Answer:
Technician B is right.
Explanation:
Air conditioning refrigerant contains Freon R22 and R410a, which have been linked to environmental damages, including ozone depletion, global warming, and energy-inefficiency. For environmentally-savvy entities and individuals, there is the modern move to a more environment-friendly refrigerant, known as R-32. Technician A's advice to vent the refrigerant outside the shop is in bad taste. He does not seem to be aware of the environmental footprint of such an action. Venting gas outside, in addition to the environmental damages, is also a waste of resources, and therefore, costly. This is why Technician B's advice should be preferred.
Answer:
E=52000Hp.h
E=38724920Wh
E=1.028x10^11 ftlb
Explanation:
To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.
Finally, when you have the result Hp h / year you convert it to Ftlb and Wh
![E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\](https://tex.z-dn.net/?f=E%3D%2812.5hp%29%28%5Cfrac%7B16h%7D%7Bday%7D%20%29%28%5Cfrac%7B5%20days%7D%7Bweek%7D%20%29%28%5Cfrac%7B52week%7D%7Byear%7D%20%29%5C%5C)
E=52000Hp.h
![E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\](https://tex.z-dn.net/?f=E%3D52000Hp.h%28%5Cfrac%7B744.71Wh%7D%7BHp.h%7D%20%29%5C%5C)
E=38724920Wh
![E=52000Hph(\frac{1977378.4 ft lb}{1Hph}](https://tex.z-dn.net/?f=E%3D52000Hph%28%5Cfrac%7B1977378.4%20%20ft%20lb%7D%7B1Hph%7D)
E=1.028x10^11 ftlb
Answer:
modern vehicles are made to crunch up a little bit so they that absorbe some of the impact instead of you
Explanation:
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
Wait why do you want me to