Answer:
(a) ------(3). (b)------(1) (c)-----(5) (d)------(2) ------ (e) -----4
Note: Kindly find an attached copy of the diagram associated with the solution to the question below.
Sources: the diagram to this question was researched from Quizlet
Explanation:
Solution
(1) Part (a)a waveform has a high frequency components compared to another waveform. the corresponding frequency components should be high.
So for the wave form a the corresponding frequency spectrum is (3)
(2) For part (b), waveform has three harmonics, the corresponding frequency spectrum is (1)
(3) The time domain waveform plot (c) is a sine wave but there exists a dc component.
Thus x[0] ≠0
For (c) the corresponding frequency spectrum is (5)
(4) For part (d) the corresponding frequency spectrum is (2)
(5) A sine wave is made of a single frequency only and its spectrum is a single point
For (e) the corresponding frequency spectrum is (4)
Answer:
λ^3 = 4.37
Explanation:
first let us to calculate the average density of the alloy
for simplicity of calculation assume a 100g alloy
80g --> Ag
20g --> Pd
ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)
= 100*10^-3/(20/11.9*10^6+80/10.44*10^6)
= 10744.62 kg/m^3
now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate
total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27
= 1.12*10^23 unit cells
mass of Pd in 1 unit cell = 20/1.12*10^23
Now,
ρ_avg= mass of unit cell/volume of unit cell
ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3
λ^3 = 4.37
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.
