The option that is not an ASE certification is . A/C and Refrigerants handling certification (609).
<h3>What is ASE certification?</h3>
The term ASE is known to be a body that tends to promotes excellence in regards to vehicle repair, service as well as parts distribution.
Note that in the world today more than a quarter of million of people are known to possess ASE certifications.
Since ASE Certified professionals work in in all areas of the transportation industry. one can say that The option that is not an ASE certification is. A/C and Refrigerants handling certification (609).
Learn more about ASE certification from
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Answer:
You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)
Step by step design:
- Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
- Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
- Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA
Our circuit meet the average voltage (Va) specification:
Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it
Answer:
The appropriate solution is "1481.76 N".
Explanation:
According to the question,
Mass,
m = 540 kg
Coefficient of static friction,
= 0.28
Now,
The applied force will be:
⇒ ![F=\mu_s mg](https://tex.z-dn.net/?f=F%3D%5Cmu_s%20mg)
By substituting the values, we get
![=0.28\times 540\times 9.8](https://tex.z-dn.net/?f=%3D0.28%5Ctimes%20540%5Ctimes%209.8)
Answer:The awnser is 5
Explanation:Just divide all of it
Answer:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density
![1\ gallon = 0.13ft^3](https://tex.z-dn.net/?f=1%5C%20gallon%20%3D%200.13ft%5E3)
Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu