I WILL GIVE BRAINLEIST!! What are some challenges did Sunita williams (astronaut) faced throughout her life and career?
Charles, Clarissa, and Francine all work in Manufacturing. The table shows the breakdown of one day for each employee.
Charles
Clarissa
Francine
9:00 a.m.–11:00 a.m.
meeting with production team and first-line supervisor of production
11:00 a.m.–12:00 p.m.
meeting with finance team
1:00 p.m.–2:00 p.m.
meeting with materials engineer for new project
2:00 p.m.–5:00 p.m.
e-mails and reports
6:00 a.m.–7:00 a.m.
review lists of damaged equipment
8:00 a.m.–11:00 a.m.
fix damaged machines in the hospital laboratory
12:00 p.m.–2:00 p.m.
fix damaged machines on the hospital floor
8:00 a.m.–9:00 a.m.
decide on which products need assessment
9:00 a.m.–12:00 p.m.
test products on factory floor against standards
1:00 p.m.–4:00 p.m.
write reports and recommendations on tested products
Which correctly identifies the career of each person?
A.)Charles is a First Line Supervisor of Production, Clarissa is an Industrial Machinery Mechanic, and Francine is a Product Safety Engineer.
B.)Charles is a Product Safety Engineer, Clarissa is a First Line Supervisor of Production, and Francine is an Industrial Machinery Mechanic.
C.)Charles is an Industrial Production Manager, Clarissa is a Medical Equipment repairer, and Francine is a Quality Control Analyst.
D.)Charles is a Quality Control Analyst, Clarissa is an Industrial Production Manager, and Francine is a Medical Equipment Repairer.
Answer:
C.)Charles is an Industrial Production Manager, Clarissa is a Medical Equipment repairer, and Francine is a Quality Control Analyst.
Explanation:
A power cycle operates between hot and cold reservoirs at 600 K and 300 K, respectively. At steady state the cycle develops a power output of 0.45 MW while receiving energy by heat transfer from the hot reservoir at the rate of 1 MW. a. Determine the thermal efficiency and the rate at which energy is rejected by heat transfer to the cold reservoir, in MW. b. Compare the results of part (a) with those of a reversible power cycle operating between these reservoirs and receiving the same rate of heat transfer from the hot reservoir
Answer:
a. The thermal efficiency of the actual cycle is 45%
ii) The rate at which energy is rejected by heat transfer to the cold reservoir is 0.55 MW
b. The (maximum) cycle efficiency, is 50%
The rate of energy rejection to the cold reservoir by heat transfer is 0.55 MW
Therefore, there is an increased efficiency and reduced heat rejection in the reversible power cycle working between the given two reservoirs
Explanation:
The given parameters are;
The temperature of the hot reservoir, [tex]T_H[/tex] = 600 K
The temperature of the hot reservoir, [tex]T_C[/tex] = 300 K
The power output, [tex]\dot W[/tex] = 0.45 MW = The cycle work per second
The heat input from the hot reservoir, [tex]\dot Q_{in}[/tex] = 1 MW
a. The thermal efficiency of the actual cycle is given by the work done divided by the heat supplied as follows;
[tex]\eta_{actual} = \dfrac{\dot W}{\dot Q_{in}}[/tex]
Therefore, we have;
[tex]\eta_{actual} = \dfrac{0.45 \ MW}{1 \ MW} = 0.45 = 45 \%[/tex]
The thermal efficiency of the actual cycle, [tex]\eta_{actual}[/tex] = 45%
ii) The rate at which energy is rejected by heat transfer to the cold reservoir, [tex]\dot Q_{c}[/tex], is given as follows;
[tex]\dot Q_{in}[/tex] = [tex]\dot Q_{c}[/tex] + [tex]\dot W[/tex]
∴ [tex]\dot Q_{c}[/tex] = [tex]\dot Q_{in}[/tex] - [tex]\dot W[/tex]
Therefore, we have;
[tex]\dot Q_{c}[/tex] = 1 MW - 0.45 MW = 0.55 MW
The rate at which energy is rejected by heat transfer to the cold reservoir, [tex]\dot Q_{c}[/tex] = 0.55 MW
b. For a reversible power cycle, we have the maximum cycle efficiency, [tex]\eta_{maximum}[/tex], given as follows;
[tex]\eta_{maximum} = \dfrac{T_H - T_C}{T_H}[/tex]
[tex]\therefore \eta_{maximum} = \dfrac{600 \, K - 300 \, K}{600 \, K} = 0.5 = 50\%[/tex]
The maximum cycle efficiency, [tex]\eta_{maximum}[/tex] = 50%
The rate of work done by the cycle, [tex]\dot W[/tex], is therefore give as follows;
[tex]\eta_{maximum} = \dfrac{\dot W}{\dot Q_{in}}[/tex]
Therefore;
[tex]\dot Q_{in}[/tex] × [tex]\eta_{maximum}[/tex] = [tex]\dot W[/tex]
1 MW × 50% = 0.5 MW = [tex]\dot W[/tex]
[tex]\dot W[/tex] = 0.5 MW
[tex]\dot Q_{c}[/tex] = [tex]\dot Q_{in}[/tex] - [tex]\dot W[/tex]
∴ [tex]\dot Q_{c}[/tex] = 1 MW - 0.5MW = 0.5MW
[tex]\dot Q_{c}[/tex] = 0.5MW
The rate of energy rejection to the cold reservoir by heat transfer, [tex]\dot Q_{c}[/tex] = 0.55 MW
Therefore the frequency and the rate at which energy is rejected by heat transfer to the cold reservoir are both higher for the reversible power cycle
This Question: 1 pt
port
The type of respirator that has its own clean air supply is the
O A. full facepiece mask
O B. self-contained breathing apparatus
OC. mouthpiece with mechanical filter
O D. half mask
This is for a carpentry class
Answer:
c
Explanation:
it has its own mechanical filter making the air clean
The what
includes all the social, physical, and
mental skills required for low-risk driving.
Answer:
id
Explanation:
How might an operations manager alter operations to meet customer demand? Name at least two ways.
Will give 50 points!!!
Pat spent 1 hour and 17 minutes playing baseball in the yard and 65 minutes playing soccer. How many more minutes did pat spend playing baseball then soccer?
Answer: Pat spent 12 more minutes playing baseball than soccer.
Explanation:
1 hour = 60 minutes
60 + 17 = 77
77 - 65 = 12
Does anyone know how to fixes rope lights? Half of the rope is not on
Someone.. Anyone..
Answer:
Step 1 - Take Down the Rope Lights. You will first have to remove the rope lights from their position.
Step 2 - Assess the Damage. Remove the insulating material from around the edge of the broken part of the rope.
Step 3 - Solder the Wire.
Step 4 - Add Insulation.
Step 5 - Test Your Repair.
Explanation:
What is the root of a fillet weld
Answer:
There are 5 pieces to each fillet weld known as the root, toe, face, leg and throat. The root of the weld is the part of deepest penetration which is the opposite angle of the hypotenuse.
While discussing the testing of a Hall-effect switch:
Technician A says that the Hall-effect switch should read zero volts with the feeler gauge inserted between the hall layer and the magnet with battery voltage applied. Technician B says that the Hall-effect switch can only be tested with a scan tool. Who is correct?
Answer:
your answer is technician A