A hot wire radiates heat at 100 Watts. If its temperature measured in degrees Kelvin is doubled then the power radiated wit be what? Select one: 1. Draw a free body diagram of a hanging mass before it is submerged in water. Make sure to label your forces.

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Answer 1

If the temperature of a hot wire measured in degrees Kelvin is doubled, the power radiated will increase by a factor of 16.

The power radiated by a hot wire is given by the Stefan-Boltzmann law:

P = σ * A * ε * T^4

where P is the power radiated, σ is the Stefan-Boltzmann constant, A is the surface area of the wire, ε is the emissivity (a measure of how effectively the wire radiates heat), and T is the temperature in Kelvin.

If the temperature T is doubled, the power radiated P' can be calculated by substituting 2T for T:

P' = σ * A * ε * (2T)^4 = σ * A * ε * 16T^4

Comparing P' to the original power P, we find that P' is 16 times greater than P:

P' = 16P

Therefore, if the temperature of the hot wire is doubled (measured in degrees Kelvin), the power radiated by the wire will increase by a factor of 16.

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Related Questions

A source emitting a sound at 300.0 Hz is moving towards a stationary observer at 25 m/s. The air temperature is 15°C. What is the frequency detected by the observer?

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The frequency detected by the observer is approximately 314.6 Hz.

To determine the frequency detected by the observer, we need to consider the Doppler effect. The formula for the observed frequency (f') in terms of the source frequency (f) and the relative velocity between the source and observer (v) is given by:

f' = f * (v + v₀) / (v + vs)

Where:

f' is the observed frequency

f is the source frequency

v is the speed of sound in air

v₀ is the velocity of the observer

vs is the velocity of the source

First, let's calculate the speed of sound in air at 15°C. The formula for the speed of sound in air is:

v = 331.4 + 0.6 * T

Where:

v is the speed of sound in m/s

T is the temperature in Celsius

Plugging in T = 15°C, we have:

v = 331.4 + 0.6 * 15

v ≈ 340.4 m/s

Now, we can calculate the observed frequency:

f' = 300.0 * (v + v₀) / (v + vs)

f' = 300.0 * (340.4 + 0) / (340.4 + (-25))

f' ≈ 314.6 Hz

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Consider an electron bound in a hydrogen atom under the influence of a homogeneous magnetic field B= z
^
B. Ignore the electron spin. The Hamiltonian of the system is H=H 0

−ωL z

with ω≡∣e∣B/2m e

c. The eigenstates ∣nℓm⟩ and eigenvalues E n
(0)

of the unperturbed hydrogen atom Hamiltonian H 0

are to be considered as known. Assume that initially (at t=0 ) the system is in the state ∣ψ(0)⟩= 2

1

(∣21−1⟩−∣211⟩) Calculate the expectation value of the magnetic dipole moment associated with the orbital angular momentum at time t.

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When a homogeneous magnetic field is applied to a hydrogen atom with an electron in the ground state, the energy levels of the electron will split into multiple sublevels. This phenomenon is known as Zeeman splitting.

In the absence of a magnetic field, the electron in the ground state occupies a single energy level. However, when the magnetic field is introduced, the electron's energy levels will split into different sublevels based on the interaction between the magnetic field and the electron's spin and orbital angular momentum.

The number of sublevels and their specific energies depend on the strength of the magnetic field and the quantum numbers associated with the electron. The splitting of the energy levels is observed due to the interaction between the magnetic field and the magnetic moment of the electron.

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--The complete Question is, Consider an electron bound in a hydrogen atom under the influence of a homogeneous magnetic field B = z. If the electron is initially in the ground state, what will happen to its energy levels when the magnetic field is applied?--

What are the expected readings of the following in the figure below? (R=9.100,ΔV=5.40 V) (i) (a) ideal ammeter (Give your answer in mA ) D ma (b) ideal voltmeter (Give your answer in volts.) (c) What Ir? How would the readings in the ammeter (in mA) and voltmeter (in volts) change if the 4.50 V. battery was filpped so that its positive rerminal was to the right? ideal ammeter A mA स V ideal voltmeter

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Similarly, the voltage measured by the voltmeter also changes sign, i.e, from 5.40V to -5.40V.

(i) (a) Ideal ammeter reading:Ammeter is connected in series with the circuit. It has very low resistance hence it can measure the current flowing through it. The ideal ammeter will have zero internal resistance and will not affect the circuit under test.

Ideal ammeter reading can be obtained using Ohm's law.i.e, V=IRWhere V= voltage, I=current and R=resistanceHere, V=5.40 V and R=9.100I=V/RI= 5.40/9.100 = 0.593 mATherefore, Ideal ammeter reading is 0.593 mA.

(b) Ideal voltmeter reading:Voltmeter is connected in parallel with the circuit. It has very high resistance hence it does not affect the circuit under test. The ideal voltmeter will have infinite internal resistance and will not allow the current to flow through it.

Ideal voltmeter reading is equal to the applied voltage. Here, the applied voltage is 5.40VTherefore, Ideal voltmeter reading is 5.40V.(c) Ir represents the current flowing through the resistor.

Using Ohm's law, we can calculate the value of current flowing through the resistor. V=IRTherefore, IR = V/RIR = 5.40/9.100IR = 0.593 mAIf the 4.50V battery is flipped,

the direction of the current flowing in the circuit gets reversed. Hence, the current measured by the ammeter gets reversed, i.e, from 0.593 mA to -0.593 mA. Similarly, the voltage measured by the voltmeter also changes sign, i.e, from 5.40V to -5.40V.

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A typical wall outlet in a place of residence in North America is RATED 120V, 60Hz. Knowing that the voltage is a sinusoidal waveform, calculate its: a. PERIOD b. PEAK VOLTAGE Sketch: c. one cycle of this waveform (using appropriate x-y axes: show the period on the y-axis and the peak voltage on the x-axis)

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The typical wall outlet in North America has a rated voltage of 120V and operates at a frequency of 60Hz. The period of the voltage waveform is 1/60 seconds, and the peak voltage is ±170V.

The frequency of the voltage waveform represents the number of complete cycles per second, which is given as 60Hz. The period of the waveform can be calculated by taking the reciprocal of the frequency: 1/60 seconds. This means that the waveform completes one cycle every 1/60 seconds.

The peak voltage refers to the maximum voltage value reached by the waveform. In this case, the rated voltage is 120V, which represents the RMS voltage. Since the waveform is sinusoidal, the peak voltage can be both positive and negative. The [tex]V_{peak} = \sqrt{2} V_{RMS} = \sqrt{2} * 120 V = 170V[/tex]. Therefore, the peak voltage is ±170V, indicating that the voltage swings from positive 170V to negative 170V during each cycle.

The cycle of wave form is given below.

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How do you get the mass of a star or planet? Kepler's third law Kepler's second law Kepler's first law

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To determine the mass of a star or planet, Kepler's third law is used. Kepler's third law states that the square of the orbital period of a planet or satellite is directly proportional to the cube of the semi-major axis of its orbit.

Kepler's third law provides a relationship between the mass of a star or planet and the orbital parameters of its satellites or planets. The law states that the square of the orbital period (T) is directly proportional to the cube of the semi-major axis (a) of the orbit. Mathematically, it can be expressed as T^2 ∝ a^3.

By measuring the orbital period and the semi-major axis of a planet or satellite, we can determine the mass of the star or planet using Kepler's third law. This is possible because the mass of the star or planet affects the gravitational force acting on the orbiting body, which in turn influences its orbital period and semi-major axis.

By observing the motion of satellites or planets around a star or planet and applying Kepler's third law, astronomers can estimate the mass of celestial objects in the universe, providing valuable information for understanding their properties and dynamics.

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Consider a point on a bicycle tire that is momentarily in contact with the ground as the bicycle rolls across the ground with constant speed. The direction for the acceleration for this point at that moment is: a. upward. b. down toward the ground. c. forward, with the direction of the bicycle's movement. d. at that moment the acceleration is zero. e. backward, against the direction of the bicycle's movement.

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So the correct option is d. At that moment, the acceleration of the point on the bicycle tire is zero. Since the bicycle is rolling with constant speed and there is no change in its motion, the point in contact with the ground.

In physics, moment refers to a turning effect or rotational force produced by a force acting on an object. It is the product of the magnitude of the force and the perpendicular distance between the line of action of the force and the pivot point or axis of rotation. Moments are measured in units of newton-meters (Nm) or foot-pounds (ft-lb) and are essential in studying rotational motion, equilibrium, and the principles of torque and angular momentum.

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A speed skater moving across frictionless ice at 8.0 m/s hits a 6.0 m -wide patch of rough ice. She slows steadily, then continues on at 6.1 m/s . Part A What is her acceleration on the rough ice? Express your answer in meters per second squared. a = m/s2

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The problem requires us to calculate the acceleration of a speed skater when she moves across a frictionless ice and hits a 6.0 m-wide patch of rough ice.

The initial velocity (u) of the speed skater = 8.0 m/s

The final velocity (v) of the speed skater = 6.1 m/s

The distance covered (s) by the speed skater = 6.0 m

The formula used here is given below:

v² = u² + 2as

where,v = final velocity

u = initial velocity

a = acceleration

and s = distance covered.

a = (v² - u²) / 2s

= (6.1² - 8.0²) / 2(6.0)a

= -2.48 m/s² [Negative sign shows the speed skater is decelerating]

Hence, the acceleration of the speed skater on the rough ice is -2.48 m/s² (rounded to two decimal places).

Note: The distance covered by the speed skater is 6.0 m only. The distance is not a factor here as the acceleration of the skater is concerned.

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A ball is dropped from rest at a height of 81 meters. What's the magnitude of the velocity of the ball as it hits the ground? (Your answer should be in units of meters per second (m/s), but just write down the number part of your answer.)

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The magnitude of the velocity of the ball as it hits the ground can be determined using the principles of motion and the equation for the velocity of a falling object. When an object falls freely under the influence of gravity, neglecting air resistance, it undergoes constant acceleration due to gravity, denoted as "g."

The value of acceleration due to gravity on Earth is approximately 9.8 m/s². To calculate the magnitude of the velocity of the ball as it hits the ground, we can use the equation:

v = [tex]\sqrt(2gh)[/tex]

where v represents the velocity, g is the acceleration due to gravity, and h is the initial height from which the ball is dropped.

In this case, the initial height (h) is given as 81 meters. By substituting this value into the equation, we can calculate the magnitude of the velocity.

The equation v = [tex]\sqrt(2gh)[/tex] represents the relationship between the velocity of a falling object and the height from which it is dropped. This equation is derived from the principles of motion and can be applied to objects falling freely under the influence of gravity.

When the ball is dropped from rest, it begins to accelerate due to gravity. As it falls, its velocity increases until it reaches the ground. The magnitude of the velocity at the moment it hits the ground is what we are interested in calculating.

By substituting the given values into the equation, we can find the magnitude of the velocity. The initial height (h) is 81 meters, and the acceleration due to gravity (g) is approximately 9.8 m/s² on Earth. Plugging these values into the equation, we can solve for the magnitude of the velocity.

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A boat whose velocity through the water is 14 km/h is moving in a river whose current is 6 km/in relative to the riverbed. The velocity of the boat relative to the riverbed must be between O 6 and 14 km/h 6 and 20 km/h and 14 km/h 8 and 20 km/h

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A boat whose velocity through the water is 14 km/h is moving in a river whose current is 6 km/h.

To determine the velocity of the boat relative to the riverbed, we need to calculate the resultant velocity of the boat. The velocity of the boat relative to the riverbed must be between 8 km/h and 20 km/h.Resolution of the velocities can be used to determine the resultant velocity. It refers to the separation of a vector quantity into two or more components. By definition, these components are scalar components.

A velocity vector's resolution into two perpendicular components is known as a rectangular resolution.

Let’s find the resultant velocity by using the formula of the Pythagorean theorem.

Velocity of the boat relative to the riverbed = Velocity of the boat in still water + velocity of the rivercurrent

= 14 km/h + 6 km/h= 20 km/h

Using the Pythagorean theorem, the resultant velocity is determined as follows:

Resolving the velocity in the x and y directions:

Velocity in the x-direction (upstream) = V × cos θ= 20 × cos 30°

= 17.32 km/h

Velocity in the y-direction (downstream) = V × sin θ= 20 × sin 30°= 10 km/h

Therefore, the boat's velocity relative to the riverbed is between 8 km/h and 20 km/h.

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Shaving/makeup mirrors typically have one flat and one concave (magnifying) surface. You find that you can project a magnified image of a lightbulb onto the wall of your bathroom if you hold the mirror 1.8 m from the bulb and 3.5 m from the wall. (a) What is the magnification of the image? (b) Is the image erect or inverted? (c) What is the focal length of the mirror?

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The magnification is approximately -1.944, indicating an inverted image. The focal length of the mirror is approximately 1.189 meters. To determine the magnification of the image formed by the magnifying mirror, we can use the mirror equation:

1/f = 1/d₀ + 1/dᵢ,

where f is the focal length of the mirror, d₀ is the object distance (distance from the bulb to the mirror), and dᵢ is the image distance (distance from the mirror to the wall).

(a) Magnification (m) is given by the ratio of the image distance to the object distance:

m = -dᵢ/d₀,

where the negative sign indicates an inverted image.

(b) The sign of the magnification tells us whether the image is erect or inverted. If the magnification is positive, the image is erect; if it is negative, the image is inverted.

(c) To find the focal length of the mirror, we can rearrange the mirror equation 1/f = 1/d₀ + 1/dᵢ, and solve for f.

d₀ = 1.8 m (object distance)

dᵢ = 3.5 m (image distance)

(a) Magnification:

m = -dᵢ/d₀ = -(3.5 m)/(1.8 m) ≈ -1.944

The magnification is approximately -1.944, indicating an inverted image.

(b)  The image is inverted.

(c) Focal length:

1/f = 1/d₀ + 1/dᵢ = 1/1.8 m + 1/3.5 m ≈ 0.5556 + 0.2857 ≈ 0.8413

Now, solving for f:

f = 1/(0.8413) ≈ 1.189 m

The focal length of the mirror is approximately 1.189 meters.

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Select the correct answer.
In which item is energy stored in the form of gravitational potential energy?
A.
a slice of bread
B.
a compressed spring
C.
an apple on a tree
D.
a stretched bow string
Reset Next

Answers

C. an apple on a tree as energy stored in the form of gravitational potential energy.

Gravitational potential energy is a form of energy that an object possesses due to its position in a gravitational field.

It is directly related to the height or vertical position of the object relative to a reference point.

Out of the given options, only the apple on a tree possesses gravitational potential energy because it is located above the ground.

As the apple is raised higher on the tree, its gravitational potential energy increases accordingly.

Thus, option C, "an apple on a tree," is the correct choice.

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A wheel with radius 37.9 cm rotates 5.77 times every second. Find the period of this motion. period: What is the tangential speed of a wad of chewing gum stuck to the rim of the wheel? tangential speed: m/s A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 26.9 m/s 2
with a beam of length 5.69 m, what rotation frequency is required? A electric model train travels at 0.317 m/s around a circular track of radius 1.79 m. How many revolutions does it perform per second (i.e, what is the motion's frequency)? frequency: Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.17 times a second. A tack is stuck in the tire at a distance of 0.351 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed. tangential speed: m/s What is the tack's centripetal acceleration? centripetal acceleration: m/s 2

Answers

Therefore, the tack's centripetal acceleration is approximately 65.2 m/s².

The given radius of a wheel is r = 37.9 cm, and it rotates 5.77 times every second. Let's find the period of this motion. The period is defined as the time taken by an object to complete one full cycle. It can be calculated using the formula: T = 1/f. where T is the period and f is the frequency. The frequency is given by: f = 5.77 rotations/sec. We can plug in the value of frequency in the above equation to get the period: T = 1/5.77 ≈ 0.173 seconds Now, let's find the tangential speed of a wad of chewing gum stuck to the rim of the wheel. The tangential speed is defined as the linear speed of an object moving along a circular path and can be calculated using the formula: v = rw where v is the tangential speed, r is the radius, and w is the angular velocity. The angular velocity can be calculated as follows: w = 2πf.

where f is the frequency. We can plug in the value of f in the above equation to get:w = 2π × 5.77 ≈ 36.24 rad/s. Now, let's plug in the values of r and w in the formula to get the tangential speed: v = rw = 37.9 × 36.24 ≈ 1374.08 cm/s = 13.74 m/s. Therefore, the tangential speed of a wad of chewing gum stuck to the rim of the wheel is approximately 13.74 m/s. Now let's find the rotation frequency that is required to achieve a radial acceleration of 26.9 m/s² with a beam of length 5.69 m. The radial acceleration is given by: a = w²rwhere w is the angular velocity and r is the radius. In this case, the radius is equal to the length of the beam, so:cr = 5.69 mWe want the radial acceleration to be 26.9 m/s², so we can plug in these values in the above formula to get:26.9 = w² × 5.69Now, let's solve for w:w² = 26.9/5.69 ≈ 4.72w ≈ 2.17 rad/s, The rotation frequency is equal to the angular velocity divided by 2π, so we can find it as follows: f = w/2π = 2.17/2π ≈ 0.345 Hz.n Therefore, the rotation frequency required to achieve a radial acceleration of 26.9 m/s² with a beam of length 5.69 m is approximately 0.345 Hz. Let's find the number of revolutions the electric model train performs per second. The speed of the train is v = 0.317 m/s, and the radius of the circular track is r = 1.79 m. The frequency is defined as the number of cycles per second, and in this case, each cycle is one full rotation around the circular track. Therefore, the frequency is equal to the number of rotations per second. The tangential speed is given by:v = rwwhere w is the angular velocity. We can rearrange this equation to get:w = v/rNow, let's plug in the values of v and r to get:w = 0.317/1.79 ≈ 0.177 rad/sThe frequency is given by:f = w/2π = 0.177/2π ≈ 0.0281 HzThe number of revolutions per second is equal to the frequency, so the train performs approximately 0.0281 revolutions per second. Finally, let's find the tack's tangential speed and centripetal acceleration. The distance between the tack and the axis of rotation is d = 0.351 m. The tangential speed is equal to the linear speed of a point on the tire at the distance d from the axis of rotation. We can find it as follows:v = rwwhere r is the radius and w is the angular velocity. The radius is equal to the distance between the tack and the axis of rotation, so:r = dNow, let's find the angular velocity. One rotation is equal to one circumference, which is equal to 2π times the radius of the tire. Therefore, the angular velocity is:w = 2πfwhere f is the frequency. We can find the frequency as follows:f = 2.17 rotations/secondThe angular velocity is:w = 2π × 2.17 ≈ 13.65 rad/sNow, let's plug in the values of r and w in the formula to get the tangential speed:v = rw = 0.351 × 13.65 ≈ 4.79 m/sTherefore, the tack's tangential speed is approximately 4.79 m/s. The centripetal acceleration is given by:a = v²/rwhere v is the tangential speed and r is the radius.We can plug in the values of v and r to get:a = v²/r = (4.79)²/0.351 ≈ 65.2 m/s². Therefore, the tack's centripetal acceleration is approximately 65.2 m/s².

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Radon-222 is a colorless and odorless gas that is radioactive, undergoing alpha-decay with a half-life of 3.8 days. What atom remains after this process? O Carbon-12 O Radium-226 O Polonium-218 O Uranium-238 O Radon-222

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Radon-222 is a radioactive, odorless and colorless gas. After undergoing alpha-decay with a half-life of 3.8 days, the atom that remains is Polonium-218.

What is radioactive? Radioactivity is the phenomenon of unstable atomic nuclei splitting or decaying spontaneously. These radioactive materials, also known as radioisotopes, are utilized in numerous applications, such as scientific study, nuclear power generation, and medical therapy. The radionuclide Radon-222 undergoes alpha decay with a half-life of 3.8 days. What happens after the alpha decay of Radon-222?Alpha decay is a type of radioactive decay that occurs when an atomic nucleus loses an alpha particle, a helium nucleus that contains two protons and two neutrons. Radon-222 emits an alpha particle and produces a new nucleus of Polonium-218 with a mass number of 218 (two less than that of the parent nucleus Radon-222). Therefore, after this process, the atom that remains is Polonium-218.

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(a) An amplitude modulated signal is given by the below equation: VAM (t) = 0.1[1 + 0.5 cos 6280t]. Sin [107t + 45°] V From the given information plot the frequency spectrum of the AM modulated signal. [7 marks] (b) The expression shown in the below equation describes the Frequency Modulated (FM) signal wave as a function of time: VFM (t) = 15 cos[2π(150 x 10³ t) + 5 cos (6 × 10³ nt)] V The carrier frequency is 150 KHz and modulating signal frequency is 3 KHz. The FM signal is coupled across a 10 2 load. Using the parameters provided, calculate maximum and minimum frequencies, modulation index and FM power that appears across the load: [12 marks] (c) Show the derivation that the general Amplitude Modulation (AM) equation has three frequencies generated from the signals below: Carrier signal, vc = Vc sinwet Message signal, um = Vm sin wmt

Answers

a) The frequency spectrum of the given AM modulated signal has the carrier frequency 6280 rad/s, upper sideband frequency 6387 rad/s, and lower sideband frequency 6173 rad/s.

b) The maximum and minimum frequencies are 150.0095 KHz and 149.9905 KHz respectively. FM power that appears across the load: 3.042 mW

c) general AM signal equation: Vm(t) = [A[tex]_{c}[/tex] cosω[tex]_{c}[/tex]t + (A[tex]_{m}[/tex]/2) cos(ω[tex]_{c}[/tex] + ω[tex]_{m}[/tex])t + (A[tex]_{m}[/tex]/2) cos(ω[tex]_{c}[/tex] - ω[tex]_{m}[/tex])t]

(a)Frequency spectrum of the AM modulated signal:

Given,

VAM (t) = 0.1[1 + 0.5 cos 6280t]. Sin [107t + 45°] V

The general form of the AM signal is given by:

Vm(t) = [A[tex]_{c}[/tex] + A[tex]_{m}[/tex] cosω[tex]_{m}[/tex]t] cosω[tex]_{c}[/tex]t

Let's compare the given signal and general form of the AM signal,

VAM (t) = 0.1[1 + 0.5 cos 6280t]. Sin [107t + 45°] V

Vm(t) = (0.5 x 0.1) cos (6280t) cos (107t + 45°)

Amplitude of carrier wave,

Ac = 0.1

Frequency of carrier wave,

ω[tex]_{c}[/tex] = 6280 rad/s

Amplitude of message signal,

A[tex]_{m}[/tex] = 0.05

Frequency of message signal,

ω[tex]_{m}[/tex] = 107 rad/s

Let's calculate the upper sideband frequency,

ω[tex]_{us}[/tex] = ω[tex]_{c}[/tex] + ω[tex]_{m}[/tex]= 6280 + 107 = 6387 rad/s

Let's calculate the lower sideband frequency,

ω[tex]_{ls}[/tex] = ω[tex]_{c}[/tex] - ω[tex]_{m}[/tex]= 6280 - 107 = 6173 rad/s

Hence, the frequency spectrum of the given AM modulated signal has the carrier frequency 6280 rad/s, upper sideband frequency 6387 rad/s, and lower sideband frequency 6173 rad/s.

(b) Calculation of maximum and minimum frequencies, modulation index, and FM power:

Given,

Carrier frequency, f[tex]_{c}[/tex] = 150 KHz

Modulating signal frequency, f[tex]_{m}[/tex] = 3 KHz

Coupling resistance, RL = 102 Ω

The general expression of FM signal is given by:

VFM (t) = A[tex]_{c}[/tex] cos[ω[tex]_{c}[/tex]t + β sin(ω[tex]_{m}[/tex]t)]

Where, A[tex]_{c}[/tex] is the amplitude of the carrier wave ω[tex]c[/tex] is the carrier angular frequency

β is the modulation index

β = (Δf / f[tex]m[/tex])Where, Δf is the frequency deviation

Maximum frequency, f[tex]max[/tex] = f[tex]m[/tex]+ Δf

Minimum frequency, f[tex]min[/tex] = f[tex]_{c}[/tex] - Δf

Maximum phase deviation, φ[tex]max[/tex] = βf[tex]m[/tex]2π

Minimum phase deviation, φ[tex]min[/tex] = - βf[tex]m[/tex]2π

Let's calculate the modulation index, β = Δf / f[tex]m[/tex]= (f[tex]max[/tex] - f[tex]min[/tex]) / f[tex]m[/tex]= (150 + 7.5 - 150 + 7.5) / 3= 5/6000= 1/1200

Let's calculate the maximum and minimum frequencies, and FM power.

The value of maximum phase deviation, φ[tex]max[/tex] = βf[tex]m[/tex]2π= (1/1200) x 6 x 103 x 2π= π/1000

The value of minimum phase deviation, φ[tex]min[/tex] = - βf[tex]m[/tex]2π= -(1/1200) x 6 x 103 x 2π= -π/1000

Let's calculate the maximum frequency,

f[tex]max[/tex] = f[tex]c[/tex] + Δf= f[tex]c[/tex] + f[tex]m[/tex] φ[tex]max[/tex] / 2π= 150 x 103 + (3 x 103 x π / 1000)= 150.0095 KHz

Let's calculate the minimum frequency,

f[tex]min[/tex] = f[tex]c[/tex]- Δf= f[tex]c[/tex] - f[tex]m[/tex]

φ[tex]max[/tex] / 2π= 150 x 103 - (3 x 103 x π / 1000)= 149.9905 KHz

Hence, the maximum and minimum frequencies are 150.0095 KHz and 149.9905 KHz respectively.

Let's calculate the FM power,

[tex]PFM = (Vm^{2} / 2) (R_{L} / (R_{L} + Rs))^2[/tex]

Where, V[tex]m[/tex] = Ac β f[tex]m[/tex]R[tex]_{L}[/tex] is the load resistance

R[tex]s[/tex] is the internal resistance of the source

PFM = (0.5 x Ac² x β² x f[tex]m[/tex]² x R[tex]_{L}[/tex]) (R[tex]_{L}[/tex] / (R[tex]_{L}[/tex] + R[tex]s[/tex]))^2

PFM = (0.5 x 15² x (1/1200)² x (3 x 10³)² x 102) (102 / (102 + 10))²

PFM = 0.003042 W = 3.042 m W

(c) Derivation of general AM signal equation:

The equation of a general AM wave is,

V m(t) = [A[tex]c[/tex] + A[tex]m[/tex] cosω[tex]m[/tex]t] cosω[tex]c[/tex]t

Where, V m(t) = instantaneous value of the modulated signal

A[tex]c[/tex] = amplitude of the carrier wave

A[tex]m[/tex] = amplitude of the message signal

ω[tex]c[/tex] = angular frequency of the carrier wave

ω[tex]m[/tex] = angular frequency of the message signal

Let's find the frequency components of the general AM wave using trigonometric identities.

cosα cosβ = (1/2) [cos(α + β) + cos(α - β)]

cosα sinβ = (1/2) [sin(α + β) - sin(α - β)]

sinα cosβ = (1/2) [sin(α + β) + sin(α - β)]

sinα sinβ = (1/2) [cos(α - β) - cos(α + β)]

Vm(t) = [Ac cosω[tex]_{c}[/tex]t + (A[tex]m[/tex]/2) cos(ω[tex]_{c}[/tex]+ ω[tex]m[/tex])t + (A[tex]m[/tex]/2) cos(ω[tex]_{c}[/tex] - ω[tex]m[/tex])t]

From the above equation, it is clear that the modulated signal consists of three frequencies,

Carrier wave frequency ω[tex]_{c}[/tex]

Lower sideband frequency (ω[tex]_{c}[/tex]- ω[tex]m[/tex])

Upper sideband frequency (ω[tex]_{c}[/tex] + ω[tex]m[/tex])

Hence, this is the derivation of the general AM signal equation which shows the generation of three frequencies from the carrier and message signals.

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Grant jumps 170 m straight up into the air to slam-dunk a basketball into the net. With what speed did he leave the floor?

Answers

The speed with which Grant left the floor was 57.7 m/s.

When Grant jumps 170m into the air to slam-dunk a basketball into the net, the speed with which he leaves the floor can be found out by using the conservation of mechanical energy, which is represented by the formula: 1/2 mvi2 + mghi = 1/2 mvf2 + mghf Here, m represents mass, vi represents the initial velocity, vf represents the final velocity, hi represents the initial height, and hf represents the final height. We can consider the initial height to be zero, so h i = 0 m. The final height will be 170 m (as he jumps 170 m high). Hence, h f = 170 m. The initial velocity can be assumed to be zero as the basketball player was on the ground before he jumped. Therefore, vi = 0 m/s. Substituting the values in the formula, we get: 1/2 mvf2 + mghf = 0 + mghf + m × g × 170 vf2 = 2 × g × hf= 2 × 9.8 × 170 vf2 = 3332vf = √3332 = 57.7 m/s. Therefore, the speed with which Grant left the floor was 57.7 m/s.

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What is the magnetic field at the center of a single (N=1 turn) circular loop of wire or radius 10 cm carrying a current of 2.5 A ? 2.41×10 −4
T 5.0×10 −6
T 1.57×10 −7
T 3.14×10 −5
T

Answers

The magnetic field at the center of a single circular loop of wire or radius 10 cm carrying a current of 2.5 A is             3.14 × 10-5 T.

Magnetic field at the center of a single circular loop of wire or radius 10 cm carrying a current of 2.5 A can be calculated using the formula;

B=μ0I/2R

where B is the magnetic field, I is the current flowing, R is the radius of the loop and μ0 is the permeability of free space.The given values are;I = 2.5 AR = 10 cm = 0.1 mμ0 = 4π × 10-7 T m/A.

Substitute the values into the formula; B = μ0I/2R = (4π × 10-7 T m/A) × (2.5 A)/2(0.1 m)= 3.14 × 10-5 T

Therefore, the magnetic field at the center of a single circular loop of wire or radius 10 cm carrying a current of 2.5 A is 3.14 × 10-5 T.

Answer: 3.14×10^−5T.

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A 2.32cm-tall object is placed 5.2 cm in front of a convex mirror with radius of curvafure 21 cm. Part (a) What is the image distance, in centimeters? Include its sign. s’ = ___________
Hints: 0% deduction per hint. Hints remaining : 2 Feedback: 0% deduction per feedback
Part (b) What is the image height, in centimeters? Include its sign.
Part (c) What is the orientation of the image relative to the object?

Answers

The image distance is + 2.00 cm and height is - 0.88 cm, inverted image.

Part (a)

Image distance, s′ = ?

We have the object distance (u) = - 5.2 cm

Radius of curvature (R) = + 21 cm (because it is a convex mirror)

We know that the mirror formula is given by:

1/f = 1/v + 1/u

where

f is the focal length of the mirror.

Putting the values of u and R, we get:

1/f = 1/v + 1/R

Since we are not given the focal length, we cannot use the above formula. However, we can use the mirror formula to calculate the image distance which is given as:

s′ = (f * u)/(u + f)s′ = - (R * u)/(u - R) [we know that for a convex mirror, the focal length is negative]

s′ = - (21 * (- 5.2))/(−5.2 − 21)s′ = 2.00 cm

Therefore, the image distance, s′ = + 2.00 cm (since the image is formed on the same side of the mirror as the object, the image distance is positive).

Part (b)

Image height, h′ = ?

The magnification of the image is given by:

- v/u,

where

v is the image distance.

Since the magnification is negative, the image is inverted with respect to the object.

Magnification, m = - v/u = h'/h

where

h' is the image height  

h is the object height

Substituting the values, we get:

m = - v/u = h'/h

2.32/h = - 2.00/(- 5.2)

h' = 0.88 cm

The image height, h′ = - 0.88 cm (because the image is inverted)

Part (c)

Orientation of the image relative to the object:

The magnification is negative, which implies that the image is inverted relative to the object.

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Vector A points in the negative z direction. Vector points at an angle of 31.0" above the positive z axis. Vector C has a magnitude of 16 m and points in a direction 42.0* below the positive x axis. Part B Express your answer using two significant figures. |B|= ________ m

Answers

Vector A points in the negative z direction. Vector points at an angle of 31.0" above the positive z axis. Vector C has a magnitude of 16 m and points in a direction 42.0* below the positive x axis.

Vector A, A = {0, 0, -a}

Vector C, C = {16 cos 42.0°, 0, - 16 sin 42.0°}

Let B = A + B + C. Hence, B = {0, 0, -a} + {B sin 31.0° cos θ, B sin 31.0° sin θ, B cos 31.0°} + {16 cos 42.0°, 0, - 16 sin 42.0°}

Then, equating the x, y, and z components of the above equation separately, we get:

B sin 31.0° cos θ = - 16 cos 42.0°B sin 31.0° sin θ = 0

B cos 31.0° = a - 16 sin 42.0°

From the second equation, we have B = 0 or sin θ = 0, we have B = 0. But, B = 0 doesn't satisfy the third equation. Hence, sin θ = 0. So, θ = 0° or θ = 180°.When θ = 0°, we get,

B sin 31.0° cos θ = - 16 cos 42.0°B sin 31.0° (1) = - 16 cos 42.0°

B = - 16 cos 42.0° / sin 31.0°

Then, |B| = 22 m (approx.)

So, the required value of |B| is 22 m (approx.)

Note: You can also solve it by using the dot product of the vectors.

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A home run is hit such a way that the baseball just clears a wall 24 m high located 135 m from home plate. The ball is hit at an angle of 38° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 2 m above the ground. The acceleration of gravity is 9.8 m/s2. What is the initial speed of the ball? Answer in units of m/s. Answer in units of m/s

Answers

The initial speed of the ball that is hit at an angle of 38° to the horizontal and air resistance is negligible found to be approximately 41.1 m/s.

To find the initial speed of the baseball, which just clears a 24 m high wall located 135 m from home plate, we can use the kinematic equations and consider the projectile motion of the ball.

In projectile motion, the vertical and horizontal components of motion are independent of each other. The vertical motion is influenced by gravity, while the horizontal motion remains constant.

Given that the ball just clears a 24 m high wall, we can use the vertical motion equation: h = v₀²sin²θ / (2g), where h is the height, v₀ is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity.

Plugging in the values, we have 24 = v₀²sin²38° / (2 * 9.8). Solving for v₀, we find v₀ ≈ 41.1 m/s.

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You are given a black box circuit and you are to apply an input vi(t)=3u(t)V. The voltage response can be described by vo(t)=(5e−8t−2e−5t)V for t≥0. What will be the steady-state response of the circuit if you apply another input voltage described by vi(t)=100cos6t V for t≥0 ?

Answers

The steady-state response of the circuit to the input voltage vi(t) = 100cos(6t) V is given by vo(t) = 100*cos(6t + φ) V

To determine the steady-state response of the circuit to the input voltage described by vi(t) = 100cos(6t) V, we need to find the response after transient effects have settled. The given voltage response vo(t) = 5e^(-8t) - 2e^(-5t) V is the transient response for the previous input.

To find the steady-state response, we need to find the particular solution that corresponds to the new input. Since the input is a sinusoidal signal, we assume the steady-state response will also be sinusoidal with the same frequency.

1. Find the steady-state response of the circuit for the new input voltage:

We assume the steady-state response will be of the form vp(t) = A*cos(6t + φ), where A is the amplitude and φ is the phase angle to be determined.

2. Apply the new input voltage to the circuit:

vi(t) = 100cos(6t) V

3. Find the output voltage in the steady-state:

vo(t) = vp(t)

4. Substitute the input and output voltages into the equation:

100cos(6t) = A*cos(6t + φ)

5. Compare the coefficients of the same terms on both sides of the equation:

100 = A  (since the cos(6t) terms are equal)

6. Solve for the amplitude A:

A = 100

7. The steady-state response of the circuit for the new input voltage is:

vo(t) = 100*cos(6t + φ) V

Therefore, the steady-state response of the circuit to the input voltage vi(t) = 100cos(6t) V is given by vo(t) = 100*cos(6t + φ) V, where φ is the phase angle that depends on the initial conditions of the circuit.

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Electrons in an x-ray machine are accelerated from rest through a potential difference of 60 000 V. What is the kinetic energy of each of these electrons in eV?
60 eV
96 eV
38 eV
60 keV
​120 eV

Answers

Electrons in an x-ray machine are accelerated from rest through a potential difference of 60 000 V. Therefore, the kinetic energy of each of these electrons is 60 keV.

Given ,Potential difference, V = 60,000 V. The energy of an electron, E = potential difference x charge of an electron (e)

The charge of an electron is e = 1.6 × 10⁻¹⁹CThe kinetic energy of an electron is calculated by using the formula, Kinetic energy = energy of an electron - energy required to remove an electron from an atom = E - ϕ where, ϕ is work function, which is the energy required to remove an electron from an atom.

This can be expressed as, Kinetic energy of an electron = eV - ϕ Now, let's find the energy of an electron.

Energy of an electron, E = potential difference x charge of an electron (e)= 60,000 V × 1.6 × 10⁻¹⁹C = 9.6 × 10⁻¹⁵ J

Now, to find the kinetic energy of each of these electrons in eV, Kinetic energy of an electron = E/e= (9.6 × 10⁻¹⁵ J) / (1.6 × 10⁻¹⁹ C) = 6 × 10⁴ eV= 60 keV

Therefore, the kinetic energy of each of these electrons in eV is 60 keV.

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Three resistors are connected in parallel. If their respective resistances are R1 = 23.0 Ω, R2 = 8.5 Ω and R3 = 31.0 Ω, then their equivalent resistance will be:
a) 5.17Ω
b) 96.97Ω
c) 0.193Ω
d) 62.5Ω

Answers

The equivalent resistance of three resistors that are connected in parallel with resistances R1 = 23.0 Ω, R2 = 8.5 Ω and R3 = 31.0 Ω is 5.17 Ω.

Therefore, the correct option is a) 5.17Ω.

How to solve for equivalent resistance?

The formula for the equivalent resistance (R) of three resistors (R1, R2, and R3) connected in parallel is given by:

1/R = 1/R1 + 1/R2 + 1/R3

Substituting the given values of R1, R2 and R3 in the above formula:

1/R = 1/23.0 + 1/8.5 + 1/31.0

Simplifying the equation by adding the fractions and then taking the reciprocal of both sides, we get:

R = 5.17 Ω

Therefore, the equivalent resistance of the three resistors connected in parallel is 5.17 Ω.

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Your brain assumes A. parallel light reflects through the focal point B. light through a focal point reflects parallel C. the angle of incidence equals the angle of reflection D. that light travels in a straight line

Answers

The correct answer is D. that light travels in a straight line. The propensity of electromagnetic waves (light) to move in a straight path is known as rectilinear propagation.

The principle that your brain assumes is known as the principle of rectilinear propagation of light. According to this principle, light travels in straight lines in a homogeneous medium unless it encounters an obstacle or undergoes a change in medium. This principle forms the basis for the behavior of light in various optical phenomena such as reflection, refraction, and image formation. When passing through a homogeneous material, which has a constant refractive index throughout, light does not deviate; otherwise, light experiences refraction. The individual rays are flowing in straight lines even if a wave front may be curved (such as the waves produced when a rock strikes a body of water). Pierre de Fermat made the discovery of rectilinear propagation.

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The burner on an electric stove has a power output of 2.0 kW. A 760 g stainless steel tea kettle is filled with 20°C water and placed on the already hot burner. If it takes 29 min for the water to reach a boil , what volume of water, in cm, was in the kettle? Stainless steel is mostly iron, so you can assume its specific heat is that of iron.

Answers

The mass of the water is 760g.

The specific heat of water is 4.18 J/gK.

To heat the water from 20 to 100°C takes 80°C.

Using Q = m x C x ΔT,

we have Q = 760 x 4.18 x 80 = 252,684 J needed to heat the water to boiling point.

The power of the stove is 2,000 W or 2,000 J/s.

Therefore the energy supplied over 29 min is 2,000 x 1,740 = 3,480,000 J.

So the volume of the water can be determined by Q = m x C x ΔT.

Rearranging, we have m = Q / C x ΔT = 3,480,000 / 4.18 x 80 = 10,486 g = 10.5 kg.

Therefore the volume of the water is V = m / ρ = 10,500 / 1 = 10,500 cm³ (since 1g = 1 cm³ for water).

Hence the volume of the water in the kettle was 10,500 cm³.

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A cannon is fired over level ground at an angle of 20 degrees to the horizontal. The initial velocity of the cannonball is 400 m/s. What are the vertical and horizontal components of the initial velocity? How long is the cannonball in the air? How far does the cannonball travel horizontally?

Answers

The vertical component of the initial velocity is 137.64 m/s, while the horizontal component is 387.88 m/s. The cannonball is in the air for approximately 81.66 seconds. It travels a horizontal distance of about 31,682.46 meters.

To determine the vertical and horizontal components of the initial velocity, we can use trigonometry. The vertical component can be calculated by multiplying the initial velocity (400 m/s) by the sine of the launch angle (20 degrees).

Thus, the vertical component is 400 m/s * sin(20 degrees) = 137.64 m/s. Similarly, the horizontal component can be found by multiplying the initial velocity by the cosine of the launch angle. Hence, the horizontal component is 400 m/s * cos(20 degrees) = 387.88 m/s.

To calculate the time the cannonball is in the air, we need to consider the vertical motion. The time of flight can be determined using the formula t = (2 * v * sinθ) / g, where v is the initial vertical velocity, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s²).

Plugging in the values, we get t = (2 * 137.64 m/s) / 9.8 m/s² = 81.66 seconds.The horizontal distance traveled can be found using the formula d = v * cosθ * t, where d is the horizontal distance, v is the initial velocity, θ is the launch angle, and t is the time of flight.

Substituting the given values, we obtain d = 387.88 m/s * cos(20 degrees) * 81.66 s = 31,682.46 meters. Therefore, the cannonball travels approximately 31,682.46 meters horizontally.

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An object is 4 cm from a converging lens with a focal length of 2.5 cm. What is the magnification, including the sign, for the image that is produced? (The sign tells if the image is inverted.) M=−1.67
M=6.67
M=−1.0
M=2.35

Answers

The magnification of the image produced by the lens is -0.38.

Magnification of an image refers to how much larger or smaller an image is than the object itself. The formula for magnification is given by;

M = -v / uwhere, M = Magnification of the imagev = Distance of the imageu = Distance of the object

To find the sign of the image, the following formula can be used:

f = Focal length of the lensIf the value of v is negative, it indicates that the image is real and inverted. If the value of v is positive, the image is virtual and erect.

A converging lens has a focal length of 2.5 cm, and the object is 4 cm away from the lens.

u = -4 cm (as the object is real) and

f = 2.5 cm (as the lens is converging)

Now, substitute the given values in the magnification formula to get the magnification.

M = -v / u

M = -(f / (f - u))

M = -(2.5 / (2.5 - (-4)))

M = -2.5 / 6.5M = -0.38

Hence, the magnification of the image produced by the lens is -0.38.

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Satellite A of mass 48.6 kg is orbiting some planet at distance 1.9 radius of planet from the surface. Satellite B of mass242.9 kg is orbiting the same planet at distance 3.4 radius of planet from the surface. What is the ratio of linear velocities of these satellites v_a/v_b?

Answers

The ratio of linear velocities of the two satellites is approximately 1.338. To find the ratio of linear velocities of the two satellites, we can use the concept of circular motion and the law of universal gravitation. The gravitational force acting on a satellite in circular orbit is given by:

F = (G * M * m) / [tex]r^2[/tex]

where F is the gravitational force, G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, and r is the distance between the satellite and the center of the planet.

In circular motion, the centripetal force required to keep the satellite in orbit is given by:

F = m * [tex](v^2 / r)[/tex]

where v is the linear velocity of the satellite.

Setting these two forces equal to each other, we can cancel out the mass of the satellite:

(G * M * m) /[tex]r^2 = m * (v^2 / r)[/tex]

Simplifying the equation, we find:

[tex]v^2[/tex] = (G * M) / r

Taking the square root of both sides gives us:

v = √[(G * M) / r]

Now, let's calculate the ratio of linear velocities[tex]v_a/v_b:[/tex]

[tex](v_a / v_b[/tex]) = [√((G * M) / [tex]r_a)[/tex]] / [√((G * M) / [tex]r_b[/tex])]

Substituting the given values:

([tex]v_a / v_b)[/tex] = [√((G * M) / (1.9 * R))] / [√((G * M) / (3.4 * R))]

Simplifying further:

([tex]v_a / v_b)[/tex] = √[(3.4 * R) / (1.9 * R)]

([tex]v_a / v_b[/tex]) = √(3.4 / 1.9)

([tex]v_a / v_b[/tex]) = √1.789

([tex]v_a / v_b[/tex]) ≈ 1.338

Therefore, the ratio of linear velocities of the two satellites is approximately 1.338.

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An electric bus operates by drawing current from two parallel overhead cables that are both at a potential difference of 380 V and are spaced 89 cm apart. The current in both cables is in the same direction. The power input (from each wire) to the bus's motor is at its maximum power of 19 kW. a. What current does the motor draw? A b. What is the magnetic force per unit length between the cables?

Answers

(a) The current that the motor draws is 100 A

(b) The magnetic force per unit length between the cables is 0.116 N/m.

The power input to the motor from each wire is maximum, i.e., P = 19 kW. Thus, the total power input to the motor is

2 × P = 38 kW.

We know that, Power (P) = V x I where V is the potential difference between the cables and I is the current flowing through them. So, the current drawn by the motor is given as

I = P / V

Substitute the given values, P = 38 kW and V = 380 V

Therefore, I = 38 x 10^3 / 380 = 100 A.

The distance between the cables is 89 cm. So, the magnetic force per unit length between the cables is given by

f = μ₀I²l / 2πd where μ₀ = 4π × 10⁻⁷ T m/A is the permeability of free space, I is the current in the cables, l is the length of the section of each cable where the magnetic field is to be calculated and d is the distance between the cables.

In this case, l = d = 89 cm = 0.89 m.

Substitute the given values,μ₀ = 4π × 10⁻⁷ T m/AI = 100 Al = d = 0.89 m

Therefore, f = μ₀I²l / 2πd= 4π × 10⁻⁷ × 100² × 0.89 / (2 × π × 0.89)= 0.116 N/m

Therefore, the magnetic force per unit length between the cables is 0.116 N/m.

Thus the current drawn by the motor is 100 A and the magnetic force per unit length between the cables is 0.116 N/m.

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What does it cost to cook a chicken for 1 hour in an oven that operates at 20 Ampere Ter 220 Volt if the electric company charge 60 fils per kWh A. 264 Fils B. 528 Fils C. 352 Fils D. 176 Fils through a surface varies with time 1 Ibr

Answers

The cost to cook a chicken for 1 hour in the given oven is 264 fils. Option A: 264 Fils. Voltage is the pressure from an electrical circuit's power source that pushes charged electrons (current) through a conducting loop, enabling them to do work such as illuminating a light. In brief, voltage = pressure, and it is measured in volts (V).

To calculate the cost of cooking a chicken for 1 hour in the given oven, we need to determine the total energy consumed by the oven during that time and then calculate the cost based on the electric company's charge.

The power consumed by the oven can be calculated using the formula:

Power (P) = Voltage (V) x Current (I)

Given:

Voltage (V) = 220 Volts

Current (I) = 20 Amperes

Using the values, we can calculate the power consumed by the oven:

P = 220 V x 20 A

P = 4400 Watts

To calculate the energy consumed, we need to convert the power from Watts to kilowatts and then multiply it by the time in hours:

Energy (E) = Power (P) x Time (t)

Given:

Time (t) = 1 hour

Converting the power from Watts to kilowatts:

Power (P) = 4400 Watts = 4.4 kilowatts

Calculating the energy consumed:

E = 4.4 kW x 1 hour

E = 4.4 kilowatt-hours (kWh)

Now we can calculate the cost using the electric company's charge:

Cost = Energy (E) x Cost per kWh

Given:

Cost per kWh = 60 fils

Calculating the cost:

Cost = 4.4 kWh x 60 fils/kWh

Cost = 264 fils

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In Circuit 64 your voltmeters were accurate in the sense that they (more or less) correctly read the actual voltages in the circuits, but they were inaccurate (for very large resistors) in that these readings are NOT the true voltage across the second resistor when the meter is not there. Now suppose you are in a different setting, with two voltmeters and a high resistance circuit. If meter A "correctly" reads 6.70 volts across a resistor in a circuit and meter B "correctly" reads 6.90V across the same resistor in the same circuit, which meter is giving you the value closest to the true value with no meters present? Explain. (4) 6. The last line of the first column (V1 reading WITHOUT the Simpson) is for the 4.7MQ. Take the value you have and use it to solve for the actual resistance of the Fluke meter. How? Suppose the resistors are both 4.70MQ and use your voltage of the power supply (if you did not write it down, use 3.00V). Remember the question that asked you to find the AV of R* when you knew IR of the other resistor? Well, here you know AV of the parallel combination of R₂ and the meter. "Reverse engineer" things to find the total current from the power supply, then the total resistance (and or you can go directly to find the Reg of the parallel combination, then solve for the meter resistance.

Answers

In the given scenario, if meter A correctly reads 6.70 volts across a resistor in a circuit and meter B correctly reads 6.90 volts across the same resistor in the same circuit, meter A is providing a value closer to the true voltage with no meters present.

When using voltmeters in high-resistance circuits, the presence of the voltmeter can affect the actual voltage across the resistor being measured. In this case, we have two voltmeters, A and B, both reading the voltage across the same resistor. If meter A reads 6.70 volts and meter B reads 6.90 volts, we need to determine which value is closer to the true voltage.

Since the voltmeters are accurate in the sense that they correctly read the actual voltages in the circuits, we can infer that the true voltage across the resistor lies between the readings of meters A and B. Considering that meter A reads 6.70 volts and meter B reads 6.90 volts, we can conclude that meter A provides a value closer to the true voltage. This is because the actual voltage is likely slightly lower than the reading on meter B, making meter A's reading more accurate in this case.

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Other Questions
Consider the crcuit shown in the diagram below. The potentiai difference across the points a and D is aV=120.0 V and the capacitors have the folowing values: C 1=13.0 jif C 2=2.00 2C 3=4.00HF, and C 4=17.0F, tnitially the cagacitors are all uncharged. mic (b) Wnat is the charge on each fully charged capacier? Q 1=Q 2=Q 3=Q 4=mcmcmcmC You are given the following program. Based on your understanding of the code, please answer the questions: (1) The output of line 18 is "1797 / 1797 correct". Please briefly explain the problem with that 100% correct output. (2) Please propose two potential solutions to that problem using 150 words maximum. (No coding required) 1# coding: utf-8 -*- 2 3 from_future import print_function, division 4 import numpy as np 5 6 from sklearn.datasets import load_digits 7 8 digits = load_digits() 9X digits.data 10 y digits.target 11 12 from sklearn.neighbors import KNeighborsClassifier 13 knn = KNeighborsClassifier (n_neighbors=1) 14 knn.fit(x, y) 15 16 y_pred = knn.predict(X) 17 == 18 print("{0} / {1} correct".format(np.sum (y 19 20 *** 21 Output: 22 1797 1797 correct 23 www 222 24 25 y_pred), len(y))) Which of the following is equal to the fraction below? (7/4)11 What is the corner frequency of the circuit below given R1=7.25kOhms,R2=9.25 kOhms, C1=7.00nF. Provide your answer in Hz. Your Answer: Answer units The BOD, of wastewater sample determine to be 35 mg/L as 20C. The K as 20C is 0.19 day . What is the BODs, if the test is run at 30C? I. Problem Solving - Design Problem 1 - A 4.2 m long restrained beam is carrying a superimposed dead load of 107 kN/m and a superimposed live load of 79 kN/m both uniformly distributed on the entire span. The beam is 400 mm wide and 650 mm deep. At the ends, it has 4-20mm main bars at top and 2-20mm main bars at bottom. At the midspan, it has 2-20mm main bars at top and 3 - $20 mm main bars at bottom. The concrete cover is 50 mm from the extreme fibers and 12 mm diameter for shear reinforcement. The beam is considered adequate against vertical shear. Given that f'c = 27.60 MPa and fy=345 MPa. Round your final answer in two decimal places. 1. Determine the design shear for the beam in kN 2. Determine the nominal shear carried by the concrete section using simplified calculation in kN 3. Determine the required spacing of shear reinforcements from simplified calculation. Express it in multiple of 10mm. 4. Determine the location of the beam from the support in which shear reinforcement are permitted not to place in the beam. Select all statements that are true tate and odor causing compounds are covered by secondary standards. Wand one must be followed by chlorination so that residual disinfectant is maintained in the distribution system OMOLG can be per than MCL Stokes Law can be used to calculate setting velocity of flocs 4 pts Solve for mEnter only the numerical value in the box. Do not enter units. Q.3 (a) The bit sequences 1001 and 0111 are to be transmitted on a communications link between two intelligent devices. For each of the methods Hamming(7,4) code and Even parity product code (a1) Calculate the transmission code-words (a2) If the most significant bit of the first bit sequence is corrupted (inverted) during the transmission, show how this error may be detected and corrected Question 2 The feasibility study by Northern Graphite Corporation for the re-start of Okanjande/Okorusu graphite producing operation indicated that Imerys did not follow Rio Tinto pilot plant design and they re-used old equipment which was unsuitable/unreliable. The design engineers are currently busy with mass balances around a hydrocyclone. The hydrocyclone overflow stream has a mass flowrate of 35t/h of solids and a pulp density of 1.35t/m3. The ore solid density was found to be 3.20t/m and the feed stream percentage solids is 35% while the pulp density of the underflow stream is 1.28t/m"". Explain how gonads are able to develop into the ovaries. Give reasons for modelling systems in state space. (6) 2.2 The closed loop transfer function of a C(s) 9s+7 system is G(s) = R(s) (s+1) (s+2) (s+3)* Find the state space representation of the system in phase variable form step by step and draw the signal-flow graph. (20) 2.3 Determine the stability of the system given in Question 2.2 using eigenvalues. (8) 2.4 For the system given in Question 2.2, if the input is a unit step signal, find the time domain response y(t). (20) 2.5 Sket ch the time domain response y(t) obtained in Question 2.4. (6) Calculate the pressure drop along 0.5 m of 0.1 m diameter horizontal steel pipe through which a fluid at 35 C is flowing at the rate of 56 m min 3 1 Viscosity of fluid at 35 C = 1156 CP Density of fluid at 35 C = 156 kg m -3 Analyse the circuit answer the questions based on Superposition theorem. (10 Marks) 30 (2 w 500 mA 60 2 50 2 2 100 2 2592 3 50 V a. The current through 100-ohm resistor due to 50v b. The current through 100 ohms due to 500mA c. The current through 100 ohms due to 50 V and 500mA source together d. The voltage across 100-ohm resistor What can be concluded about the values of H and S from this graph? (A) H>0,S>0 (B) H>0,S 3.00 kilograms of hydrogen are converted to helium by nuclear fusion. How much of it, in kilograms, remains as matter (and is thus not converted to energy)? ke Question has to be executed using the commands provided in command prompt (Windows) and be done using scrapy shellGo to the given Stackoverflow (jobs) page and extract the titles/role of all the jobs listed on the page, request the page in (or use the same shell), fetch the location of all the jobs posted on the given page.url = https://stackoverflow.com/jobs/companies Write "TRUE" if the statement is correct, otherwise write "FALSE".___________ Narratives are more credible sources than Relics.___________ If it can be demonstrated that the witness or source has no direct interest in creating bias, then the credibility of the message decreases.___________ If a number of credible sources contain the same message, the credibility of the message strongly increases.___________ When two sources disagree on a particular point, the historian will prefer the source with most "authority"that is the source created by the expert or by the eyewitness.___________ When two sources disagree and there is no other means of evaluation, then historians take the source which benefit the best interest of the historian.___________ If a source is an oral transmission of information, then, automatically, it is not a legitimate source.___________ If the person who personally witness a murder but has personal grudge on the murderer but was made a witness still, his testimony has unquestionable credibility.___________ If the source is the pronouncement of government official whose basis is his own words of assurance, then, therefore, it is a reliable source.___________ If it can be demonstrated that the witness or source has no direct interest in creating bias then the credibility of the message decreases.___________ If a person identifies another as a criminal, then he tells the authority about the matter. Automatically, the person is really a criminal, and thereby, should be arrested. A reaction mixture initially contains 1.12 M COCI. Determine the equilibrium concentration of CO if Kc for the reaction at this temperature is 8.33 x 10 Calculate this based on the assumption that the answer is negligible compared to 1.12. COCCO+ Cla Calculate the angular momenta of the earth due to its rotational motion about its own axis (effect days and nights) and due to its rotational motion around the sun (effect season change).