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3 years ago

6

Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average valu

e. For LaTeX: R=1kΩR = 1 k Ω, and assuming that the diode can be represented by a constant voltage drop model with LaTeX: V_D=0.7VV D = 0.7 V.

1 answer:

Varvara68 [4.7K]3 years ago

8 0

The question is not complete. We are supposed to find the average value of v_o.

Answer:

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

3/(T/4) = 0.7/t1

So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)

v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)

v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

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thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat

eduard

Answer:

$q_{in} = 528.6\,\frac{kJ}{kg}$

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

$q_{in} = h_{g} - h_{mix}$

The specific enthalpies are:

Liquid-Vapor Mixture:

$h_{mix} = 2217.2\,\frac{kJ}{kg}$

Saturated Vapor:

$h_{g} = 2745.8\,\frac{kJ}{kg}$

The thermal energy per unit mass required to heat the steam is:

$q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}$

$q_{in} = 528.6\,\frac{kJ}{kg}$

7 0

3 years ago

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. A

irina [24]

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.

So; we can say :

$D_{pursuit} =D_{police}$

By using the second equation of motion to find the distance S;

$S= ut + \dfrac{1}{2}at^2$

$D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)$

$D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)$

$D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)$

$D_{police} = ut _P + \dfrac{1}{2}at_p^2$

where ;

u = 0

$D_{police} = \dfrac{1}{2}at_p^2$

$D_{police} = \dfrac{1}{2}*(1.96)*(t+12)^2$

$D_{police} = 0.98*(t+12)^2$

$D_{police} = 0.98*(t^2 + 144 + 24t)$

$D_{police} = 0.98t^2 + 141.12 + 23.52t$

Recall that:

$D_{pursuit} =D_{police}$

$(187.8+15.65 \ t-1.525 t^2)= 0.98t^2 + 141.12 + 23.52t$

$(187.8 - 141.12) + (15.65 \ t - 23.52t) -( 1.525 t^2 - 0.98t^2) = 0$

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

t = -6.16 OR t = 3.02

Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs

From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;

Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s

Total time required for the police car to over take the automobile = 15.02 sec

8 0

3 years ago

Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical

Olin [163]

Answer:

The theoretical density for Niobium is $1.87 g/cm^3$ .

Explanation:

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density of the unit cell

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

$a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm$

$1 nm = 10^{-7} cm$

On substituting all the given values , we will get the value of 'a'.

$\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}$

$\rho =1.87 g/cm^3$

The theoretical density for Niobium is $1.87 g/cm^3$ .

6 0

3 years ago

Read 2 more answers

A driver counts 21 other vehicles using 3 EB lanes on one section of I-80 between her rented car and an overpass ahead. It turne

SVEN [57.7K]

Answer:

vehicle density = 28.205 veh/mile

flow rate = 0.909 veh/hr

Explanation:

given data

count n = 21

distance = 0.78 miles

speed = 52 mph

solution

we get here vehicle density that is express as

vehicle density = n ÷ distance ...............1

vehicle density = ( 21 + 1 ) ÷ 0.78

vehicle density k = 28.205 veh/mile

and

now we get here flow rate that is express as

flow rate = k × vs .................2

flow rate = 28.205 × ( 52 × 0.00062 ÷ 1m )

flow rate = 0.909 veh/hr

6 0

2 years ago

Education is the foundation of profession. Justify

ololo11 [35]

Answer:

Explanation:

Foundations of Education refers to a broadly-conceived field of educational and profoundly important academic and professional purpose unifies persons Education as a profession in remarks made in relation to the expertise has long been the foundation of free and open academic inquiry.Setting goals for schools and students is an important process. It helps to motivate people when there is something specific to aim for. It is also a good way to show the clear path to success. The principal should work with the teachers to develop a plan and mission for the school

plz mark as brainliest

8 0

3 years ago