can be referred to or considered as the types of brakes that are useful in application of brakes to an object, such as wheels, in motion. To understand better, it can be stated that the system of braking under drum brakes is completely in contrast to that of disc brakes.
Drum brakes have a hydraulic pressure, which means that if the condition of lining is to be checked, removal of drums becomes essential. If the drums are not removed, correction or alignment of wheels cannot be performed.
Therefore, the significance regarding drum brakes has been aforementioned, and the statement given above with respect to their removal also holds false.
Learn more about drum brakes here:
https://brainly.com/question/14937026
#SPJ2
Consider a Rankine cycle with reheat using water (steam) as the working fluid. Saturated liquid at 100C exits the condenser. The fluid exits the boiler at 6500C, 20000 kPa. The cycle uses a single reheater with operating pressure 200 kPa and exit temperature 5000C. The mass flow rate of water (steam) in the cycle is 40 kg/s. The operation of the turbines and pump are adiabatic and have isentropic efficiency 0.90. As usual, neglect pressure losses in piping, boiler, reheater, and condenser. Compute: quality at the exit of the high-pressure turbine.
identify which country has an absolute advantage in production of cookies and which has the absolute advantage in production of milk
a) Question Completion:
INPUT HOURS OF LABOR
Country Cookies Milk
Atlantis 2 hours 1 hour
Neverland 4 hours 1 hour
Answer:
1. Atlantis has the absolute advantage in the production of cookies.
2. No country has the absolute advantage in the production of milk.
Explanation:
Absolute advantage refers to superior production capability. It is determined when a country, for example, has the ability to produce a particular good or service at lower cost or more efficiently (i.e. with less resources) than the other country. In the scenario above, Atlantis has an absolute advantage in the production of cookies because it can produce the same quantity of cookies using 2 labor hours that Neverland can produce using 4 labor hours. But for the production of milk, Atlantis and Neverland share the same comparative advantage less they can use less labor hours to produce milk than they can produce cookies.
A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account for 36%, 47% and 17% of water flow respectively. How much water is discharged in 30 minutes from tributary B?
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
When an electron in a valence band is raised to a conduction band by sufficient light energy, semiconductors start conducting ________.
Answer:
This band gap also allows semiconductors to convert light into electricity in photovoltaic cells and to emit light as LEDs when made into certain types of diodes. Both these processes rely on the energy absorbed or released by electrons moving between the conduction and valence bands.
Explanation:
On the internet
Conduction is thermal energy transfer by:_______.
a. molecular interactions.
b. bulk (macroscopic) particle motion.
c. electromagnetic waves inside the solid.
d. a combination of molecular interactions and macroscopic particle motion.
Answer:
a. molecular interactions.
Explanation:
Conduction is thermal energy transfer by molecular interactions. Therefore, conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.
In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.
Some examples of conductors include metal, steel, aluminum, copper, graphite, etc.
Hence, conduction is thermal energy transfer as a result of the movement of electrons and collision between the molecules of an object.
The steel framework is used to support the reinforced stone concrete slab that is used for an office. The slab is 200 mm thick. Calculate the value of the distributed load on BE. Take a = 2 m, b = 5 m. Assume density for concrete slab = 23.6 kN/m3. Hint: Look at Example 2 from class notes.
Answer:
Total distributed load on BE = 5 m²
Explanation:
The first process is to get the value for the Dead load (DL) on the slab;
This is determined by using the formula:
DL = ρ × t
DL = 23.6 kK/m³ × 0.2 m
DL = 4.72 kN/m²
From table 1.4 which relates to the office buildings, we derive the value for the minimum live load (LL) = 2.40 kN/m³
Hence the total load TL = Dead Load DL) + Live loaf (LL)
DL = (4.72 + 2.40) kN/m³
DL = 7.12 kN/m³
Now; from the imaginative view of the information given; the member BE which get the load from half area of the panel BEDC & half the area of BEFA panel parallel to member BE; then the tributary area on member BE can be calculated as;
[tex]A_{BE} = ( \dfrac{a}{2}+\dfrac{a}{2}) \times width[/tex]
[tex]A_{BE} = ( \dfrac{2}{2}+\dfrac{2}{2}) \times 1[/tex]
[tex]A_{BE} =2\times 1[/tex]
[tex]A_{BE} =2 m^2/m[/tex]
The total distributed load acting on BE is:
[tex]Total \ load = TL \times A_{BE}[/tex]
[tex]Total \ load = 7.12 \ \dfrac{kN}{m^2 }\times (2 \times \dfrac{m^2}{m})[/tex]
Total load = 2.5 × 2
Total load = 5 m²
Perform the following unit conversions. Please do not use an on-line unit converter since this problem is given to you as practice in preparation for what you need to be proficient in:
a. 180 in^3 to L
b. 750 ft-lbf to kJ
c. 75.0 hp to kW
d. 2500.0 lb/h to kg/s
e. 120 psia to kPa
f. 120 psig to kPa
g. 300 ft/min to m/s
h. 125 km/h to miles/h
i. 6000 N to Ibf
j. 6000 N to ton
Answer:
The answers are below
Explanation:
a. 180 in^3 to L
1 in³ = 0.0164L
180 in³ = [tex]180\ in^3*\frac{0.0164\ L}{1\ in^3}= 2.95\ L[/tex]
b. 750 ft-lbf to kJ
1 ft-lbf = 0.00136 kJ
750 ft-lbf = [tex]750\ ft-lbf *\frac{0.00136\ kJ}{1\ ft-lbf} =1.02\ kJ[/tex]
c. 75.0 hp to kW
1 hp = 0.746 kW
75 hp = [tex]75\ hp*\frac{0.746\ kW}{1\ hp}=55.95\ kW[/tex]
d. 2500.0 lb/h to kg/s
1 lb/h = 0.000126 kg/s
2500.0 lb/h = [tex]2500.0\ lb/h*\frac{0.000126\ kg/s}{1\ lb/h} =0.315\ kg/s[/tex]
e. 120 psia to kPa
1 psia = 6.89 kPa
120 psia = [tex]120\ psia*\frac{6.89\ kPa}{1\ psia} =826.8\ kPa[/tex]
f. 120 psig to kPa
1 psig = 6.89 kPa
120 psig = [tex]120\ psia*\frac{6.89\ kPa}{1\ psig} =826.8\ kPa[/tex]
g. 300 ft/min to m/s
1 ft/min = 0.005 m/s
300 ft/min = [tex]300\ ft/min*\frac{0.005\ m/s}{1\ ft/min} = 1.5\ m/s\\[/tex]
h. 125 km/h to miles/h
1 km/h = 0.62 mph
125 km/h = [tex]125\ km/h*\frac{0.62\ mph}{1\ km/h} =77.5\ mph[/tex]
i) 6000 N to Ibf
1 N = 0.2248 lbf
6000 N = [tex]6000\ N*\frac{ 0.2248\ lbf}{1\ N}=1348.8\ N[/tex]
j. 6000 N to ton
1 N = 0.000102 Ton-force
6000 N = [tex]6000\ N*\frac{ 0.000102\ Ton-force}{1\ N}=0.612\ N[/tex]
64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump efficiency of 74 percent, determine the required power input to the pump. Disregard frictional losses in the pipes, and assume the geothermal water at 200m depth to be exposed to the atmosphere.
Answer:
835,175.68W
Explanation:
Calculation to determine the required power input to the pump
First step is to calculate the power needed
Using this formula
P=V*p*g*h
Where,
P represent power
V represent Volume flow rate =0.3 m³/s
p represent brine density=1050 kg/m³
g represent gravity=9.81m/s²
h represent height=200m
Let plug in the formula
P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m
P=618,030 W
Now let calculate the required power input to the pump
Using this formula
Required power input=P/μ
Where,
P represent power=618,030 W
μ represent pump efficiency=74%
Let plug in the formula
Required power input=618,030W/0.74
Required power input=835,175.68W
Therefore the required power input to the pump will be 835,175.68W
Axial forces in a column due to service loads are as follows (assume the live load used in these calculationsis less than 100 psf):Dead:150 kcompressionLive:280 kcompressionRoof Live:40 kcompressionSnow:50 kcompressionWind:120 kcompression or tensionEarthquake:200 kcompression or tension1. Compute the required axial strength,TuandPu, for this column in tension and compression usingLRFD load combinations. (Neglect self-weight.)2. Describe the loading scenario that represents the worst case tension and compression loading for thecolumn (remember, wind and earthquake loading can act in either direction).
Answer:
a) attached below
b) The worst case in tension is case ( 9-7 ) b which is
= -65k
The worst case in compression is case ( 9-5 ) a which is
= 670 k
Explanation:
Given data :
D = 150k , L = 280k , Lr = 40k , s = 50k , w = ± 120k
E = ± 200k
attached below is a detailed solution to the given problem ( problem 1 )
A) attached below
b) The worst case in tension is case ( 9-7 ) b which is
= -65k
The worst case in compression is case ( 9-5 ) a which is
= 670 k
How is varnished timber shaped and cut?