Answer:
The low side pressure of an A/C system losing vacuum and the pressure rising above zero indicates that there is too much refrigerant in the system.
Explanation:
Considering an A/C system, the condenser fan might be malfunctioning if the low side pressure of the air conditioner is excessive. On the other hand, it's also conceivable that the system has been overcharged with refrigerant.
Stated the scenario that the refrigerant of the system was being recovered, it is an indication that the system is merely overcharged. Even with the engine off, you will notice high pressures.
Either too much oil is present, or there is too much refrigerant in the air conditioning system. In either case, until you let some of that pressure out—ideally, a mechanic should do this—the issue won't go away on its own.
To know more about the pressure scenarios related to AC systems, refer to:
brainly.com/question/17072827
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Answer: The volume of gas expands because of the decrease in pressure as he tries to exit the water body, therefore he must take necessary precaution.
Explanation:
Using Boyle's law which states that the the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature
ie P1VI=P2V2
A diver absorbs compressed nitrogen gas when he dives into the water body, As he ascends out of the water body having less pressure, the volume of nitrogen gas which he absorbs will tend to expand following Boyle's Law. Therefore a scuba driver should not rises quickly but slowly to the surface or else the expanding nitrogen gas can cause tiny bubbles in his blood and tissue to form together with joints pains and eventually cause decompression sickness needing medical attention.
Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as
Here,
P = Pressure
d = Diameter
t = Thickness
At the same time the longitudinal stress is given as,
The letters have the same meaning as before.
Then he hoop stress would be,
And the longitudinal stress would be
The Mohr's circle is attached in a image to find the maximum shear stress, which is given as
Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi
C.
The range of temperatures on Earth allows water to exist in all of its states.
