Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation
![2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]](https://tex.z-dn.net/?f=2850.6%20%2B%20%5B%5Cfrac%7B50%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D%20%3D%20h2%20%2B%5B%20%5Cfrac%7B600%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D)
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99
Answer:
a) S = 1.69 10⁹ W/m², b) P = 5.63 Pa
, c) F = 20.6 10⁻¹² N
Explanation:
a) The intensity defined as the energy per unit area
S = U / A
Area of a circle is
W = 6.2 mw = 6.2 10-3 W
R = 1080 nm = 1080 10⁻⁹ m = 1.080 10⁻⁶ m
A = π R2
A = π (1,080 10⁻⁶)²
A = 3.66 10 -12 m²
S = 6.2 10-3 / 3.66 10-12
S = 1.69 10⁹ W / m²
b) The radiation pressure
P = 1 / c (dU / dt) / A
S = (dU / dt) / A
P = S / c
P = 1.69 10 9 / 3. 108
P = 5.63 Pa
c) the definition of pressure is force over area
P = F / A
F = P A
F = 5.63 3.66 10⁻¹²
F = 20.6 10⁻¹² N
d) for this we use Newton's second law
F = ma
a = F / m
Answer:
i don't know if this is right but here,9900 kgm/s
Explanation:
calculator UwU
Answer:
a. A list of the names of each student present today. (microstate)
b. The number of students in attendance. (macrostate)
Explanation:
You can fins the answer to this question by comparing the situation of the problem with a system of molecules with discrete energy.
Without importance of which molecules have a specific energy, but rather, what is the total amount of energy, you can get for different configurations of energy the same amount of the total energy. If different configurations of the energies of the molecules give you the same total energy of the system, you say that the macrostate is the same. In the case of the classroom, it does not matter how are distributed the students in the class, the total number of students is always the same. The macrostate is the same for what ever organization of the students in the class.
If you would interested in the energy of each molecules, you will obtain different configurations. In the case of the classroom. The names of the student will define a microstate because in this case there are many configurations.
a. A list of the names of each student present today. (microstate)
b. The number of students in attendance. (macrostate)
