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A 100 MHz generator with Vg= 10/00 v and internal resistance 50 ohms air line that is 3.6m and terminated in a 25+j25 ohm load

Lilit [14]

The value of Vz at point z from the generator = 17.748 ∠ -107.62° V

Given data :

Internal resistance = 50 ohms

Vg = 10 ∠ 0° v

length of air line = 3.6 m

Terminating resistance = 25 + j25 Ω

First step : Determine the Total impedance 

Total Impedance ( z ) = Rin + Rline + Rl + jXl

= 50 + 50 + 25 + j25

= 125 + j25 ≈ 127.47 ∠ 11.3°

Next step : Determine the current in the circuit

current ( I ) = Vg / z

= ( 10∠0° v ) / ( 127.47 ∠ 11.3° )

= 0.0784 ∠ -11.3 amp

Final step : Determine the value of Vz at point z from the generator

Vz = ( Vg + I * Ri ) - ( RI * I + Rline * I )

= ( 10∠0° + 0.0784 ∠ -11.3 * 50 ) - ( 25 + j25 + 50 * 0.0784 ∠ -11.3 )

= -5.37 - j16.91 ≈ 17.748 ∠ -107.62° V

Hence we can conclude that the value of Vz at point z form the generator 17.748 ∠ -107.62° V

Learn more about voltage : brainly.com/question/11288970

Hello your question is incomplete below is the complete question

A 100 MHz generator with Vg =10< 0 degree (v) and internal resistance 50-ohm is connected to a lossless 50-ohm air line that is 3.6m long and terminated in a 25+j25 (ohm) load.

Find (a) V(z) at a location z from the generate.

4 0

3 years ago

The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate

Sonja [21]

Answer:

a) Mechanical efficiency ( $\varepsilon$ )=63.15% b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, $Wmec=0.95*6kW=5.7 kW$ .

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump $\Delta P$ . So $P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W$ .

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is $\varepsilon=3.6kW/5.7kW=0.6315=63.15\%$

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling H to the heat transfered to the fluid, the specif heat of the water and $\rho$ the density of the water:

$[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T$ [/tex]

Finally, the temperature rise is:

$\Delta T=2100/75348 \ºC=0.028 \ºC$

7 0

3 years ago

Oliver is designing a new children's slide to increase speed at which a child can descend.His first design involved steel becaus

Ede4ka [16]

Answer:

Steel can rusts easily in the presence of moisture and oxygen, and tarnishes as the rust progresses.

Explanation:

Steel is an alloy of carbon and iron. it is a very useful alloy that is found in almost any engineered piece. The problem with steel is that it is very susceptible to rust, and the cost of maintaining it in order to prevent rust is very high. Steel rusts in the presence of moisture and oxygen (rust is an oxidation-reduction process). Using it as a water slide exposes it constantly to moisture, and to prevent it from rusting in this case will involve a lot of maintenance cost, which is why steel is not advisable to be used in this case.

4 0

3 years ago

Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street r

irakobra [83]

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area = 1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000 × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

4 0

3 years ago

Find the velocity and rate of flow of water through a rectangular channel of 6m wide and 3m deep when it's running full. The cha

Elza [17]

Answer:

V = 1.5062 m/s

Explanation:

look to the photos

5 0

3 years ago