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DerKrebs [107]
3 years ago
8

g Nuclear power supplies about _______ of all electricity in Europe. Nuclear power supplies about _______ of all electricity in

Europe. two-thirds one-third one-fifth one-half three-fourths
Engineering
1 answer:
Alex_Xolod [135]3 years ago
5 0

Answer:

The answer is not in the options. It is one-fourth.

Explanation:

As of 2017, it was recorded that nuclear power supplies 25% of electricity in Europe. That's 1/4 of the total electrical power supply.

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garik1379 [7]

????? what do u mean tier 2

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Discuss the environmental concerns regarding microfibre products
zlopas [31]

I don't know

Explanation:

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High-mount stop lamps are typically attached to the exterior of the vehicle using:
I am Lyosha [343]

Answer:

Option A

Explanation:

Nylon retainers are knocking screws that have nylon coating on their thread which prevents them from loosening. It generally has two heads Hex Key an star screw. For high mount fixing, star screw is used as it is less prone to stripping. The nylon insert of a Nylock nuts enters into the screw and prevent it from loosening. It is useful in areas where there is high vibration.

Hence, option A is correct

8 0
3 years ago
4. Which type of duct undergoes more rigorous testing before it's labeled?
Alenkasestr [34]

Answer:

Flexible duct

Explanation:

it is flexible duct due to current flow

7 0
3 years ago
A 75 ! coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 j75 !. If the relative p
Cerrena [4.2K]

Answer:

4.26

Explanation:

The wavelength λ is given by:

\lambda=v/f=c/nf\\c=speed\ of\ light=3*10^8m/s,f=frequency=3*10^9Hz,n=permittivity=2.56\\\\\lambda=3*10^8/(2.56*3*10^9)=0.0625\ m\\

Phase constant (β) = 2π/λ

βl = 2π/λ × l

l = 2 cm = 0.02 m

βl = 2π/0.0625 × 0.02=2.01 rad = 115.3°

1 rad = 180/π degrees

Z_L=load\ impedance=37.5+j75\\\\Z_o=characteristic impedance = 75\ ohm\\\\\tilde {Z_L}=Z_L/Z_o=37.5+j75/75=0.5+j

\tilde {Z_{in}}=\frac{\tilde {Z_{L}}+jtan\beta l}{1+j\tilde {Z_{L}}tan\beta l}=\frac{0.5+j+jtan(115.2)}{1+j(0.5+j)tan(115.2)}=0.253-j0.274\\  \\Z_{in}=Z_o\tilde {Z_{in}}=75(0.253-j0.274)=19-j20.5\\\\\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_o}=\frac{37.5+j75-75}{37.5+j75+75}=0.62\angle 83^o\\\\\Gamma_{in}=\frac{Z_{in}-Z_0}{Z_{in}+Z_o}=\frac{19-j20.5-75}{19-j20.5+75}=0.62\angle -147^o\\\\VSWR=\frac{1+\rho}{1-\rho} =\frac{1+0.62}{1-0.62}=4.26

6 0
3 years ago
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