2. Testing Encryption Validity Encryption is the process of converting information algorithmically so that only a valid recipien
1 answer:
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Answer:
The number of germanium atoms per cubic centimeter for this germanium-silicon alloy is 3.16 x 10²¹ atoms/cm³.
Explanation:
Concentration of Ge () = 15%
Concentration of Si (C) = 85%
Density of Germanium (ρ) = 5.32 g/cm³
Density of Silicon (ρ) = 2.33 g/cm³
Atomic mass of Ge (A)= 72.64 g/mol
To calculate the number of Ge atoms per cubic centimeter for the alloy, we will use the formula:
No of Ge atoms/cm³=[Avogadro's Number*]/([
*A
/ρ
)+(C
*A
/ρ
)]
= (6.02x10²³ * 15%) / [(15% * 72.64/5.32)+(72.64*85%/2.33)]
= (9.03x10²²)/(2.048+26.499)
= (9.03x10²²)/(28.547)
No of Ge atoms/cm³ = 3.16 x 10²¹ atoms/cm³
Answer:
A) while (num >= 0)
Explanation:
To understand why we need to focus on the module and division operation inside the loop. num % 10 divide the number by ten and take its remainder to then add this remainder to sum, the important here is that we are adding up the number in reverse order and wee need to repeat this process until we get the first number (1%10 = 1), therefore, num need to be one to compute the last operation.
A) this is the correct option because num = 1 > 0 and the last operation will be performed, and after the last operation, num = 1 will be divided by 10 resulting in 0 and 0 is not greater than 0, therefore, the cycle end and the result will be printed.
B) This can not be the option because this way the program will never ends -> 0%10 = 0 and num = 0/10 = 0
C) This can not be the option because num = 1 > 1 will produce an early end of the loop printing an incomplete result
D) The same problem than C
E) There is a point, before the operations finish, where sum > num, this will produce an early end of the loop, printing an incomplete result
Answer:
Explanation:
The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.
The answer to the above question is
The pressure at point 2 = 75.959 kPa
Explanation:
Bernoulli's equation with losses gives
hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)
Between points 1 and 2, z₁ = z₃ + 0.6 m therefore
hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)
hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)
But v = Q/A
or since A = π×D²/4 we have
A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²
Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃ = 1.315 m/s from there we find the friction coefficient from Moody Diagram as follows
ε = Which gives
the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175
Substituting he above values into the equation we get
= 19.761 m
Combined head loss = 19.761 m
Hence 19.743 m = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)
or 260 kPa-18.82 m × 9.81 m/s²×ρ= P₃
Where ρ = density of water, we have
260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa