Answer:
a) ![(dP)_{v} = 9.692 kPa](https://tex.z-dn.net/?f=%28dP%29_%7Bv%7D%20%3D%209.692%20kPa)
b) ![(dP)_{T} = -9.692 kPa](https://tex.z-dn.net/?f=%28dP%29_%7BT%7D%20%3D%20-9.692%20kPa)
c) dP = 0 Pa
Explanation:
The specifies equation is :
![dz = (\frac{\delta z}{\delta x}) _{y} dx + (\frac{\delta z}{\delta y}) _{x} dy](https://tex.z-dn.net/?f=dz%20%3D%20%28%5Cfrac%7B%5Cdelta%20z%7D%7B%5Cdelta%20x%7D%29%20_%7By%7D%20dx%20%2B%20%28%5Cfrac%7B%5Cdelta%20z%7D%7B%5Cdelta%20y%7D%29%20_%7Bx%7D%20dy)
Note that:
![dP = \frac{R}{v} dT - \frac{RT}{v^{2} } dV](https://tex.z-dn.net/?f=dP%20%3D%20%5Cfrac%7BR%7D%7Bv%7D%20dT%20-%20%5Cfrac%7BRT%7D%7Bv%5E%7B2%7D%20%7D%20dV)
1% increase in temperature at specific volume:
![dT = \frac{0.01}{1} *350\\dT = 3.5 K](https://tex.z-dn.net/?f=dT%20%3D%20%5Cfrac%7B0.01%7D%7B1%7D%20%2A350%5C%5CdT%20%3D%203.5%20K)
a) Change in pressure of helium at constant volume:
![(dP)_{v} = \frac{R}{v} dT](https://tex.z-dn.net/?f=%28dP%29_%7Bv%7D%20%3D%20%5Cfrac%7BR%7D%7Bv%7D%20dT)
R = 2.0769 kJ/kg-K
dT = 3.5 K
v = 0.75 m³/kg
![(dP)_{v} = \frac{2.0769}{0.75} * 3.5\\(dP)_{v} = 9.692 kPa](https://tex.z-dn.net/?f=%28dP%29_%7Bv%7D%20%3D%20%5Cfrac%7B2.0769%7D%7B0.75%7D%20%2A%203.5%5C%5C%28dP%29_%7Bv%7D%20%3D%209.692%20kPa)
b)
dv = (1%/100%) *0.75
dv = 0.0075 m³/kg
Change in pressure of helium at constant temperature:
![(dP)_{T} = \frac{-RT}{v^{2} } dv](https://tex.z-dn.net/?f=%28dP%29_%7BT%7D%20%3D%20%5Cfrac%7B-RT%7D%7Bv%5E%7B2%7D%20%7D%20dv)
R = 2.0769 kJ/kg-K
T = 350 K
v = 0.75 m³/kg
dv = 0.0075 m³/kg
![(dP)_{T} = \frac{-(2.0769*350)}{0.75^{2} } *0.0075\\(dP)_{T} = -9.692 kPa](https://tex.z-dn.net/?f=%28dP%29_%7BT%7D%20%3D%20%5Cfrac%7B-%282.0769%2A350%29%7D%7B0.75%5E%7B2%7D%20%7D%20%2A0.0075%5C%5C%28dP%29_%7BT%7D%20%3D%20-9.692%20kPa)
c) The change in pressure of helium :
![dP = (dP)_{v} + (dP)_{T}](https://tex.z-dn.net/?f=dP%20%3D%20%28dP%29_%7Bv%7D%20%2B%20%28dP%29_%7BT%7D)
dP = 9.692 - 9.692
dP = 0
Answer:
C. Yes, the water could be changing the phase.
Explanation:
Answer: 1.57
Explanation:
This described situation is known as Refraction, a phenomenon in which light bends or changes its direction when passing through a medium with a index of refraction different from the other medium.
In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.
According to Snell’s Law:
(1)
Where:
is the first medium index of refraction (the value we want to know)
is the second medium index of refraction (air)
is the angle of incidence
is the angle of refraction
Now, let's find
from (1):
(2)
Substituting the known values:
Finally: