The two signals x1 and x2 are periodic signals with different periods.
Signal x1 is a periodic signal with a period of 9 samples, and each sample is a cosine wave with a frequency of 2π/9 radians per sample. Signal x2 is a periodic signal with a period of 10 samples, and each sample is a cosine wave with a frequency of 2π/10 radians per sample.
The DFT of each signal X1 and X2 is a set of complex numbers that represent the frequency content of each signal. The DFT of x1 shows a single non-zero frequency component at index 1, while the DFT of x2 shows two non-zero frequency components at indices 1 and 9.
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Select the correct answer.
Alan is walking on a street and passes by a very busy construction site. He sees engineers hanging one end of various objects to the ceiling and the other end to heavy
weights. They observe the objects and then then take down notes. What are the engineers doing
Answer: the engineers are performing a tensile test
Explanation:
i took the test and got it right
Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and (v) time invariant. For invertibility, either find an inverse system or an example of two inputs that lead to the same output. Note that y[n] denotes the system output and x[n] denotes the system input.
a. y[n] = x[n] x[n-1] + [n+1]
b. y[n] = cos(x[n])
Answer:
a.
y[n] = x[n] x[n-1] x[n+1]
(i) Memory-less - It is not memory-less because the given system is depend on past or future values.
(ii) Causal - It is non-casual because the present value of output depend on the future value of input.
(iii) Invertible - It is invertible and the inverse of the given system is [tex]\frac{1}{x[n] . x[n-1] x[n+1]}[/tex]
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.
b.
y[n] = cos(x[n])
(i) Memory-less - It is memory-less because the given system is not depend on past or future values.
(ii) Causal - It is casual because the present value of output does not depend on the future value of input.
(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .
For example - for x[n] = 0 , y[n] = cos(0) = 1
for x[n] = 2[tex]\pi[/tex] , y[n] = cos(2[tex]\pi[/tex]) = 1
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.
Where does Burj Khalifa located? and how many meters?
Answer:
Burj Khalifa is located in dubai UAE at over 828m
Explanation:
828 metres
Answer:
The Burji Khalifa, known as the Burj Dubai prior to its inauguration in 2010, is a skyscraper in Dubai, United Arab Emirates. With a total hight of 829.8 m and a roof hight of 828 m, the Burji Khalifa has been the tallest structure and building in the world since its topping out in 2009.
A hoop of mass m and radius r starts from rest and rolls down an incline at an angle θ. The hoop’s inertia is given by IG = mr 2. The static friction coefficient is μs. Determine the acceleration of the center of mass aGx and the angular acceleration α. Assume that the hoop rolls without bouncing or slipping. Use two approaches to solve the problem: (a) Use the moment equation about the mass center G and (b) use the moment equation about the contact point P. (c) Obtain the frictional condition required for the hoop to roll without slipping.
The acceleration of the center of mass aGx is gsinθ/(1+I/mr^{2}), and the angular acceleration α is gsinθ/r(1+I/mr^{2}).
To find the acceleration of the center of mass aGx and the angular acceleration α of a hoop rolling down an incline at an angle θ, we can use two approaches. The first approach is to use the moment equation about the mass center G, which gives us aGx = gsinθ/(1+I/mr^{2}) and α = gsinθ/r(1+I/mr^{2}). The second approach is to use the moment equation about the contact point P, which gives us the same results. To ensure that the hoop rolls without slipping, we need to have a frictional force that is greater than or equal to the static friction coefficient μs times the normal force, which is equal to mgcosθ. Therefore, the required frictional condition is μs ≥ gcosθ/(1+I/mr^{2}).
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The three sub regions of South America are the Andes Mountains, the Amazon Rainforest, and the Eastern Highlands. The Atacama Desert is the driest place on Earth.
Answer:
<:
Explanation:
Answer:
d
Explanation:
Experiment with a simple derivation relationship between two classes. Put println statements in constructors of both the parent and child classes. Do not explicitly call the constructor of the parent in the child classes. Do not explicitly call the constructor of teh parent in the child. What happens? Why? Change the child's constructor to explicitly call the constructor of the parent. Now what happens?
I need an example program in java, because I can't visualize what I am supposed to do, and I do need help with that, if you could send me the sample programs I would be grateful thank you.
In Java, when a class is derived from another class, a relationship is formed between them. This relationship is known as the inheritance relationship, where the derived class inherits properties and methods from the parent class.
When you create a new object of the child class, the constructor of the parent class is automatically called before the constructor of the child class is executed. This is because the child class needs to initialize all the properties that it inherited from the parent class.
Now, if you experiment with a simple derivation relationship between two classes and put println statements in constructors of both the parent and child classes, but do not explicitly call the constructor of the parent in the child classes, you will see that the parent class constructor is still called before the child class constructor. This is because it is done implicitly by Java.
Here is a sample program that demonstrates this:
```
class Parent {
public Parent() {
System.out.println("Parent constructor called");
}
}
class Child extends Parent {
public Child() {
System.out.println("Child constructor called");
}
}
public class Main {
public static void main(String[] args) {
Child childObj = new Child();
}
}
```
If you run this program, you will see the output as:
```
Parent constructor called
Child constructor called
```
Now, if you change the child's constructor to explicitly call the constructor of the parent, you will see that the output remains the same. However, the parent constructor is called explicitly this time.
Here is the modified sample program:
```
class Parent {
public Parent() {
System.out.println("Parent constructor called");
}
}
class Child extends Parent {
public Child() {
super();
System.out.println("Child constructor called");
}
}
public class Main {
public static void main(String[] args) {
Child childObj = new Child();
}
}
```
If you run this program, you will still see the output as:
```
Parent constructor called
Child constructor called
```
But this time, the parent constructor is called explicitly using the `super()` keyword inside the child's constructor.
I hope this helps you understand the inheritance relationship and how constructors work in Java. Let me know if you have any more questions.
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the "Parent" constructor was specifically called from the "Child" constructor. This is such that each constructor can either call a constructor in the superclass (called "super()") or a constructor in the same class (called "this()").
```
class Parent {
public Parent() {
System.out.println("Parent constructor called");
}
}
class Child extends Parent {
public Child() {
System.out.println("Child constructor called");
}
}
public class Main {
public static void main(String[] args) {
Child c = new Child();
}
}
```
In this program, we have two classes: `Parent` and `Child`. `Child` is a subclass of `Parent`, meaning it inherits all of `Parent`'s methods and fields.
In the `Parent` constructor, we simply print a message saying that the constructor was called. Similarly, in the `Child` constructor, we also print a message saying that the constructor was called.
In the `Main` class, we create an instance of `Child` by calling its constructor. Notice that we do not explicitly call the `Parent` constructor in the `Child` constructor.
If you run this program, you'll see the following output:
```
Parent constructor called
Child constructor called
```
This is because when we create a new `Child` object, the `Child` constructor is called first. Since the `Child` constructor does not explicitly call the `Parent` constructor, the `Parent` constructor is automatically called for us.
Now, let's change the `Child` constructor to explicitly call the `Parent` constructor:
```
class Child extends Parent {
public Child() {
super();
System. out.println("Child constructor called");
}
}
```
Notice that we added the `super()` statement, which calls the `Parent` constructor.
If you run this program now, you'll see the following output:
```
Parent constructor called
Child constructor called
```
The output is the same as before. However, this time, we explicitly called the `Parent` constructor in the `Child` constructor. This is because every constructor must call either another constructor in the same class (`this()`) or a constructor in the superclass (`super()`).
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an undisturbed soil sample has a void ratio of 0.56, water content of 15 nd a specific gravity of soils of 2.64. find the wet and dry unit weights in lb/ft3 , porosity and degree of saturation.
The wet unit weight is 106.5 lb/ft3, the dry unit weight is 97.3 lb/ft3, the porosity is 35.9%, and the degree of saturation is 23.3%.
To solve this problem, we need to use the following equations:
Void ratio (e) = Volume of voids (Vv) / Volume of solids (Vs)
Porosity (n) = Vv / Vt, where Vt is the total volume of the soil sample (Vt = Vv + Vs)
Degree of saturation (Sr) = (Vw / Vv) x 100, where Vw is the volume of water in the soil sample
Dry unit weight ([tex]γd[/tex]) = (Gs / (1 + e)) x [tex]γw[/tex], where Gs is the specific gravity of the soil and [tex]γw[/tex] is the unit weight of water (62.4 lb/ft3)
Wet unit weight [tex](γw[/tex]) = [tex]γd[/tex] + (w x [tex]γw[/tex]), where w is the water content of the soil sample
Given data:
Void ratio (e) = 0.56
Water content (w) = 15%
Specific gravity of soil (Gs) = 2.64
First, we need to calculate the dry unit weight:
[tex]γd[/tex] = (Gs / (1 + e)) x [tex]γw[/tex]
[tex]γd[/tex] = (2.64 / (1 + 0.56)) x 62.4
[tex]γd[/tex]= 97.3 lb/ft3
Next, we can calculate the wet unit weight:
[tex]γw[/tex] = [tex]γd[/tex] + (w x [tex]γw[/tex])
[tex]γw[/tex] = 97.3 + (0.15 x 62.4)
[tex]γw[/tex] = 106.5 lb/ft3
Now we can calculate the porosity:
n = Vv / Vt
n = e / (1 + e)
n = 0.56 / (1 + 0.56)
n = 0.359 or 35.9%
Finally, we can calculate the degree of saturation:
Sr = (Vw / Vv) x 100
Sr = (0.15 x Vt) / Vv
Sr = (0.15 x (Vv + Vs)) / Vv
Sr = (0.15 / (1 - n)) x 100
Sr = (0.15 / (1 - 0.359)) x 100
Sr = 23.3%
Therefore, the wet unit weight is 106.5 lb/ft3, the dry unit weight is 97.3 lb/ft3, the porosity is 35.9%, and the degree of saturation is 23.3%.
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Suppose you are choosing between four different desktop computers: one is an Apple Mac Intosh and the other three are PC-compatible computers that use a Pentium 4, an AMD processor (using the same compiler as the Pentium 4), and a Pentium 5 (which does not yet exist in 2004 but has the same architecture as the Pentium 4 and uses the same compiler). Which of the following statements are true?
a. The fastest computer will be the one with the highest clock rate.
b. Since all PCs use the same Intel-compatible instruction set and execute the same number of instructions for a program, the fastest PC will be the one with the highest clock rate.
c. Since AMD uses different techniques than Intel to execute instructions,they may have different CPIs. But, you can still tell which of the two Pentium-based PCs is fastest by looking at the clock rate.
d. Only by looking at the results of benchmarks for tasks similar to your workload can you get an accurate picture of likely performance.
Answer:
d.
Explanation:
An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter D=1 mm, electrical resistivity rhoe=10−6Ω⋅m, thermal conductivity k=25W/m⋅K, and emissivity ε=0.20. The heater is designed to deliver air at a temperature of T[infinity]=50∘C under flow conditions that provide a convection coefficient of h=250W/m2⋅K for the wire. The temperature of the housing that encloses the wire and through which the air flows is Tsur=50∘C. If the maximum allowable temperature of the wire is Tmax=1200∘C, what is the maximum allowable electric current I? If the maximum available voltage is ΔE=110V, what is the corresponding length L of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.
Solution :
Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :
[tex]$h_T= \epsilon \sigma (T_s+T_{surr})(T_s^2 +T^2_{surr})$[/tex]
[tex]$h_T= 0.20 \times 5.67 \times 10^{-8} (1473+323)(1473^2 +323^2)$[/tex]
[tex]$h_T=46.3 \ W/m^2 .K$[/tex]
The total heat transfer coefficient will be :
[tex]$h_T=(250+46.3) \ W/m^2 .K$[/tex]
[tex]$=296.3 \ W/m^2 .K$[/tex]
Now calculating the maximum volumetric heat generation :
[tex]$\dot {q}_{max}=\frac{2h_t}{r_0}(T_s-T_{\infty})$[/tex]
[tex]$\dot {q}_{max}=\frac{2\times 296.3}{0.0005}(1200-50)$[/tex]
[tex]$= 1.362 \times 10^{9} \ W/m^3$[/tex]
The heat generation inside the wire is given as :
[tex]$\dot{q} = \frac{I^2R}{V}$[/tex]
Here, R is the resistance of the wire
V is the volume of the wire
∴ [tex]$\dot{q} = \frac{I^2\left( \rho \times \frac{L}{A} \right)}{A \times L}$[/tex]
[tex]$=\frac{I^2 \rho}{\left(\frac{\pi}{4}D^2 \right)}$[/tex]
where, ρ is the resistivity.
[tex]$I_{max}= \left(\frac{\dot{q}_{max}}{\rho} \right)^{1/2} \times \frac{\pi}{4}D^2$[/tex]
[tex]$I_{max}= \left(\frac{1.36 \times 10^9}{10^{-6}} \right)^{1/2} \times \frac{3.14}{4}(1 \times 10^{-3})^2$[/tex]
= 28.96 A
Now considering the relation for the current flow through the finite potential difference.
[tex]$E=I_{max} \times R$[/tex]
[tex]$E=I_{max} \times \rho \times \frac{L}{A}$[/tex]
[tex]$L=\frac{AE}{I_{max} \ \rho}$[/tex]
[tex]$L=\frac{\frac{\pi}{4} \times (1 \times 10^{-3})^2 \times 110}{28.96 \times 10^{-6}}$[/tex]
= 2.983 m
Now calculating the power rating of the heater:
[tex]$P= E \times I_{max}$[/tex]
[tex]$P= 110 \times 28.96}$[/tex]
= 3185.6 W
= 3.1856 kW
a given project activity has the following time estimates: a = 8 b = 27 m = 16 what is the variance ( ) of this project activity's estimated duration? (round to 2 decimal places)
To calculate the variance of this project activity's estimated duration, we can use the formula:
Variance = [(b-a)/6]^2
where a is the optimistic time estimate, b is the pessimistic time estimate, and m is the most likely time estimate.
In this case, the optimistic time estimate (a) is 8, the pessimistic time estimate (b) is 27, and the most likely time estimate (m) is 16.
So, plugging these values into the formula:
Variance = [(27-8)/6]^2
Variance = 3.08
Therefore, the variance of this project activity's estimated duration is 3.08 (rounded to 2 decimal places).
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An insulated closed piston–cylinder device initially contains 0.3 m3 of carbon dioxide at 200 kPa
and 27°C. A resistance heater inside the cylinder is turned on and supplied heat to the gas. As a
result, the gas expanded by pushing the piston up, until the volume doubled. During this process,
6
the pressure changed according to = 4, in which the constant 6 has units of kPa.m
a) Find the mass of the hydrogen in the tank in kg.
b) Determine the work done by the gas in kJ.
To solve this problem, we can use the ideal gas law and the equation for polytropic process.
What is ideal gas law ?The ideal gas law is a fundamental law of physics that describes the behavior of an ideal gas. It relates the pressure, volume, temperature, and number of particles of a gas using the following equation:
PV = nRT
a) First, we need to find the mass of the carbon dioxide in the tank. The ideal gas law is:
PV = mRT
where P is the pressure, V is the volume, m is the mass, R is the universal gas constant, and T is the temperature. Rearranging for the mass, we get:
m = PV / RT
Substituting the given values, we have:
m = (200 kPa)(0.3 m3) / [(0.287 kPam3/kgK)(27°C + 273.15)] = 3.87 kg
So the mass of the carbon dioxide in the tank is 3.87 kg.
b) To determine the work done by the gas during the process, we can use the equation for polytropic process:
P1V1^n = P2V2^n
where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and n is the polytropic index. Substituting the given values, we have:
(200 kPa)(0.3 m3)^n = (4)(0.6 m3)^n
Dividing both sides by (0.3 m3)^n and taking the logarithm of both sides, we get:
log(200) + nlog(0.3) = log(4) + nlog(0.6)
Solving for n, we get:
n = log(4/200) / log(0.6/0.3) ≈ 1.235
Using the polytropic work equation:
W = (P2V2 - P1V1) / (1 - n)
Substituting the given values, we have:
W = [(4 kPa)(0.6 m3) - (200 kPa)(0.3 m3)] / (1 - 1.235) = 233.7 kJ
So the work done by the gas during the process is 233.7 kJ.
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a load of 12tonnes is put along a horizontal plane by a force at 30°to and above the flat. if the coefficient of sliding friction is 0.2 find the frictional force
Answer:
20368.917N
Explanation:
Frictional force (F) is the product of the Coefficient of friction and the normal reaction.
F = μN
Coefficient of friction, μ = 0.2
Normal reaction = MgCosθ
Mass, m = 12 tonnes = 12 * 1000 = 12000 kg
N = 12000 * 9.8 * cos30
N = 101844.58
F = 0.2 * 101844.58
F = 20368.917N
Trace the following tree scan function and describe its action. template int treeFunc (tnode *t) int n = 0, left, right; if (t != NULL) { if (t->left != NULL) n++; if (t->right != NULL) n++; left = treeFunc (t->left); right = treeFunc (t->right); return n + left + right; else return 0; (a) identifies the number of leaf nodes in the tree (c) identifies the number of edges in the tree (b) identifies the number of nodes in the tree (d) identifies the depth of the tree.
The treeFunc function counts the number of nodes in a binary tree.
The function takes a pointer to the root of a binary tree as input.
It initializes a counter variable "n" to zero, and two variables "left" and "right" to store the results of recursive calls on the left and right subtrees, respectively.
If the root is not NULL, it checks whether the left and right children are NULL or not, and increments the counter "n" accordingly (if either child is not NULL, the node is not a leaf).
The function then recursively calls itself on the left and right children, storing the results in the variables "left" and "right".
Finally, it returns the sum of "n", "left", and "right", which gives the total number of nodes in the tree.
If the root is NULL, the function immediately returns 0.
Therefore, the correct answer is (b) identifies the number of nodes in the tree.
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Convert an acceleration of 12m/s² to km/h²
Complete the function ConvertToPennies() so that the function returns the total number of pennies given a number of dollars and (optionally) a number of pennies.
Ex: ConvertToPennies(5 , 6) returns 506 and ConvertToPennies(8) returns 800.
complete the code:
function totalPennies = ConvertToPennies(numDollars, numPennies)
% numDollars: Number of dollars
% numPennies: Number of pennies (optional)
% Function output: Total number of pennies
To complete the ConvertToPennies function, include a conditional statement that checks if numPennies is provided and calculate the total pennies accordingly.
To complete the ConvertToPennies() function, follow these steps:
1. Add an 'if' statement to check if the 'numPennies' input is provided by using the 'nargin' function, which returns the number of function input arguments.
2. If 'numPennies' is provided (nargin == 2), calculate the total pennies by multiplying 'numDollars' by 100 and adding 'numPennies'.
3. If 'numPennies' is not provided (nargin == 1), calculate the total pennies by simply multiplying 'numDollars' by 100.
Here's the completed code:
function totalPennies = ConvertToPennies(numDollars, numPennies)
% numDollars: Number of dollars
% numPennies: Number of pennies (optional)
% Function output: Total number of pennies
if nargin == 2
totalPennies = (numDollars * 100) + numPennies;
else
totalPennies = numDollars * 100;
end
end
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2. generate a fm signal using fm modulator passband block, where the message signal is 10 hz and carrier frequency is 100 hz. demodulate the fm signal using pll (phase locked loop)
To generate a FM signal using FM modulator passband block, we need to first generate the message signal of 10 Hz and the carrier frequency of 100 Hz. The FM modulator passband block is used to modulate the message signal onto the carrier signal.
This can be achieved using a frequency modulation process, where the amplitude of the carrier signal is kept constant, while its frequency is varied according to the message signal.Once we have generated the FM signal, we need to demodulate it using a PLL (Phase Locked Loop). A PLL is a closed-loop feedback system that compares the phase and frequency of the input signal with the output signal to produce a stable output signal with a fixed frequency. In FM demodulation, the PLL is used to extract the modulating signal from the FM signal.To demodulate the FM signal using a PLL, we first need to tune the PLL to the carrier frequency of the FM signal. This can be done by adjusting the frequency of the PLL's voltage-controlled oscillator (VCO) to match the frequency of the carrier signal. Once the PLL is locked onto the carrier frequency, any changes in the frequency of the FM signal will result in a corresponding change in the phase of the output signal.By measuring the phase difference between the output signal of the PLL and the original carrier signal, we can extract the modulating signal from the FM signal. This is because the phase difference between the output signal and the carrier signal is proportional to the frequency deviation of the FM signal. Therefore, by integrating the phase difference over time, we can recover the original message signal that was modulated onto the carrier signal.In conclusion, the process of generating and demodulating an FM signal involves using an FM modulator passband block to modulate the message signal onto the carrier signal and a PLL to extract the modulating signal from the FM signal.For such more question on frequency
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To generate an FM signal using an FM modulator passband block with a message signal of 10 Hz and a carrier frequency of 100 Hz, we can use the following steps:
Create the message signal with a frequency of 10 Hz and a maximum amplitude of 1 using a waveform generator.
Create the carrier signal with a frequency of 100 Hz and a maximum amplitude of 1 using a waveform generator.
Use an FM modulator passband block to modulate the carrier signal with the message signal. The modulation index should be set to a value that provides the desired frequency deviation.
The FM modulated signal can be expressed mathematically as:
s(t) = Ac * cos[2pifct + KfInt(m(t)*dt)]
where s(t) is the FM modulated signal, Ac is the carrier amplitude, fc is the carrier frequency, Kf is the frequency sensitivity of the modulator, and m(t) is the message signal.
Apply a low-pass filter to the modulated signal to remove any high-frequency components that may have been introduced during modulation.
To demodulate the FM signal using a PLL, we can use the following steps:
Create a reference signal with the same frequency as the carrier signal using a waveform generator.
Use a mixer to multiply the FM modulated signal by the reference signal.
Pass the output of the mixer through a low-pass filter to obtain the error signal.
Feed the error signal into a phase detector.
Use a voltage-controlled oscillator (VCO) to generate a signal with a frequency that is proportional to the error signal.
Use a loop filter to adjust the output of the phase detector to control the frequency of the VCO.
Use a buffer to amplify the output of the VCO to obtain the demodulated message signal.
The output of the buffer will be the demodulated message signal, which should be a replica of the original message signal.
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State model can be utilized to validate the execution of object-oriented software. Which one is invalid?
Group of answer choices
The final state should only have inbound transitions.
The initial state should only have outbound transitions.
The final state is reachable from all other states.
There is an event and an action associated with a state transition.
Every state is reachable from any state.
The statement that is invalid is: "The final state is reachable from all other states."
This statement is incorrect because the final state is a state in which the object-oriented software has completed its execution, and it should not have any outbound transitions. Therefore, it should not be possible to reach the final state from any other state. The final state should only have inbound transitions from other states. On the other hand, the other statements are valid. The initial state should only have outbound transitions because it is the starting point of the object-oriented software. There should be an event and an action associated with a state transition to specify what triggers the transition and what happens during the transition. Every state should be reachable from any state to ensure that all possible paths through the software can be tested.
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If the sum of the two numbers is 4 and the sum of their squares minus three times their product is 76,find the number
Answer:
-2 and 6
Explanation:
Let "x" and "y" be 2 numbers.
The sum of the two numbers is 4. The mathematical expression is:
x + y = 4
y = 4 - x [1]
The sum of their squares minus three times their product is 76. The mathematical expression is:
x² + y² - 3 x y = 76 [2]
If we substitute [1] in [2], we get:
x² + (4 - x)² - 3 x (4 - x) = 76
x² + 16 - 8 x + x² - 12 x + 3 x² = 76
5 x² - 20 x - 60 = 0
We apply the solving formula for second order equations and we get x₁ = 6 and x₂ = -2.
If we replace these x values in [1], we get:
y₁ = 4 - x₁ = 4 - 6 = -2
y₂ = 4 - x₂ = 4 - (-2) = 6
As a consequence, one of the numbers is 6 and the other is -2.
Which option identifies what is missing in the categorical syllogistic argument below? Quadrupeds have four legs so horses are quadrupeds.
Conclusion
Minor premise
Inference
Major premise
1. What is the productivity rate using cycle time for the following information:
I
Type of Work – Hauling
Average Cycle Time – 35 Minutes
Truck Capacity – 25 Tons
Crew - One Driver
Productivity Factor - 0.85
System Efficiency – 55 Minutes
per
Hour
... is an actual sequence of interactions (i.e., an instance) describing one specific situation; a ... is a general sequence of interactions (i.e., a class) describing all possible ... associated with a situation. ... are used as examples and for clarifying details with the client. ... are used as complete descriptions to specify a user task or a set of related system features.
Answer:
ScenarioUse caseScenariosScenariosUse caseExplanation:
A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.
A storm with a duration of about 24 hours resultsin the following hydrograph at a gaging station on a river. The flow was 52 cubic meters per second (cms) before the rain began. The drainage area above the gaging station is 1,450 square kilometers. Use the observed hydrograph to develop a 24-hour rainfall duration unit hydrograph for this watershed.
Time 0 6 12 18 24 30 36 42 48 54 60 66 72 78 84
(Hours)
Flow 52 52 55 56 97 176 349 450 442 370 328 307 280 259 238
(m3/s)
Time 90 66 102 108 114 120 126 132 138 144 155 156 162 168 176
(Hours)
(m3/s) 214 193 173 145 135 124 114 107 97 86 79 66 62 58 52
Answer:
attached below
Explanation:
Given data:
Gaged flow = 52 m^3 / sec
Depth covering drainage area = 1450 km^2
Develop a 24-hour rainfall duration unit hydrograph for the watershed using observed hydorgraph
Runoff flow = gaged flow - base flows
= 52 - 52 = 0 m^3/sec
For 18 hours time duration
Direct runoff volume ( Vr ) = ∑ ( QΔt1 )
where Δt = 6
∑Q = 3666 m^3/sec
hence Vr = Δt * ∑Q = 79185600 m^3
Next we will convert the Direct runoff volume to its equivalent depth covering the drainage area
= Vr / drainage area depth
= 79185600 / 1450000000 = 5.46 cm
Next we will find the unit hydrograph flows by applying the relation below
[tex]Q_{1.0cm} = Q_{5.46cm} (\frac{1.0cm}{5.46cm} )[/tex]
where 14m^3/sec = [tex]Q_{5.6cm}[/tex]
input value back to the above relation
[tex]Q_{1.0cm} = 2.57 m^3/sec[/tex]
Attached below is The remaining part of the solution tabulated below
and A graph of the unit hydorgraph for the given watershed
Note :
Base flow total = 1560
UH total = 671.30
Branch circuits shall be __________ in accordance with the maximum permitted.
210. 3
Branch circuits shall be sized in accordance with the maximum permitted.
What is the sizing requirement for branch circuits?The sizing of branch circuits ensure that they can safely and effectively carry the electrical load that will be placed on them. The National Electrical Code specifies the maximum ampacit or current-carrying capacity of branch circuits based on the size and type of wire being used.
So, it important to size branch circuits correctly to avoid overloading the circuit which can result in overheating, fires, and other hazards. The contractors and other professionals responsible for installing and maintaining electrical systems should be familiar with the NEC requirements for branch circuit sizing and ensure that all installations comply with these standards.
Read more about Branch circuits
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Now, you get a turn to practice writing a short program in Scratch. Try to re-create the program that was shown that turns the sprite in a circle. After you have completed that activity, see if you can make one of the improvements suggested. For example, you can try adding a sound. If you run into problems, think about some of the creative problem-solving techniques that were discussed.
When complete, briefly comment on challenges or breakthroughs you encountered while completing the guided practice activity.
Pls help im giving 100 points for this i have this due in minutes
Answer:
u need to plan it out
Explanation:
u need to plan it out
Answer:
use the turn 1 degrees option and put a repeat loop on it
Explanation:
u can add sound in ur loop
During a shrinkage limit test, a 19.3 cm3 saturated clay sample with a mass of 37 g was placed in a porcelain dish and dried in the oven. The oven-dried sample had a mass of 28 g with a final volume of 16 cm3 . Determine the shrinkage limit and the shrinkage ratio.
Answer:
shrinkae limit = 20.35%
shrinkage ratio = 1.45
Explanation:
1. to get the shrinkage limit we would first calculate the moisture content w.
w = (37-28)/28
= 9/28
= 0.3214
then the formula for shrinkage limit is
[tex][w-\frac{(V-Vd}{wd} ]*100[/tex]
w = 0.3214
V = 19.3
Vd = 16
Wd = 28
when we put these values into the formula:
[tex][0.3214-\frac{(19.3-16)}{28} ]*100\\[/tex]
= 20.35%
2. the shrinkage limit = Wd/V
= 28/19.3
= 1.45
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 2.00 MPa, calculate the magnitude of applied stress necessary to cause slip to occur on the (111) plane in the direction.
Answer:
Explanation:
From the given information:
The equation for applied stress can be expressed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda}[/tex]
where;
[tex]\phi[/tex] = angle between the applied stress [100] and [111]
To determine the [tex]\phi[/tex] and [tex]\lambda[/tex] for the system
Using the equation:
[tex]\phi= cos^{-1}\Big [\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}}\Big][/tex]
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [111]
[tex]l_1 = 1 , m_1 = 1, n_1 = 1[/tex]
Thus;
[tex]\phi= cos^{-1}\Big [\dfrac{1*1+0*1+0*1}{\sqrt{(1^2+0^2+0^2)(1^2+1^2+1^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(3)}}\Big][/tex]
[tex]\phi= 54.74^0[/tex]
To determine [tex]\lambda[/tex] for [tex][1 \overline 1 0][/tex]
where;
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [tex][1 \overline 1 0][/tex]
[tex]l_1 = 1 , m_1 = -1, n_1 = 0[/tex]
Thus;
[tex]\lambda= cos^{-1}\Big [\dfrac{1*1+0*1+0*0}{\sqrt{(1^2+0^2+0^2)(1^2+(-1)^2+0^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(2)}}\Big][/tex]
[tex]\phi= 45^0[/tex]
Thus, the magnitude of the applied stress can be computed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda }[/tex]
[tex]\sigma_{app} = \dfrac{2.00}{cos (54.74) \ cos (45) }[/tex]
[tex]\mathbf{\sigma_{app} =4.89 \ MPa}[/tex]
Travel Time Problem: Compute the time of concentration using the Velocity, Sheet Flow Method for Non-Mountainous Orange County and SCS method at a 25 year storm evert.
Location Slope (%) Length (ft) Land Use
1 4.5 1000 Forest light underbrush with herbaceous fair cover.
2 2.5 750 Alluvial Fans (eg. Natural desert landscaping)
3 1.5 500 Open Space with short grasses and good cover
4 0.5 250 Paved Areas (1/4 acre urban lots)
Answer:
Total time taken = 0.769 hour
Explanation:
using the velocity method
for sheet flow ;
Tt = [tex]\frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4} }[/tex]
Tt = travel time
n = manning CaH
Pl = 25years
L = how length ( ft )
s = slope
For Location ( 1 )
s = 0.045
L = 1000 ft
n = 0.06 ( from manning's coefficient table )
Tt1 = 0.128 hour
For Location ( 2 )
s = 2.5 %
L= 750
n = 0.13
Tt2 = 0.239 hour
For Location ( 3 )
s = 1.5%
L = 500 ft
n = 0.15
Tt3 = 0.237 hour
For Location (4)
s = 0.5 %
L = 250 ft
n = 0.011
Tt4 = 0.165 hour
hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4
= 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour
The movement of the piston and connecting rod assembly driven by the force of combustion turns the ________.
Answer:
the answer. for this is Crankshaft
Which gas is released in the SMAW process causing a
shielding affect on the molten weld pool?
•nitrogen
•carbon dioxide
•argon
•hydrogen
Which of the following devices is a simple machine?
A.
an engine
B.
a pulley
C.
a motor
D.
a bicycle
E.
a crane
Answer:
A PULLY
Explanation:
HAD THIS ONE THAT IS THE CORRECT ANWSER
Answer:
The answer is B. a pulley
Explanation:
I hope I answered your question:)