1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vlada [557]
3 years ago
6

Can someone help me plz!!! It’s 25 points

Engineering
2 answers:
Dmitrij [34]3 years ago
8 0
The modus of electricity is probably around 25 but I’m not too sure
lora16 [44]3 years ago
3 0
Where’s the question at ???
You might be interested in
A 75 ohm coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 + j75 Ohm. If the diel
Hatshy [7]

Answer:

The load reflection coefficient, \Gamma =0.62\angle 82.875^{\circ} \Omega

Reflection coefficient at input,  \Gamma = 0.62\angle - 147.518^{\circ} \Omega

SWR = 4.26

Given:

Characteristic impedance of the co-axial cable, Z_{c} = 75 \Omega

Length of the cable, L = 2.0 cm = 0.02 m

Z_{Load} = 37.5 + j75 \Omega

Dielectric constant, K = 2.56

frequency, f = 3.0 GHz = 3.0 \times 10^{9} Hz

Explanation:

In order to calculate the reflection coefficient at load, we first calculate these:

The line input impedance Z_{i} is given by:

Z_{i} = Z_{c}\frac{Z_{Load} + jZ_{c} tan(\beta L)}{Z_{c} + jZ_{Load} tan (\beat L)}                     (1)

Now, we calculate the value of \beta:

\beta = \frac{2\pi}{\lambda'} = \farc{2\pi f\sqrt{K}}{c}

(since, \lambda' = \farc{c}{f\sqrt{K}})

\beta = \farc{2\pi f\sqrt{2.56}}{3\times 10^{8}} = 100.53

Now, Substituting the value in eqn (1):

Z_{i} = 75\frac{37.5 + j75 + j75 tan(100.53\times 0.02)}{75 + j(37.5 + j75) tan ( 100.53\times 0.02)} = 18.99 - j20.55 \Omega = 27.98\angle - 47.257^{\circ} \Omega    

Now, the load reflection coefficient is given by:

\Gamma = \frac{Z_{Load} - Z_{c}}{Z_{c} + Z_{Load}}}

Thus

\Gamma = \frac{37.5 + j75 - 75}{75 + 37.5 + j75}} = 0.077 + j0.615 = 0.62\angle 82.875^{\circ} \Omega

Similarly,

Reflection coefficient at input:

\Gamma' = \frac{Z_{i} - Z_{c}}{Z_{c} + Z_{i}}}

\Gamma' = \frac{18.99 - j20.55 - 75}{75 + 18.99 - j20.55}} = - 0.523 - j0.334 = 0.62\angle - 147.518^{\circ} \Omega

Now, the SWR is given by:

SWR, Standing Wave Ratio = \frac{1 +|\Gamma|}{1 - |\Gamma|}

SWR = \frac{1 +|0.62|}{1 - |0.62|} = 4.26

8 0
3 years ago
Cual es el impacto medioambiental del mortero?
ddd [48]

Explanation:

La industria del acero desecha gran cantidad de polvos, los que acopiados al aire libre, sobre el suelo y sometidos a ¡a acción de los agentes climáticos, contaminan el medio ambiente. El propósito del presente trabajo es evaluar el impacto ambiental que produce la incorporación de polvos de acería a matrices de cemento portland, sometidas a condiciones de humedad, presión y temperatura. Se caracterizó químicamente el material de desecho de una acería de Argentina, para identificar los elementos presentes. Por difractometría de rayos X (XRD) se determinaron las estructuras de las sustancias y su grado de oxidación. Se utilizó SEM-EDAX con el propósito de observar forma, tamaño y composición química de las partículas. Se elaboraron diferentes morteros de cemento Portland y se incorporó este material en diferentes porcentajes. Se moldearon probetas normalizadas para realizar ensayos mecánicos, de lixiviación y estudios petrográficos. Se analizó la composición química del agua de curado de las probetas y del líquido obtenido en ensayos de lixiviación. Los resultados permitieron identificar y cuantificar los elementos peligrosos que pueden se incorporados al ambiente y sugerir medidas de prevención.

3 0
3 years ago
What is over head line
CaHeK987 [17]

Explanation:

An overhead power line is a structure used in electric power transmission and distribution to transmit electrical energy across large distances. It consists of one or more conductors (commonly multiples of three) suspended by towers or poles. Since most of the insulation is provided by air, overhead power lines are generally the lowest-cost method of power transmission for large quantities of electric energy.

<h3><em><u>Constr</u></em><em><u>uction</u></em></h3>

Towers for support of the lines are made of wood (as-grown or laminated), steel or aluminum (either lattice structures or tubular poles), concrete, and occasionally reinforced plastics. The bare wire conductors on the line are generally made of aluminum (either plain or reinforced with steel or composite materials such as carbon and glass fiber), though some copper wires are used in medium-voltage distribution and low-voltage connections to customer premises. A major goal of overhead power line design is to maintain adequate clearance between energized conductors and the ground so as to prevent dangerous contact with the line, and to provide reliable support for the conductors, resilience to storms, ice loads, earthquakes and other potential damage causes. Today overhead lines are routinely operated at voltages exceeding 765,000 volts between conductors.

<em>Please</em><em> </em><em>mark</em><em> </em><em>it</em><em> </em><em>as</em><em> </em><em><u>brainliest</u></em><em>. </em><em>Follow</em><em> </em><em>me</em><em> </em><em>I </em><em>w</em><em>ill</em><em> </em><em>fo</em><em>llow</em><em> you</em><em> back</em><em>. </em>

6 0
4 years ago
Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of "84" m3/min and exits at 12
Nina [5.8K]

Answer:

W = - 184.8 kW

Explanation:

Given data:

P_1 = 1.05 bar

T_1 = 300K

\dot V_1 = 84 m^3/min

P_2 = 12 bar

T_2 = 400 K

We know that work is done as

W = - [ Q + \dor m[h_2 - h_1]]

forP_1 = 1.05 bar,  T_1 = 300K

density of air is 1.22 kg/m^3 and h_1 = 300 kJ/kg

for P_2 = 12 bar, T_2 = 400 K

h_2 = 400 kj/kg

\dot m = \rho \times \dor v_1 = 1.22 \frac{84}{60} =1.708 kg/s

W = -[14 + 1.708[400-300]]

W = - 184.8 kW

8 0
3 years ago
Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part
ra1l [238]

Answer:

h1 = 290.16kj/kg

P = 1.2311

Prandil expression at 8

P=p1/p7×pr

=8(1.2311)

=9.85

Enthalpy state at 8 corresponding to 9.85

h1 = 526.13kj/kg

Now prandtl state at 9 that correspond to 1400k.

h9 = 1515.42kj/kg

Pr = 450.5

Prandtl expression at state 10

P= p10/p9×pr

=1/8(450.5)

=56.31

Enthalpy at state 10 corresponding to prandtl 56.31

h10 = 860.39kj/kg

At 520k

h11 = 523.63kj/kg

4 0
3 years ago
Other questions:
  • Please perform 100 searches on Twitter using keyword "Lebron". Please print out the the number of tweets returned that were in E
    9·1 answer
  • Which of the following is used as part of a four-wheel drive system? A. Front drive axles B. Transfer Case C. Front Drive Shaft
    10·1 answer
  • The three areas of knowledge that are required for a product safety engineer are:
    6·1 answer
  • The ________ is the part of the drill press that holds and rotates the cutting tool.
    7·2 answers
  • Write a simple calculator program. Your program should ask for three things two whole numbers and an operator in the form of an
    10·1 answer
  • Refrigerant 134a at p1 = 30 lbf/in2, T1 = 40oF enters a compressor operating at steady state with a mass flow rate of 200 lb/h a
    14·1 answer
  • A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons
    14·1 answer
  • If a wire has 23.6 A of current and a resistance of 33.3Ω how much voltage is present?
    8·1 answer
  • Describe some typical pairs of entities that you think might be common in business, and describe their relationships, whether ma
    14·1 answer
  • Type the correct answer in the box. Spell all words correctly.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!