What are the important factors needed to be considered while selecting a brake or clutch?
Answers
Answer:
The correct answer is: the following factors are needed to properly consider while selecting a brake or clutch:
-Engagement
-Friction
-Electromagnetic
-Mechanical
-Actuation
-Electric
-Fluid power
-Self-actuation
-Key concepts
-Application notes
-Selection criteria
Explanation:
Clutches and brakes are important devices in many rotating drive systems, it is very important to guarantee the security and the proper function of them accomplishing a high quality parameters in those factors.
Related Questions
The following program includes fictional sets of the top 10 male and female baby names for the current year. Write a program that creates: A set all_names that contains all of the top 10 male and all of the top 10 female names. A set neutral_names that contains only names found in both male_names and female_names. A set specific_names that contains only gender specific names. Sample output for all_names: {'Michael', 'Henry', 'Jayden', 'Bailey', 'Lucas', 'Chuck', 'Aiden', 'Khloe', 'Elizabeth', 'Maria', 'Veronica', 'Meghan', 'John', 'Samuel', 'Britney', 'Charlie', 'Kim'}
8. 15 A manual arc welding cell uses a welder and a fitter. The cell operates 2,000 hriyr. The welder is paid $30/hr and the fitter is paid $25/hr. Both rates include applicable overheads. The cycle time to complete one welded assembly is 15. 4 min. Of this time, the arc-on time is 25%, and the fitter's participation in the cycle is 30% of the cycle time. A robotic arc welding cell is being considered to replace this manual cell. The new cell would have one robot, one fitter, and two workstations, so that while the robot is working at the first sta tion, the fitter is unloading the other station and loading it with new components. The fitter's rate would remain at $25/hr. For the new cell, the production rate would be eight welded assemblies per hour. The arc-on time would increase to almost 52%, and the fitter's participation in the cycle would be about 62%. The installed cost of the robot and worksta tions is $158,000. Power and other utilities to operate the robot and arc welding equipment will be $3. 80/hr, and annual maintenance costs are $3,500. Given a 3-year service life, 15% rate of return, and no salvage value, (a) determine the annual quantity of welded assem blies that would have to be produced to reach the break-even point for the two methods. (b) What is the annual quantity of welded assemblies produced by the two methods work. Ing 2,000 hryr?
With the aid of a labbled diagram describe the operation of a core type single phase transformer
How to find magnitude of net electric charge when current flows through a capacitor with initial charge?
Who is right?
Answers
Technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.
What are hybrid vehicle?
Hybrid vehicle are defined as a powered by a combustion engine and/or a number of electric motors that draw power from batteries. A gas-powered car simply has a traditional gas engine, but a hybrid car also features an electric motor.
One important advantage of hybrid cars is their capacity to reduce the size of the main engine, which improves fuel efficiency. Many hybrid vehicles employ electric motors to accelerate slowly at first until they reach higher speeds. They then use gasoline-powered engines to increase fuel efficiency.
Thus, technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.
To learn more about hybrid vehicle, refer to the link below:
#SPJ5
Fluorescent lamps
Mercury-containing lamps
All of the above
Answers
Answer: D all above
Explanation:
Jus done it
b. The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.
Answers
Answer:
1)ΔL = 0.616 mm
2)Δd = 0.00194 mm
Explanation:
We are given;
Force; F = 52900 N
Initial length; L_o = 207 mm = 0.207 m
Diameter; d_o = 19.2 mm = 0.0192 m
Elastic modulus; E = 61.4 GPa = 61.4 × 10^(9) N/m²
Now, from Hooke's law;
E = σ/ε
Where; σ is stress = force/area = F/A
A = πd²/4 = π × 0.0192²/4
A = 0.00009216π
σ = 52900/0.00009216π
ε = ΔL/L_o
ε = ΔL/0.207
Thus,from E = σ/ε, we have;
61.4 × 10^(9) = (52900/0.00009216π) ÷ (ΔL/0.207)
Making ΔL the subject, we have;
ΔL = (52900 × 0.207)/(61.4 × 10^(9) × 0.00009216π)
ΔL = 0.616 × 10^(-3) m
ΔL = 0.616 mm
B) Poisson's ratio is given as;
υ = ε_x/ε_z
ε_x = Δd/d_o
ε_z = ΔL/L_o
Thus;
υ = (Δd/d_o) ÷ (ΔL/L_o)
Making Δd the subject gives;
Δd = (υ × d_o × ΔL)/L_o
We are given Poisson's ratio to be 0.34.
Thus;
Δd = (0.34 × 19.2 × 0.616)/207
Δd = 0.00194 mm
Answers
Answer:
Explanation:
A multipurpose transformer can act as step up as well as step down transformer according to the desired setting by a user.
When the voltage at the output is greater than the voltage at the input of the transformer then it acts as step-up transformer and vice-versa acting is a step down transformer.
Given that:
input (primary) voltage of the transformer,
no. of turns in the primary coil,
- When the output voltage is 5.60 V:
turns compensating the losses
- When the output voltage is 12.0 V:
turns compensating the losses
- When the output voltage is 480 V:
turns compensating the losses
Answers
Answer:
The problem is that the pumps would consume more energy than the generators would produce.
Explanation:
Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.
A pump uses electricity to add energy to the water to send it to a higher potential energy state.
Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.
What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.
Answers
Answer:
#include <iostream>
#include <iomanip>
using namespace std;
int main(){
//declare arrays
string ItemName[100];
double ItemCost[100]={0.0};
//declare variable
string name="";
double Total=0.0;
int NumItems=0;
cout<<"Enter number of grocery item you will be entering";
cin>>NumItems
cout<<"\n";
if (NumItems>100)
{
cout<<"Enter grocery items less than 100"
cin>>NumItems
cout<<"\n";
}
for (int i=1;i>=NumItems;i++)
{
cout<<"Please enter the item name in one word only, Example: icecream\n";
cin>>ItemName[i]
cout<<"Please enter the cost as a decimal number, Example: 2.05\n\n";
cin>>ItemCost[i];
}
cout<<"Items"<<" "<<""$""<<"Cost"<<endl
for (int i=1;i>=NumItems;i++)
{
cout<<ItemName[i]<<" "<<"$"<<ItemCost[i]<<endl
}
for (int i=1;i>=NumItems;i++)
{
Total=Total+ItemCost[i];
}
cout<<"Total:$"<<Total<<endl;
system("PAUSE");
return 0;
}