Power electronics and motion control system

A single-phase full-bridge uncontrolled (diode) rectifier is supplied by 220 V, 50 Hz source. Neglecting the diodes volt-drops,

a. Calculate the Average and rms values of the Output Voltage, Output (load) Current, the Ripple and Form Factors, when load is pure resistive R=10 Ohm.

b. Assume that load has inductive nature and L>> R and load current is flat and equal to 12 Ampere. Calculate the input Active Power, input Apparent Power and Power Factor (neglect diode losses)

Hello!!

For calculate the GPE let's applicate the formula:

\(\boxed{GPE = m g h}\)

\(\textbf{Being:}\)

\(\sqrt{}\) GPE = Gravitational potential energy = ?

\(\sqrt{}\) m = Mass = 100 kg

\(\sqrt{}\) g = Gravity = 10 m/s²

\(\sqrt{}\) h = Height = 40 m

⇒ \(\text{Then let's \textbf{replace it according} we information:}\)

\(GPE = 100 \ kg * 10 \ m / s ^{2} * 40 \ m\)

⇒ \(\text{Let's resolve it: }\)

\(GPE = 40000 \ J\)

\(\textbf{Result:}\\\text{The gravitational potential energy is \textbf{40 000 Joules}}\)

Answer:

What ever, thanks for the points

Explanation:

The  agents of physical weathering that might occur in a mountainous region are: Water and  changes in temperature.

Particularly in their settings with little soil and limited plant life, like mountainous areas and harsh deserts, physical weathering is one that do occurs.

Therefore, some Agents of weathering include water, ice, acids, salts, plants, animals, and variations in temperature. After a rock has been fractured, the fragments of rock and mineral are carried away by a process known as erosion. The forces of weathering and erosion cannot be withstood by any rock on Earth.

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Answer:

Her computer is producing thermal energy, not heat.

Explanation:

Answer:

Her computer is producing thermal energy,not heat

Explanation:

Writing in an online journal is different from writing in a paper one in the following way: it is easier to communicate online (option B).

A journal is a newspaper or magazine dealing with a particular subject.

A journal is an efficient medium to communicate the findings or results of an investigation to the public.

However, a journal can be virtual (online) or paper (hard copy). In this 21st century, it is easier to communicate to the masses online because more audience will be captured.

Therefore, writing in an online journal is different from writing in a paper one in the following way: it is easier to communicate online.

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Answer:

  F = 69.3 N

Explanation:

For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

               fr = μ N

We define a reference system parallel to the floor

block B  ( lower)

Y axis  

            N - W₁-W₂ = 0

            N = W₂ + W₂

            N = (M + m) g

X axis

              F -fr = M a

for block A (upper)

X axis

              fr = m a                 (2)

so that the blocks do not slide, the acceleration in both must be the same.

Let's solve the system by adding the two equations

             F = (M + m) a          (3)

             a =\(\frac{F}{ M+m}\)

the friction force has the formula

            fr = μ N

             fr = μ (M + m) g

let's calculate

            fr = 0.34 (2.0 + 0.250) 9.8

            fr = 7.7 N

we substitute in equation 2

             fr = m a

             a = fr / m

             a = 7.7 / 0.250

             a = 30.8 m / s²

we substitute in equation 3

             F = (2.0 + 0.250) 30.8

             F = 69.3 N

\({ \qquad\qquad\huge\underline{{\sf Answer}}} \)

Here we go ~

So,

[ taking unit vector along north be \({ \sf \hat{j}} \) ]

[ taking unit vector along west be \({ \sf \hat{i}} \) ]

\( \qquad \dashrightarrow\sf \: V_{wr}= V_{w} - V_{r }\)

[ where \({ \sf V_{wr}= } \) Relative velocity of wind with respect to runner ]

\( \qquad \dashrightarrow\sf \: V_{wr}= - 3 \hat{ j} - ( - 4 \hat{i})\)

\( \qquad \dashrightarrow\sf \: V_{wr}= - 3 \hat{ j} + 4 \hat{i}\)

That is : 3 m/s towards south and 4 m/s towards east.

So, it makes an angle of 37° with south, and 53° with East.

[ The resultant is depicted by yellow arrow in attachment ]

\( \qquad \dashrightarrow\sf \: |V_{wr} | = \sqrt{3 {}^{2} + 4 {}^{2} }\)

\( \qquad \dashrightarrow\sf \: |V_{wr} | = \sqrt{9 + 16 }\)

\( \qquad \dashrightarrow\sf \: |V_{wr} | = \sqrt{25 }\)

\( \qquad \dashrightarrow\sf \: |V_{wr} | = 5 \: \: m/s\)

\( \dashrightarrow \sf \tan( \alpha) = \dfrac{b\: \sin \theta}{b + a \sin( \theta) } \)

\({\sf \theta }\) = angle between the two vectors ( along east and south) i.e 90°

\( \dashrightarrow \sf \tan( \alpha) = \dfrac{4\: \sin (90 \degree)}{3+ 4 \cos( 90 \degree) } \)

\( \dashrightarrow \sf \tan( \alpha) = \dfrac{4 \times 1}{3+( 4 \times 0)} \)

\( \dashrightarrow \sf \tan( \alpha) = \dfrac{4}{3} \)

\( \dashrightarrow \sf \alpha = \tan {}^{ - 1} \bigg( \dfrac{4}{3 } \bigg)\)

\( \dashrightarrow \sf \alpha = 53 \degree\)

So, it's direction is 53° east from south ~

Answer:

Explanation:

Let v be the velocity acquired by electron in electric field

V q = 1/2 m v²

V is potential difference applied on charge q , m is mass of charge , v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .

Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron

= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

= 79.02 x 10⁻¹³ N .

Minimum force will be zero when electron moves along the direction of magnetic field .

(a) The maximum force on the  electron due to the magnetic field will be  F= 79.02 x 10⁻¹³ N .

(b) Minimum force will be zero when electron moves along the direction of magnetic field .

Whenever a current is passes through a wire then the magnetic fields are generated around the wire and if any other charged particle comes under the influence of this magnetic field then the magnetic force is applied in the charge.

Let v be the velocity acquired by an electron in electric field

\(Vq=\dfrac{1}{2}mv^2\)

V is potential difference applied on charge q ,

m is mass of charge ,

v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

Maximum force will be exerted on the moving electron when it moves perpendicular to the magnetic field .

Maximum force = Bqv , where B is magnetic field , q is charge on an electron and v is velocity of electron

F=Bqv

F= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

F= 79.02 x 10⁻¹³ N .

Minimum force will be zero when electron moves along the direction of magnetic field .

Hence the maximum force on the  electron due to the magnetic field will be  F= 79.02 x 10⁻¹³ N and the Minimum force will be zero when electron moves along the direction of magnetic field .

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The body of a jellyfish can be  excluded due to the absence of fossil records and is denoted as option D.

This is referred to as the remains of plants and animals which have been preserved in the earth crust and it is important as it helps to give more information about life history of different organisms.

Examples of fossils include bones, shells etc as they are the parts of the organism which are preserved and not doesn't decay. The body of a jellyfish will decay and not form a fossilized structure which is therefore the reason why option D was chosen.

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Answer: B

Explanation: B is the correct statement describing the positions of the Moon, the Sun, and the Earth during a new moon. The Moon is between Earth and the Sun.

During a new moon, the Moon is positioned between the Sun and the Earth, with the illuminated side of the Moon facing away from the Earth. This means that the side of the Moon that faces the Earth is not receiving any sunlight, making it invisible to us from Earth. The new moon is the first phase of the lunar cycle and occurs roughly every 29.5 days.

Answer:

1min since there is no gravity on the moon so it will take time to drop on the moon.

Explanation:

Answer:

a) interior,  b) inside,   c) minor

Explanation:

In this exercise you are asked to select the correct words so that the statements have been correct

Electric charges always repel each other when they are of the same sign, in conductors this has the consequence that charges accumulate on the surface and the interior remains without electric charges.  with this we analyze the statements

a) interior

b) inside

c) minor

therefore the phrase would be:

(a) The net charge is always zero ---INTERIOR--- the surface of an isolated conductor.

(b) The electric field is always zero ---INSIDE--- a perfect conductor.

(c) The charge density on the surface of an isolated, charged conductor is highest where the surface is ---MINOR---to)

The circuit consists of a series of number of pairs of parallel and series

arranged resistors.

Correct responses:

The total resistance can be calculated as follows;

Resistance between point C to D and from D to E are in series, therefore;

We have;

\(R_{CD}\) = 5 Ω and \(R_{DE}\) = 9 Ω are parallel

Therefore;

\(R_{TCE} = \mathbf{ \dfrac{1}{\dfrac{1}{R_{CE}} + \dfrac{1}{R_{CD} +R_{DE}} }}\)

\(R_{TCE} = \dfrac{1}{\dfrac{1}{14} + \dfrac{1}{5+9} } = 7\)

\(R_{TCE}\) = 7 Ω

\(R_{BC}\) = 11 Ω is in series with \(R_{CE}\) both of which are parallel to \(R_{BE}\) = 18 Ω

\(R_{TBE} = \mathbf{\dfrac{1}{\dfrac{1}{R_{BE}} + \dfrac{1}{R_{BC} + R_{TCE}} }}\)

Therefore;

\(R_{TBE} = \mathbf{\dfrac{1}{\dfrac{1}{18} + \dfrac{1}{11 + 7} }} = 9\)

\(R_{TBE}\) = 9 Ω

\(R_{TAE} = \mathbf{\dfrac{1}{\dfrac{1}{R_{AE}} + \dfrac{1}{R_{AB} + R_{TBE}} }}\)

Therefore;

\(R_{TAE} = \dfrac{1}{\dfrac{1}{22} + \dfrac{1}{13 + 9} } = 11\)

\(R_{TAE}\) = 11 Ω

\(R_T = \mathbf{R_{TAE} + R_{A_F}}\)

Therefore;

The current in the circuit, I, is therefore;

\(\displaystyle I = \frac{24 \ v}{12 \ \Omega} = \mathbf{2 \ A}\)

By current divider rule, due to the equality of the parallel resistances we have;

Current through AB, \(I_{AB}\) = \(\frac{1}{2} \times 2 \ A\) = 1 A

Similarly;

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Answer:

um

Explanation:

Answer: Dang thats tough g but u did it

Explanation:

Answer:

A frequency by 2 seonds is  0.5 Hertz.

Explanation:

The minimum coefficient of static friction with the floor is 0.3846.

To find the minimum coefficient of static friction with the floor, we need to consider the forces acting on the couch. In this case, the force of gravity is pulling the couch downward with a magnitude of mg, where m is the mass of the couch (40 kg) and g is the acceleration due to gravity (9.8 m/s²).

Since the couch does not move, the force of static friction between the couch and the floor must be equal in magnitude but opposite in direction to the horizontal pushing force of 150 N.

Therefore, we have the equation F_friction = F_push, where F_friction is the force of static friction.

The force of static friction can be calculated using the formula F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.

Since the couch is on a level floor and is not accelerating vertically, the normal force N is equal in magnitude but opposite in direction to the force of gravity, which is mg.

Substituting the values into the equation, we have μs * mg = 150 N.
Solving for μs, we get μs = 150 N / (mg).
Substituting the given values, we have μ_s = 150 N / (40 kg * 9.8 m/s²).
Simplifying, we find that μs = 0.3846.

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Answer:

1. True

2. False

3. True

4. False

Explanation:

a) In this case, length of the copper rod is 3.6 m which is much larger than the distance 4.9 cm to the point at which electric field is to be determined. Therefore, yes, cylindrical symmetry holds.

b) In this case, length of the copper rod is 8.9 cm which is of the same order of magnitudes the distance 4.9 cm to the point at which electric field is to be determined. Therefore, no, cylindrical symmetry does not hold.

c) In this case, length of the copper rod is 3.8 m which is much larger than the distance 4.9 cm to the point at which electric field is to be determined. Therefore, yes, cylindrical symmetry holds.

d) In this case, length of the copper rod is 3.6 m which is of the same order of magnitudes the distance 4.9 cm to the point at which electric field is to be determined. Therefore, no, cylindrical symmetry does not hold.

The two displacements that move from (5,5) to (-5,-5)

(5,5) → ( 5,-5) [10 units down]

(5,-5) → (-5,5) [ 10 units left ]

The single displacement that connects the 2 points is the hypotenuse of the formed triangle where each side is 10 m long.

Apply Pythagorean theorem:

c2 = 10^2+10^2

c^2 = 100 + 100

c^2 = 200

c =√200

c= 14.14

Compared to the simple displacement (14.14) that connects both points, it is greater.

20m > 14.14 m

Answer:

a) ΔV ’= 1.66 10¹ V= 16.6 V,  b)  U = 55.64 10⁻¹² J,  c)    U_f = 1.18 10⁻¹⁰ J

d)     W = 6.236 10⁻¹¹ J

Explanation:

Capacitance can be found for a parallel plate capacitor

          C = ε₀  \(\frac{A}{d}\)  

Let's reduce the magnitudes to the SI system

           A = 9.30 cm² (1 m / 10² cm) 2 = 9.30 10⁻⁴ m²

           c = 4.50 mm (1 m / 1000 mm) = 4.50 10⁻³ m

          Co = 8.85 10⁻¹²    9.30 10⁻⁴ /4.50 10⁻³

          Co = 1.829 10⁻¹² F

when the plates separate at d = 9.60 10⁻³ m, the capcitance changes to

          C = ε₀ \frac{A}{d_1}

          C = 8.85 10⁻¹² 9.30 10⁻⁴/9.60 10⁻³

          C = 8.57 10⁻¹³ F

a) the potential difference

            C =

since the capacitor is not discharged, let's look for the initial charge

            Co = \frac{Q}{ \Delta V}

             Q = C₀ ΔV

              Q = 1.829 10⁻¹² 7.80

             Q = 14.2662 10⁻¹² C

when the condensate plates are separated

             C = \frac{Q}{ \Delta V' }

              ΔV ’= Q / C

              ΔV ’= 14.266 10⁻¹² / 8.57 10⁻¹³

              ΔV ’= 1.66 10¹ V= 16.6 V

b) the stored energy is

             U = ½ C ΔV²

for initial separation

              U = ½ C₀ ΔV²

             U = ½ 1.829 10⁻¹² 7.80²

              U = 55.64 10⁻¹² J

c) The energy for end separation;

               U_f = ½ C DV’2

                U_f = ½ 8.57 10⁻¹³ 16,6²2

                U_f = 1.18 10⁻¹⁰ J

d) The work

as there are no losses, the work is equal to the variation of the energy

                W = ΔU = U_f -U₀

                 W = 1.18 10⁻¹⁰ - 55.64 10-12

                 W = 6.236 10⁻¹¹ J

Answer:

m = 312.5

Explanation:

Given that,

The focal length of the objecive lens, \(f_o=1\ cm\)

The focal length of eye piece, \(f_e=2\ cm\)

length of the tube, L = 25 cm

We need to find the magnitude of the overall magnification of the microscope. It is given by the formula as follows :

\(m=\dfrac{L}{f_o}\times \dfrac{D}{f_e}\)

D = 25 cm

So,

\(m=\dfrac{25}{1}\times \dfrac{25}{2}\\\\m=312.5\)

So, the overall magnification of the microscope is 312.5.

Quaynisha Perry's strategy was interesting: ask everyone in the room to be part of the decision-making process.

The statement has been corrected of the capitalization error.

Capitalization errors is said to take place when the writer capitalizes a word that does not require a capital letter or when the writer does not capitalize a word the requires one.

In all cases, Capitalization errors hinder the reader's experience with the writing and must  be avoided at all cost.

It is pertinent to note that you should always capitalize the first letter of the first word in a sentence, no matter what the word is.

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Answer: The Boat will rise

Explanation: Because high amplitude means high in heights.

The final temperature of the water is approximately 19.8°C.First, we need to calculate the heat lost by the water as it cools down from 80°C to the final temperature.

We can use the formula: Q = m × c × ΔT where Q is the heat lost, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

The mass of water in the calorimeter is:

90 ml = 90 g

The heat lost by the water is:

Q1 = 90 g × 4.18 J/g°C × (80°C - T)

Q = m × Lf + m × c × ΔT

The mass of ice added to the water is 10 g. The heat gained by the ice is:

Q2 = 10 g × 334 J/g + 10 g × 4.18 J/g°C × (T - (-20°C))

where 334 J/g is the heat of fusion of water.

Since the calorimeter is insulated, we know that the heat lost by the water must equal the heat gained by the ice. Therefore:

Q1 = Q2

90 g × 4.18 J/g°C × (80°C - T) = 10 g × 334 J/g + 10 g × 4.18 J/g°C × (T - (-20°C))

Simplifying and solving for T, we get:

T = 19.8°C

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I dont know but I am messaging because maybe someone else see it and help u :)

Radius of the planet has the greatest influence on the width of the light curve of a transiting planet.

"Radius is a straight line from the centre to the circumference of a circle or sphere."

"Planets are the largest objects in the solar system after the Sun."

"A light curve is a graph of light intensity of a celestial object or region, as a function of time."

"A transit occurs when a planet passes between a star and its observer."

According to Kepler, size of the planet is proportional to how much light is blocked . Hence , more deeper the transits are , more the size of the planet will be. This implies radius has the greatest influence on the width of the light curve of a transiting planet.

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Answer:

6.25%

Explanation:

20/5=4

half-life=1/2 of previous life

1/2=50%

50%=0.5

(0.5)^4 =  ==> get the 4 from dividing total years(20) by the half-life(5): 20/5

0.0625

0.0625*100%=6.25%

The speed of the truck just before braking is 23.24 m/s.

The speed of the truck before braking is calculated by applying the third kinematic equation as shown below.

v² = u² - 2as

where;

When the truck stops, the final velocity = 0

0 = u² - 2as

u²  = 2as

u = √2as

u = √ ( 2 x 6 x 45 )

u = 23.24 m/s

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Answer:

\(1.125\; {\rm m}\).

Explanation:

The torque that Sarah and Sally exert on this seesaw need to be equal in magnitude.

The weight of Sarah is \(m(\text{Sarah})\, g\), where \(m(\text{Sarah}) = 40\; {\rm kg}\).

The weight of Sally is \(m(\text{Sally})\, g\), where \(m(\text{Sally}) = 30\; {\rm kg}\).

Assuming that the seesaw is level. The force that Sarah exerts on the seesaw will be perpendicular to the seesaw. The resultant torque will be of magnitude \(\tau(\text{Sarah}) = m(\text{Sally})\, g\, l(\text{Sarah})\), where \(l(\text{Sarah})\) is the distance between Sarah and the center of the seesaw (the fulcrum).

Similarly, the torque from Sally will have a magnitude of \(\tau(\text{Sally}) = m(\text{Sally})\, g\, l(\text{Sally})\), where \(l(\text{Sally}) = (3.0 / 2)\; {\rm m} = 1.5\; {\rm m}\) is the distance between Sally and the center of the seesaw.

The magnitude of the two torques should be equal. Thus:

\(m(\text{Sarah})\, g\, l(\text{Sarah}) = m(\text{Sally})\, g\, l(\text{Sally})\).

Rewrite this equation and solve for \(l(\text{Sarah})\):

\(\begin{aligned}l(\text{Sarah}) &= \frac{m(\text{Sally})\, g\, l(\text{Sally})}{m(\text{Sarah})\, g} \\ &= \frac{m(\text{Sally})}{m(\text{Sarah})}\, l(\text{Sally})\\ &= \frac{30\; {\rm kg}}{40\; {\rm kg}}\, (1.5\; {\rm m}) \\ &= 1.125\; {\rm m}\end{aligned}\).