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3 years ago

I don’t understand this

Engineering

1 answer:

blondinia [14]3 years ago

7 0

Answer:

Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of $160cm^{2}$ by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.

Have an amazing day and I hope this can somewhat help :)

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Q1-A Lossless transmission line is 80 cm long and operates at a frequency of 500 MHz. Two line parameters are L=0.15μH/m and C=9

hjlf

The characteristic impedance, the phase constant, the velocity on the line, and the input impedance are respectively;

A) Z_o = 40.82 Ω

B) β = 11.543 rad/m

C) v_p = 2.72 × 10^(8) m/s

D) Z_in = 115.91 Ω

We are given;

Inductance; L = 0.15 μH/m = 0.15 × 10^(-6) H/m

Capacitance; C = 90 pf/m = 90 × 10^(-12) f/m

Frequency; f = 500 MHz = 500 × 10^(6) Hz

Load impedance; Z_l = 80 Ω

Length of transmission line; l = 80cm = 0.8m

A) Formula for characteristic impedance is;

Z_o = √(L/C)

Thus;

Z_o = √((0.15 × 10^(-6))/(90 × 10^(-12)))

Z_o = 40.82 Ω

B) Formula for the phase constant is;

β = ω√(LC)

Where;

ω = 2πf

Thus;

β = (2π × 500 × 10^(6))√(0.15 × 10^(-6) × 90 × 10^(-12))

β = 11.543 rad/m

C) Formula for phase velocity is;

v_p = ω/β

v_p = (2π × 500 × 10^(6))/11.543

v_p = 2.72 × 10^(8) m/s

D) Formula for the input impedance is;

Z_in = Z_o[(Z_l + Z_o*tanβl)/(Z_o + Z_l*tanβl)]

Z_in = 40.82[(80 + 40.82*tan(11.543*0.8))/(40.82 + 80*tan(11.543*0.8))]

Z_in = 40.82(72.1335/25.4031)

Z_in = 115.91 Ω

Read more about impedance at; brainly.com/question/25153504

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2 years ago

Seperate real and imaginary parts tan(2x+i3y)

ahrayia [7]

Answer:

First you have to separate real and imaginary parts of Tan(x+iy)=Tan(z)=sin(z)/cos(z)

sinz=sin(x+iy)=sinxcos(iy)+cosxsin(iy)=sinxcoshy-icosx sinhy

cosz=cos(x+iy)=cosxcos(iy)-sinxsin(iy)=cosxcoshy−isinxsinhy

Now if you plug in Tan(z) and simplify (it is easy!) you get

Tan(z)=(sin(2x)+isinh(2y))/(cos(2x)+cosh(2y))= A+iB.

This means that

A=sin(2x)/(cos(2x)+cosh(2y)) and B= sinh(2y)/(cos(2x)+cosh(2y))

Now,

A/B=sin(2x)/sinh(2y)

If any questions, let me know.

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3 years ago

Which of the following is NOT an ASE certification? Select one:

stiks02 [169]

The option that is not an ASE certification is . A/C and Refrigerants handling certification (609).

<h3>What is ASE certification?</h3>

The term ASE is known to be a body that tends to promotes excellence in regards to vehicle repair, service as well as parts distribution.

Note that in the world today more than a quarter of million of people are known to possess ASE certifications.

Since ASE Certified professionals work in in all areas of the transportation industry. one can say that The option that is not an ASE certification is. A/C and Refrigerants handling certification (609).

Learn more about ASE certification from

brainly.com/question/5533417

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1 year ago

If the old radiator is replaced with a new one that has longer tubes made of the same material and same thickness as those in th

Nookie1986 [14]

Answer: hello some parts of your question is missing attached below is the missing information

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube

answer : Total surface area = 3/2 * area of old radiator

Explanation:

we will use this relation

K = $\frac{Qd }{A* change in T }$

change in T = ΔT

therefore New Area ( A ) = 3/2 * area of old radiator

Given that the thermal conductivity is the same in the new and old radiators

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What should you release to re-establish vehicle control and tire traction?

MrMuchimi

Answer: The accelerator and the brakes.

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