Answer:
Yes, it can can be completed adiabatically
Explanation:
To solve the problem we will resort to the theory of thermodynamics,
It is necessary to develop this problem to resort to the A-11E tables in English Units for R134a (since the problem requires it, if it were SI just to change to that table)
State 1 indicates that the refrigerant is at 60 ° F,
In the first table (attached image of the value taken) the value of the entropy is
![S_f=S_1= 0.06675Btu/lbm.R](https://tex.z-dn.net/?f=S_f%3DS_1%3D%200.06675Btu%2Flbm.R)
For State 2 the refrigerant is at 50% quality and at a pressure of ![140lbf / in ^ 2](https://tex.z-dn.net/?f=140lbf%20%2F%20in%20%5E%202)
In table 2 of the refrigerant (for the pressure values) we perform the reading and we have to
![Sf= 0.09214](https://tex.z-dn.net/?f=Sf%3D%200.09214)
![Sg=0.21879](https://tex.z-dn.net/?f=Sg%3D0.21879)
We know that,
![S_2 = Sf +X_{quality}(S_g-S_f)](https://tex.z-dn.net/?f=S_2%20%3D%20Sf%20%2BX_%7Bquality%7D%28S_g-S_f%29)
![S_2 = 0.09214+0.5(0.21979-0.09214)](https://tex.z-dn.net/?f=S_2%20%3D%200.09214%2B0.5%280.21979-0.09214%29)
![S_2 = 0.155965BTU/Lb.R](https://tex.z-dn.net/?f=S_2%20%3D%200.155965BTU%2FLb.R)
The change in enthalpy would be given by
![\Delta S = S_2-S_1 = 0.155965BTU/Lb.R- 0.06675Btu/lbm.R](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%20S_2-S_1%20%3D%200.155965BTU%2FLb.R-%200.06675Btu%2Flbm.R)
![\Delta S = 0.089215Btu/lbm.R](https://tex.z-dn.net/?f=%5CDelta%20S%20%3D%200.089215Btu%2Flbm.R)
<em>The change in enthalpy is positive, so the process can be completed adiabatically</em>
The <span>Atomic theory that states that atoms have three fundamental parts, and that electrons orbit the nucleus is the kinetic molecular theory of gases.</span>
Explanation:
V = 3 cm/s = 0.03 m/s. BY THE FORmULA OF K.E. K.E = 1/2 mV^2. 300 =1/2 m (0.03)^2. m = 300 x 2/0.0009.
Answer:
speed when it reaches y = 4.00cm is
v = 14.9 g.m/s
Explanation:
given
q₁=q₂ =2.00 ×10⁻⁶
distance along x = 3.00cm= 3×10⁻²
q₃= 4×10⁻⁶C
mass= 10×10 ⁻³g
distance along y = 4×10⁻²m
r₁ =
=
= 3.61cm = 0.036m
r₂ =
=
= 5cm = 0.05m
electric potential V = ![\frac{kq}{r}](https://tex.z-dn.net/?f=%5Cfrac%7Bkq%7D%7Br%7D)
change in potential ΔV = ![V_{1} - V_{2}](https://tex.z-dn.net/?f=V_%7B1%7D%20-%20V_%7B2%7D)
ΔV =
, where
2.00μC
ΔV = ![2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })](https://tex.z-dn.net/?f=2kq%28%5Cfrac%7B1%7D%7Br_%7B1%7D%7D%20-%20%5Cfrac%7B1%7D%7Br_%7B2%7D%20%7D%29)
ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × ![(\frac{1}{0.036} - \frac{1}{0.05} )](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B0.036%7D%20-%20%5Cfrac%7B1%7D%7B0.05%7D%20%29)
ΔV= 2.789×10⁵
= ΔV × q₃
ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶
v² = 223.12 g.m/s
v = 14.9 g.m/s
The surface of the earth changes. Some changes are due to slow processes, such as erosion and weathering, and some changes are due to rapid processes, such as landslides, volcanic eruptions, and earthquakes.