Find the value of R2 required if vs = 18 V , R1 = 50 kΩ , and the desired no-load output voltage is vo = 6 VPart b)Several different loads are going to be used with the voltage divider from Part A. If the load resistances are 300 kΩ , 200 kΩ , and 100 kΩ , what is the output voltage that is the most different from the design output voltage vo = 6 V ?part c)The circuit designer wants to change the values of R1 and R2 so that the design output voltage vo = 6 V is achieved when the load resistance is RL = 200 kΩ rather than at no-load. The actual output voltage must not drop below 5.4 V when RL = 100 kΩ . What is the smallest resistor value that can be used forR1?
Answers
Answer:
Explanation:
Check attachment for solution
a. The value of R2 required for the desired no-load output voltage R2 = 10 kΩ
b. The most different output voltage is 5.8 V, which occurs when the load resistance is 300 kΩ.
c. The smallest resistor value that can be used for R1 is 45 kΩ.
Part A:
To find the value of R2 required for the desired no-load output voltage, we can use the following voltage divider equation:
vo = vs * R2 / (R1 + R2)
Substituting in the known values, we get:
6 V = 18 V * R2 / (50 kΩ + R2)
Solving for R2, we get:
R2 = 18 V * 6 V - 6 V * 50 kΩ / 6 V - 18 V
R2 = 10 kΩ
Part B:
To find the output voltage that is the most different from the design output voltage, we need to calculate the output voltage for each load resistance. We can use the following voltage divider equation:
vo = vs * (R2 / (R1 + R2) || RL)
Substituting in the known values for each load resistance, we get:
vo (RL = 300 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 300 kΩ)
vo (RL = 300 kΩ) = 5.8 V
vo (RL = 200 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 200 kΩ)
vo (RL = 200 kΩ) = 7.2 V
vo (RL = 100 kΩ) = 18 V * (10 kΩ / (50 kΩ + 10 kΩ) || 100 kΩ)
vo (RL = 100 kΩ) = 8.6 V
Therefore, the output voltage that is the most different from the design output voltage is 5.8 V, which occurs when the load resistance is 300 kΩ.
Part C:
To change the values of R1 and R2 so that the design output voltage vo = 6 V is achieved when the load resistance is RL = 200 kΩ rather than at no-load, we can use the following voltage divider equation:
vo = vs * R2 / (R1 + R2)
Substituting in the known values, we get:
6 V = 18 V * R2 / (R1 + 200 kΩ)
Solving for R2, we get:
R2 = 30 kΩ
Next, we need to calculate the value of R1 to ensure that the actual output voltage does not drop below 5.4 V when RL = 100 kΩ. We can use the following voltage divider equation:
vo = vs * (R2 / (R1 + R2) || RL)
Substituting in the known values, we get:
5.4 V = 18 V * (30 kΩ / (R1 + 30 kΩ) || 100 kΩ)
Solving for R1, we get:
R1 = 45 kΩ
Therefore, the smallest resistor value that can be used for R1 is 45 kΩ.
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Related Questions
1. A priority queue is an abstract data type which is like a regular queue or some other data structures, but where additionally each element has a "priority" associated with it. In a priority queue, an element with high priority is served before an element with low priority like scheduler. If two elements have the same priority, they are served according to their order in the queue.
Answers
Answer:
The code is as below whereas the output is attached herewith
Explanation:
package brainly.priorityQueue;
class PriorityJobQueue {
Job[] arr;
int size;
int count;
PriorityJobQueue(int size){
this.size = size;
arr = new Job[size];
count = 0;
}
// Function to insert an element into the priority queue
void insert(Job value){
if(count == size){
System.out.println("Cannot insert the key");
return;
}
arr[count++] = value;
heapifyUpwards(count);
}
// Function to heapify an element upwards
void heapifyUpwards(int x){
if(x<=0)
return;
int par = (x-1)/2;
Job temp;
if(arr[x-1].getPriority() < arr[par].getPriority()){
temp = arr[par];
arr[par] = arr[x-1];
arr[x-1] = temp;
heapifyUpwards(par+1);
}
}
// Function to extract the minimum value from the priority queue
Job extractMin(){
Job rvalue = null;
try {
rvalue = arr[0].clone();
} catch (CloneNotSupportedException e) {
e.printStackTrace();
}
arr[0].setPriority(Integer.MAX_VALUE);
heapifyDownwards(0);
return rvalue;
}
// Function to heapify an element downwards
void heapifyDownwards(int index){
if(index >=arr.length)
return;
Job temp;
int min = index;
int left,right;
left = 2*index;
right = left+1;
if(left<arr.length && arr[index].getPriority() > arr[left].getPriority()){
min =left;
}
if(right <arr.length && arr[min].getPriority() > arr[right].getPriority()){
min = right;
}
if(min!=index) {
temp = arr[min];
arr[min] = arr[index];
arr[index] = temp;
heapifyDownwards(min);
}
}
// Function to implement the heapsort using priority queue
static void heapSort(Job[] array){
PriorityJobQueue object = new PriorityJobQueue(array.length);
int i;
for(i=0; i<array.length; i++){
object.insert(array[i]);
}
for(i=0; i<array.length; i++){
array[i] = object.extractMin();
}
}
}
package brainly.priorityQueue;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
public class PriorityJobQueueTest {
// Function to read user input
public static void main(String[] args) {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n;
System.out.println("Enter the number of elements in the array");
try{
n = Integer.parseInt(br.readLine());
}catch (IOException e){
System.out.println("An error occurred");
return;
}
System.out.println("Enter array elements");
Job[] array = new Job[n];
int i;
for(i=0; i<array.length; i++){
Job job =new Job();
try{
job.setJobId(i);
System.out.println("Element "+i +"priority:");
job.setJobName("Name"+i);
job.setSubmitterName("SubmitterName"+i);
job.setPriority(Integer.parseInt(br.readLine()));
array[i] = job;
}catch (IOException e){
System.out.println("An error occurred");
}
}
System.out.println("The initial array is");
System.out.println(Arrays.toString(array));
PriorityJobQueue.heapSort(array);
System.out.println("The sorted array is");
System.out.println(Arrays.toString(array));
Job[] readyQueue =new Job[4];
}
}
Please write the following code in Python 3. Also please show all output(s) and share your code.
Below is a for loop that works. Underneath the for loop, rewrite the problem so that it does the same thing, but using a while loop instead of a for loop. Assign the accumulated total in the while loop code to the variable sum2. Once complete, sum2 should equal sum1.
sum1 = 0
lst = [65, 78, 21, 33]
for x in lst:
sum1 = sum1 + x
Answers
Explanation of rewriting code from for loop to while loop in Python.
Explanation:To rewrite the given code using a while loop instead of a for loop, you can iterate over the list indices and manually accumulate the total.
Here is the code:
sum1 = 0lst = [65, 78, 21, 33]
index = 0
sum2 = 0
while index < len(lst):
sum2 += lst[index]
index += 1
By using this code snippet, the sum accumulated using the while loop will be stored in the variable sum2, which should equal the sum1 calculated using the for loop.
The elastic cords used for bungee jumping are designed to endure large strains. Consider a bungee cord that stretches to a maximum length 3.85 times the original length. There are different ways to report this extensional deformation. Calculate how 'wrong' the engineering strain is compared to the true strain by evaluating the ratio:
εtrue / εengr = __________
Answers
Answer:
(εtrue/εengr) = (1.3481/2.85) = 0.473
This shows that the engineering strain is truly a bit far off the true strain.
Explanation:
εengr
Engineering strain is a measure of how much a material deforms under a particular load. It is the amount of deformation in the direction of the applied force divided by the initial length of the material.
ε(engineering) = ΔL/L₀
Lf = final length = 3.85 L₀
L₀ = original length = L₀
ΔL = Lf - L₀ = 3.85 L₀ - L₀ = 2.85 L₀
ε(engineering) = ΔL/L₀ = (2.85L₀)/L₀ = 2.85
εtrue
True Strain measures instantaneous deformation. It is obtained mathematically by integrating strain over small time periods and Running them up. Hence,
ε(true) = In (Lf/L₀)
Lf = 3.85L₀
L₀ = L₀
ε(true) = In (Lf/L₀) = In (3.85L₀/L₀) = In 3.85 = 1.3481
(εtrue/εengr) = (1.3481/2.85) = 0.473
(a) Write the RTL specification for an arithmetic shift right on an eight-bit cell. (b) Write the RTL specification for an arithmetic shift left on an eight-bit cell.
Answers
(a) The RTL specification for an arithmetic shift right on an eight-bit cell can be described as follows:
1. Take the eight-bit input value and assign it to a variable, let's call it "input".
2. Create a temporary variable, let's call it "shifted".
3. Assign the most significant bit (MSB) of the "input" to the least significant bit (LSB) of the "shifted" variable.
4. Shift the remaining seven bits of the "input" one position to the right, discarding the least significant bit (LSB) and shifting in the sign bit to the most significant bit (MSB).
5. Repeat step 3 and step 4 seven more times to shift all the bits in the "input" variable to the right.
6. Output the final value of the "shifted" variable, which represents the result of the arithmetic shift right.
Here's an example to illustrate the RTL specification:
Input: 10111010
Shifted: 11011101 (after one shift)
Shifted: 11101110 (after two shifts)
Shifted: 11110111 (after three shifts)
Shifted: 11111011 (after four shifts)
Shifted: 11111101 (after five shifts)
Shifted: 11111110 (after six shifts)
Shifted: 11111111 (after seven shifts)
Shifted: 11111111 (after eight shifts)
(b) The RTL specification for an arithmetic shift left on an eight-bit cell can be described as follows:
1. Take the eight-bit input value and assign it to a variable, let's call it "input".
2. Create a temporary variable, let's call it "shifted".
3. Assign the least significant bit (LSB) of the "input" to the most significant bit (MSB) of the "shifted" variable.
4. Shift the remaining seven bits of the "input" one position to the left, discarding the most significant bit (MSB) and shifting in a zero to the least significant bit (LSB).
5. Repeat step 3 and step 4 seven more times to shift all the bits in the "input" variable to the left.
6. Output the final value of the "shifted" variable, which represents the result of the arithmetic shift left.
Here's an example to illustrate the RTL specification:
Input: 10111010
Shifted: 01110100 (after one shift)
Shifted: 11101000 (after two shifts)
Shifted: 11010000 (after three shifts)
Shifted: 10100000 (after four shifts)
Shifted: 01000000 (after five shifts)
Shifted: 10000000 (after six shifts)
Shifted: 00000000 (after seven shifts)
Shifted: 00000000 (after eight shifts)
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Under standard conditions, a given reaction is endergonic (i.e., DG >0). Which of the following can render this reaction favorable: using the product immediately in the next step, maintaining a high starting-material concentration, or keeping a high product concentration
Answers
Answer:
Explanation:using the product immediately in the next step
The girl has a mass of 45kg and center of mass at G. (Figure 1) If she is swinging to a maximum height defined by theta = 60, determine the force developed along each of the four supporting posts such as AB at the instant theta = 0.
Answers
Answer and Explanation:
the answer is attached below
A 750-turn solenoid, 24 cm long, has a diameter of 2.3 cm . A 19-turn coil is wound tightly around the center of the solenoid. Part A If the current in the solenoid increases uniformly from 0 to 5.1 A in 0.70 s , what will be the induced emf in the short coil during this time? Express your answer to two significant figures and include the appropriate units. |E| = nothing nothing Request Answer Provide Feedback
Answers
Answer: 2.26x10^-4 v
Explanation:
Lenght of the selonoid = 24x10^-2m
Diameter of the selonoid = 2.3cm
The radius will then be = 1.15cm = 1.15x10^-2m
The area of the selonoid = ¶r^2 = 3.142 x (1.15x10^-2)^2 = 0.000415m^2.
Number of turns on selonoid N1 is 750
For the small center coil, number of turns N2 is 19.
There is a change in current dI/dt from 0 to 5.1 in 0.7s, dI/dt = (5.1-0)/0.7
dI/dt = 7.29A/s.
Induced EMF on selonoid due to magnetic Flux due to changing current in small coil is given as;
E = -M(dI/dt), where M is the mutual inductance of the coils.
but M = (u°AN1N2)/L, where u°= 4¶x10^-7,
A = area of selonoid,
L = Lenght of selonoid.
M = (4¶X10^-7X0.000415X750X19)/(24X10^-2)
M = 3.096X10^-5H
Induced EMF E = 3.096X10^-5 x 7.29
E = 2.26x10^-4V
This report contains three fields: the label of the vending machine, what percentages of the beverages it was last stocked with are sold, and how many total dollars of sales has this generated. You will need to create a new dictionary where the keys are the vending machine labels, and the values are a new type of object called a `MachineStatus`. For each instance, the `MachineStatus` class should store:
Answers
Answer:
the label of a vending machinethe total amount of beverages the vending machine was previously stocked withthe total amount of beverages currently in stock in the vending machinethe total income of the machine from the last time it was stocked until now (note: beverages have different prices, so you cannot simply multiply the change in stock times $1.50 to get the total income)Explanation:
For each instance, the `MachineStatus` class should store: the label of a vending machine the total amount of beverages the vending machine was previously stocked with the total amount of beverages currently in stock in the vending machine the total income of the machine from the last time it was stocked until now.
A cylindrical specimen of a metal alloy 10 mm (0.4 in.) in diameter is stressed elastically in tension. A force of 15,000 N (3370 lbf) produces a reduction in specimen diameter of 7 * 10-3 mm (2.8 * 10-4 in.). Compute Poisson’s ratio for this material if its elastic modulus is 100 GPa (14.5 * 106 psi).
Answers
The Poisson's ratio is 0.36
Explanation:
Given-
Diameter = 10 mm = 10 X 10⁻³m
Force, F = 15000 N
Diameter reduction = 7 X 10⁻3 mm = 7 X 10⁻⁶ mm
The equation for Poisson's ratio:
ν = [tex]\frac{-E_{x} }{E_{z} }[/tex]
and
εₓ = Δd / d₀
εz = σ / E
= F / AE
We know,
Area of circle = π (d₀/2)²
[tex]E_{z}[/tex]= F / π (d₀/2)² E
εz = 4F / πd₀²E
Therefore, Poisson's ratio is
ν = - Δd ÷ d / 4F ÷ πd₀²E
ν = -d₀ΔDπE / 4F
ν = (10 X 10⁻³) (-7 X 10⁻⁶) (3.14) (100 X 10⁻⁹) / 4 (15000)
ν = 0.36
Therefore, the Poisson's ratio is 0.36
Pro-Cut Rotor Matching Systems provide a non-directional finish without performing additional swirl sanding — true or false?
Answers
Answer:True
Explanation:Rotor matching is a term used to describe the process through a brake rotor is aligned to the hub and bearing assembly to produce a smooth, flat, and straight friction surface.
Rotor matching is an essential process which is required to prevent swirling of a break when a vehicle is stopping, rotor matching is needed for efficient and effective break system performance as it is one of the main determinant factors for accidents arising from break failures.
A wind chill factor is defined as the temperature in still air required for a human to suffer the same heat loss as he does for the actual air temperature with the wind blowing. On a very cold morning on the ski slopes at Big Bear, the outside temperature is 15 ℉ and the wind chill factor is-30 °F. Find the wind speed in miles/hour. Assumptions: For a person fully clothed in ski gear, assume that temperature on the outside surface of the clothes is 40 °F and the heat transfer coefficient in sl air is 4 Btu/hr ft2 °F. For simplicity, model the person as a cylinder 1 ft in diameter by 6 ft tall. Use the correlation for forced convection past an upright cylinder: Nu 0.0239 ReD 0805 . Properties of air: thermal conductivity 0.0 134 Btu/hrft。F, density 0.0845 lbm/ft, dynamic viscosity 3.996×10-2 bm/ft hr 2.0
Answers
Answer: V = 208514.156 ft/hr
Explanation:
we will begin by giving a step by step order of answering.
given that
A = area available for convection which will be same for both cases
h (still) = heat transfer coefficient in still air
h(blowing) = heat transfer coefficient in blowing air.
Therefore,
h(still) A [40-(-30)] = h(blowing) A (40-15)
canceling out we have
h(still) (70) = h(blowing) (25)
where h(still) = 4 Btu/hr.ft².°F
4 × 70 = h(blowing) (25)
h (blowing) = 11.2 Btu/hr.ft².°F
Also, we have that NUD = 0.0239 ReD 0805
h (blowing) D / k = 0.0239 (ρVD/μ)˄0.805
where from our data,
D = diameter = 1 ft
ρ = density = 0.0845 lbm/ft
μ = viscosity = 3.996×10-2 bm/ft hr
K = 0.0134 Btu/hr.ft²°F
So from
h (blowing) D / k = 0.0239 (ρVD/μ)˄0.805
we have;
11.2 × 1 / K = 0.0239 (0.0845×V×1 / 3.996×10-2 )˄0.805
where K = 0.0134
V = 208514.156 ft/hr
cheers i hope this helps
1. A wood board is one of a dozen different parts in a homemade robot kit. The width, depth, and height dimensions of the board are 7.5 x 14 x 1.75 inches, respectively. The board is made from southern yellow pine, which has an air dry weight density of .025 lb/in.3. a. What is the volume of the wood board? Precision = 0.00
Answers
Answer:
183.75 cubic inches.
Explanation:
The volume of the wood board is determine by means of this expression:
[tex]V = w \cdot h \cdot l[/tex]
By replacing variables:
[tex]V = (7.5 in) \cdot (14 in) \cdot (1.75 in)\\V = 183.75 in^{3}[/tex]
A thick oak wall (rho = 545 kg/m3 , Cp = 2385 J/kgK, and k = 0.17 W/mK) initially at 25°C is suddenly exposed to combustion products at 800°C. Determine the time of exposure necessary for the surface to reach the ignition temperature of 400°C, assuming the convection heat transfer coefficient between the wall and the products to be 20 W/m2 K. At that time, what is the temperature 1 cm below the surface? (Note: use an appropriate equation for the semi-infinite wall case; compare equations 18.20 and 18.21 in the text).
Answers
Answer:
Explanation:
The detailed calculation and appropriate equation with substitution is as shown in the attached file.
If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed of the pulse if you stretch the hose more tightly
Answers
Answer:
Explanation:if you stretch the hose more tightly the speed of the pulse will reduce..
Which has the capability to produce the most work in a closed system;
1 kg of steam at 800 kPa and 180°C or 1 kg of R–134a at 800 kPa and 180°C? Take T0 = 25°C and P0 = 100 kPa.
Answers
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
The amount of work produced by 1 kg of steam and 1 kg of R–134a would be the same in a closed system at the given conditions.
Explanation:In a closed system, the amount of work produced depends on the change in internal energy of the system. The change in internal energy is given by the formula ΔU = Q - W, where Q is the heat added to the system and W is the work done by the system. Since both 1 kg of steam and 1 kg of R–134a are at the same temperature and pressure, their internal energy changes would be the same for the same amount of heat added. Therefore, the amount of work produced by both substances would also be the same in a closed system.
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A 14-cm-radius, 90-cm-high vertical cylindrical container is partially filled with 60-cm-high water. Now, the cylinder is rotated at a constant angular speed of 180 rpm. Determine how much the liquid level at the center of the cylinder will drop as a result of this rotational motion.
Answers
The student's question involves calculating the water level drop at the center of a rotating cylindrical container, which requires understanding of rotational motion and equilibrium in physics.
Explanation:The student is asking about the change in water level in a rotating cylindrical container due to the centrifugal force. In a rotating frame, the water surface forms a parabolic shape and the level at the center drops. To solve this problem, principles of rotational motion and physics are applied. The task is to calculate the drop in the center of the water surface within a cylindrical container rotating at a constant angular speed. This can be determined by setting up the equilibrium of forces acting on a particle of the water in the radial direction and using the relationship between the angular speed, radius, and acceleration in a rotating system.
Design a 10-to-4 encoder with inputs in the l-out-of-10 code and outputs in a code like normal BCD except that input lines 8 and 9 are encoded into "E" and " F", respectively.
Answers
Answer:
See image attached.
Explanation:
This device features priority encoding of the inputs
to ensure that only the highest order data line is en-
coded. Nine input lines are encoded to a four line
BCD output. The implied decimal zero condition re-
quires no input condition as zero is encoded when
all nine datalinesare athigh logic level. Alldata input
and outputs are active at the low logic level. All in-
puts are equipped with protection circuits against
static discharge and transient excess voltage.
A 10-to-4 encoder is requested to be designed for mapping 1-out-of-10 input code to a modified BCD output, where the digits 8 and 9 are encoded as 'E' and 'F'. The implementation involves the use of digital logic gates and could also reference quantum gates depending on the context of the course.
Explanation:The student is asking to design a 10-to-4 encoder with a specific bit pattern for the numbers 8 and 9. This encoder takes a l-out-of-10 input code and produces a BCD-like output with special cases for inputs 8 ('E') and 9 ('F'). The design would necessitate the use of digital logic gates to map each of the 10 inputs to correspondent 4-bit BCD outputs, with the additional requirement to encode the inputs representing the decimal numbers 8 and 9 into the hexadecimal digits 'E' and 'F', respectively.
In terms of circuitry, the encoder might use a combination of logical gates such as AND, OR, and NOT to create the desired output for each input. For example, when the ninth input is active (representing the number 8), the output should be '1110', which signifies 'E' in hexadecimal notation. Similarly, an active tenth input (representing the number 9) should produce '1111', corresponding to 'F' in hexadecimal.
The encoding and decoding elements are described using terms such as CnNOT, CNOT, I (Identity), and Toffoli gates, which suggests a more sophisticated setup, possibly involving quantum computing principles, as these terms relate to quantum gates.
The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur.
Assuming the punch applies an average force of 2 kN over a time of 2 ms to the 200-g implant, determine (a) the velocity of the implant immediately after impact, (b) the average resistance of the implant to penetration if the implant moves 1 mm before coming to rest.
Answers
Answer:
a) the velocity of the implant immediately after impact is 20 m/s
b) the average resistance of the implant is 40000 N
Explanation:
a) The impulse momentum is:
mv1 + ∑Imp(1---->2) = mv2
According the exercise:
v1=0
∑Imp(1---->2) = F(t2-t1)
m=0.2 kg
Replacing:
[tex]0+F(t_{2} -t_{1} )=0.2v_{2}[/tex]
if F=2 kN and t2-t1=2x10^-3 s. Replacing
[tex]0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s[/tex]
b) Work and energy in the system is:
T2 - U(2----->3) = T3
where T2 and T3 are the kinetic energy and U(2----->3) is the work.
[tex]T_{2} =\frac{1}{2} mv_{2}^{2} \\T_{3} =0\\U_{2---3} =-F_{res} x[/tex]
Replacing:
[tex]\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N[/tex]
A Diesel cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and cutoff ratio (rc) determine the efficiency and other values listed below.
Note: The gas constant for air is R=0.287 kJ/kg-K.
--Given Values--
T1 (K) = 306
P1 (kPa) = 140
r = 10
rc = 1.75
a) Determine the specific internal energy (kJ/kg) at state 1.
Your Answer =
b) Determine the relative specific volume at state 1.
Your Answer =
c) Determine the relative specific volume at state 2.
Your Answer =
d) Determine the temperature (K) at state 2.
Your Answer =
e) Determine the pressure (kPa) at state 2.
Your Answer =
f) Determine the specific enthalpy (kJ/kg) at state 2.
Your Answer =
g) Determine the temperature (K) at state 3.
Your Answer =
h) Determine the pressure (kPa) at state 3.
Your Answer =
i) Determine the specific enthalpy (kJ/kg) at state 3.
Your Answer =
j) Determine the relative specific volume at state 3.
Your Answer =
k) Determine the relative specific volume at state 4.
Your Answer =
l) Determine the temperature (K) at state 4.
Your Answer =
m) Determine the pressure (kPa) at state 4.
Your Answer =
n) Determine the specific internal energy (kJ/kg) at state 4.
Your Answer =
o) Determine the net work per cycle (kJ/kg) of the engine.
Your Answer =
p) Determine the heat addition per cycle (kJ/kg) of the engine.
Your Answer =
q) Determine the efficiency (%) of the engine.
Your Answer =
help A through Q
Answers
Answer:
Explanation: see attachment
//This method uses the newly added parameter Club object //to create a CheckBox and add it to a pane created in the constructor //Such check box needs to be linked to its handler class
Answers
Comments on Question
Your questions is incomplete.
I'll provide a general answer to create a checkbox programmatically
Answer:
// Comments are used for explanatory purpose
// Function to create a checkbox programmatically
public class CreateCheckBox extends Application {
// launch the application
public void ClubObject(Stage chk)
{
// Title
chk.setTitle("CheckBox Title");
// Set Tile pane
TilePane tp = new TilePane();
// Add Labels
Label lbl = new Label("Creating a Checkbox Object");
// Create an array to hold 2 checkbox labels
String chklbl[] = { "Checkbox 1", "Checkbox 2" };
// Add checkbox labels
tp.getChildren().add(lbl);
// Add checkbox items
for (int i = 0; i < chklbl.length; i++) {
// Create checkbox object
CheckBox cbk = new CheckBox(chklbl[i]);
// add label
tp.getChildren().add(cbk);
// set IndeterMinate to true
cbk.setIndeterminate(true);
}
// create a scene to accommodate the title pane
Scene sc = new Scene(tp, 150, 200);
// set the scene
chk.setScene(sc);
chk.show();
}
Blocks A and B are able to slide on a smooth surface. Block A has a mass of 30 kg. Block B has a mass of 60 kg. At a given instant, block A is moving to the right at 0.5 m/s, block B is moving to the left at 0.3 m/s, and the spring connected between them is stretched 1.5 m. Determine the speed of both blocks at the instant the spring becomes unstretched. 9.
Answers
Answer and Explanation:
The answer is attached below
The flying boom B is used with a crane to position construction materials in coves and underhangs. The horizontal "balance" of the boom is controlled by a 250-kg block D, which has a center of gravity at G and moves by internal sensing devices along the bottom flange F of the beam. Determine the position x of the block when the boom is used to lift the stone S, which has a mass of 60 kg. The boom is uniform and has a mass of 80 kg.
Answers
Answer:
0.34
Explanation:
See the attached picture.
If, for a particular junction, the acceptor concentration is 1017/cm3 and the donor concentration is 1016/cm3 , find the junction built-in voltage. Assume ni = 1.5 × 1010/cm3 . Also, find the width of the depletion region (W) and its extent in each of the p and n regions when the junction terminals are left open. Calculate the magnitude of the charge stored on either side of the junction. Assume that the junction area is 100 μm2 .
Answers
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
the spring mass system has an attached mass of 10g the spring constant is 30g/s^2. A dashpot mechanism is attached. which has a damping coefficient of 40g/s. The mass is pulled down and released. At time t=0, the mass is 3cm below the rest position and moving upward at 5cm/s
solve the differential equation. state whether the motion of the sping-mass system is harmonic, damped oscillation, critically damped or overdamped.
Answers
Answer:
Explanation: see attachment
A heat engine operates between a source at 477°C and a sink at 27°C. If heat is supplied to the heat engine at a steady rate of 65,000 kJ/min, determine the maximum power output of this heat engine.
Answers
Answer:
[tex] T_C = 27+273.15 = 300.15 K[/tex]
[tex] T_H = 477+273.15 = 750.15 K[/tex]
And replacing in the Carnot efficiency we got:
[tex] e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%[/tex]
[tex] W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}[/tex]
Explanation:
For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:
[tex] e = 1 -frac{T_C}{T_H}[/tex]
We have on this case after convert the temperatures in kelvin this:
[tex] T_C = 27+273.15 = 300.15 K[/tex]
[tex] T_H = 477+273.15 = 750.15 K[/tex]
And replacing in the Carnot efficiency we got:
[tex] e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%[/tex]
And the maximum power output on this case would be defined as:
[tex] W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}[/tex]
Where [tex] Q_H[/tex] represent the heat associated to the deposit with higher temperature.
The value read at an analog input pin using analogRead() is returned as a binary number between 0 and the maximum value that can be stored in [X] bits. 1. This binary number is directly linearly proportional to the input voltage at the analog pin, with the smallest and largest numbers returned corresponding to the minimum and maximum ADC input values, respectively. At a Vcc of 3.3 V, analogRead(A0) returns a value of 1023. Approximately what voltage is present at the pin A0 on the MSP430F5529?
Answers
Answer: The approximate voltage would be the result from computing analogRead(A0)*3.3 V / 1023
Explanation:
Depending on the binary read of the function analogRead(A0) we would get a binary value between 0 to 1023, being 0 associated to 0V and 1023 to 3.3V, then we can use a three rule to get the X voltage corresponding to the binary readings as follows:
3.3 V ---------> 1023
X V------------>analogRead(A0)
Then [tex]X=\frac{3.3 \times analogRead(A0)}{1023} Volts[/tex]
Thus depending on the valule analogRead(A0) has in bits we get an approximate value of the voltage at pin A0, with a precission of 3,2mV approximately (3.3v/1023).
For this problem, you may not look at any other code or pseudo-code (even if it is on the internet), other than what is on our website or in our book. You may discuss general ideas with other people. Assume A[1. . . n] is a heap, except that the element at index i might be too large. For the following parts, you should create a method that inputs A, n, and i, and makes A into a heap.
Answers
Answer:
(a)
(i) pseudo code :-
current = i
// assuming parent of root is -1
while A[parent] < A[current] && parent != -1 do,
if A[parent] < A[current] // if current element is bigger than parent then shift it up
swap(A[current],A[parent])
current = parent
(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))
(b)
(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.
If above condition is not true then we will check for parent node and if it is less than current node then swap these.
pseudo code :-
current = i
// assuming parent of root is -1
parent is parent node of current node
while A[parent] < A[current] && parent != -1 do,
if A[grandparent] < A[current] // if current element is bigger than parent then shift it up
swap(A[current],A[parent])
swap(A[grandparent],A[parent])
current = grandparent
else if A[parent] < A[current]
swap(A[parent],A[current])
current = parent
(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)
so order would be log(n)/2
(iii) C++ code :-
#include<bits/stdc++.h>
using namespace std;
// function to return index of parent node
int parent(int i)
{
if(i == 0)
return -1;
return (i-1)/2;
}
// function to return index of grandparent node
int grandparent(int i)
{
int p = parent(i);
if(p == -1)
return -1;
else
return parent(p);
}
void shift_up(int A[], int n, int ind)
{
int curr = ind-1; // because array is 0-indexed
while(parent(curr) != -1 && A[parent(curr)] < A[curr])
{
int g = grandparent(curr);
int p = parent(curr);
if(g != -1 && A[g] < A[curr])
{
swap(A[curr],A[p]);
swap(A[p],A[g]);
curr = g;
}
else if(A[p] < A[curr])
{
swap(A[p],A[curr]);
curr = p;
}
}
}
int main()
{
int n;
cout<<"enter the number of elements :-\n";
cin>>n;
int A[n];
cout<<"enter the elements of array :-\n";
for(int i=0;i<n;i++)
cin>>A[i];
int ind;
cout<<"enter the index value :- \n";
cin>>ind;
shift_up(A,n,ind);
cout<<"array after shift up :-\n";
for(int i=0;i<n;i++)
cout<<A[i]<<" ";
cout<<endl;
}
Explanation:
Assume that a specific hard disk drive has an average access time of 16ms (i.e. the seek and rotational delay sums to 16ms) and a throughput or transfer rate of 185MBytes/s, where a megabyte is measured as 1024.
Required:
What is the average access time for a hard disk spinning at 360 revolutions per second with a seek time of 10 milliseconds?
Answers
Answer:
Average access time for a hard disk = 11.38 ms
Explanation:
See attached pictures for step by step explanation.
This assignment covers the sequential circuit component: Register and ALU. In this assignment you are supposed to create your own storage component for two numbers using registers. Those two numbers are then passed into a custom ALU that calculates the result of one of four possible operations. Key aspect of this assignment is to understand how to control registers, how to route signals and how to design a custom ALU.
Answers
Answer:
The part I called command in the first diagram has been renamed to opcode, or operation code. This is a set of bits (a number) that will tell the ALU which action to perform. I can get the LC-3 opcodes for ADD and NOT and ADD from the book, so I'm not too worried.
Note the #? comment by the switch above opcode. This means I'm not sure how many switches I will need. How many bits do I need to perform all the operations I want? The textbook will tell me.
Materials
Now I make a list of all the materials you have accumulated so far. This list is just an example; yours may be different.
Two 4-bit inputs
One 4-bit output
Two keypads for 4-bit input
Three 7-segment displays (2 for input, 1 for output)
A bunch of switches for opcode (could use a keypad, I guess, but switches are so much more geeky)
A bunch of lights too
The "is zero" LED
One button for clock
One button for reset
One switch for carry-in
Include logic to perform a SUB instruction. That is, subtract the second operand from the first (out = in1 - in2). All three values -- both inputs and the output -- must be two's complement numbers (negative numbers must be represented). Your design may work in one (8 points) or two (4 points) clock cycles.
Explanation:
You are allowed to use the Logisim built-in registers.
The clear input of the register should not be used (do not connect anything to
them).
Custom ALU
Use the provided subcircuit in the template to implement your ALU. You do not have to create additional subcircuits to do this. The ALU has a total of three inputs: First number, second number and select operation input. And one output: Result. The first and second number are used as input for the operations the ALU performs. The select input decides which operation result will be on the single output of the ALU. The ALU is supposed to calculate: NumberA OPERATION NumberB. Register 1 of the storage contains NumberA and Register 2 contains NumberB. The ALU must be able to compute signals with a 4-bit width. Make sure to add labels to all inputs and outputs.
The following operations should be performed for each select input combination (s1s0): • 00: Logic Bitwise XOR
• 01: Multiplication
• 10: Division
• 11: Addition Notes:
You can change the inputs bit width / data bits of any gate to more that 1-bit.
The Logic Bitwise XOR operation can be done with a single XOR gate.
You are also allowed to use the built-in arithmetic logic components and multi- plexer provided by Logisim.
If the result is larger than 4 bits, it will be truncated (only 4 LSB will be shown). This behavior is intended for this assignment. Also, negative results do not have to be considered.
Once you have implemented the ALU circuit, connect the wires in the main circuit properly and test all four operations of your ALU in combination with the storage component.
The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upper surface: Cp constant at – 0.8 from the leading edge to 60% chord, then increasing linearly to +0.1 at the trailing edge. Lower surface: Cp constant at – 0.4 from the leading edge to 60% chord, then increasing linearly to + 0.1 at the trailing edge. Estimate the lift coefficient and the pitching moment coefficient about the leading edge due to lift.
Answers
Answer:
The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.
Explanation:
The Upper Surface Cp is given as
[tex]Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1[/tex]
The Lower Surface Cp is given as
[tex]Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1[/tex]
The difference of the Cp over the airfoil is given as
[tex]\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32[/tex]
Now the Lift Coefficient is given as
[tex]C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192[/tex]
Now the coefficient of moment about the leading edge is given as
[tex]C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306[/tex]
So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.
The lift coefficient and the pitching moment coefficient about the leading edge due to lift are respectively; 0.3192 and -0.13
Aerodynamics engineeringWe are given;
Distance of upper surface from leading edge to percentage of chord = -0.8
Percentage of chord for both surfaces = 60% = 0.6
Rate of increase at trailing edge for both surfaces = +0.1
Distance of lower surface from leading edge to percentage of chord = -0.6
angle of incidence; α_i = 4° = 4π/180 rad
Let us first calculate the Cp constant for both the upper and lower surface.
Cp for upper surface is;
Cp_u = (-0.8 × 0.6) - 0.1∫¹₀.₆ dx
Solving this integral gives;
Cp_u = (-0.8 × 0.6) - (0.1 × 0.4)
Cp_u = -0.52
Cp for lower surface is;
Cp_l = (-0.4 × 0.6) + 0.1∫¹₀.₆ dx
Solving this integral gives;
Cp_l = (-0.4 × 0.6) + (0.1 × 0.4)
Cp_l = -0.2
Change in Cp across the foil is;
ΔCp = Cp_l - Cp_u
ΔCp = -0.2 - (-0.52)
ΔCp = 0.32
Formula for the lift coefficient is;
C_L = ΔCp * cosα_i
C_L = 0.32 * cos (4π/180)
C_L = 0.3192
Formula for the pitching moment coefficient is;
(-0.3 * 0.4 * 0.6) - ((0.6 + (0.4/3)) * 0.2 * 0.4)
C_m,p = -0.072 - 0.059
C_mp ≈ -0.13
Read more about aerodynamics at; https://brainly.com/question/6672263