Therefore, the point's angular acceleration when t = 2.13s is 99.589 rad/s2.(e) Angular acceleration constantSince the angular acceleration is not constant, the question of angular acceleration constant does not apply.
Given equation: $$\theta = 7.85t * 2.85t^2 + 1.77t^3$$where $\theta$ is in radians and t is in seconds.(a) The point's angular position when t = 0.Substitute t = 0 in the above equation,$$\theta = 7.85(0) * 2.85(0)^2 + 1.77(0)^3$$$\theta = 0$ radians(b) The point's angular velocityTo find the angular velocity, differentiate the equation with respect to time.$$ \begin{aligned} \frac{d\theta}{dt} &= \frac{d}{dt}(7.85t * 2.85t^2 + 1.77t^3) \\ &= 7.85 * 2.85t^2 + 7.08t^2 \\ &= 7.08t^2(1 + 2.85) \\ &= 23.352t^2 \end{aligned} $$Substitute t = 0 to find the point's angular velocity at t = 0.$$ \begin{aligned} \frac{d\theta}{dt} &= 23.352t^2 \\ &= 23.352(0)^2 \\ &= 0 \end{aligned} $$Therefore, the point's angular velocity when t = 0 is zero.(c) The point's angular velocity when t = 6.29sSubstitute t = 6.29 in the equation for angular velocity.$$ \begin{aligned} \frac{d\theta}{dt} &= 23.352t^2 \\ &= 23.352(6.29)^2 \\ &= 926.089 \ rad/s \end{aligned} $$Therefore, the point's angular velocity at t = 6.29s is 926.089 rad/s.(d) The point's angular acceleration when t = 2.13sTo find the angular acceleration, differentiate the angular velocity with respect to time.$$ \begin{aligned} \frac{d^2\theta}{dt^2} &= \frac{d}{dt}(23.352t^2) \\ &= 46.704t \\ &= 46.704(2.13) \\ &= 99.589 \ rad/s^2 \end{aligned} $$
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A heat exchanger tube with an outside diameter of 3 inches and a wall thickness of 0.05 inches has a temperature difference of 47C between the inside and outside surfaces. If the tube is made of steel (k = 50 W/mC) and is 0.96 m long, what is the heat transfer rate through the tube
Using these values in the above formula, we get:Q = (2π × 50 × 0.96 / 4.094) × 47Q = 1122.12 WThe heat transfer rate through the tube is 1122.12 W. Therefore, the correct option is (c) 1122.12 W.
Given data: Outside diameter of the heat exchanger tube (D0)
= 3 inches Wall thickness of the tube (δ)
= 0.05 inches Length of the tube (L)
= 0.96 m Temperature difference between inside and outside surfaces of the tube (ΔT)
= 47°C Thermal conductivity of steel (k)
= 50 W/m°C The heat transfer rate through the tube can be calculated using the formula given below:Q
= (2πkL / ln (D0 / δ)) × ΔTWhere,Q
= Heat transfer rate through the tubeπ
= 3.14L
= Length of the tubeΔT
= Temperature difference between inside and outside surfaces of the tubek
= Thermal conductivity of steel D0
= Outside diameter of the heat exchanger tubed
= Inside diameter of the heat exchanger tube
= (D0 - 2 × δ)ln
= Natural logarithmδ
= Wall thickness of the tubeLet us calculate the inside diameter of the heat exchanger tube,d
= (D0 - 2 × δ)d
= (3 - 2 × 0.05)d
= 2.9 inches 1 inch
= 0.0254 mSo, d
= 2.9 × 0.0254
= 0.07366 mln (D0 / δ)
= ln (3/0.05)ln (60)
= 4.094.Using these values in the above formula, we get:Q
= (2π × 50 × 0.96 / 4.094) × 47Q
= 1122.12 W
The heat transfer rate through the tube is 1122.12 W. Therefore, the correct option is (c) 1122.12 W.
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The wavelength range of the visible spectrum is approximately 400-700 nm. White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in the first order. (Calculate your angle to the nearest 0.1 deg)
The angular width of the visible spectrum in the first order is approximately 1142.9 deg to 2000 deg.
To find the angular width of the visible spectrum in the first order, we can use the formula:
Δθ = λ / d
Where,
Δθ is the angular width
λ is the wavelength of light
d is the slit spacing of the diffraction grating
Given,
Wavelength range of visible spectrum: 400-700 nm
Slit spacing of the diffraction grating: 350 slits/mm = 0.35 slits/μm
For the shortest wavelength (λ = 400 nm):
Δθ = 400 nm / (0.35 slits/μm) = 1142.9 μm/μm = 1142.9 deg
For the longest wavelength (λ = 700 nm):
Δθ = 700 nm / (0.35 slits/μm) = 2000 μm/μm = 2000 deg
Therefore, the angular width of the visible spectrum in the first order is approximately 1142.9 deg to 2000 deg (rounded to the nearest 0.1 deg).
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The height of a moving object is given by
h(t)=2.69sin(3.90t+8.0)
where h is in feet and t is in seconds. Find the velocity at the instant t=2 seconds. Be accurate to 3 decimal places and include correct units.
To find the velocity at the instant t = 2 seconds, we need to take the derivative of the height function h(t) with respect to time.
Given:
h(t) = 2.69sin(3.90t + 8.0)
Taking the derivative of h(t) with respect to t
h'(t) = 2.69 * (3.90) * cos(3.90t + 8.0)
Now we can evaluate the velocity at t = 2 seconds by substituting t = 2 into the derivative expression:
h'(2) = 2.69 * (3.90) * cos(3.90 * 2 + 8.0)
Calculating the value:
h'(2) = 2.69 * (3.90) * cos(7.80 + 8.0)
≈ 2.69 * (3.90) * cos(15.80)
≈ 2.69 * (3.90) * (-0.759)
≈ -7.76 ft/s
Therefore, at t = 2 seconds, the velocity of the object is approximately -7.76 ft/s. The negative sign indicates that the object is moving downwards.
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The pavement compactor is traveling down the incline at vG=5
ft/s when the motor is disengaged. The body of the compactor,
excluding the rollers, has a weight of 8000 lb and a center of
gravity at G.
When the motor is disengaged, the pavement compactor travels down the slope at 5 feet per second, which implies that its initial velocity is 5 feet per second. The pavement compactor's weight is 8000 pounds, and its center of gravity is located at G. Let us assume that the slope's incline angle is θ.
Let's make some further assumptions. Let us assume that there is no rolling friction, that the rollers' moment of inertia is negligible, and that the pavement compactor's center of gravity moves in a straight line throughout the slope.The force acting on the pavement compactor is the gravitational force component parallel to the slope, and its magnitude is 8000 pounds multiplied by the sine of the incline angle. The acceleration of the pavement compactor equals the gravitational force's parallel component divided by the pavement compactor's mass, or 8000 pounds divided by 32.174 feet per second squared, multiplied by the cosine of the incline angle.
The velocity of the pavement compactor at any point down the slope is equal to the square root of twice the distance down the slope multiplied by the acceleration. The distance down the slope is equal to the slope's length multiplied by the sine of the angle of inclination.
Therefore, the velocity of the pavement compactor at any point down the slope is as follows:
v = √[2gs sin(θ)cos(θ)]
Where,
v = Velocity of the pavement compactor (ft/s)
g = Acceleration due to gravity (32.174 ft/s²)
s = Distance travelled by the pavement compactor down the slope (ft)θ = Angle of inclination of the slope (radians)It is worth noting that this formula only works if the slope's length is far greater than the pavement compactor's length.
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What is the phasor representation of 1(t) = locos(wt) at half-period? a. ← b. → c. Arrow up d. Arrow down
the phasor representation will be a vector pointing upward or in the direction of the arrow up symbol.
The phasor representation of 1(t) = locos(wt) at half-period is represented by the arrow up symbol.
Let's break down the problem,
First, let's determine what a phasor is. A phasor is a vector that rotates with the same frequency as a sinusoidal function. It helps in representing the sinusoidal function as a sum of cosine and sine components.
Now let's determine the phasor representation of the given equation:1(t) = locos(wt)
The phasor representation of a cosine function is a vector rotating in a counterclockwise direction. In this case, the cosine function is at a half-period. Therefore, the phasor representation will be a vector pointing upward or in the direction of the arrow up symbol.
Hence, the correct option is c. Arrow up.
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0 [8] In the circuit shown below: (a) [5] i) If the load \( Z_{L} \) consists of a pure resistance \( R_{L} \), find the value of \( R_{L} \) for which the source delivers maximum power to the load. i
The given circuit diagram can be shown as below:We can find the value of RL for which the source delivers maximum power to the load by using the following steps:Step 1: We need to find the expression for the power delivered to the load (PL). We know that, Power, P = I2R
Therefore, the power delivered to the load can be written as,PL = IL2RL ---------(1)Step 2: Now, we need to find the expression for the current through the load (IL).Using the current divider rule, the current through the load can be written as,IL = VS / (R + ZL) ----------(2)Where, ZL is the impedance of the load, R is the resistance of the circuit, and VS is the source voltage.Step 3: Now, we need to substitute the value of IL from equation (2) into equation (1), to get the expression for power delivered to the load in terms of RL.
PL = (VS / (R + RL))2RLPL = (VS2 RL) / ((R + RL)2) ----------(3)
Step 4: We need to differentiate equation (3) w.r.t RL to get the value of RL for which PL is maximum. Therefore, we get,dPL / dRL = (VS2 (R - RL)) / ((R + RL)3)We need to equate the above equation to zero to find the value of RL for which PL is maximum. Hence,0 = (VS2 (R - RL)) / ((R + RL)3)VS2 (R - RL) = 0R - RL = 0RL = RThe value of RL for which the source delivers maximum power to the load is R. The power delivered to the load can be calculated using equation (3), as follows,
PL = (VS2 R) / (4R2)PL = (VS2) / (4R)
Therefore, the value of RL for which the source delivers maximum power to the load is R.
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The equation for calculating how much energy (E in units of Joules) is required to heat an object is E=CmΔT. If we are heating water, the value for C (the specific heat content) is 4100 Joules per kg per Kelvin (or "J/kg/K"). If the water we are heating is 0.1 kg and we heat it 100 degrees, how much energy (E) does it require?
• 41000
• 41
• 0.41
• 4100000000000
The amount of energy (E) required to heat 0.1 kg of water by 100 degrees is 4100 Joules.
The equation for calculating the energy required to heat an object is E = CmΔT, where E represents the energy in Joules, C is the specific heat content in J/kg/K, m is the mass of the object in kg, and ΔT is the change in temperature in Kelvin. For water, the specific heat content (C) is 4100 J/kg/K. In this case, we are heating 0.1 kg of water with a temperature change (ΔT) of 100 degrees. Plugging these values into the equation, we get E = (4100 J/kg/K) * (0.1 kg) * (100 K) = 4100 Joules. Therefore, it requires 4100 Joules of energy to heat 0.1 kg of water by 100 degrees. The specific heat content of water indicates that it takes a relatively high amount of energy to raise its temperature compared to other substances. This property is why water is often used as a coolant or heat transfer medium in various applications. Understanding the energy requirements for heating substances is crucial in fields such as engineering, physics, and chemistry, where precise control and calculations of heat transfer are necessary.
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A spring with an unstretched length of 40 cm and a k value of
120 N/cm is used to lift a 0.5 kilogram box from a height of 20 cm
to a height of 30 cm. If the box starts at rest, what would you
expect
According to the law of conservation of energy, the total initial energy should be equal to the final energy.
Based on the given information, we can analyze the situation using principles of energy conservation and Hooke's Law for the spring.
Potential Energy:
The potential energy of the box can be calculated using the formula:
Potential Energy = m * g * h,
where m is the mass of the box (0.5 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the change in height (30 cm - 20 cm = 10 cm = 0.1 m).
Potential Energy = 0.5 kg * 9.8 m/s² * 0.1 m = 0.49 J.
Spring Potential Energy:
The spring potential energy can be calculated using the formula:
Spring Potential Energy = (1/2) * k * x²,
where k is the spring constant (120 N/cm = 120 N/m = 12,000 N/m) and x is the change in length of the spring.
Change in length of the spring, x = final length - initial length = (30 cm - 40 cm) = -10 cm = -0.1 m (negative sign indicates compression).
Spring Potential Energy = (1/2) * 12,000 N/m * (-0.1 m)² = 60 J.
Total Initial Energy:
The total initial energy of the system is the sum of the potential energy and the spring potential energy when the box is at rest:
Total Initial Energy = Potential Energy + Spring Potential Energy = 0 + 60 J = 60 J.
Final Energy:
The final energy of the system is the potential energy when the box reaches the new height:
Final Energy = Potential Energy = 0.49 J.
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Complete Answer:
A Spring With An Unstretched Length Of 40 Cm And A K Value Of 120 N/Cm Is Used To Lift A 0.5 Kilogram Box From A Height Of 20 Cm To A Height Of 30 Cm. If The Box Starts At Rest, What Would You Expect The Final Velocity To Be?
A spring with an unstretched length of 40 cm and a k value of 120 N/cm is used to lift a 0.5 kilogram box from a height of 20 cm to a height of 30 cm. If the box starts at rest, what would you expect the final velocity to be?
A 3-phase 4-pole ac machine has double-layer stator windings and 12 slots per pole. Each stator coil has 2 turns, and the coil pitch is y,=10 slot pitch. Each winding has 2 parallel circuits. If balanced 3-phase currents of 60 Hz and 30 A are injected to the stator windings, find the magnitude and the speed of the fundamental, the 5th, and the 7th harmonics of total mmf.
A 3-phase 4-pole ac machine has double-layer stator windings and 12 slots per pole. Each stator coil has 2 turns, and the coil pitch is y,=10 slot pitch. The magnitude of the 7th harmonic component of the mmf is given by 17.5 A.
Each winding has 2 parallel circuits. If balanced 3-phase currents of 60 Hz and 30 A are injected to the stator windings, the magnitude and the speed of the fundamental, the 5th, and the 7th harmonics of total mmf can be found as follows: Calculation of fundamental frequency
From the given problem, the total number of stator slots = 12 × 4 = 48 and the number of poles = 4.
Thus, the synchronous speed Ns is given by: [tex]Ns = 120f / p = 120 × 60 / 4 = 1800 rpm[/tex]
The fundamental component of the mmf wave rotates in synchronism with the rotor at a speed of 1800 rpm. The fundamental frequency f1 is given by: [tex]f 1 = ns / 120 = 1800 / 120 = 15 Hz[/tex]
Magnitude of the fundamental frequency of mmf
The magnitude of the fundamental component of the mmf is given by: [tex]Mf = 1.5× √2 × 2 × 30 = 127.3 A[/tex]
Now, let's calculate the harmonic frequencies of the mmf wave. The harmonic frequencies in an AC machine are given by the formula: nf = nf1, where n is an integer
Calculation of 5th harmonic frequency
The frequency of the 5th harmonic of the mmf wave is given by:
n5 = 5f1
= 5 × 15
= 75 Hz
Speed of 5th harmonic
The speed of the 5th harmonic of the mmf wave is given by:
N5 = 120f / p
= 120 × 75 / 4
= 2250 rpm
Magnitude of 5th harmonic frequency of mmf
The magnitude of the 5th harmonic component of the mmf is given by:
M5 = (1/5) × 1.5 × √2 × 2 × 30
= 25.45 A
Calculation of 7th harmonic frequency
The frequency of the 7th harmonic of the mmf wave is given by:
n7 = 7f1
= 7 × 15
= 105 Hz
Speed of 7th harmonic
The speed of the 7th harmonic of the mmf wave is given by: N7 = 120f / p
= 120 × 105 / 4
= 3150 rpm
Magnitude of 7th harmonic frequency of mmf
The magnitude of the 7th harmonic component of the mmf is given by: M7 = (1/7) × 1.5 × √2 × 2 × 30 = 17.5 A
Thus, the fundamental frequency, the 5th, and the 7th harmonics of total mmf of the given ac machine have been calculated.
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A 1 m x 2 m glass window that is in your room at __18__°C, while the temperature of the inside surface of the window is _15___°C. The heat transfer coefficient between the room and in window is 10 W/m2K. Find the rate of heat flow from the room to the surface of the window.
The rate of heat flow from the room to the surface of the window can be calculated using the formula; Q = U*A*ΔT, where
Q = rate of heat flow,
U = heat transfer coefficient,
A = surface area,
ΔT = temperature difference between the two sides.
The values are as follows:
A = 1 m x 2 m
= 2 m²
ΔT = (18°C - 15°C)
= 3°C
U = 10 W/m²K
Substituting these values in the formula:
Q = U*A*ΔT
= 10 * 2 * 3
= 60 W
The rate of heat flow from the room to the surface of the window is 60 W.
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The process where a photon comes into an atom and increases the energy of an electron is known as...
absorption
emission
fluorescence
reflection
The process where a photon comes into an atom and bounces off an electron is known as...
absorption
emission
fluorescence
reflection
The process where a photon is given off by an electron dropping to a lower energy state is known as...
absorption
emission
fluorescence
reflection
The process where a photon comes into an atom and increases the energy of an electron is known as absorption.
Absorption is the process in which light photons are absorbed by atoms or molecules when they pass through a medium. The photons' energy is transferred to the absorbing material in this process, typically elevating one or more of the material's electrons to higher energy states. When an electron moves from a lower energy level to a higher one, it absorbs energy.
Electrons release energy when they move from a higher energy level to a lower one in emission. Reflection is the phenomenon in which a light wave incident on a boundary is sent back into the same medium from which it came. Emission is the opposite of absorption, and it is the process where the energy of an electron increases, and a photon is emitted as it moves to a lower energy level.
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PRACTICE IT Use the worked example above to help you solve this problem. A diverging lens of focal length f = -9.9 cm forms images of an object situated at various distances. (a) If the object is placed p₁ = 29.7 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. 9 = -7.42 cm M = 0.25 (b) Repeat the problem when the object is at p₂ = 9.9 cm. 9 = -4.95 cm 0.17 M X Your response differs from the correct answer by more than 10%. Double check your calculations. (c) Repeat the problem again when the object is 4.95 cm from the lens. a -3.3 cm -0.11 X M Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. EXERCISE HINTS: GETTING STARTED I I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. Repeat the calculation, finding the position of the image and the magnification if the object is 20.6 cm from the lens. q = -6.69 cm 0.23 X M = What factors affect the magnification of an image?
a) the magnification is 0.17.
b) the magnification is 0.5.
c) the magnification is 0.67.
(a) The given information is:focal length,
f = -9.9 cm
p₁ = 29.7 cm
9 = -7.42 cm
M = 0.25
The object is placed at a distance of 29.7 cm from the lens. The image is formed at a distance of 9 cm from the lens.
Using the lens formula,
1/f = 1/v - 1/u
where,
u = -29.7 cm,
f = -9.9 cm
On substituting the values, we get
1/v = 1/-9.9 - 1/-29.7
v = -6.633 cm
The image is formed at a distance of 6.633 cm from the lens.
Since the image is formed on the same side as the object, the image is virtual. Magnification is given by,
|m| = v/u
|0.25| = -6.633/-29.7
On simplifying,
|m| = 0.17
Therefore, the magnification is 0.17.
(b) When the object is placed at a distance of 9.9 cm from the lens, then,
u = -9.9 cm,
f = -9.9 cm
The lens formula is given as,
1/f = 1/v - 1/u
On substituting the values, we get,
1/v = 1/-9.9 - 1/-9.9
v = -4.95 cm
The image is formed at a distance of 4.95 cm from the lens. Since the image is formed on the same side as the object, the image is virtual.
Magnification is given by,
|m| = v/u
|0.25| = -4.95/-9.9
On simplifying,
|m| = 0.5
Therefore, the magnification is 0.5.
(c) When the object is placed at a distance of 4.95 cm from the lens, then,
u = -4.95 cm,
f = -9.9 cm
The lens formula is given as,
1/f = 1/v - 1/u
On substituting the values, we get,
1/v = 1/-9.9 - 1/-4.95
v = -3.3 cm
The image is formed at a distance of 3.3 cm from the lens. Since the image is formed on the same side as the object, the image is virtual.
Magnification is given by,
|m| = v/u
|0.25| = -3.3/-4.95
On simplifying,
|m| = 0.67
Therefore, the magnification is 0.67.
Factors affecting the magnification of an image are:
i) the focal length of the lens
ii) the distance between the lens and the object
iii) the distance between the lens and the image.
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7. Write the complete a decay equation for Ra. (b) Find the energy released in the decay.
Write the complete a decay equation for 249 Cf. (b) Find the energy released in the decay.
9. Write the complete / decay equation for 90 Sr . a major waste product of nuclear reactors.
b) Find the energy released in the decay.
The complete alpha decay equation for Ra is ²²⁵Ra -> ²²¹Rn + α particle. The energy released in the decay is approximately 4.87 MeV.
Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, which consists of two protons and two neutrons. For Ra (radium), the complete alpha decay equation is as follows:
²²⁵Ra -> ²²¹Rn + α
In this equation, the parent isotope Ra undergoes alpha decay and transforms into the daughter isotope Rn, while emitting an alpha particle. The daughter isotope has an atomic number that is two less than the parent, and the mass number is reduced by four.
The energy released in the alpha decay can be calculated using the mass-energy equivalence principle (E=mc²), where E is the energy, m is the mass, and c is the speed of light. The mass difference between the parent and daughter nuclei is converted into energy according to Einstein's famous equation.
The energy released in the alpha decay of Ra is approximately 4.87 MeV (million electron volts). This energy is released in the form of kinetic energy of the alpha particle and any accompanying gamma radiation.
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9. While normal load is being supplied, an open circuit takes place in one of the pilot
wires. What will be the consequences as far as the busbar differential relay is
concerned?
10. Suggest an add-on to the differential relay, to avert a possible maloperation in the above scenario.
11. Sketch the high impedance busbar differential protection for a three-phase busbar having three incoming and two outgoing feeders.
The consequences for the busbar differential relay can vary depending on the specific configuration and design of the relay. One such add-on is the use of voltage supervision or voltage restraint features. The voltage supervision feature provides an additional layer of security and helps maintain the integrity of the busbar differential protection.
In the event of an open circuit in one of the pilot wires, while the normal load is being supplied, the consequences for the busbar differential relay can vary depending on the specific configuration and design of the relay. However, generally, an open circuit in one of the pilot wires can lead to a loss of communication or signal transmission between the relay and the associated current transformers (CTs) or other devices connected to the pilot wires. This loss of communication can potentially cause the busbar differential relay to operate falsely or fail to operate when a fault occurs, compromising the protection of the busbar.
To avert possible maloperation of the busbar differential relay in the scenario described above, an add-on or additional protection scheme can be implemented. One such add-on is the use of voltage supervision or voltage restraint features. This feature monitors the voltage across the pilot wires and ensures that a sufficient voltage is present for proper relay operation. If the voltage falls below a certain threshold, indicating an open circuit or communication failure, the differential relay can be blocked from operation to prevent false tripping or loss of protection. The voltage supervision feature provides an additional layer of security and helps maintain the integrity of the busbar differential protection.
Sketching a complete high-impedance busbar differential protection scheme for a three-phase busbar with three incoming and two outgoing feeders would require a more detailed understanding of the specific system configuration, CT locations, and associated relay settings. However, I can provide a general overview of the components involved in such a protection scheme.
In a high-impedance busbar differential protection, each incoming and outgoing feeder is equipped with a current transformer (CT) that measures the current flowing in and out of the busbar. The secondary side of the CTs is connected to high-impedance differential relays. The relay outputs are interconnected and connected to a tripping circuit that can trip the relevant circuit breakers in case of a fault.
The differential relays compare the currents from the CTs to detect any imbalance or fault current flowing into or out of the busbar. A differential current exceeding a set threshold indicates a fault within the protected zone, and the relay initiates tripping actions to isolate the faulted section.
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If there is a wave function going in the positive x direction at y1(x,t) = 1.60 cos (3.31x - 25.9t) and a second wave function also going in the positive x direction at y2(x,t) = 2.55 cos (14.7x - wt) but this second wave function moves energy 12 times faster than the first wave. Where x is in meters and t is in seconds. What is the frequency of the second wave in hertz?
The frequency of the second wave in hertz is 2.341 Hz.
Wave functions:
y₁(x,t) = 1.60 cos (3.31x - 25.9t)y₂(x,t) = 2.55 cos (14.7x - wt) the frequency of the second wave in hertz. To calculate the frequency of the second wave in the heart.
The angular frequency of the second wave.y_2(x,t)=2.55\cos (14.7x-wt) .The angular frequency is given by:
omega=2\pi f Here, w is the angular frequency. Frequency is f.w=14.7.
The frequency of the second wave in hertz, f is given by the relation: f=w/2\pi Substitute the value of w to calculate the frequency of the second wave in hertz. f=14.7/(2\pi).
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i.
Determine the rms current of the periodic function.
ii. When a 100-ohm resistor is connected in this periodic
function, what will be the average power?
Given: The periodic function is v(t) = 40sin(100πt) + 60cos(100πt) RMS Current of the periodic functionThe RMS current of the periodic function is given by the formula:
[tex]Irms=√((I1² +I2² +....+ In² )/ n )[/tex]where I1, I2, .....In are the instantaneous currents at the time t1, t2, ......tn respectively. And n is the number of instantaneous currents.The current in the circuit can be calculated using Ohm’s Law.[tex]i(t) = v(t) / R = v(t) / 100 1 = 40sin(100πt) / 100 = 0.4sin(100πt)I2 = 60cos(100πt) / 100 = 0.6cos(100πt)[/tex]Therefore, [tex]Irms = √[(0.4)² + (0.6)²]/√2= 0.7071 or 0.71 A[/tex] (approx) When a 100-ohm resistor is connected in this periodic function,
We know that Average power = (Irms)²RThe value of Irms is 0.71 A and the value of resistance R is 100 Ohm.Average power = (0.71)² x 100= 50.41 W (approx)Therefore, the average power in the given periodic function is 50.41 W (approx).
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Saku is a competitor in the world strongman competition this year Premise 2. Every competitor in the history of the competition has weighed more than 100 kg Conclusion. Therefore, Saku weighs more than 100 kg Inductive, Weak and Uncogent Deductive, Valid and Unsound Deductive, Invalid and Unsound Inductive, Strong and Cogent Question 12 (1 point) What did Karl Popper see as the defining characteristic of the scientific process? Falsifiability Experimental design Prediction testing Randomization
The argument is deductive, valid, but unsound.
The argument follows a deductive reasoning pattern where the conclusion is derived from the premise. It is valid because if the premise is true, the conclusion logically follows. However, the argument is unsound because the premise itself is not necessarily true.
It claims that every competitor in the history of the competition has weighed more than 100 kg, but there is no evidence or guarantee that this premise is accurate or universally applicable. Therefore, the conclusion that Saku weighs more than 100 kg cannot be considered reliable based solely on the given argument.
Karl Popper saw falsifiability as the defining characteristic of the scientific process. According to Popper, scientific theories should be formulated in a way that allows for the possibility of being disproven or falsified through empirical observations or experiments. The ability to make predictions and subject those predictions to testing is crucial for scientific theories to be considered valid. Randomization and experimental design are important components within the scientific process, but Popper emphasized that the core principle is the ability to potentially refute or disprove theories through empirical evidence.
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What is the proper interpretation of E=mc
2
in the position-electron pair production experiment? no energy was created or lost because the positron and the electron cancel each other in electric charge. kinetic energy and mass are created simultaneously. the kinetic energy created is equal in quantity to the mass created. the kinetic energy lost ended up as mass created.
E=mc^2 states that energy and mass are interconvertible, with no energy created or lost in the process.
The proper interpretation of E=mc^2 in the position-electron pair production experiment involves the conversion of energy into mass and vice versa.
According to the equation, energy (E) is equal to mass (m) multiplied by the square of the speed of light (c^2). In this experiment, a high-energy photon interacts with the electric field of an atomic nucleus, resulting in the creation of an electron-positron pair.
No energy is created or lost in this process, as energy is conserved. The positron and electron do cancel each other in terms of their electric charge, but they possess equal and opposite amounts of kinetic energy.
The energy that was lost during the creation of the positron-electron pair is transformed into mass. This means that the kinetic energy lost is exactly equal to the mass created, demonstrating the equivalence of energy and mass.
Overall, the experiment highlights the profound connection between energy and mass, where both are interconvertible and conserved in accordance with the principles of E=mc^2.
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210Pb (half life = 22.3 years) decays by beta decay to 210Po (half life = 139 days). If the concentration of 21° Po is initially = zero, how long must decay take place before the activity of 21°Po equals half that of parent 210Pb? =
Decay should take place for 8.5 years before the activity of 210Po equals half that of parent 210Pb.
Let the initial activity of 210Pb be A1 and the initial activity of 210Po be A2.T1/2 of 210Pb = 22.3 years.
So, the decay constant of 210Pb can be given by:
λ1 = (0.693/T1/2)1/λ1 = (0.693/22.3)
1/λ1 = 0.03106 y-1
Now, T1/2 of 210
Po = 139 days = 0.38 years
So, the decay constant of 210Po can be given by:λ2 = (0.693/T1/2)2/λ2 = (0.693/0.38)2/λ2 = 1.83 y-1
The rate of decay of 210Pb is given by: dN1/dt = - λ1N1
The rate of decay of 210Po is given by: dN2/dt = λ1N1 - λ2N2Where N1 and N2 are the number of nuclei of 210Pb and 210Po respectively.
The general solution to the second differential equation is given by:
N2 = {(A1/λ1) - [(A1/λ1) + (A2/λ2)]e-λ2t }e-λ1t
The time at which the activity of 210Po becomes half of the activity of 210Pb can be obtained by equating the activity of 210Po to half of the activity of 210Pb.
So we get: A2 = (1/2)A1
The above equation can be written as: (A1/λ1) - [(A1/λ1) + (A2/λ2)]e-λ2t = 0.5A1
Simplifying, we get:e-λ1t - [1 + (λ1/λ2) (0.5)] e-λ2t = 0
Using a graph or trial and error, we can find out that the time at which the above equation is satisfied is t = 8.5 years.
Therefore, decay should take place for 8.5 years before the activity of 210Po equals half that of parent 210Pb.
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4 20 the Fart. (d) What is the mass"s velocify along the y-axis, in meters per second, time t
1
=0.15 s? w(t
j
)=−2.1392 23 205 - Part (e). What is the magaitude of the mass"s maximum acceleration, in meters per second syuared? (11\%) Problem 2: A mass m=15 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k= 75 N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by y(t)=Acos(ωt−ϕ), with the positive y-axis pointing upward. At time t=0 the mass is observed to be at a distance d=0.35 melow its equilibrium height with an upward speed of v
0
=4 m/s.
The velocity along the y-axis, at time t1 = 0.15 s, is -1.533 m/s. the magnitude of the maximum acceleration of the mass is approximately 187.9 m/s².
Part (d):
We have the following equation of motion for the simple harmonic motion:
y(t) = A cos(ωt - ϕ)
From this equation, we can find the velocity along the y-axis as follows:
dy(t)/dt = -Aωsin(ωt - ϕ)
We know that at time t1 = 0.15 s, w(t1) = -2.139 m
Therefore,
ω = 23.205 rad/s
A = d = 0.35 mϕ = 0
(as we have been given that the positive y-axis points upward)
Thus,
vy = -0.35*23.205*sin(23.205*0.15)
≈ -1.533 m/s
Hence, the velocity along the y-axis, at time t1 = 0.15 s, is -1.533 m/s.
Part (e):
The maximum acceleration of the mass can be found as follows:
a_max = ω^2A
From the given values,
ω = 23.205 rad/s
A = d = 0.35 m
Therefore,
a_max = (23.205)^2*0.35
≈ 187.9 m/s²
Hence, the magnitude of the maximum acceleration of the mass is approximately 187.9 m/s².
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Which of the following describes the relationship between the distance between charges and electrical force? directly proportional relationship inversely proportional relationship direct square relationship inverse square relationship Newton's universal law of gravitation describes forces that are , while Coulomb's law describes forces that are
The electrical force is directly proportional to the product of charges and inversely proportional to the square of the distance between charges.
Therefore, the correct option is an inverse square relationship. Newton's universal law of gravitation describes forces that are gravitational, while Coulomb's law describes forces that are electrical.
Coulomb's law is a mathematical equation that describes the interactions between electric charges. It quantifies the amount of electrical force that two charged objects exert on each other based on their distance and charge. The equation can be used to calculate the force between two point charges, which are charged particles that have a negligible size and shape relative to the distance between them.
Newton's law of gravitation is a mathematical equation that describes the force of gravity between two objects with mass. It states that any two objects with mass exert an attractive force on each other that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
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A 0.7 kg aluminum pan, c
al
=900, on a stove is used to heat 0.25 liters of water from 19
∘
C to 788
∘
C. (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan % and (c) the water?
(a) The amount of heat required is 3.1333 x 10⁵ J. (b) The percentage of the heat that is used to raise the temperature of the pan is 4.43%. (c) The percentage of the heat that is used to raise the temperature of the water is 95.57%.
Given,
Mass of aluminum pan (m) = 0.7 kg
Specific heat of aluminum (c) = 900 J/kg∘C
(a) To find the heat required to heat the water, we use the specific heat of water. Specific heat of water (c) = 4186 J/kg∘C Volume of water (V) = 0.25 L = 0.25 x 10⁻³ m³
Increase in temperature of water (ΔT1) = 788 - 19 = 769∘C
The mass of water (m1) is given by:
mass = density x volume
Density of water (ρ) = 1000 kg/m³ mass = 1000 x 0.25 x 10⁻³ = 0.25 kg
The amount of heat required to heat the water is given by:
Q1 = m1 x c x ΔT1 Q1
= 0.25 x 4186 x 769 Q1
= 7.82 x 10⁵ J
(b) To find the percentage of heat used to raise the temperature of the pan, we use the formula: percentage of heat used to raise the temperature of the pan
= Q2 / Q x 100
where Q2 is the heat used to raise the temperature of the pan. The amount of heat used to raise the temperature of the pan is given by:
Q2 = m2 x c x ΔT2
m2 is the mass of the pan. ΔT2 is the increase in temperature of the pan. The initial temperature of the pan is 19°C. The final temperature of the pan is the same as the final temperature of the water, which is 788°C.
ΔT2 = 788 - 19 = 769°C
m2 = 0.7 kg
Q2 = 0.7 x 900 x 769
Q2 = 4.14 x 10⁵ J
The total amount of heat required is given by:
Q = Q1 + Q2
Q = 7.82 x 10⁵ + 4.14 x 10⁵
Q = 1.20 x 10⁶ J
(c) To find the percentage of heat used to raise the temperature of the water, we use the formula: percentage of heat used to raise the temperature of the water
= Q1 / Q x 100
The percentage of heat used to raise the temperature of the water is given by: percentage of heat used to raise the temperature of the water
= 7.82 x 10⁵ / 1.20 x 10⁶ x 100
percentage of heat used to raise the temperature of the water
= 95.57%
The amount of heat required to heat the water is 7.82 x 10⁵ J. The percentage of heat used to raise the temperature of the pan is 4.43%. The percentage of heat used to raise the temperature of the water is 95.57%.
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The effective potential corresponding to a pair of particles interacting through a central force is given by the expression La U₁rs(r)= +C where C>0 and the parameters have their usual meaning. What is the radial component of force? Is it repulsive or Zur attractive? O a. f(r)--30r, attractive O b. (r)--4Cr¹, attractive O c f(r)=-3Cr, repulsive Od. f(r)=3Cr, repulsive
The radial component of force can be calculated by taking the negative gradient of the effective potential. The effective potential is given by
U₁rs(r) = +C.
So the radial component of force can be expressed as follows:
f(r) = -dU₁rs/dr
The negative gradient of U₁rs with respect to r results in the radial component of force.Therefore, the radial component of force is given by;
f(r) = -dU₁rs/dr= -d/dru(+C)=-0
The radial component of force is zero, which indicates that there is no force acting in the radial direction. This means that there is no repulsive or attractive force acting between the particles. In conclusion, the radial component of force is zero. Thus, the correct answer is option E.
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(a) What are the main limitations of the wheeled linear induction motor for vehicle applications? Why do you think magnetically levitated induction motor is advantageous compared to the wheeled linear induction motor for vehicle applications?
The main limitations of wheeled linear induction motors (LIMs) for vehicle applications are traction efficiency and wear and tear. Wheeled LIMs face challenges in maintaining high traction efficiency due to the friction between the wheels and the track.
Wheeled LIMs face challenges in maintaining high traction efficiency due to the friction between the wheels and the track. The contact between the wheels and the track leads to energy losses, reducing the overall efficiency of the motor. Additionally, this friction causes wear and tear on the wheels, requiring frequent maintenance and replacement.
On the other hand, magnetically levitated induction motors (MLIMs) offer several advantages over wheeled LIMs for vehicle applications. MLIMs utilize magnetic levitation to suspend the vehicle, eliminating the need for wheels and physical contact with the track. This leads to reduced friction, significantly improving traction efficiency and reducing wear and tear.
Furthermore, MLIMs provide a smoother and quieter ride as there are no physical wheels or track vibrations. The absence of mechanical components, such as wheels and axles, also reduces the weight of the vehicle, improving energy efficiency and maneuverability.
Moreover, MLIMs offer the potential for higher speeds, better acceleration, and regenerative braking. Magnetic levitation allows for more precise control over the vehicle's movement and enables dynamic stabilization, enhancing safety.
In conclusion, the magnetically levitated induction motor (MLIM) overcomes the limitations of the wheeled linear induction motor (LIM) by providing higher traction efficiency, reduced wear and tear, smoother ride, quieter operation, improved energy efficiency, and enhanced control and safety features.
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The units of the time variable "r" and angular frequency "o" in this IRA are in seconds and rad/second, respectively. IRA#6_1. An ideal highpass filter has a cutoff angular frequencies of 5 rad/sec and a passband gain of 1 (i.e. frequency response in the passband is one). If this filter is used to filter the input signal x(t)=2cos(31)-3sin(4t), then the output of the filter is:_________
The output of the filter can be found out by first calculating the Fourier transform of the input signal x(t) and then multiplying it with the frequency response of the filter
Y(jω) = 0.3π(δ(ω - 31) + δ(ω + 31)) - 0.3jπ(δ(ω - 31) + δ(ω + 31)) - 0.15[δ(ω - 4) - δ(ω + 4)]
The input signal
x(t) = 2cos(31t) - 3sin(4t)
is to be filtered using an ideal high pass filter that has a cutoff angular frequency of 5 rad/sec and a passband gain of 1, and the output of the filter is to be found out. The units of the time variable r and angular frequency ω in this IRA are in seconds and rad/second, respectively. IRA#6_1.
The highpass filter can be defined as having the frequency response
H(jω) = (jω/5 + 1) / (jω + 5).
Here, j is the imaginary unit, ω is the angular frequency in rad/sec, and 5 is the cutoff angular frequency of the filter, which is 5 rad/sec. Since this is an ideal highpass filter, its gain is unity in the passband (angular frequencies greater than 5 rad/sec) and zero in the stopband (angular frequencies less than 5 rad/sec).
The output of the filter can be found out by first calculating the Fourier transform of the input signal x(t) and then multiplying it with the frequency response of the filter
H(jω).x(t) = 2cos(31t) - 3sin(4t)X(jω) = [π(δ(ω - 31) + δ(ω + 31))] / 2j - 1.5[δ(ω - 4) - δ(ω + 4)]
Now, the output of the filter Y(jω) can be obtained as follows.
Y(jω) = H(jω)X(jω)
= [(jω/5 + 1) / (jω + 5)][π(δ(ω - 31) + δ(ω + 31))] / 2j - 1.5[δ(ω - 4) - δ(ω + 4)]
The final answer is:
Y(jω) = 0.3π(δ(ω - 31) + δ(ω + 31)) - 0.3jπ(δ(ω - 31) + δ(ω + 31)) - 0.15[δ(ω - 4) - δ(ω + 4)]
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Transmission Line Dispersion
A transmission line with no leakage (Go = 0) is carrying a signal with angular frequency
ω = 105 rad s−1. The capacitance per unit length is Co = 10−7 Fd m−1 and the inductance per
unit length is 10−5 H m−1, and the length of the line is 100 m.
A. If the resistance per unit length Ro = 0, how long does it take the signal to travel from
one end of the line to the other?
B. If there is some resistance per unit length, Ro = 1 Ω m−1, then the propagation constant
γ will be a function of frequency and the line becomes dispersive.
What is the propagation constant in this case?
C. In the case of part B, how long does it take the signal to get from one end of the line to
the other?
D. At what angular frequency, ω, will the time needed to go from one end to the other be
two times the result in part A?
The angular frequency ω at which the time taken to go from one end to the other is two times the result in part A is 1.65 × 109 rad/s.
A) If the resistance per unit length Ro = 0, then the characteristic impedance and the propagation constant will become
\[{Z_c} = \sqrt {\frac{L}{{C}}}
= 1000\Omega \& \& {\gamma _o}
= j\sqrt {\omega LC}
= j1\;
{\rm{rad/m}}\]
The velocity of propagation on the line is
v = ω/γo
= 105/1
= 105 m/s.
The time taken for the signal to travel from one end of the line to the other can be calculated as
t = L/v
= 100/105
= 0.95 s.
B) If Ro = 1 Ωm−1, then the propagation constant becomes
\[\gamma = \sqrt {j\omega \left( {L + R\Delta x} \right)\left( {C + \frac{\Delta x}{R}} \right)}
= j0.9949\;
{\rm{rad/m}}\]
C) The time taken for the signal to travel from one end of the line to the other can be calculated as
t = L/v
= L/ωIm[γ]
= L/ωβ,
where β is the phase constant.
Thus, t = 100/(105 × 0.9949)
= 0.952 s.
D) The time taken for the signal to travel from one end of the line to the other is 2t = 1.9 s.
Thus, using the relation obtained in part C, we have
\[2t = \frac{2L}{{\omega \beta }}
= \frac{{2L}}{{\omega \sqrt {{{\left( {L + R\Delta x} \right)}\left( {C + \frac{\Delta x}{R}} \right)}} }}\]
Rearranging the above equation gives
\[{\omega ^2} = \frac{{4{L^2}}}{{{{\left( {2t\sqrt {{\rm{LC}}} } \right)}^2} + {L^2}{\rm{R}}\Delta x}}
= 1.65 \times {10^9}\;
{\rm{rad}}{{\rm{s}}^{ - 1}}\]
Therefore, the angular frequency ω at which the time taken to go from one end to the other is two times the result in part A is 1.65 × 109 rad/s.
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Q1: Consider the vectors A = j - 5k and B = -2î + 5j – 2ť. a. Calculate the dot product between the vectors. b. Find the angle between the vectors.
The dot product between the vectors is 5. The angle between vectors A and B is approximately 106.9 degrees.
Given vectors, A = j - 5k and B = -2î + 5j – 2ť.
To calculate the dot product between vectors A and B, we use the formula, A . B = |A||B| cos θ, where |A| and |B| are magnitudes of vectors A and B and θ is the angle between them. (Note that since A and B have different units, we can't calculate their magnitudes without knowing what those units are. But we can still find the dot product and angle between them.)
a. To calculate the dot product between vectors A and B, we need to take the dot product of their respective components:
A . B = (0)(-2) + (1)(5) + (-5)(0) = 5
So, A . B = 5
b. To find the angle between vectors A and B, we can rearrange the formula we used above:
cos θ = (A . B) / (|A||B|)θ = cos⁻¹((A . B) / (|A||B|))
Substituting the values of A . B, |A|, and |B|,θ = cos⁻¹(5 / (√(1² + (-5)² + 0²) × √((-2)² + 5² + (-2)²)))θ ≈ 106.9°
So, the angle between vectors A and B is approximately 106.9 degrees.
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An object is moving in a circular motion law: s(t)=2t^3=3t^2+4. In t=2s, the module of its total acceleration is a=40m/s^2
Compute the Radius R of the circle. Compute the module of the acceleration in t=1s.
The module of the acceleration in t = 1 is 18 m/s².
:Radius of the circle = 16m
Module of the acceleration in t = 1 is 18 m/s².
Given:An object is moving in a circular motion law:
s(t) = 2t³ = 3t² + 4.
In t = 2s, the module of its total acceleration is
a = 40m/s²
To Find: The Radius R of the circle. Compute the module of the acceleration in t=1s.
We are given the equation of the motion as follows,
s(t) = 2t³ = 3t² + 4
Differentiating the equation twice, we get v(t) and a(t).
v(t) = s'(t)
= 6t² + 6ta(t)
= v'(t)
= 12t + 6
Now, we have to find out the radius R of the circle.
Substituting the value of t = 2 in the equation s(t), we have,
s(2) = 2 x 2³ - 3 x 2² + 4
= 16 m
If R be the radius of the circle, then we have,
R = s(2) = 16 m
Also, we have to find the module of the acceleration in t = 1.
s(t) = 2t³ = 3t² + 4,
we have to find out the values of s(1), s'(1), and s''(1) by putting the value of t = 1.
s(1) = 2 x 1³ - 3 x 1² + 4 = 3 m
Now, we can calculate v(1) and a(1) by putting t = 1 in the equations v(t) and a(t).
v(1) = 6 x 1² + 6 x 1 = 12 m/sa(1) = 12 x 1 + 6 = 18 m/s²
Hence, the module of the acceleration in t = 1 is 18 m/s².
:Radius of the circle = 16m
Module of the acceleration in t = 1 is 18 m/s².
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A Volt is defined as the potential difference between two points of a conducting wire carrying a constant current of 1 ampere when the power dissipated between these points is 1 watt O a. True O b. False
False. A volt is defined as the potential difference when one joule of work is done per coulomb of charge moved, not specifically related to a conducting wire carrying a constant current and power dissipation.
A volt is defined as the unit of electric potential difference or voltage. It is not specifically tied to a conducting wire carrying a constant current of 1 ampere and power dissipation of 1 watt. The volt is defined as the potential difference between two points when one joule of work is done per coulomb of charge moved between those points.
This definition holds true in various electrical contexts, not limited to a specific current or power dissipation scenario. Therefore, the statement that a volt is defined based on a conducting wire with a constant current and power dissipation of 1 watt is incorrect.
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A 1.70 m cylindrical rod of diameter 0.450 cm is connected to a power supply that maintains a constant potential difference of 13.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.3 A , while at 92.0 ∘C it reads 17.0 A . You can ignore any thermal expansion of the rod.
a) Find the resistivity and for the material of the rod at 20 ∘C.
b) Find the temperature coefficient of resistivity at 20 ∘Cfor the material of the rod.
a)the resistivity and for the material of the rod at 20 ∘C is 1.53 × 10⁻⁷ Ω m.
b) the temperature coefficient of resistivity at 20 ∘Cfor the material of the rod is 7.29 × 10⁻³ K⁻¹.
a) Resistivity is defined as the resistance offered by a wire of unit length and unit area of cross-section. Its SI unit is Ω m.
It depends on temperature and is represented by the symbol ρ. Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
Hence the formula for resistivity is given by:
ρ = RA / L
Where,ρ = Resistivity of the material.
A = Area of cross-section of the rod
L = Length of the rod
R = Resistance
We can calculate R from the following equation:
R = V / I
Where, V = Potential difference across the rod
I = Current flowing through the rod.
The resistivity and the material of the rod at 20 °C are given by:ρ = RA / L= [(D/2)²π] [V/I] / L= [(0.0045/2)²π] [13/18.3] / 1.7= 1.53 × 10⁻⁷ Ω m.
b) Temperature coefficient of resistivity is defined as the change in resistivity per degree change in temperature. It is given by:
α = (ρ₂ - ρ₁) / ρ₁ (T₂ - T₁)
Where,α = Temperature coefficient of resistivity.
ρ₂ = Resistivity at 92 °C.
ρ₁ = Resistivity at 20 °C.T₂ = 92 + 273 = 365 K.T₁ = 20 + 273 = 293 K.
Substituting the values of ρ₂, ρ₁, T₂, and T₁ in the formula, we get:
α = (1.57 × 10⁻⁷ - 1.53 × 10⁻⁷) / (1.53 × 10⁻⁷) (365 - 293)= 3.88 × 10⁻⁴ / 0.0531= 7.29 × 10⁻³ K⁻¹.
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