Answer: Let us use the pickled file - DeckOfCardsList.dat.
Explanation: So that our possible outcome becomes
7♥, A♦, Q♠, 4♣, 8♠, 8♥, K♠, 2♦, 10♦, 9♦, K♥, Q♦, Q♣
HPC (High Point Count) = 16
Answer:
mass flow rate = 0.0534 kg/sec
velocity at exit = 29.34 m/sec
Explanation:
From the information given:
Inlet:
Temperature ![T_1 = -16^0\ C](https://tex.z-dn.net/?f=T_1%20%3D%20-16%5E0%5C%20C)
Quality ![x_1 = 0.2](https://tex.z-dn.net/?f=x_1%20%3D%200.2)
Outlet:
Temperature ![T_2 = -16^0 C](https://tex.z-dn.net/?f=T_2%20%3D%20-16%5E0%20C)
Quality ![x_2 = 1](https://tex.z-dn.net/?f=x_2%20%3D%201)
The following data were obtained at saturation properties of R134a at the temperature of -16° C
![v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg](https://tex.z-dn.net/?f=v_f%3D%200.7428%20%5Ctimes%2010%5E%7B-3%7D%20%5C%20m%5E3%2Fkg%20%5C%5C%20%5C%5C%20%20v_g%20%3D%200.1247%20%5C%20m%5E3%20%2Fkg)
![v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg](https://tex.z-dn.net/?f=v_1%20%3D%20v_f%20%2B%20x_1%20%28%20vg%20-%20%28%20v_f%29%29%20%5C%5C%20%5C%5C%20v_1%20%3D%200.7428%20%5Ctimes%2010%5E%7B-3%7D%20%2B%200.2%20%280.1247%20-%280.7428%20%5Ctimes%2010%5E%7B-3%7D%29%29%20%5C%5C%20%5C%5C%20%20v_1%20%3D%200.0255%20%5C%20m%5E3%2Fkg%20%5C%5C%20%5C%5C%20%5C%5C%20%20v_2%20%3D%20v_g%20%3D%200.1247%20%5C%20m%5E3%2Fkg)
![m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}](https://tex.z-dn.net/?f=m%20%3D%20%5Crho_1A_1v_1%20%3D%20%5Crho_2A_2v_2%20%5C%5C%20%5C%5C%20%20m%20%3D%20%5Cdfrac%7B1%7D%7B0.0255%7D%20%5Ctimes%20%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Ctimes%20%281.7%20%5Ctimes%2010%5E%7B-2%7D%29%5E2%5Ctimes%206%20%20%5C%5C%20%5C%5C%20%5Cmathbf%7Bm%20%3D%200.0534%20%5C%20kg%2Fsec%7D)
![\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}](https://tex.z-dn.net/?f=%5Crho_1A_1v_1%20%3D%20%5Crho_2A_2v_2%20%5C%5C%20%5C%5C%20A_1%20%3DA_2%20%20%5C%5C%20%5C%5C%20%20%5Crho_1v_1%20%3D%20%5Crho_2v_2%20%20%20%5C%5C%20%5C%5C%20%5Cimplies%20%5Cdfrac%7B1%7D%7B0.0255%7D%20%5Ctimes6%20%3D%20%5Cdfrac%7B1%7D%7B0.1247%7D%5Ctimes%20%28v_2%29%5C%5C%20%5C%5C%20%5C%5C%5Cmathbf%7B%5C%5C%20v_2%20%3D%2029.34%20%5C%20m%2Fsec%7D)
Answer:
Using the formula
V =20y/(x^2+y^2)^1/2 - 20x/(x^2+y^2)^1/2
Hence fluid speed at x axis =20x/(x^2+y^2)^1/2
While the fluid speed at y axis =20y/(x^2+y^2)^1/2
Now the angle at 1, 5
We substitute into the formula above
V= 20×5/(1+25)^1/2= 19.61
For x we have
V = 20× 1/(1+25)^1/2= 3.92
Angle = 19.61/3.92= 5.0degrees
Angel at 5, and 2
We substitute still
V = 20×5/(2+25)^1/2=19.24
At 2 we get
V= 20×2/(2+25)^1/2=7.69
Dividing we get 19.24/7.69=2.5degrees
At 1 and 0
V = 20/(1)^1/2=20
At 0, v =0
Angel at 2 and 0 = 20degrees
At 5 and 2
V = 100/(25+ 4)^1/2=18.56
At x = 2
40/(√29)=7.43
Angle =18.56/7.43 = 2.49degrees.
Answer: National anthem contributed to the development of unity diversity in nepal; The Sayaun Thunga Phool Ka was officially adopted as national anthem of Nepal on 3rd August 2007. This anthem which means made of hundred flowers strongly portrays the great unity and diversity of Nepalese people.
Explanation: