Please what is the name of this tool
2 answers:
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Answer:
Glass: Low-Loss dielectric
α = 8.42*10^-11 Np/m
β = 468.3 rad/m
λ = 1.34 cm
up = 1.34*10^8 m/s
ηc = 168.5 Ω
Tissue: Quasi-Conductor
α = 9.75 Np/m
β = 12.16 rad/m
λ = 51.69 cm
up = 0.52*10^8 m/s
ηc = 39.54 + j 31.72 Ω
Wood: Good conductor
α = 6.3*10^-4 Np/m
β = 6.3*10^-4 Np/m
λ = 10 km
up = 0.1*10^8 m/s
ηc = 6.28*( 1 + j )
Explanation:
Given:
Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz
Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.
Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz
Find:
Determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, up, and ηc:
Solution:
- We need to determine the loss tangent to determine category of the medium as follows:
σ / w*εr*εo
Where, w is the angular speed of wave
εo is the permittivity of free space = 10^-9 / 36*pi
- Now we classify as follows:
- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.
Glass: Low-Loss dielectric
Tissue: Quasi-Conductor
Wood: Good conductor
- Now we will use categorized material base equations from Table 17-1 as follows:
Glass: Low-Loss dielectric
α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)
α = 8.42*10^-11 Np/m
β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)
β = 468.3 rad/m
λ = 2*pi / β = 2*pi / 468.3
λ = 1.34 cm
up = λ*f = 0.0134*10^10
up = 1.34*10^8 m/s
ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)
ηc = 168.5 Ω
Tissue: Quasi-Conductor
α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)
α = 9.75 Np/m
β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)
β = 12.16 rad/m
λ = 2*pi / β = 2*pi / 12.16
λ = 51.69 cm
up = λ*f = 0.5169*100*10^6
up = 0.52*10^8 m/s
ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5
ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5
ηc = 39.54 + j 31.72 Ω
Wood: Good conductor
α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )
β = α = 6.3*10^-4 Np/m
λ = 2*pi / β = 2*pi / 6.3*10^-4
λ = 10 km
up = λ*f = 10,000*1*10^3
up = 0.1*10^8 m/s
ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4
ηc = 6.28*( 1 + j )
Answer:
a.)
US Sieve no. % finer (C₅ )
4 100
10 95.61
20 82.98
40 61.50
60 42.08
100 20.19
200 6.3
Pan 0
b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4
c.) Cu = 3.33
d.) Cc = 1
Explanation:
As given ,
US Sieve no. Mass of soil retained (C₂ )
4 0
10 18.5
20 53.2
40 90.5
60 81.8
100 92.2
200 58.5
Pan 26.5
Now,
Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g
⇒ w = 421.2 g
As we know that ,
% Retained = C₃ = C₂×
∴ we get
US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)
4 0 0
10 4.39 4.39
20 12.63 17.02
40 21.48 38.50
60 19.42 57.92
100 21.89 79.81
200 13.89 93.70
Pan 6.30 100
Now,
% finer = C₅ = 100 - C₄
∴ we get
US Sieve no. Cummulative % retained (C₄) % finer (C₅ )
4 0 100
10 4.39 95.61
20 17.02 82.98
40 38.50 61.50
60 57.92 42.08
100 79.81 20.19
200 93.70 6.3
Pan 100 0
The grain-size distribution is :
b.)
From the diagram , we can see that
D10 = 0.12
D30 = 0.22
D60 = 0.12
c.)
Uniformity Coefficient = Cu =
⇒ Cu =
d.)
Coefficient of Graduation = Cc =
⇒ Cc = = 1
