Answer:
Technician A only
Explanation:
Both high-side pressures and low-side pressures are low with the engine running and the selector set to the air-conditioning position. Technician A says that the system is undercharged. Technician B says the cooling fan could be inoperative. Which technician is correct?
usually . An overcharged system will result in lower than normal low side pressures
An undercharged system will not enable the compressor to create pressure. As a result of the low amount of refrigerant, the cooling ability is reduced. When we say undercharged, we mean the refrigerant in the system is low, so the both the high side pressures and low side pressures will be low.
It is not possible to see the other waves on the electromagnetic spectrum because only other species can see the other parts of the spectrum because they have different components in their eyes than we do, therefore, only allowing us to see a
portion of the spectrum, which is visible light.
It would be a good game for you but if I get a pic I don’t want you can you come to my crib I just
Answer:
explained
Explanation:
Yes, the heating of filament is what causes the light production (photon emission), and this heating is caused because of current in the light bulb
(H= i^2*R*t i=current, H= heat, t= time and R= resistance).But using constant current source is not a good idea because in constant current source resistance is very low that can cause short circuit and ultimately fusing it. Whereas in constant voltage source current adjusts itself and prevents fusing because of high resistance in the circuit.
Answer:
W= -2.5 (p₁*0.0012) joules
Explanation:
Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then
In adiabatic compression, work done by mixture during compression is
W= 
where f= final volume and i =initial volume, p=pressure
p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ
W= 
W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)
W=1/1-γ (p₁Vf-p₀Vi)
W= 1/1-1.40 (p₁*6/5 -p₀*0)
W= -2.5 (p₁*6/5*0.001) changing liters to m³
W= -2.5 (p₁*0.0012) joules
