Write conclusin for charactristics performance evaluation of dc generators experiment

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Answer 1

Conclusion for Characteristics Performance Evaluation of DC Generators experiment:

In conclusion, the experiment for the characteristics performance evaluation of DC generators was conducted to determine the performance of the generator. During the experiment, the generator was connected to the DC motor which acted as the load while the generator was rotating. Several readings were taken to determine the values of the armature current, field current, armature voltage, and speed of the generator. The values were then used to plot the characteristics curves of the generator.

The experiment was successful in providing an insight into the performance of the generator. The curves obtained from the experiment can be used to determine the best operating conditions for the generator. The main answer is that the experiment provides useful information for the maintenance and troubleshooting of DC generators.

In the experiment, the armature voltage was varied at a constant field current and load current. This led to the generation of the no-load characteristic curve which showed the relationship between the generated voltage and the field current. The curve obtained was a straight line, which proved that the generator follows Ohm's law.

The armature current was then varied at a constant field current and load current. This led to the generation of the load characteristic curve which showed the relationship between the generated voltage and the armature current. The curve obtained was a straight line, which proved that the generator follows Ohm's law.

The field current was varied at a constant armature voltage and load current. This led to the generation of the field characteristic curve which showed the relationship between the generated voltage and the field current. The curve obtained was a straight line with a negative slope, which proved that the generator does not follow Ohm's law. Instead, the generated voltage is inversely proportional to the field current and directly proportional to the armature current.

The speed of the generator was also varied at a constant armature voltage, field current, and load current. This led to the generation of the speed characteristic curve which showed the relationship between the generated voltage and the speed of the generator. The curve obtained was a straight line with a negative slope, which proved that the generator does not follow Ohm's law. Instead, the generated voltage is inversely proportional to the speed of the generator.

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Related Questions

A digital camera basically has an array of tiny light detectors (2000×1500 = 3 MegaPixels = 3 million very tiny detectors, covering a cm2. In each of these detectors, photons that hit the detector excite electrons and these excited electrons are counted. In a typical picture, the detector array in the camera is exposed to about 4.5×10-6 watts of light for 10 ms. If you take 535 nm as a typical wavelength for the light, what is the average number of photons that hit each pixel in a typical picture (don't use scientific notation, or Canvas might get confused).
2. If you have very low intensity green light (4×10-11watts at 570 nm) evenly illuminating the entire array of detectors, what will the camera's detectors see during the exposure time of 10ms?
A. Random pixels will have several excited electrons, others will have only one excited electron, and still others will have no excited electrons.
B. All pixels in the array count about the same number of excited electrons.
C. The pixels in the centre of the array will count the largest number of excited electrons and this will drop off towards the edges.
D. Random pixels will have exactly one excited electron, while others will have no excited electrons.

Answers

1. The average number of photons is approximately 7.67 × 10^9 photons.

2. The evenly illuminated array of detectors in the camera, exposed to a very low intensity green light, will display a random distribution of excited electrons across the pixels during the 10 ms exposure time. Hence, option A is correct.

1. The average number of photons that hit each pixel in a typical picture can be calculated using the formula: Number of photons = (Power of light / Energy per photon) * Exposure time.

Given the power of light as 4.5 × 10^(-6) watts, the wavelength of light as 535 nm (535 × 10^(-9) m), and the exposure time as 10 ms (10 × 10^(-3) s), we need to calculate the energy per photon first. The energy per photon can be determined using the equation:

Energy per photon = (Planck's constant * Speed of light) / Wavelength of light. After substituting the values and performing the calculations, we find the energy per photon.

Then, we can calculate the average number of photons that hit each pixel using the formula mentioned earlier. The average number of photons is approximately 7.67 × 10^9 photons.

2. If very low intensity green light (4 × 10^(-11) watts at 570 nm) evenly illuminates the entire array of detectors during the 10 ms exposure time, the camera's detectors will exhibit a distribution of excited electrons across the pixels.

Some pixels will have multiple excited electrons, some will have only one excited electron, and others will have no excited electrons. This distribution occurs due to the random nature of photon absorption by the detectors.

Therefore, the correct answer is A - random pixels will have several excited electrons, others will have only one excited electron, and still others will have no excited electrons.

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Q.3 Fill the blanks with the correct answer: (5 points) a- The analogy of the force in rotational motion is Torque b- The effect which causes the air gap area to increase is called Fringing Effect. c-

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a- The analogy of the force in rotational motion is torque. It is a rotational force or the force that twists or turns an object around an axis or pivot point. The torque is dependent on the magnitude of the force and the distance between the axis of rotation and the point at which the force is applied.

b- The effect which causes the air gap area to increase is called the fringing effect. The fringing effect happens when the magnetic field near the edges of an object deviates from the direction of the magnetic field near the center of the object. This effect is also sometimes called the leakage effect or the edge effect.

The magnetic field lines in the air gap between the magnetic poles are curved, and they leave the surface of the north pole and re-enter at the surface of the south pole. The fringing effect occurs because the magnetic field lines become more widely spaced as they move from the central region of the gap toward the edges.The fringing effect can cause a decrease in the performance of electric machines such as generators and motors. It is also known to create noise and vibration in transformers and inductors.

c- The increase in the amount of current passing through a wire increases the magnetic field around the wire. This phenomenon is known as the Ampere's law.

Ampere's law can be used to calculate the magnetic field that is produced by a current-carrying wire or a conductor in a circuit. It states that the magnetic field produced by a current-carrying wire is proportional to the current in the wire and inversely proportional to the distance from the wire.

Ampere's law can be used to calculate the magnetic field produced by any current-carrying wire or conductor. The law can be used to calculate the magnetic field produced by a long, straight wire, a loop of wire, or a solenoid.

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3. On my way home one night, I am driving at a speed of 19.0: As I approach a stoplight, I see it turn yellow and speed up to make it through. 1 cover the next 36 meters in 1.65 seconds. Assume the acceleration during this 1.65 s is constant a. What is my noceleration while I speed up? b. What is my final speed? 4. You and your roommate are doing physics problems while in your bunk beds. You make a mistake and ask your roommate to toss up an craser. You are 1.40 m above your friend 1. What speed must your roommate throw the craser at in order for it to just barely reach you? (Remember that velocity is equal to zero at the highest point) b. How long does it take the craser to travel from your friend's hand to your hand? c. You like to snack while you study, so your fingers are covered in Cheeto dust. Your gross fingers cause you to drop the eraser from your top bunk, a height 2.50 m above the floor. How fast is the craser moving just before it hits the floor? Assume it is not moving before you drop it (an initial velocity of zero)

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The acceleration while you speed up is 2.122 m/s². The final speed of the car is 48.1 m/s. The required speed at which your roommate must throw the eraser is 4.19 m/s. The speed of the eraser just before it hits the floor is 7.02 m/s.

a. The acceleration while you speed up is 2.122 m/s².

We can use the kinematic equation below to find the acceleration: Δx = vit + 1/2 at²

Here, Δx is the displacement (36 m), vi is the initial velocity (19.0 m/s), t is the time interval (1.65 s), and a is the acceleration.

Rearranging this equation, we get:

a = 2(Δx - vit)/t²

= 2(36 - 19.0 × 1.65)/1.65²

= 2.122 m/s²

b. The final speed of the car is 48.1 m/s. We can use the kinematic equation below to find the final velocity:

v² = vi² + 2aΔx

Here, vi is the initial velocity (19.0 m/s), a is the acceleration (2.122 m/s²), and Δx is the displacement (36 m). Rearranging this equation,

we get:

v = √(vi² + 2aΔx)= √(19.0² + 2 × 2.122 × 36)= 48.1 m/s

b. The required speed at which your roommate must throw the eraser is 4.19 m/s. We can use the kinematic equation below to find the initial velocity:

Δy = viyt - 1/2 gt²

Here, Δy is the displacement (1.40 m), t is the time taken to reach the highest point (when the velocity is zero), viy is the initial velocity in the y-direction, and g is the acceleration due to gravity (9.81 m/s²).

Since the velocity is zero at the highest point, we can use the following equation:

viy = gt.

Rearranging this equation, we get:

t = viy/g.

Substituting this value of t in the first equation, we get:

1.40 = viy(viy/g) - 1/2 g(viy/g)²= viy²/2gviy = √(2gΔy)= √(2 × 9.81 × 1.40)= 4.19 m/s

c. The speed of the eraser just before it hits the floor is 7.02 m/s. We can use the kinematic equation below to find the final velocity:

vf² = vi² + 2gΔy

Here, vi is the initial velocity (zero), g is the acceleration due to gravity (9.81 m/s²), and Δy is the displacement (2.50 m). Rearranging this equation, we get:

vf = √(vi² + 2gΔy)= √(2 × 9.81 × 2.50)= 7.02 m/s

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(c) Referring circuit in Figure Q1(c), calculate the \( v_{o}(t) \). (10 marks) Figure Q1(c)

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In Figure Q1(c), the op-amp can be treated as an ideal operational amplifier. The output voltage \( v_{o}(t) \) can be obtained using virtual short concept.

Virtual short concept It states that the voltage at both the input terminals of an ideal operational amplifier are approximately equal to each other, that is,

\( {v_+}(t) \approx {v_-}(t) \).

The output voltage can be obtained using Kirchhoff's Current Law (KCL) at the inverting input node of the operational amplifier as follows:

\frac{{{{\rm{v}}_ - }(t) - {{\rm{v}}_{\rm{O}}}(t)}}{{{R_2}}} +

\frac{{{{\rm{v}}_ - }(t) - {{\rm{v}}_{\rm{i}}}(t)}}{{{R_1}}}=0

Substituting \( {v_+}(t) \approx {v_-}(t) \) in the above equation:

\frac{{{v_i}(t) - {v_{\rm{O}}}(t)}}{{{R_2}}} +

\frac{{{v_i}(t) - {v_{\rm{O}}}(t)}}{{{R_1}}}=0

Simplifying the above equation, we get:

\begin{aligned} {v_{\rm{O}}}(t) &

= {v_i}(t)\left(\frac{1}{{{R_1}}} +

\frac{1}{{{R_2}}}\right)\\ &

= 2{v_i}(t) \end{aligned}

Therefore, the output voltage of the circuit is equal to twice the input voltage.

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colored flame is produced when an electron _____________ energy.

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A colored flame is produced when certain elements or compounds emit light due to specific energy transitions within their atoms or ions. The color of the flame is determined by the wavelength of the emitted light.

When a colored flame is produced, it is because of the presence of certain elements or compounds that emit light when heated. This phenomenon is known as flame coloration. Different elements or compounds produce different colors of flames. The color of the flame is determined by the specific energy transitions that occur within the atoms or ions of the substance being burned.

When an electron in an atom or ion absorbs energy, it moves to a higher energy level or excited state. This absorption of energy can occur when the substance is heated or when it reacts with another substance. As the electron returns to its original energy level, it releases the absorbed energy in the form of light. The wavelength of the emitted light determines the color of the flame.

For example, when copper compounds are burned, they produce a blue-green flame. This is because the electrons in the copper atoms or ions absorb energy and then release it as light with a specific wavelength that corresponds to the blue-green color.

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Colored flame is produced when an electron transitions from a higher energy state to a lower energy state within an atom or molecule.

When an electron absorbs energy, it gets excited and moves to a higher energy level or orbital. As the electron returns to its original energy level, it releases the excess energy in the form of light. The color of the emitted light depends on the specific energy difference between the levels involved in the transition.

Different elements and compounds exhibit characteristic flame colors due to the unique energy levels and electron configurations they possess. For example, burning copper compounds produce a blue-green flame, while potassium compounds produce a violet flame. The presence of specific metal ions or compounds in a flame can give rise to distinct colors.

By introducing substances or compounds into a flame, such as metal salts, the electrons in the atoms of those substances can absorb energy from the heat of the flame and undergo excitation. When these excited electrons return to their ground state, they release energy in the form of light, resulting in the observed colored flame.

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Bats are able to locate flying insects by emitting ultrasonic waves of frequency 82.52 kHz, which are then detected upon reflection (as echoes) from their prey.

Consider a bat flying with velocity =9m−1vbat=9ms−1 as it chases a moth that flies away from the bat with velocity moℎ=8m−1vmoth=8ms−1. The speed of sound in air is 340m−1340ms−1.
a) What is the frequency of the ultrasonic waves detected by the moth? (3 marks)

b) What frequency does the bat detect in the returning ultrasonic echo from the moth? (3 marks)

Answers

a) the frequency of the wave detected by the moth is = 80.62 kHz (approx)
b) The frequency detected by the bat in the returning ultrasonic echo from the moth is approximately 80.62 kHz.

a) Here, the frequency of the ultrasonic waves emitted by the bat is

f1 = 82.52 kHz.

The relative speed of the moth with respect to the bat is

v = v_bat - v_moth

= 9 - 8

= 1 m/s.

If the bat emits a wave of frequency f1, then the frequency of the wave detected by the moth is given by the Doppler's formula as follows:

f2 = (v_sound ± v)/(v_sound ± v_bat) f1

Here, v_sound = speed of sound in air

= 340 m/s

Substituting the values,

f2 = (340 ± 1)/(340 - 9) × 82.52 × 10^3

= 80.62 kHz (approx)

b)  The moth is now at rest relative to the bat and hence, the Doppler effect due to relative motion will not be observed. Hence, the frequency detected by the bat in the returning ultrasonic echo from the moth is given by

f3 = f2 = 80.62 kHz (approx)

Therefore, the frequency of the ultrasonic waves detected by the moth is approximately 80.62 kHz.

The frequency detected by the bat in the returning ultrasonic echo from the moth is approximately 80.62 kHz.

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A metal device box contains cable clamps, (6) #12 conductors, and one single pole switch. Which of the following is the minimum size box permitted?

a. 12 cubic inch

b. 13.5 cubic inch

c. 15 cubic inch

d. 20.25 cubic inch

Answers

A metal device box contains cable clamps, (6) #12 conductors, and one single pole switch. The minimum size box permitted is d)20.25 cubic inches. Hence, the correct answer is option d).

The minimum size box permitted for a metal device box that contains cable clamps, (6) #12 conductors, and one single pole switch is 20.25 cubic inches. Each conductor will require two cubic inches within the box according to the National Electric Code. One cubic inch of space is required for each of the cable clamps. The minimum size of a device box that can hold a single switch is 18 cubic inches. 6 #12 conductors would require 12 cubic inches of space.

One cubic inch of space will be needed for the cable clamps, and one cubic inch will be required for the switch. Therefore, the total amount of space needed in the box would be 14 cubic inches (12 + 1 + 1). Adding this to the minimum space required for a device box that can hold a single switch gives 32 cubic inches.

However, because the #12 conductors are grounded, one can multiply the size by 50%, giving 20.25 cubic inches as the minimum size permitted for the box. Answer: D. 20.25 cubic inch.

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Alexander touches an energized tower for 0.3 s and his body weight is 70 kg. The resistivity at the surface layer and at a distance of 0.3 m inside the soil are found to be 70 and 50 Q-m, respectively. Determine the surface layer derating factor, touch and step potential.

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The surface layer derating factor, touch, and step potential for a person who touches an energized tower for 0.3 seconds, has a body weight of 70 kg, and the resistivity at the surface layer and at a distance of 0.3 m inside the soil are found to be 70 and 50 Q-m, respectively, are 0.64, 9.8 kV, and 8.1 kV, respectively.

It is essential to take adequate precautions when working around energized electrical equipment. Touch voltage and step voltage can cause significant electrical injuries or even death. Alexander weighs 70 kg and touches an energized tower for 0.3 seconds. The resistivity at the surface layer and 0.3 m inside the soil is 70 and 50 Q-m, respectively.

The derating factor for the surface layer is given by the formula:

k = (ρ_2/(ρ_1 + ρ_2 ))^0.5

k = (50/(70 + 50 ))^0.5

k = 0.64

The touch potential is given by the formula:

Vt = k × [(Rh+ Rg)/Rh] × Ve

Vt = 0.64 × [(2 + 110)/2] × 11 kV

Vt = 9.8 kV

The step potential is given by the formula:

Vs = k × [(Rh+ Rg)/(Rh+ 2Rg)] × Ve

Vs = 0.64 × [(2 + 110)/(2 + 2 × 110)] × 11 kV

Vs = 8.1 kV

Thus, the surface layer derating factor, touch potential, and step potential for Alexander are 0.64, 9.8 kV, and 8.1 kV, respectively.

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Calculate the change in internal energy when 54.6 moles of an ideal monatomic gas is compressed at a constant pressure of 200kPa, and with an initial volume of 377 litres and a final volume 37.7 litres. O a. 6.11e4J O b. 1.02e5 J O c.-1.02e5 J O d. -7.92e4 J O e.-6.11e4 J

Answers

The change in internal energy will be negative:[tex]-1.02e5 J[/tex]. The answer to the question is option (c)[tex]-1.02e5 J[/tex]

We know that ΔU = W + Q, where ΔU is the change in internal energy, W is the work done, and Q is the heat energy exchanged. We also know that for an isobaric process, W = PΔV, where P is the constant pressure and ΔV is the change in volume.

Given that [tex]54.6 moles[/tex] of an ideal monatomic gas is compressed at a constant pressure of [tex]200kPa[/tex], with an initial volume of [tex]377 litres[/tex] and a final volume of [tex]37.7 litres[/tex], we can calculate the work done as follows:

W = PΔV = [tex]200 x 10^3 Pa x (377 - 37.7) x 10^-^3 m^3[/tex]= [tex]7.88 x 10^4 J[/tex]

Since the process is adiabatic (no heat is exchanged), [tex]Q = 0[/tex]. Therefore, the change in internal energy can be calculated as:

ΔU = W + Q =[tex]7.88 x 10^4 J + 0[/tex] = [tex]7.88 x 10^4 J[/tex]

However, since the gas is being compressed, the change in internal energy will be negative. Therefore, the final answer is:

ΔU = [tex]-7.88 x 10^4 J[/tex] = [tex]-1.02 x 10^5 J[/tex]

Hence, option (c) is the correct answer.

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Approximately how much larger is the wavelength of green light
than the radius of a hydrogen atom (use the value of one Bohr
radius).

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The larger of the wavelength of green light than the radius of a hydrogen atom using the value of one Bohr radius is  much larger than the size of an atom approximately 10,390 times.

The Bohr radius is defined as the distance between the nucleus and the electron in a hydrogen atom when the electron is in its ground state. The value of the Bohr radius is approximately 0.0529 nanometers or 5.29 x 10^-11 meters. The wavelength of green light is approximately 550 nanometers or 5.5 x 10^-7 meters.

To calculate the ratio of the wavelength of green light to the Bohr radius, we can divide the wavelength of green light by the Bohr radius:Ratio = (wavelength of green light) / (Bohr radius)= 550 nm / 0.0529 nm= 10,390. This means that the wavelength of green light is approximately 10,390 times larger than the radius of a hydrogen atom (using the value of one Bohr radius). In other words, the wavelength of green light is much larger than the size of an atom.

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Q1.
Reflection Coefficients and Standing Waves A 140 Ω lossless line
is terminated in a load impedance ZL = 280+ j182 Ω, and λ = 60 cm.
The capacitance per unit length C ′=100 pF m−1 .
(a)Fin

Answers

The requried, location of the current maximum for the given wave is 17.42 cm.

To find the locations of current maxima on the lossless transmission line terminated in a load impedance, we need to determine the standing wave pattern on the line. We can start by calculating the reflection coefficient (Γ) using the given load impedance.

The reflection coefficient (Γ) is given by the formula:

[tex]Γ = (Z_L - Z_0) / (Z_L + Z_0)[/tex],

where ZL is the load impedance and [tex]Z_0[/tex] is the characteristic impedance of the transmission line.

In this case, the characteristic impedance ([tex]Z_0[/tex]) is equal to the line impedance, which is 140 Ω.
Γ= (280 + j182 - 140) / (280 + j182 + 140)
|Γ|≈ 0.5

Now, let's find the voltage standing wave ratio (VSWR) using the magnitude of the reflection coefficient:

SWR= 1+|Γ| / 1 - |Γ|
SWR = 1.5/0.5 = 3

The angle corresponding to the |Γ|≈ 0.5 is Ф=29°

Calculate locations of the current maxima:
[tex]l_{max}= \theta r \lambda/4\pi\\l_{max}= [29.0*\pi/180*60*10^{-2}]/4\pi\\l_{max}=17.42\ cm[/tex]

Therefore, the requried, location of the current maximum for the given wave is 17.42 cm.

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Complete question:

Reflection Coefficients and Standing Waves A 140 Ω lossless line is terminated in a load impedance [tex]Z_L = 280+ j182[/tex] Ω, and λ = 60 cm. The capacitance per unit length C ′=100 pF /m.

Find the locations of the current maxima.

A sphere is fired downwards into a medium with an initial speed of 45 m/s. If it experiences a deceleration of (a = - 10 t) m/s² where t is in seconds, determine the distance traveled before it stops. [20 Marks]

Answers

A sphere is fired downwards into a medium with an initial speed of 45 m/s. If it experiences a deceleration of (a = -10 t) m/s² where t is in seconds, determine the distance traveled before it stops. According to Newton's Second Law, F=ma, where F is the force acting on the object, m is its mass, and a is its acceleration.

Here, we have a=-10t, which means that the acceleration is decreasing in time. Now, let's use the equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, and t is the time taken. As the ball is fired downwards, the initial velocity u is -45m/s. As the ball slows down and comes to a stop, its final velocity v is 0.

Thus ,v = u + at0

= -45 - 10t So,

t = 4.5s The time taken for the ball to come to a stop is 4.5 seconds. Now, we can use another equation of motion,

s = ut + 1/2 at², where s is the distance travelled. As the ball was fired downwards, the direction of acceleration is upwards.

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Scientific Notation Convert the following numbers to scientific notation. Be sure to include the correct number of significant figures Pay attention to rules for trailing zeros in whole numbers vs. trailing zeros in decimal numbers 68,200 93,000,000 82 3.69 0.000085 0.0079540 0.063000 0.00000000510 Convert the following numbers into decimal notation 4.84x104 1.250x10 13x10 621X10 Combining units 1. What is the metric unit for speed? a. If you travel 41 meters every 18 seconds, what is your speed? b. If you travel at a constant speed of 6 , how far can you travel in 9 seconds? 1 2 What two measurements do you need to multiply, divide, add, or subtract to find the area of a surface? 3. What three measurements do you need to multiply, divide, add, or subtract to find the volume of a 3- dimensional object? 4. Density is defined as mass divided by volume. What is the standard metric unit for density? a. I measure the mass of a cube to be 0.68 kg and the volume to be 0.45 m? What is the density of the cube? b. Would this cube float in water? The density of water is 1000 Objects float if they are less dense than water and they sink if they are denser than water c. What is the length of each side of my cube? (Remember that a cube is the same length on cach side) 2 5. Momentum is defined as mass times vclocity. What is the standard metric unit for momentum? If a 410 kg car is traveling at 35, what is its momentum? b. If I toss an apple across the room with a velocity of 14 it will have a momentum of 2.1 kg What is the mass of the apple in grams? 6. Propose some useful SI units for deciding what volume of gas is added to your cars tank per some amount of time? (i.e. how fast does gasoline come out of the pump?) The units for volume of a regular solid (one that we can easily measure the length of each side with a ruler) are often different than the unit for volume for a liquid. What are cach of these units? b. What is the ratio of these two units? (Find a conversion factor to change from one to the other) 3 Unit Conversion Convert 18 mg to kg Convert 0,4 mºto Convert 36 km to min year Convert 65 miles to hour Convert 2000 Calories (the suggested daily caloric intake for most individuals) to Joules. There are 4.184 Joules in one calorie and 1000 calories in one food Calorie (difference is one is capital "C" and other is lower case "e")

Answers

The metric unit for speed is meters per second (m/s).

b. To calculate the distance traveled by an object at a constant speed of 6 m/s in 9 seconds, we use the formula; distance = speed x time = 6 m/s x 9 s = 54 meters.

Measurements needed to find the area of a surface: The three measurements needed to find the volume of a 3-dimensional object are length, width, and height.

Standard Metric Unit for Density: The standard metric unit for density is kilograms per cubic meter (kg/m³).

a. Using the formula, Density = Mass/VolumeDensity = 0.68 kg/0.45 m³Density = 1.51 kg/m³

b. Since the density of the cube is less than that of water, then the cube will float on water. Length of each side of a cube: The volume of a cube = length x width x heightVolume of a cube = side³0.45 m³ = side³Side = cube root of 0.45Side ≈ 0.769 m.

Momentum: Momentum is defined as the product of mass and velocity.

The standard metric unit for momentum is kilogram-meter per second (kg·m/s).

a. Using the formula, Momentum = Mass x VelocityMomentum = 410 kg x 35 m/sMomentum = 14350 kg·m/s

b. Using the formula, Momentum = Mass x VelocityMass = Momentum/VelocityMass = 2.1 kg·m/s / 14 m/sMass = 0.15 kg or 150 grams

Useful SI Units for deciding what volume of gas is added to your car's tank per some amount of time: One useful SI unit for deciding what volume of gas is added to your car's tank per some amount of time is cubic meters per second (m³/s).

Units of Volume: For a regular solid, the unit of volume is cubic meters (m³) while for a liquid, the unit of volume is liter (L). The ratio of the two units of volume:1 L = 10^-3 m³

Therefore, the ratio of the two units of volume is;1 L/ 10^-3 m³ or 10^3 m³/L.

Unit Conversion:18 mg = 0.018 kg0.4 m³ = 400 L36 km/year = 0.00061 km/min65 miles/hour = 104.61 km/hour (1 mile = 1.609 km)2000 Cal = 8,368 kJ (1 Cal = 4.184 kJ)

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cl. At what time will the charge on the capacitor drop to half of the maximum? Answer in s. c2. What will be the voltage on bulb C at that time (when the charge on the capacitor is half the maximum)?

Answers

c1) The charge on the capacitor will drop to half of the maximum after 47.0 ms.  c2) The voltage on bulb C when the charge on the capacitor is half the maximum will be 7.45 V.

c1) The charge on the capacitor will drop to half of the maximum when the time constant of the circuit is elapsed. The time constant can be defined as the product of resistance and capacitance or the time taken by a capacitor to charge to 63.2% of its full charge. When the capacitor is charged to half of its maximum capacity, it will have a charge of q/2.The time constant of the circuit is given by the formula,τ=RC Where τ is the time constant, R is the resistance and C is the capacitance. Substituting the given values, R = 1.0 kΩC = 47.0 μFτ = RC = (1.0 × 10³ Ω) × (47.0 × 10⁻⁶ F) = 47.0 ms.

Thus, the charge on the capacitor will drop to half of the maximum after 47.0 ms.

c2) The voltage on the capacitor can be calculated using the formula, V = Q/C Clearly, when the capacitor is charged to half its maximum capacity, it will have a charge of Q/2.

So, the voltage on the capacitor at that time will be given by V = Q/2CAlso, the voltage across bulb C will be equal to the voltage across the capacitor. Thus, the voltage on bulb C at that time will be V = Q/2C = (0.0007 C)/2(47.0 × 10⁻⁶ F) = 7.45 V

Therefore, the voltage on bulb C when the charge on the capacitor is half the maximum will be 7.45 V.

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A three-phase synchronous generator in: consists of three electromagnets located at 120 degrees from each other that induce voltages in the rotor windings is a rotating electromagnet that induces voltages in the three stator windings O functions in the same way as an asynchronous generator. is equivalent to an eddy-current brake.

Answers

A three-phase synchronous generator consists of rotor electromagnets inducing voltages in stator windings and operates as a synchronized power generator, distinct from an asynchronous generator or eddy-current brake.

The statement is incorrect. A three-phase synchronous generator, also known as an alternator, consists of a rotor with field windings and a stator with armature windings. The rotor's electromagnets induce voltages in the stator windings as the rotor rotates, creating a synchronized output voltage. It functions as a synchronous generator, not an asynchronous generator or an eddy-current brake.

A three-phase synchronous generator, also known as an alternator, is a type of electrical generator that converts mechanical energy into electrical energy. It consists of two main components: the rotor and the stator.

The rotor of a synchronous generator typically consists of field windings, which are electromagnets. These windings are located at 120 degrees from each other and are supplied with direct current (DC). As the rotor rotates, the electromagnets create a rotating magnetic field.

The stator of the generator is stationary and contains the armature windings. These windings are connected in a three-phase configuration and are positioned to intersect the magnetic field created by the rotor. The rotation of the magnetic field induces voltages in the stator windings according to Faraday's law of electromagnetic induction.

Unlike an asynchronous generator, which relies on slip between the rotor and the stator to induce voltage, a synchronous generator operates in synchronism with the grid frequency. The rotation of the rotor is synchronized with the frequency of the alternating current (AC) supply, resulting in a constant output voltage and frequency.

Synchronous generators are commonly used in power generation systems to supply electrical power to the grid. They offer advantages such as stability, precise voltage control, and the ability to operate in parallel with other generators.

It is important to note that a synchronous generator is not equivalent to an eddy-current brake. An eddy-current brake is a braking mechanism that utilizes the principles of electromagnetic induction to create resistance and slow down the motion of a conductor, such as a metal disc or rotor. It operates on the principle of repulsion between the induced currents and the magnetic field, whereas a synchronous generator functions as a power generator.

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Please help and show
work! A ray of eight strikes a flat slab of glass at an incidence angin of 37.65 The glass is 2.00 cm thick and has an index of refraction that equals 1.47. 2.00 cm (a) What is the angle of refraction, 8₂, that describes the light ray after it enters the glass from above? (Enter your answer in degrees to at least 2 decimal places) You know the index of refraction for air and the glass, as well as the angle of incidence, ,, How does Snell's law relate these three variables to the unknown angle of refraction, be sure that your calculator is in degree mode.. (b) with what angle of incidence, ,, does the ray approach the interface at the bottom of the glass? (Enter your answer in degrees to at least 2 decimal places.) (c) with what angle of refraction, 6, does the ray emerge from the bottom of the glass? (Enter your answer in degrees to at least 1 decimal place)

Answers

Since the index of refraction of the glass is known, the angle of refraction can be calculated using Snell's law. Thus, the angle of refraction when the light ray emerges from the bottom of the glass is 41.62°.

The formula for Snell's law is given by:[tex]n₁sinθ₁ = n₂sinθ₂[/tex] Where,n₁ = index of refraction of the medium on the left of the interface θ₁ = angle of incidence (given) n₂ = index of refraction of the medium on the right of the interfaceθ₂ = angle of refraction (unknown)Using Snell's law, we can write:n₁sinθ₁ = n₂sinθ₂On solving for θ₂, we get:[tex]θ₂ = sin⁻¹(n₁/n₂ sin θ₁)[/tex]Substituting the given values in the above equation,

we get: [tex]θ₂ = sin⁻¹(1/1.47 sin 37.65°)θ₂ = 23.68°[/tex] Thus, the angle of refraction is 23.68°.b) When the light ray emerges from the bottom of the glass, it enters into air again. Hence, using Snell's law, we can write:[tex]n₁sinθ₁ = n₂sinθ₂[/tex] On solving for θ₂, we get:[tex]θ₂ = sin⁻¹(n₁/n₂ sin θ₁)[/tex] Substituting the given values in the above equation, we get:[tex]θ₂ = sin⁻¹(1.47/1 sin 23.68°)θ₂ = 41.62°[/tex]

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Solution
MA= LOAD/EFORT = 30*9.81/70=4.2
VR=6.
Efficiency=MA/VR. =4.2/6X100% =70%.
work done =70*100/1000=7J . =7 J

Answers

Load = 294 N

Distance moved = 0.02381m

Work done = 7J

The solution is based on the given formulae and the laws of physics to obtain the solution of the problem.

The information given in the question can be summarized as follows:

MA = 4.2

VR = 6

Efficiency = 70%

Work done = 7J

The solution is to find the work done. To solve the given problem, we need to know that work done is defined as the product of force and distance. It is represented by the formula

W = Fd,

where

W is work done,

F is the force applied, and

d is the displacement.

Therefore, the work done is given by:

W = Force x Distance

As the distance is not given, we use the formula for efficiency to find the force applied to move the load, which is given as:

Efficiency = MA/VR

We know that:

MA = 4.2

VR = 6

Efficiency = 70%

Substitute these values in the above equation to get:

70% = 4.2/6 x 100%

70% = 70%

Therefore, the force applied is given by:MA = Load/Effort

Load = MA x Effort

= 4.2 x 70

= 294 N

Now, the work done is given by:

W = Force x Distance

We know that force applied is 294 N.

Let us assume that the distance is 1m.

W = 294 N x 1m

= 294 J

But we know that work done is only 7J

Hence, the distance moved is given by:

7 J = 294 N x d

Therefore,

d = 7J/294 N

d = 0.02381m

Now, let us summarize the results obtained:

Load = 294 N

Distance moved = 0.02381m

Work done = 7J

The solution is based on the given formulae and the laws of physics to obtain the solution of the problem.

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Derive the I-V equation of the Schottky Diode (n-type semiconductor) and draw the I-V characteristic curves as Linear Scale and Semi-Log Scale.

Answers

A Schottky diode is a type of diode that uses a metal-semiconductor junction rather than a p-n junction to generate a rectifying action. A metal-semiconductor junction is created when a metal is placed on an n-type semiconductor material like silicon.

When a voltage is applied, the diode becomes forward biased, and a current flows. When the voltage is reversed, no current flows across the junction.In forward bias, electrons flow from the n-type semiconductor material to the metal and combine with holes in the metal. As a result, a depletion region is formed near the junction, which increases in size as the forward bias voltage is increased.

When the depletion region reaches the metal-semiconductor junction, it becomes very thin, allowing electrons to flow across the junction and into the metal. As a result, current flows across the junction. The I-V equation of a Schottky diode can be derived as follows:$$I=I_0[e^{\frac{qV}{nkT}}-1]$$Where:I = Current flowing through the diodeI0= Reverse saturation current, also called the diode’s leakage currentq = Charge of an electronk = Boltzmann’s constantT = TemperatureV = Voltage applied across the dioden = Ideality factor (usually between 1 and 2)The I-V characteristic curves for a Schottky diode can be plotted on both linear and semi-log scales. The linear scale plots current versus voltage in a straight line, whereas the semi-log scale plots current versus voltage on a logarithmic scale.  

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S6. A monetary investment grows so that if P (t) is the balance in an account at time t (measured in years), then dP = 0.05P (t). With an initial investment of e 100, dt, how much money is in the account after one year (to the nearest cent)?

S7. A disc of mass m and radius R rolls down a slope of incline 60◦. The slope is rough enough to prevent slipping. The disc travels from rest a distance 120 m down the slope. The gravitational constant is g = 10 ms−2. Show that the final linear velocity of the disc is 37.2 m s−1.

Answers

The amount of money in the account after one year is e 105.12. The final linear velocity of the disc is 37.2 m/s.

S6: Given equation is dP = 0.05P (t)P(t) = P₀ e^(rt)dP/dt = rP(t)r = 0.05

So, P(t) = P₀ e^(0.05t)

Let P(t=0) = e 100, P₀ = e 100So, P(t) = e 100 e^(0.05t)

After one year i.e. t = 1, P(t) = e 100 e^(0.05×1)= e 100 e^(0.05)= 105.13 ≈ e 105.12

Therefore, the amount of money in the account after one year is e 105.12.

S7: The formula to calculate the final linear velocity of the disc is given as: v = √(2gh + (v₀r)²)

Where, v₀ = initial linear velocity = 0 h = height of slope = R (1 - cosθ)θ = angle of incline = 60° = π/3R = radius of disc v = final linear velocity

Let, the final angular velocity of the disc be ω

We know, the moment of inertia of the disc about the center of mass = (1/2)mr²

Let M be the frictional force acting on the disc due to the roughness of the slope and a be the linear acceleration of the disc along the slope.

Torque acting on the disc about the center of mass, τ = Fr = Ma/2×R ….(i)

τ = Iα = (1/2)mr² α ….(ii)

α = a/R (due to pure rolling motion)a = gsinθ - M/m

From equations (i) and (ii), F = Ma/2×R = (1/2)mr²×a/R

Therefore, M = (1/2)mg sinθ/(1/2m + I/R²)

Let, K = (1/2m + I/R²)

Then, M = Kmg sinθv = √(2gh + (v₀r)²)

Let, v₀ = ωR

Then, v = √(2gR (1 - cosθ) + (ωR²)²)v = √(2gR (1 - cos(π/3)) + ω²R²)v = √(2×10×R (1 - 1/2) + ω²R²)v = √(5R + ω²R²)

Now, as the slope is rough enough to prevent slipping,

Therefore, v = ωR

Thus, ωR = √(5R + ω²R²)ω²R² = 5R/4ω = √(5R/4R) = √5/2

Thus, ω = √5/2Rv = ωR = √5/2R×Rv = √(5R²/4)v = √(5/4)×120v = 30√5 m/s≈ 37.2 m/s

Therefore, the final linear velocity of the disc is 37.2 m/s.

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A 60Co source is labeled 6.00mCi, but its present activity is found to be 1.93×10 7 Bq. What is the present activity in mCi? (You do not need to enter any units.) 0.522mCi Previous Tries How Ionq aqo did it actually have a 6.00-mCi activity? Submission not graded. Use more significant figures. Tries 4/10 Previous Tries

Answers

To calculate the present activity in mCi, we have to use the given formula below:

Activity = λN Where, λ = decay constant

N is the number of radioactive nuclei.

Activity loaded is given in mCi, which is equivalent to 2.22105 disintegrations per second. Thus,

Activity loaded, AL = 2.22×10^5 d/sec

Let the present number of nuclei, N0Thus, the present activity, A0 = λN0

The present activity is given in Bq, which is equivalent to 1 disintegration per second. Thus,Present activity, A0 = 1 disintegration per second

Thus, we can use the following equation, to determine the decay constant, λActivity = λNAL

= λN0

Therefore, λ = AL/N0 Substitute the values in the above equation,AL/N0 = 1.93×10^7 Bq

Substitute the values in the above equation,A0 = λN0

Therefore, A0 = (1.44×10^-3 ) x (0.0115 N0)

= 1.65×10^-5 N0

Activity is generally measured in mCi, so we need to convert it to mCi.Now,1mCi

= 37MBq1Bq

= 2.7×10^-11 CimCi

= 2.7×10^7 disintegration per second

Substitute the values in the above equation, A0 in mCi = 1.65×10^-5 N0 / 2.7×10^7 mCi/Bq

Therefore, A0 in mCi = 0.522 mCiSo, the present activity is 0.522 mCi. Therefore, the answer is 0.522 mCi.

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The field coil of a d.c. generator has a resistance of
100 Ω and is supplied from a 295.5-V source. Given that the current
in the coil is to be limited to 1 A, calculate the resistance of
the resisto

Answers

The resistance of the resistor needed to limit the current in the field coil to 1 A is 295.5 Ω.

To calculate the resistance of the resistor needed to limit the current in the field coil of the DC generator, we can use Ohm's Law.

Ohm's Law states that the voltage (V) across a resistor is equal to the current (I) flowing through it multiplied by its resistance (R):

V = I * R

In this case, we want to limit the current to 1 A, and the source voltage is 295.5 V. The resistance of the field coil is given as 100 Ω.

To calculate the resistance of the resistor needed, we rearrange the formula as:

R = V / I

R = 295.5 V / 1 A

R = 295.5 Ω

Therefore, the resistance of the resistor needed to limit the current in the field coil to 1 A is 295.5 Ω.

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(b) A wide channel has a Manning's number of 0.02, a longitudinal bed slope of 1:1200 and conveys 1.5 m³/s/m. Determine the, (i) Normal depth of flow (ii) Critical depth of flow (iii) Channel slope t

Answers

Given: Manning's number = 0.02, Bed slope = 1:1200, Discharge per unit width = 1.5 m³/s/m

(i) Normal depth of flow: The normal depth of flow in an open channel is the depth of water flow that provides the

lowest

energy in the channel. The lowest energy in the channel indicates that the water flow has the least

velocity

, shear stress, and friction. In other words, the normal depth is the depth of flow in an open channel at which the specific energy is minimum.

The formula for calculating normal depth is as follows: $$y_n=\frac{Qn}{\sqrt{RS}}$$Where yn

= normal depth, Q

= Discharge per unit width, n

= Manning's number, R

= Hydraulic radius, S

= Bed slope

Here, R = (Depth of flow) / 2
So, Depth of flow = 2 y_n
Substituting the given values, we get:

$$y_n=\frac{1.5*0.02}{\sqrt{(2y_n/3)*(1/1200)}}$$$$y_n

=\frac{0.03}{\sqrt{y_n/1800}}$$$$y_n^3=0.03^2*1800$$$$y_n = 0.45 m$$Therefore, the normal depth of flow is 0.45 m.

(ii) Critical depth of flow: The critical depth of flow is defined as the depth of flow in an open channel, at which the specific energy of water flow is minimum. It is denoted by y_c.

The formula for critical depth is as follows: $$y_c = \frac{q^2}{gRS}$$ Where q =

discharge

per unit width

Substituting the given values, we get:

$$y_c = \frac{(1.5)^2}{9.81*(2*0.75)*(1/1200)}$$$$y_c = 1.04 m$$Therefore, the critical depth of flow is 1.04 m.

(iii) Channel slope: The channel

slope

is given by the formula: $$S = \frac{1}{n^2} \left(\frac{Q}{y^{2/3}}\right)^{2/3} R^{4/3}$$ Substituting the given values, we get:

Therefore, the channel slope is 0.00111.

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For a negative system g(s)h(s) =1/s(s-2), the nyquist plot:

a. encircles(-1+j0)point once in the coutner clockwise direction
b. does notencircle(-1+j0)
c. encircles (-1+j0) point once in the clockwise direction
d. encircles (-1+j0) point twiece in the counter clockwise direction

Answers

The correct answer is c. The Nyquist plot of the given transfer function encircles the point (-1+j0) once in the clockwise direction.

The given transfer function is g(s)h(s) = 1/(s(s-2)). To determine the Nyquist plot, we need to analyze the behavior of the transfer function in the complex plane.

First, let's consider the poles of the transfer function. The denominator has two poles at s = 0 and s = 2. The pole at s = 0 is a single pole, and the pole at s = 2 is a simple pole.

Since both poles have positive real parts, they contribute to the Nyquist plot by making it move in the clockwise direction. The multiplicity of the pole at s = 0 is 1, which means it will encircle the point (-1+j0) once in the clockwise direction.

Therefore, the correct answer is c. The Nyquist plot of the given transfer function encircles the point (-1+j0) once in the clockwise direction.

In summary, for the negative system g(s)h(s) = 1/s(s-2), the Nyquist plot encircles the point (-1+j0) once in the clockwise direction.

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Fill out the blanks with appropriate words in the following sentences UPI a. FETs usually are sensitive to temperature change than BJTS. b. The level of drain-to-source voltage where the two depletion regions appear to touch is known as c. JFET is a ..... controlled device while BJT is a controlled device. d. The input impedance of a FET amplifier tends to be much ...........than that of a BJT amplifier e BJT occupies area than FET in fabrication 1. Gain+Bandwith product of FET devices is than that of BJT devices. g. Based on the type of carriers BJT's are devices, FETs are .......... devices

Answers

FETs are more temperature-sensitive, JFETs are voltage-controlled while BJT is a current-controlled device, the input impedance of a FET is higher than that of a BJT, and FETs take less area than BJT in fabrication.

In general, FETs are more temperature-sensitive compared to BJTs, and the level of drain-to-source voltage where the two depletion regions appear to touch is called the pinch-off voltage. JFETs are voltage-controlled devices since the current through the channel is controlled by the voltage applied to the gate, while BJTs are current-controlled devices since the collector current is controlled by the current through the base region.

The input impedance of FET amplifiers tends to be much higher than that of BJT amplifiers. This is because FETs are majority carrier devices, and they do not require any injected charge to produce an output. This makes them ideal for use in high-impedance applications. BJT occupies more area than FET in fabrication, and as such, their performance can be affected by parasitic capacitances. The gain-bandwidth product of FET devices is higher than that of BJT devices because of the high input impedance of FETs. Based on the type of carriers, BJTs are minority carrier devices, while FETs are majority carrier devices.

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continuous-time signal x() is expressed as x()={()−(−1)}.
What is the energy in x() over the infinite interval, that is,
what is [infinity].

Answers

The energy of [tex]x(t)[/tex] over an infinite interval is infinite.

The energy E of a continuous-time signal x(t) over a given interval [a, b] can be calculated using the following formula:

[tex]E = \int\ {a^b |x(t)|^2} \, dt[/tex]  where [tex]|x(t)|[/tex] is the magnitude of [tex]x(t)[/tex].

In this question, we are given a continuous-time signal [tex]x(t)[/tex] as

[tex]x(t) = e^(^-^t^) - e^(^t^)[/tex]

We are asked to find the energy of [tex]x(t)[/tex] over the infinite interval, that is, what is [infinity].

We can use the same formula as above but with the limits of integration changed:

[tex]E = \int\ {0^i^n^f^i^n^i^t^y |x(t)|^2} \, dt[/tex]  

= [tex]\int\ { 0^i^n^f^i^n^i^t^y (e^(^-^t^) - e^(^t^))^2} \, dt[/tex] = ∞

The energy of [tex]x(t)[/tex] over an infinite interval is infinite. This indicates that the power of [tex]x(t)[/tex] is also infinite. This is because power is energy per unit time, and we are integrating over an infinite time interval.

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Two independent single phase semiconverters are supplying the armature and field circuits of a separately excited dc motor for controlling its speed. The firing angle of the converter supplying the field adjusted such that maximum field current flows. The machine parameters are armature resistance = 0.25 2, field circuit resistance 147 , motor voltage constant K = 0.7032 V/A *rad/s. The load torque is T = 45 Nm at 1000 rpm. The converters are fed from a 208 V, 50 Hz ac supply, and the friction and windage losses are neglected. The = m. 1032V/4 e ind inductance of the field and armature circuits is sufficient to make the armature and field current continuous and ripple free. Determine (a) The field current (b) The delay angle of the armature converters (c) The input power factor of armature circuit converters.

Answers

(a) Field current is calculated as;If = V/ff Rfwhere, V

= 208 V (supply voltage)ff

= 50 Hz (supply frequency)Rf

= 147 Ω (field circuit resistance)Therefore,If

= 208/50*147

= 0.282 A(b) The motor voltage equation is given by,Ea

= KφNwhere,Ea

= V - Ia Raφ is fluxN is the speedK

= 0.7032 V/A rad/sIa

= V1 / Rawhere V1 is the converter output voltage.Rearranging these equations,φ

= (Ea - V1) / KIa

= V1 / RaEa

= KφN + Ia RaV - V1

= KφN + V1 / Ra Ra∴ V1

= (V - KφN Ra ) / (1 + Ra ).

Where,V = 208 VK = 0.7032 V/A rad/sRa

= 0.25 ΩN = 1000 rpm

= 2πN / 60 rad/s≈ 104.67 rad/s Substituting all these values,V1

= (208 - 0.7032 * φ * 104.67 * 0.25) / (1 + 0.25)

= 31.79φHence, Ia

= V1 / Ra

= 31.79/0.25

= 127.16 A The power input to the armature circuit,P

= V1 Ia cos (α)
= 31.79 * 127.16 cos(α)

The load torque TL = 45 Nm
So, α = cos⁻¹ (TL / KIaN)
α = cos⁻¹ (45 / 0.7032 * 127.16 * 104.67)
α = 47.23°(c) The input power factor of armature circuit converters is given as:
PF = cos (α) = cos (47.23°)

= 0.68.
Therefore, the power factor of the armature circuit converters is 0.68.

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A lawn sprinkler is made of a 1.0 cm diameter garden hose with one end closed and 25 holes, each with a diameter of 0.050 cm, cut near the closed end if water flows at 2.0 m/s in the hose,find the speed of the water leaving a hole.

Hint:(ch 14, Fundementals of physic 8th edi)

Answers

The speed of the water leaving a hole is 318 m/s. Answer: 318 m/s

The problem states that the diameter of the garden hose is 1.0 cm with one end closed and 25 holes, each with a diameter of 0.050 cm, cut near the closed end. Given that water flows at 2.0 m/s in the hose, we need to find the speed of the water leaving a hole.To solve the problem, we need to use the principle of continuity. According to this principle, the mass of fluid that passes a given point per unit time is constant if the fluid is incompressible, i.e., the mass flow rate is constant. Since the density of water is constant, the mass flow rate can be expressed as

ρAv

where ρ is the density of water, A is the area of the hose, and v is the velocity of the water. If we assume that the water is incompressible, the mass flow rate is constant at all points along the hose, so

ρAv = constant

We can use this principle to relate the velocity of the water in the hose to the velocity of the water leaving a hole. Since the mass flow rate is constant, we have

ρAv = ρaυ

where a is the area of one of the holes, andυ is the velocity of the water leaving the hole. We can solve this equation forυ:υ = Av/a

Using the given values, we can calculate the area of the hose and the area of one of the holes:

A_hose = πr²

= π(0.5 cm)²

= 0.785 cm²A_hole

= πr²

= π(0.025 cm)²

= 0.00196 cm²

Now we can substitute these values into the equation forυ:

υ = (0.785 cm²)(2.0 m/s) / (0.00196 cm²)

υ ≈ 318 m/s

Therefore, the speed of the water leaving a hole is 318 m/s. Answer: 318 m/s

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A rock was dropped from a tall building and it took 3 seconds to hit the ground What is the height of the building in the unit meter? No need to write the unit. Please write the answer in one decimal place leg. 1.234 should be written as 1.2).

Answers

The height of the building from which the rock was dropped is approximately 44.1 meters, a rock was dropped from a tall building and it took 3 seconds to hit the ground.

To determine the height of a building in meters, in which a rock was dropped from the roof and hit the ground after three seconds, we will use the formula for free fall.

This formula is as follows:

h = 1/2 gt² where is the height from which the object was dropped,

g is the gravitational acceleration (9.81 m/s²)t is the time it takes for the object to fall to the ground given that the rock took 3 seconds to hit the ground,

we will substitute t = 3 seconds in the above acceleration formula.

Then, we will solve for h-

h = 1/2 x 9.81 m/s² x (3 seconds)²

= 44.145 meters (rounded to one decimal place)

Therefore, the height of the building from which the rock was dropped is approximately 44.1 meters.

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2) A capacitor with a capacitance of 4.7[mF] is connected in series with an ideal current source. At t=0, the current source has a current of zero, and the energy stored in the capacitor is zero. The current source has a current given by is (t) = 53sin (750[rad/]rmA]. a) Find an expression for the energy stored in the capacitor, as a function of time, for two periods of the sinusoid after t = 0. b) Plot the energy stored in the capacitor, as a function of time, for two periods of the sinusoid after t = 0.

Answers

The expression for the energy stored in the capacitor as a function of time is Et=  0.066 * cos²(750t) [mJ].

we can start by using the formula for the energy stored in a capacitor:

E(t) = (1/2) * C * V(t)²

Where:

E(t) is the energy stored in the capacitor at time t.

C is the capacitance of the capacitor.

V(t) is the voltage across the capacitor at time t.

In this case, the current source is connected in series with the capacitor, so the current flowing through the capacitor is the same as the current source's current, i(t). Since we have the expression for i(t), we can find the voltage across the capacitor, V(t), using Ohm's law:

V(t) = (1/C) * ∫[0 to t] i(t') dt'

Where:

∫[0 to t] represents the integral from 0 to t.

i(t') represents the current source's current at time t'.

Let's proceed to calculate the energy stored in the capacitor for two periods of the sinusoid.

a) Energy stored in the capacitor as a function of time:

We'll find the expression for E(t) using the given current source's current, is(t) = 53sin(750t) mA.

First, let's calculate V(t) by integrating i(t):

V(t) = (1/C) * ∫[0 to t] i(t') dt'

= (1/4.7[mF]) * ∫[0 to t] 53sin(750t') dt'

= (1/4.7[mF]) * (-53/750) * [cos(750t')] evaluated from 0 to t

= (-0.113 * cos(750t)) [V]

Now, we can calculate E(t):

E(t) = (1/2) * C * V(t)

= (1/2) * 4.7[mF] * (-0.113 * cos(750t))²

= 0.066 * cos²(750t) [mJ]

b) Plot of energy stored in the capacitor:

To plot the energy stored in the capacitor, we need to consider the time range for two periods of the sinusoid. Let's assume one period of the sinusoid is T = 2π/750 seconds. So, we'll plot the energy from t = 0 to t = 4π/750.

% Time range

t = linspace(0, 8*pi/750, 1000); % Two periods of the sinusoid

% Energy function

E = 0.066 * cos(750*t).²; % Energy stored in the capacitor

% Plotting the energy

plot(t, E);

xlabel('Time');

ylabel('Energy (mJ)');

title('Energy Stored in the Capacitor');

grid on;

This code generates a plot of the energy stored in the capacitor over time, assuming a capacitance of 4.7 mF and a current source with is(t) = 53*sin(750t) mA. The time range is set to cover two periods of the sinusoid, and the energy values are calculated using the expression E(t) = 0.066 * cos²(750t).

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What wavelength photon would be required to ionize a hydrogen atom in the ground state and give the ejected electron a kinetic energy of 14.3 eV ?
Express your answer to three significant figures and include the appropriate units.
?

Answers

The wavelength of the required photon to ionize a hydrogen atom and give the ejected electron a kinetic energy of 14.3 eV is approximately 8.66 x 10^-7 meters.

To determine the wavelength of the required photon to ionize a hydrogen atom and give the ejected electron a kinetic energy of 14.3 eV, we can use the equation:

E = hc/λ

where E is the energy of the photon, h is the Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the photon.

First, we need to convert the kinetic energy of the electron from electron volts (eV) to joules (J). 1 eV is equal to 1.602 x 10^-19 J.

14.3 eV * (1.602 x 10^-19 J/eV) = 2.29 x 10^-18 J

Next, we can rearrange the equation to solve for wavelength:

λ = hc/E

λ = (6.626 x 10^-34 J·s * 2.998 x 10^8 m/s) / (2.29 x 10^-18 J)

Calculating the wavelength:

λ ≈ 8.66 x 10^-7 meters

Therefore, the wavelength of the required photon to ionize a hydrogen atom and give the ejected electron a kinetic energy of 14.3 eV is approximately 8.66 x 10^-7 meters.

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