It will take approximately 7.96 seconds for the tire to accelerate from 33.3 revolutions per minute to 109 revolutions per minute with a rotational acceleration of 1.01 rad/s².
To solve this problem, we need to find the time it takes for the tire to accelerate from 33.3 revolutions per minute to 109 revolutions per minute, given its rotational acceleration.
First, let's convert the given rotational velocities to radians per second:
Initial rotational velocity (ω1) = 33.3 revolutions per minute
Final rotational velocity (ω2) = 109 revolutions per minute
To convert revolutions per minute to radians per second, we can use the conversion factor:
1 revolution = 2π radians
1 minute = 60 seconds
So, we have:
ω1 = 33.3 revolutions per minute × (2π radians / 1 revolution) × (1 minute / 60 seconds)
= 3.49 radians per second
ω2 = 109 revolutions per minute ×(2π radians / 1 revolution) × (1 minute / 60 seconds)
= 11.45 radians per second
Now, we can use the rotational acceleration and the initial and final velocities to find the time (t) using the following equation:
ω2 = ω1 + α × t
Where:
ω1 = initial rotational velocity
ω2 = final rotational velocity
α = rotational acceleration
t = time
Rearranging the equation to solve for t:
t = (ω2 - ω1) / α
Substituting the given values:
t = (11.45 radians per second - 3.49 radians per second) / 1.01 rad/s²
t ≈ 7.96 seconds
Therefore, it will take approximately 7.96 seconds for the tire to accelerate from 33.3 revolutions per minute to 109 revolutions per minute with a rotational acceleration of 1.01 rad/s².
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A circular region 8.00 cm in radius is filled with an electric field perpendicular to the face of the circle. The magnitude of the field in the circle varies with time as E(t)=E0cos(ωt) where E0=10.V/m and ω=6.00×109 s−1. What is the maximum value of the magnetic field at the edge of the region? T
Therefore, the maximum value of the magnetic field at the edge of the region is 6.37×10−7 T. Answer: 6.37×10−7 T.
The time-varying electric field produces a time-varying magnetic field according to Faraday's law. The maximum magnetic field on the edge of the circular region can be determined using the equation for the magnetic field: B = μ0ωE0r / (2c) where μ0 is the permeability of free space, ω is the angular frequency, E0 is the amplitude of the electric field, r is the radius of the circular region, and c is the speed of light.
This equation applies when the radius of the region is much smaller than the wavelength of the electromagnetic wave. Here, the radius is only 8.00 cm, whereas the wavelength is λ = 2πc / ω = 5.24×10−3 cm. Therefore, the equation is valid. We can substitute the given values to get: Bmax = μ0ωE0r / (2c) = (4π×10−7 T m A−1)(6.00×109 s−1)(10. V/m)(8.00×10−2 m) / (2 × 3.00×108 m/s) = 6.37×10−7 T.
Therefore, the maximum value of the magnetic field at the edge of the region is 6.37×10−7 T. Answer: 6.37×10−7 T.
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What is a convergent lens?
What is a divergent lens?
How do we find out the focal length of a convergent lens?
How does a refractive telescope work?
Explain the physical characteristics of the images the form by the refractive and reflective telescopes.
A convergent lens is a lens that refracts light and forms a real or virtual image of an object. This type of lens is thicker at the center than at the edges and is also known as a convex lens. Light passing through a convergent lens is brought to a focal point, causing the rays to converge on a single point.
A divergent lens is a lens that spreads out the light rays that enter it and forms a virtual image. This type of lens is thinner at the center than at the edges and is also known as a concave lens. The light passing through the lens is bent in a way that causes the rays to diverge away from a single point.
The focal length of a convergent lens can be found using the lens equation, which is given as:
1/f = 1/v - 1/u
Where f is the focal length of the lens, v is the distance from the lens to the image, and u is the distance from the lens to the object.
A refractive telescope works by using two lenses, a convergent lens to collect light and a divergent lens to magnify the image. The light enters the telescope and is collected by the convergent lens, which focuses the light to a point. The light then passes through the divergent lens, which magnifies the image and forms a virtual image for the observer to see.
The physical characteristics of the images formed by refractive and reflective telescopes are different. Refractive telescopes produce images that are chromatic, meaning they have different colors around the edges of the image. They also produce images that are slightly distorted due to the lens being curved. Reflective telescopes produce images that are not chromatic and are free of the distortion that is produced by the curvature of the lens.
A convergent lens refracts light and forms a real or virtual image, while a divergent lens spreads out the light rays and forms a virtual image. The focal length of a convergent lens can be found using the lens equation. Refractive telescopes use lenses to collect and magnify light, while reflective telescopes use mirrors to reflect the light and form an image. The physical characteristics of the images formed by refractive and reflective telescopes are different.
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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
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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a) A Hall-effect probe operates with a 107mA current. When the probe is placed in a uniform magnetic field with a magnitude of 0.0806T, it produces a Hall voltage of 0.689 μV. When it is measuring an unknown magnetic field, the Hall voltage is 0.352 μV. What is the unknown magnitude of the field?
b) If the thickness of the probe in the direction of B is 1.94mm, calculate the charge-carrier density (each of charge e).
(a) The unknown magnitude of the field is 0.00506 T.
(b) The charge-carrier density is 495 × 1019 m⁻³.
a) The Hall coefficient for the probe can be calculated using the equation: RH = VHB/I = 0.689μV/(107mA × 0.0806T) = 8.12×10⁻⁷ m³/C
The unknown magnetic field's magnitude can be determined using the equation: VB = RH × I × B0.352 × 10-6 V = 8.12 × 10⁻⁷ m³/C × 107 mA × BUnknown magnetic field, B = 0.00506 T
b) The charge-carrier density (n) can be calculated using the equation:n = 1/Re × e × μn, Where Re is the resistance of the material, e is the charge of an electron, and μn is the mobility of the material.
The resistance of the probe can be calculated using the equation: Re = l/(σt)where l is the length of the probe, t is the thickness of the probe in the direction of B, and σ is the conductivity of the material. Assuming the probe is rectangular in shape, we can use the equation: Re = w × h/(σt)where w is the width of the probe, and h is the height of the probe.
The area of the probe can be calculated using the equation:
A = w × h = t × w = 1.94 × 10⁻³ m²
The conductivity of the material can be calculated using the equation:σ = n × e2 × μ
The mobility of the material is given by the Hall coefficient equation:
RH = 1/ne = 1/Re × B
The charge-carrier density can now be calculated using the equation:n = 1/Re × e × μn = (B/Re × RH) × e × μn = (0.00506 T/Re × 8.12 × 10⁻⁷ m³/C) × 1.6 × 10⁻¹⁹ C × 0.001 m2/Vs = 495 × 1019 m⁻³
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A bug of mass 0.026 kg is at rest on the edge of a solid cylindrical disk (M=0.10 kg,R=0.13 m) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 14.5rad/s. The bug crawls to the center of the disk. (a) What is the new angular velocity of the disk (in rad/s)? (Enter the magnitude. Round your answer to at least one decimal place.) rad/s (b) What is the change in the kinetic energy of the system (in J)? स ] (c) If the bug crawls back to the outer edge of the disk, what is the angular velocity of the disk (in rad/s) then? (Enter the magnitude.) rad/s (d) What is the new kinetic energy of the system (in J)? J (e) What is the cause of the increase and decrease of kinetic energy? The work of the bug crawling on the disk causes the kinetic energy to increase or decrease. Score: 1 out of 1 Comment:
a)The new angular velocity of the disk is 1.45 rad/s.b)Change in the kinetic energy of the system is given by:ΔK=-0.592 J.c)The new angular velocity of the disk is 1.45 rad/s.d)New kinetic energy of the system is given by:Kf = 0.385 J.e)The cause of the increase and decrease of kinetic energy is the work done by the bug.
Given data: Mass of the bug = m₁ = 0.026 kgMass of the disk = M = 0.10 kgRadius of the disk = R = 0.13 mInitial angular velocity of the disk = ω₁ = 14.5 rad/sInitial moment of inertia of the disk = I₁ = (1/2)MR²Final moment of inertia of the disk = I₂ = (1/2)M(R/2)² + M(3R/2)² = 5MR²/4 = 0.08125 kg-m².
Let the new angular velocity of the disk be ω₂. Then, using the law of conservation of angular momentum, we get:I₁ω₁ = I₂ω₂ω₂ = I₁ω₁/I₂ω₂ = (0.5 × 0.10 × (0.13)² × 14.5)/(0.08125 × 14.5) = 1.45 rad/sTherefore, the new angular velocity of the disk is 1.45 rad/s.
Change in the kinetic energy of the system is given by:ΔK = Kf - Ki = (1/2)I₂ω₂² - (1/2)I₁ω₁² = (1/2)(0.08125)(1.45² - 14.5²) J= -0.592 J (negative sign indicates decrease in kinetic energy)If the bug crawls back to the outer edge of the disk, then the new angular velocity of the disk is the same as the initial angular velocity (since the angular momentum is conserved):ω₃ = ω₁ = 14.5 rad/s.
New kinetic energy of the system is given by:Kf = (1/2)I₁ω₁² = (1/2)(0.10)(0.13)²(14.5)² J= 0.385 J.
The cause of the increase and decrease of kinetic energy is the work done by the bug. When the bug crawls towards the center of the disk, it does negative work (i.e. work done by external force is negative) and the kinetic energy of the system decreases.
When the bug crawls towards the outer edge of the disk, it does positive work (i.e. work done by external force is positive) and the kinetic energy of the system increases.
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A crate with a mass of 193.5 kg is suspended from the end of a uniform boom with a mass of 90.3 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall. Calculate the tension in the cable.
To calculate the tension in the cable supporting the boom and the crate, we need to consider the equilibrium of forces acting on the system.
The crate has a mass of 193.5 kg, while the boom itself has a mass of 90.3 kg. The upper end of the boom is supported by the cable attached to the wall, and the lower end is supported by a pivot on the same wall.
In this situation, we can start by considering the forces acting on the boom. The downward force of gravity acting on the boom is equal to the sum of the weight of the crate and the weight of the boom itself. This force acts at the center of mass of the boom. To maintain equilibrium, the tension in the cable must balance this downward force.
By summing the forces acting vertically, we can set up the equation: Tension - Weight of crate - Weight of boom = 0. The weight of the crate is given by the mass of the crate multiplied by the acceleration due to gravity (9.8 m/s^2). The weight of the boom is calculated similarly using its mass.
Solving the equation, we can find the tension in the cable by rearranging terms: Tension = Weight of crate + Weight of boom.
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A 30.4 cm diameter coil consists of 23 turns of circular copper wire 1.80 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.70E-3 T/s. Determine the current in the loop.
The current in a 30.4 cm diameter coil with 23 turns of circular copper wire can be determined by calculating the rate of change of a uniform magnetic field perpendicular to the coil's plane, which is 8.70E-3 T/s. The current is found to be 0.0979 A.
To find the current in the loop, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in a closed loop is equal to the negative rate of change of magnetic flux through the loop. In this case, the loop is a coil with 23 turns, and the diameter of the coil is given as 30.4 cm. The magnetic field is changing at a rate of 8.70E-3 T/s.
First, we calculate the area of the coil. The radius of the coil can be determined by dividing the diameter by 2, giving us a radius of 15.2 cm (0.152 m). The area of the coil is then calculated using the formula for the area of a circle: [tex]A = \pi r^2[/tex]. Plugging in the value, we find [tex]A = 0.07292 m^2[/tex].
Next, we calculate the rate of change of magnetic flux through the coil by multiplying the magnetic field change rate (8.70E-3 T/s) by the area of the coil ([tex]A = 0.07292 m^2[/tex]). The result is 6.349E-4 Wb/s (webers per second).
Finally, we use Ohm's law, V = IR, to find the current in the loop. The induced EMF is equal to the voltage, so we have EMF = IR. Rearranging the formula, we get I = EMF/R. Substituting the values, we find I = 6.349E-4 Wb/s divided by the resistance of the loop.
To determine the resistance, we need the length of the wire. The length can be calculated by multiplying the circumference of the coil by the number of turns. The circumference is given by the formula [tex]C = 2\pi r[/tex], where r is the radius of the coil. Substituting the values, we find C = 0.957 m. Multiplying the circumference by the number of turns (23), we get the length of the wire as 22.01 m.
Using the formula for the resistance of a wire, R = ρL/A, where ρ is the resistivity of copper ([tex]1.72 * 10^-^8[/tex] Ωm), L is the length of the wire, and A is the cross-sectional area of the wire, we can calculate the resistance. Substituting the values, we find [tex]R = 3.59 * 10^-^4[/tex] Ω.
Now, we can calculate the current using the formula I = EMF/R. Substituting the values, we find I = 6.349E-4 Wb/s divided by [tex]3.59 *10^-^4[/tex] Ω, which equals 0.0979 A.
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A 35 kg student bounces up from a trampoline with a speed of 7.4 m/s.
(a) Determine the work done on the student by the force of gravity when she is 1.3 m above the trampoline.
(b) Determine her speed at 1.3 m above the trampoline.
(a) Therefore, Wg = -(1/2)mv^2where Wg is the work done by gravity on the student. Wg = - (1/2)mv^2 = - (1/2)(35 kg)(7.4 m/s)^2 = -1368.5 J. (b) Therefore, Work done by gravity on the student = -1368.5 J, Speed of the student at 1.3 m above the trampoline = 0 m/s.
Mass of student, m = 35 kg, Speed of the student when he leaves the trampoline, v = 7.4 m/s
Distance between the student and the trampoline, h = 1.3 m
(a) Work done on the student by the force of gravity
when she is 1.3 m above the trampoline. Work done by gravity on the student will be equal to the decrease in the student's kinetic energy. Initial kinetic energy of the student, K1 = (1/2)mv1^2Where v1 is the initial velocity of the student
Final kinetic energy of the student, K2 = 0 (At the highest point, velocity becomes zero)
The work done by gravity, Wg = K1 - K2 = (1/2)mv1^2 – 0 = (1/2)mv1^2The gravitational potential energy of the student at a height h above the trampoline, U1 = mgh
The gravitational potential energy of the student at the highest point, U2 = 0
Therefore, the decrease in gravitational potential energy of the student, U1 - U2 = mgh
Joule’s Law of Work and Energy states that the total work done on an object is equal to the change in its kinetic energy.
The work done by gravity on the student must be equal to the decrease in his kinetic energy.
Wg = -ΔKWhere ΔK is the change in kinetic energy of the student.This equation can be written as follows: Wg = - (Kf - Ki)Where Ki is the initial kinetic energy of the student, and Kf is the final kinetic energy of the student.
The final kinetic energy of the student is zero since he stops at the highest point.
The initial kinetic energy of the student is (1/2)mv^2, where m is the mass of the student and v is his speed just before leaving the trampoline. Therefore, Wg = -(1/2)mv^2where Wg is the work done by gravity on the student. Wg = - (1/2)mv^2 = - (1/2)(35 kg)(7.4 m/s)^2 = -1368.5 J (Negative sign indicates the work done by gravity is in opposite direction to the motion of the student)
(b) Determine her speed at 1.3 m above the trampoline. The speed of the student just before he leaves the trampoline is 7.4 m/s.
When he reaches a height of 1.3 m above the trampoline, his speed will be zero.
This is because at the highest point, the velocity of the student is zero. So, the speed of the student when he is 1.3 m above the trampoline is zero.
Therefore, Work done by gravity on the student = -1368.5 JSpeed of the student at 1.3 m above the trampoline = 0 m/s.
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Snell's Law: Light enters air from an ice cube. The angle of refraction will be... o less than the angle of incidence greater than the angle of incidence equal to the angle of incidence
The angle of refraction when light enters air from an ice cube will be greater than the angle of incidence.
Snell's law describes the relationship between the angles of incidence and refraction when light passes through the interface between two different media.
It states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of light in the two media. In this case, as light travels from the denser medium (ice) to the less dense medium (air), it undergoes refraction.
When light passes from a denser medium to a less dense medium, such as from ice to air, the angle of refraction is always greater than the angle of incidence.
This phenomenon is due to the change in the speed of light as it enters the new medium. As light enters air from an ice cube, it speeds up since the refractive index of air is lower than that of ice.
This increase in speed causes the light rays to bend away from the normal, resulting in a greater angle of refraction compared to the angle of incidence.
Therefore, the angle of refraction when light enters air from an ice cube will be greater than the angle of incidence, according to Snell's law.
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I₁ = 102 - 32° Arms I2 = 184 + 49° Arms 13 = = 172 + 155° Arms ZA = 3 + j2 Ω Zg = 4 - j4 Ω ZA Zc = 10-j3 n Ω 13 The average power absorbed by impedance Z, in the circuit above is closest to... The reactive power absorbed by impedance Zc in the circuit above is closest to... I₁ ZB Zc
Average power absorbed by impedance Z: 10404 * Re(Z)
Reactive power absorbed by impedance Zc: 29584 * Im(Zc)
To calculate the average power absorbed by impedance Z and the reactive power absorbed by impedance Zc in the given circuit, we can use the formulas for power calculations in AC circuits.
Given values:
I₁ = 102 ∠ -32° A
I₂ = 184 ∠ 49° A
I₃ = 172 ∠ 155° A
ZA = 3 + j2 Ω
Zg = 4 - j4 Ω
Zc = 10 - j3 Ω
Average Power Absorbed by Impedance Z:
The average power (P) absorbed by an impedance Z can be calculated using the formula:
P = |I|² * Re(Z)
Where |I| is the magnitude of the current and Re(Z) is the real part of the impedance.
In this case, the impedance Z is not directly given, but we can calculate it by adding the parallel combination of ZA and Zg:
Z = (ZA * Zg) / (ZA + Zg)
Calculating Z:
Z = (3 + j2) * (4 - j4) / (3 + j2 + 4 - j4)
= (12 + j12 + j8 - j8) / (7 - j2)
= (12 + j20) / (7 - j2)
Now, we can calculate the average power absorbed by impedance Z:
P = |I₁|² * Re(Z)
= |102 ∠ -32°|² * Re(Z)
= (102)² * Re(Z)
= 10404 * Re(Z)
Reactive Power Absorbed by Impedance Zc:
The reactive power (Q) absorbed by an impedance Zc can be calculated using the formula:
Q = |I|² * Im(Zc)
Where |I| is the magnitude of the current and Im(Zc) is the imaginary part of the impedance Zc.
Now, we can calculate the reactive power absorbed by impedance Zc:
Q = |I₃|² * Im(Zc)
= |172 ∠ 155°|² * Im(Zc)
= (172)² * Im(Zc)
= 29584 * Im(Zc)
Therefore, the closest values for the average power absorbed by impedance Z and the reactive power absorbed by impedance Zc are:
Average power absorbed by impedance Z: 10404 * Re(Z)
Reactive power absorbed by impedance Zc: 29584 * Im(Zc)
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A(n) ultraviolet photon has a wavelength of 0.00900 cm. Find the momentum, the frequency, and the energy of the photon in electron volts. (a) the momentum kg · m/s (b) the frequency Hz (c) the energy of the photon in electron volts eV Need Help? Read It
A(n) ultraviolet photon has a wavelength of 0.00900 cm.(a)Frequency ≈ 3.33 x 10^12 Hz.(b)Energy ≈ 1.366 eV.(c) Energy of the photon: 1.366 eV
To find the momentum of a photon, we can use the formula:
Momentum = (Planck's constant) / (wavelength)
The Planck's constant, denoted as h, is approximately 6.626 x 10^-34 J·s.
Given the wavelength of the ultraviolet photon as 0.00900 cm (or 0.0000900 m), we have:
Momentum = (6.626 x 10^-34 J·s) / (0.0000900 m)
Momentum ≈ 7.362 x 10^-30 kg·m/s
(a) Momentum: 7.362 x 10^-30 kg·m/s
To find the frequency of the photon, we can use the formula:
Frequency = (speed of light) / (wavelength)
The speed of light, denoted as c, is approximately 3.00 x 10^8 m/s.
Using the wavelength of the photon as 0.00900 cm (or 0.0000900 m), we have:
Frequency = (3.00 x 10^8 m/s) / (0.0000900 m)
Frequency ≈ 3.33 x 10^12 Hz
(b) Frequency: 3.33 x 10^12 Hz
To find the energy of the photon in electron volts (eV), we can use the formula:
Energy = (Planck's constant) ×(frequency) / (electron charge)
The electron charge, denoted as e, is approximately 1.602 x 10^-19 C.
Substituting the values, we have:
Energy = (6.626 x 10^-34 J·s)× (3.33 x 10^12 Hz) / (1.602 x 10^-19 C)
Energy ≈ 1.366 eV
(c) Energy of the photon: 1.366 eV
Note: 1 electron volt (eV) is defined as the energy gained or lost by an electron when it moves through a potential difference of 1 volt.
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A transformer transfers electrical energy from primary to secondary usually with a change in a) Frequency b) power c) time period d) none of the previous the 2- The voltage per turn of the high voltage winding of a transformer is voltage per turn of the low voltage winding. a) more than b) less than c) the same as d) none of the previous 3- A single phase transformer, 50 Hz, core-type transformer has square core of 24.5 cm side. The permissible flux density is 1 Wb/m². if the iron factor is 0.9, the induced voltage per turn is -----------. a) 12 b) 6 c) 11 d) none of the previous. 4- A transformer takes a current of 0.5A and absorbs 60 W when the primary is connected to its normal supply of 220 V, 50 Hz; the secondary being on open circuit. The magnetizing current is --‒‒‒‒‒‒‒‒ a) 0.42 A b) 0.22 A c) 0.3 A d) none of the previous. 5- A transformer will have maximum efficiency at --- a) No-load. b) full-load. c) if W₁ = WcuFL. d) none of the previous.
1) b) power. 2) c) the same as. 3) b) 6. 4) a) 0.42 A. 5) b) full-load.
1) The correct answer is b) power. A transformer transfers electrical energy from the primary winding to the secondary winding, resulting in a change in power. The primary coil converts the incoming electrical power into a magnetic field, which induces a corresponding voltage in the secondary coil. While the voltage and current may change in the transformation process, the power remains constant (ideally), disregarding losses.
2) The voltage per turn of the high voltage winding of a transformer is the same as the voltage per turn of the low voltage winding. This relationship is based on the turns ratio of the transformer. The turns ratio determines the voltage transformation between the primary and secondary windings. If the turns ratio is, for example, 1:2, the high voltage winding will have twice as many turns as the low voltage winding, resulting in the same voltage per turn for both windings.
3) In this case, the induced voltage per turn of the transformer can be calculated by dividing the permissible flux density (1 Wb/m²) by the iron factor (0.9) and multiplying it by the area of the square core (24.5 cm × 24.5 cm). The result is 6.
4) The magnetizing current of a transformer is the current required to establish the magnetic field in the core. In this scenario, when the primary is connected to its normal supply of 220 V, 50 Hz, and the secondary is on open circuit, the magnetizing current is 0.42 A.
5) A transformer achieves its maximum efficiency at full-load. At full-load, the power output of the transformer is closest to the power input, resulting in the highest efficiency. At no-load or other partial loads, the efficiency of the transformer decreases due to various losses such as core losses and copper losses. Therefore, the transformer operates most efficiently when operating at its designed full-load capacity.
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The switch opens at t = 0 after a very long time. Find v(t) for t > 0. Draw circuits clearly for each step using 4-step approach to illustrate the situation when t<0 and t>0 when doing circuit analysis for full credit. Write final answers in the box provided. [10 pts] 6 V 30 k 0 47 (1) 60 k 5 μF 60 k
The voltage of switch function v(t) for t > 0 is approximately 5.992 V. The 5 μF capacitor does not affect the voltage at steady-state.
To analyze the circuit and find the voltage function v(t) for t > 0, let's go through the 4-step approach and consider the circuit at t < 0 and t > 0 separately.
Step 1: Circuit at t < 0 (before the switch opens)
At t < 0, the switch is closed, and the capacitor is assumed to have been charged to a steady-state. In this case, the capacitor behaves like an open circuit, and the 60 kΩ resistor is effectively disconnected.
The circuit at t < 0 can be represented as follows:
Step 2: Circuit at t = 0 (when the switch opens)
At t = 0, the switch opens. The capacitor retains its voltage, and the voltage across it remains constant. However, the circuit topology changes as the capacitor now acts as a voltage source with an initial voltage of 6 V.
The circuit at t = 0 can be represented as follows:
Step 3: Circuit at t > 0 (after the switch opens)
At t > 0, the switch remains open, and the circuit reaches a new steady-state. The capacitor acts like an open circuit in the steady-state, and the 60 kΩ resistor is effectively disconnected.
The circuit at t > 0 can be represented as follows:
Step 4: Solving for v(t) for t > 0
To find the voltage function v(t) for t > 0, we can use the voltage divider rule to determine the voltage across the 30 kΩ resistor.
The voltage across the 30 kΩ resistor is given by:
v(t) = (30 kΩ / (30 kΩ + 47 Ω)) * 6 V
Simplifying the equation:
v(t) = (30000 / 30047) * 6 V
v(t) ≈ 5.992 V (approximately)
Therefore, the voltage function v(t) for t > 0 is approximately 5.992 V.
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A moving particle encounters an external electric field that decreases its kinetic energy from 9320 eV to 6600 eV as the particle moves from position A to position B. The electric potential at A is -65.0 V, and that at B is +15.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.
A moving particle encounters an external electric field that decreases its kinetic energy from 9320 eV to 6600 eV as the particle moves from position A to position B. The electric potential at A is -65.0 V, and that at B is +15.0 V.
We need to determine the charge of the particle.
The work done on the charged particle as it moves from point A to point B is
W = q (Vb - Va)
As the charged particle moves from point A to point B, the potential difference is,
Vb - Va = (+15 V) - (-65 V) = 80 V
Work done on the charged particle, W is,
W = q (Vb - Va) = (1.6 × 10^-19 C) × (80 V) = 1.28 × 10^-17 J
Kinetic energy of the charged particle at position A is,
KEA = 9320 eV = 1.495 × 10^-15 J
And the kinetic energy of the charged particle at position B is,
KEB = 6600 eV = 1.061 × 10^-15 J
The loss of kinetic energy of the charged particle from position A to position B is
W = KEA - KEB1.28 × 10^-17 J = (1.495 × 10^-15 J) - (1.061 × 10^-15 J)1.28 × 10^-17 J = 0.434 × 10^-15 J
Therefore, charge of the particle is,
q = W / (Vb - Va) = 1.28 × 10^-17 C / 80 V = 1.6 × 10^-19 C
As work done on the charged particle is negative, the algebraic sign of charge is also negative. Therefore, the charge of the particle is -1.6 × 10^-19 C.
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Q5. Solve the equation for temperature distribution in a rod d²T T(0) = 0 and T(1)-100°C, take dx-0.25, To=30°C 7 Marks dxi (T-To)
The temperature gradient is constant throughout the length of the rod. Thus, the temperature distribution in the rod is linear and is given by T=100x.
Given equation is d²T/dx²=0 (using equation for heat conduction in one direction)According to the question, the rod is of length 1m. So, let the length of the elemental segment of the rod is dx. Since we know that the thermal conductivity is constant then: $\frac {d²T}{dx²}$= k $\frac {d²T}{dt²}$=0 (Since k is constant).So, $\frac{dT}{dx}$=c₁, integrating both sides with respect to x gives T=c₁x + c₂.The boundary conditions are, T(0)=0 and T(1)=100°CPutting T(0)=0, we get c₂=0Putting T(1)=100, we get c₁=100Therefore, T=100xTaking dx=0.25, To=30°CThe temperature distribution in the rod is: x 0.00 0.25 0.50 0.75 1.00T(x) 0.00 25.00 50.00 75.00 100.00Hence, the temperature of the rod at various segments are as follows: At x = 0.25m, T = 25°CAt x = 0.50m, T = 50°CAt x = 0.75m, T = 75°CAt x = 1m, T = 100°CThe temperature of the rod is increasing linearly from 0 to 100°C. The gradient of the line represents the rate of increase of temperature. The temperature gradient is constant throughout the length of the rod. Thus, the temperature distribution in the rod is linear and is given by T=100x.
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An ideal Carnot engine operates between a high temperature reservoir at 219°C and a river with water at 17°C. If it absorbs 4000 J of heat each cycle, how much work per cycle does it perform?
The work per cycle that is performed by the is Carnot engine -42382.4 J.
The Carnot engine is an ideal reversible engine that is used to understand the working of heat engines. It works between two temperatures, namely the high temperature and low temperature to extract work from heat. It is based on the concept of the second law of thermodynamics. It is used to establish the maximum efficiency of the engines.
The work per cycle that is performed by an ideal Carnot engine operating between a high temperature reservoir at 219°C and a river with water at 17°C and it absorbs 4000 J of heat each cycle can be calculated as:
Wcycle = QH - QL
where
Wcycle is the work per cycle,
QH is the heat absorbed per cycle,
QL is the heat rejected per cycle
The heat rejected per cycle QL can be calculated as:
QL = TH / (TH - TL) * QH
where
TH is the temperature of the high temperature reservoir,
TL is the temperature of the low-temperature reservoir
Substituting the given values in the above formula,
QL = 219 / (219 - 17) * 4000= 46382.4 J
The work per cycle can be calculated by substituting the values in the formula:
Wcycle = QH - QL= 4000 - 46382.4= -42382.4 J (Negative sign indicates that work is done on the engine rather than by the engine)
Therefore, the work per cycle that is performed by the Carnot engine is -42382.4 J.
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Which of the following is correct in AC circuits? For a given peak voltage, the peak current is inversely proportional to capacitance, inversely proportional to inductance, and directly proportional to resistance. For a given peak voltage, the peak current is directly proportional to resistance, directly proportional to capacitance, and inversely proportional to inductance. For a given peak voltage, the peak current is inversely proportional to resistance, inversely proportional to capacitance, and inversely proportional to inductance. For a given peak voltage, the peak current is directly proportional to capacitance, inversely proportional to inductance, and inversely proportional to resistance.
For a given peak voltage, the peak current in an AC circuit is directly proportional to resistance, inversely proportional to capacitance, and inversely proportional to inductance.
In an AC circuit, the relationship between peak voltage (Vp), peak current (Ip), resistance (R), capacitance (C), and inductance (L) can be described using Ohm's Law and the formulas for capacitive reactance (Xc) and inductive reactance (Xl).
Ohm's Law states that Vp = Ip * R, where Vp is the peak voltage and R is the resistance. According to Ohm's Law, the peak current is directly proportional to resistance. Therefore, for a given peak voltage, the peak current is directly proportional to resistance.
In a capacitive circuit, the capacitive reactance (Xc) is given by Xc = 1 / (2πfC), where f is the frequency of the AC signal and C is the capacitance. The higher the capacitance, the lower the capacitive reactance. Therefore, for a given peak voltage, the peak current is inversely proportional to capacitance.
In an inductive circuit, the inductive reactance (Xl) is given by Xl = 2πfL, where L is the inductance. The higher the inductance, the higher the inductive reactance. Therefore, for a given peak voltage, the peak current is inversely proportional to inductance.
Thus, the correct statement is: For a given peak voltage, the peak current is directly proportional to resistance, inversely proportional to capacitance, and inversely proportional to inductance.
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The Twisti In The Wring (The Nolks Slides To The Right In The Diagram Below.)
Always wear gloves and eye protection when handling wire ropes. In conclusion, the Twisti in the wring technique is a very effective method for unraveling the twisted ropes. It is easy to use and requires minimal effort.
Twisti in the wring refers to the act of unraveling the twisted ropes. The Nolks Slides to the Right in the Diagram Below is a type of the Twisti in the wring technique. In this technique, we use two strands of wire ropes to form the twist.
The twist can be easily undone by simply sliding the nolks or the kinks in the ropes. This technique is commonly used in the shipping industry to unravel the twisted ropes.However, before you start unraveling the ropes, you need to check the strength and the tensile strength of the wire ropes. The strength of the wire ropes depends on the size, grade, and construction of the wire ropes.
The tensile strength of the wire ropes is measured in pounds per square inch (psi).The Twisti in the wring technique is a very effective method for unraveling the twisted ropes. It is easy to use and requires minimal effort. The technique is commonly used in the shipping industry to unravel the twisted ropes. It is important to follow proper safety precautions when using this technique.
Always wear gloves and eye protection when handling wire ropes. In conclusion, the Twisti in the wring technique is a very effective method for unraveling the twisted ropes. It is easy to use and requires minimal effort.
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Tarzan wishes to save Jane from the jaws of a large Tyrannosaurus Rex. He deftly throws a rope upwards, catching it on a lower tooth which is at a height of 150 m above the ground. He knows that jungle vines can withstand a tension force of 1.5 times his weight. If he has a mass of 200 kg find a. the maximum acceleration of Tarzan up the vine. b. the length of time required to climb the vine
Tarzan, with a mass of 200 kg, aims to rescue Jane from a Tyrannosaurus Rex by using a rope. The rope is attached to a tooth 150 m above the ground. The maximum acceleration of Tarzan up the vine is 1.5m/s^2. The length of time required to climb the vine is 17.32 seconds.
To calculate the maximum acceleration of Tarzan up the vine, we need to consider the tension force the vine can withstand. Since the vine can endure a tension force of 1.5 times Tarzan's weight, we multiply his mass (200 kg) by 1.5 to find the maximum tension force: 200 kg * 1.5 = 300 kg.
a)To calculate the maximum acceleration, we can use Newton's second law of motion, which states that force equals mass multiplied by acceleration (F = m * a). In this case, the force is the tension force (300 kg), and the mass is Tarzan's weight (200 kg). Rearranging the formula, we get [tex]a = F / m = 300 kg / 200 kg = 1.5 m/s^2[/tex].
b)To find the time required to climb the vine, we need to determine the distance Tarzan needs to cover. This distance is equal to the height of the tooth, which is 150 m. We can use the equation of motion, [tex]s = ut + (1/2)at^2[/tex], where s is the distance, u is the initial velocity (which is zero in this case), a is the acceleration ([tex]1.5 m/s^2[/tex]), and t is the time we want to find. Rearranging the formula, we get [tex]t = \sqrt(2s / a) = \sqrt(2 * 150 m / 1.5 m/s^2) = 17.32 seconds.[/tex]
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Sound waves entering human ear first pass through the auditory canal before reaching the eardrum. If a typical adult has an auditory canal of 2.5cm long and 7.0mm in diameter, suppose that when you listen to ordinary conversation, the intensity of sound waves is about 3.2 × 10−6W/m2 ; a) What is the average power delivered to the eardrum?
Sound waves entering human ear first pass through the auditory canal before reaching the eardrum. If a typical adult has an auditory canal of 2.5cm long and 7.0mm in diameter, suppose that when you listen to ordinary conversation, the intensity of sound waves is about 3.2 × 10−6W/m^2 ,, the average power delivered to the eardrum when listening to ordinary conversation is approximately 1.23 × 10^(-10) Watts.
To calculate the average power delivered to the eardrum, we can use the formula:
Power = Intensity× Area
Given:
Intensity (I) = 3.2 × 10^(-6) W/m^2
Auditory canal length (L) = 2.5 cm = 0.025 m
Auditory canal diameter (d) = 7.0 mm = 0.007 m
First, we need to find the area of the cross-section of the auditory canal. Since the canal has a circular cross-section, the area can be calculated using the formula:
Area = π × (d/2)^2
Substituting the given values:
Area = π × (0.007/2)^2
Area ≈ 3.85 × 10^(-5) m^2
Now we can calculate the power delivered to the eardrum:
Power = Intensity × Area
Power = (3.2 × 10^(-6) W/m^2) * (3.85 × 10^(-5) m^2)
Power ≈ 1.23 × 10^(-10) W
Therefore, the average power delivered to the eardrum when listening to ordinary conversation is approximately 1.23 × 10^(-10) Watts.
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What is the critical angle for light traveling from crown glass (n=1.52) into water ( n=1.33) ? Just two significant digits please.
The critical angle is 61°. The critical angle is the angle of incidence in the first medium such that the angle of refraction in the second medium is 90 degrees.
Using Snell's law, we have:
n1 sin θc = n2
where
n1 is the refractive index of the first medium (crown glass)
n2 is the refractive index of the second medium (water)
θc is the critical angle
Plugging in the values, we get:
1.52 sin θc = 1.33
θc = sin⁻¹ (1.33/1.52) ≈ 61.1°
To two significant digits, the critical angle is 61°.
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A rock is suspended by a light string. When the rock is in air, the tension in the string is 41.5 NN . When the rock is totally immersed in water, the tension is 25.1 NN . When the rock is totally immersed in an unknown liquid, the tension is 20.0 NN .
Part A
What is the density of the unknown liquid?
Let's determine the density of the unknown liquid.
Step 1:Calculate the weight of the rockThe tension in the string while the rock is in the air= 41.5 NFrom the Newton's second law,Force= mass x accelerationTherefore, Force due to gravity= mass x acceleration due to gravity= weight of rock= mgwhere m is mass of rockg is the acceleration due to gravity= 9.8 m/s²Weight of rock= 41.5 N (given)
Step 2:Calculate the weight of the rock when it is in waterThe tension in the string when the rock is in water= 25.1 NThe weight of rock when it is in water= (Tension in string while in air) - (Tension in string while in water)= 41.5 - 25.1= 16.4 N
Step 3:Calculate the weight of the rock when it is immersed in the unknown liquidThe tension in the string when the rock is immersed in unknown liquid= 20.0 NThe weight of rock when it is in the unknown liquid= (Tension in string while in air) - (Tension in string while in unknown liquid)= 41.5 - 20.0= 21.5 N
Step 4:Calculate the buoyant force on the rock when it is in waterThe buoyant force on rock when it is in water= Weight of rock when it is in air - weight of rock when it is in water= 41.5 - 16.4= 25.1 NThe buoyant force on the rock when it is in the unknown liquid= Weight of rock when it is in air - weight of rock when it is in unknown liquid= 41.5 - 21.5= 20.0 N
Step 5:Calculate the volume of the rockTo calculate the density of the unknown liquid, we need to calculate the volume of the rock. For this, we can use Archimedes' principle.Archimedes' principle: The buoyant force on a body equals the weight of the fluid it displaces. We know the buoyant force on the rock when it is in water and in the unknown liquid, respectively.
Buoyant force= weight of displaced liquidVolume of rock = (Buoyant force in air)/ (Density of air) = (Weight of rock in air)/ (Density of air) = (41.5 N)/(1.29 kg/m³) = 32.2 x 10⁻³ m³
Step 6:Calculate the density of the unknown liquidDensity of the unknown liquid= (Weight of rock in air - weight of rock in unknown liquid)/ (Weight of fluid displaced)= (41.5 N - 21.5 N)/ (25.1 N - 20.0 N)= 20.0 N/ 5.1 N= 3.92 kg/m³The density of the unknown liquid is 3.92 kg/m³.
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A student pushes on a 5-kg box with a force of 20 N forward. The force of sliding friction is 10 N backward. What’s the acceleration of the box?
The acceleration of the box is 2 m/s².
To determine the acceleration of the box, we need to consider the net force acting on it. The net force is the vector sum of all the forces acting on the box.
In this case, the force applied by the student is 20 N forward, while the force of sliding friction is 10 N backward. Since the forces are in opposite directions, we need to subtract the frictional force from the applied force to find the net force:
Net force = Applied force - Frictional force
= 20 N - 10 N
= 10 N
Now, we can apply Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied and inversely proportional to its mass.
Net force = mass * acceleration
Rearranging the equation to solve for acceleration, we have:
Acceleration = Net force / mass
= 10 N / 5 kg
= 2 m/s²
Therefore, the acceleration of the box is 2 m/s².
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A train is moving West at 25 m/s and blows its horn which has a frequency of 256 Hz according to the train driver. A car is 500 m West of the train and is moving East at 35 m/s. If it is a hot day with a temperature of 30oC then what is frequency of the train horn observed by the car driver?
The car driver, moving towards the train, would observe a higher frequency of the train horn compared to its actual frequency due to the Doppler effect. The observed frequency can be calculated using the Doppler effect equation. The frequency of the train horn observed by the car driver is approximately 278.84 Hz.
The Doppler effect is the change in frequency or wavelength of a wave observed by an observer moving relative to the source of the wave. In this case, the car is moving towards the train, causing a shift in the frequency of the train horn observed by the car driver.
The Doppler effect equation for sound is given by:
f' = f((v + v₀) / (v + vₛ))
Where:
f' is the observed frequency,
f is the actual frequency of the sound source,
v is the speed of sound,
v₀ is the velocity of the observer (car driver), and
vₛ is the velocity of the source (train).
Given that the car is moving towards the train, its velocity (v₀) would be positive, while the velocity of the train (vₛ) would be negative.
Substituting the given values:
f' = 256 Hz * ((343 m/s + 35 m/s) / (343 m/s - 25 m/s))
By evaluating the above expression, the frequency of the train horn observed by the car driver is approximately 278.84 Hz. Thus, the car driver would hear a higher frequency compared to the actual frequency of the train horn due to the Doppler effect.
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A neutron star results when a star in its final stages collapses due to gravitational pressure, forcing the electrons to combine with the protons in the nucleus and converting them into neutrons. (a) Assuming that a neutron star has a mass of 3.00×10 30
kg and a radius of 1.20×10 3
m, determine the density of a neutron star. ×10 20
kg/m 3
(b) How much would 1.0 cm 3
(the size of a sugar cube) of this material weigh at Earth's surface? ×10 15
N
(a) Density of neutron star = 3.27 × 10¹⁷ kg/m³
(b) Weight of 1.0 cm³ neutron star at Earth's surface = 3.21 × 10¹⁵ N
(a) Density of neutron star:
Given,Mass of neutron star = 3.00 × 10³⁰ kg
Radius of neutron star = 1.20 × 10³ m
Density = Mass / Volume
Volume of neutron star = (4/3)πr³
Volume of neutron star = (4/3) × π × (1.20 × 10³)³m³
Volume of neutron star = 9.16 × 10⁹ m³
Density of neutron star = 3.00 × 10³⁰ / 9.16 × 10⁹
Density of neutron star = 3.27 × 10¹⁷ kg/m³
(b) Weight of 1.0 cm³ neutron star at Earth's surface:
We can calculate the weight using the formula;
W = mg
where, W = weight, m = mass, g = acceleration due to gravity at earth's surface
g = 9.8 m/s²
Let's convert the density into g/cm³1 kg/m³ = 10⁻⁶ g/cm³
Density = 3.27 × 10¹⁷ kg/m³
Density = 3.27 × 10¹¹ g/cm³
Mass of 1.0 cm³ neutron star = density × volume
Mass of 1.0 cm³ neutron star = 3.27 × 10¹¹ g/cm³ × 1.0 cm³
Mass of 1.0 cm³ neutron star = 3.27 × 10¹¹ g
Weight of 1.0 cm³ neutron star = mass × acceleration due to gravity
Weight of 1.0 cm³ neutron star = 3.27 × 10¹¹ g × 9.8 m/s²
Weight of 1.0 cm³ neutron star = 3.21 × 10¹² N
Weight of 1.0 cm³ neutron star = 3.21 × 10¹⁵ nN
The weight of a 1.0 cm³ neutron star at Earth's surface is 3.21 × 10¹⁵ N. Therefore, the answer is (a) Density of neutron star = 3.27 × 10¹⁷ kg/m³(b) Weight of 1.0 cm³ neutron star at Earth's surface = 3.21 × 10¹⁵ N.
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A chair of mass 15.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F=42.0 N that is directed at an angle of 43.0 ∘
below the horizontal and the chair slides along the floor. Use Newton's laws to calculate the normal force that the floor exerts on the chair
The normal force that the floor exerts on the chair is approximately 107.9 N.
Mass of chair, m = 15 kgForce, F = 42.0 NAngle, θ = 43°Normal force, N is given by,Newton’s second law of motion states that the force acting on an object is directly proportional to the acceleration produced in it and inversely proportional to its mass.
It is given by, `F = ma`Where, F is the net force applied on the object, m is the mass of the object and a is the acceleration produced in the object. When an object is in contact with a surface, it experiences two types of forces:Normal force (N)Frictional force (f)According to Newton’s third law of motion, the normal force acting on an object is equal in magnitude and opposite in direction to the force applied by the object on the surface in contact.
Hence,Normal force, N = Force applied by the object on the surface in contactLet N1 be the normal force acting on the chair. From the free-body diagram of the chair, we can write,N1 + Fsinθ = mgwhere, m is the mass of the chair, g is the acceleration due to gravity and Fsinθ is the component of force F acting parallel to the surface.
Substituting the given values in the above equation, we getN1 = mg - Fsinθ= (15 kg) × (9.8 m/s²) - (42 N) × sin 43°≈ 107.9 N.
Therefore, the normal force that the floor exerts on the chair is approximately 107.9 N.
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Drag each tile to the correct box. Arrange the letters to show the path of the light ray as it travels from the object to the viewer’s eye. An illustration depicts the passage of light ray through four positions labeled A on the top, B on the top right, C on the right middle and E on the left middle in an object. A B C D E → → → →
Answer:
Explanation:
To arrange the letters to show the path of the light ray as it travels from the object to the viewer's eye, the correct order is:
D → C → E → B → A
This sequence represents the path of the light ray starting from position D, then moving to position C, followed by E, B, and finally A.
To meet the hot water requirements of a family in summer, it is necessary to use two glass solar collectors (transmittance 0.9, emissivity 0.88), each one 1.4 m high and 2 m wide. The two collectors join each other on one of their sides so that they give the appearance of being a single collector with a size of 1.4 m x 4 m. The temperature of the glass cover is 31 °C while the surrounding air is at 22 °C and the wind is blowing at 32 km/h. The effective sky temperature for radiation exchange between the glass cover and the open sky is –46 °C. Water enters the tubes attached to the absorber plate at a rate of 0.5 kg/min. If the back surface of the absorber plate is heavily insulated and the only heat loss is through the glass cover, determine: a) the total rate of heat loss from the collector. b) If the efficiency of the collector is 21%, what will be the value of the incident solar radiation on the collector [W/m2]? Note: Efficiency is defined as the ratio of the amount of heat transferred to the water to the incident solar energy on the collector.
a)The total rate of heat loss from the collector is 12,776.99 W.b). The value of the incident solar radiation on the collector is 905.76 W/m2.
a) Total rate of heat loss from the collector:The total rate of heat loss from the collector can be determined using the following expression:Q=α * F * (Ts-Tsur),Where Q is the total rate of heat loss, α is the heat transfer coefficient, F is the area of the glass cover, Ts is the temperature of the glass cover, and Tsur is the effective sky temperature for radiation exchange between the glass cover and the open sky.
The heat transfer coefficient can be calculated as follows:α = 5.7 + 3.8V,Where V is the wind speed. The value of V is given to be 32 km/h. Converting km/h to m/s, we get:V = (32 * 1000) / (60 * 60) = 8.89 m/sSubstituting the values, we get:α = 5.7 + 3.8(8.89)α = 39.17 W/m2KThe area of the glass cover can be calculated as follows:A = 2 * 1.4 * 2A = 5.6 m2Substituting the values, we get:Q=α * F * (Ts-Tsur)Q = 39.17 * 5.6 * (31 + 273) - (-46 + 273)Q = 12, 776.99 WTherefore, the total rate of heat loss from the collector is 12,776.99 W.
b) Value of the incident solar radiation on the collector:We can use the definition of efficiency to calculate the value of the incident solar radiation on the collector.Efficiency = (Heat transferred to water / Incident solar energy) * 100Given that the efficiency is 21%, we can rearrange the above expression to calculate the incident solar energy.Incident solar energy = Heat transferred to water / (Efficiency / 100).
Substituting the values, we get:Heat transferred to water = m * Cp * ΔT,Where m is the mass flow rate, Cp is the specific heat of water, and ΔT is the temperature difference between the inlet and outlet of the absorber plate.The mass flow rate is given to be 0.5 kg/min. Converting kg/min to kg/s, we get:m = 0.5 / 60 = 0.0083 kg/sThe specific heat of water is 4.18 kJ/kgK. The temperature difference can be calculated as:T = m * Cp * ΔT / P,Where P is the power generated by the collector.
The power generated can be calculated as:P = Efficiency * Incident solar energy * FSubstituting the values, we get:T = m * Cp * ΔT / (Efficiency * Incident solar energy * F).
Rearranging the expression, we get:Incident solar energy = m * Cp * ΔT / (Efficiency * F * (Tout - Tin))Substituting the values, we get:Incident solar energy = 0.0083 * 4.18 * (60 - 22) / (0.21 * 5.6 * (60 - 31))Incident solar energy = 905.76 W/m2Therefore, the value of the incident solar radiation on the collector is 905.76 W/m2.
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A DVD is initially at rest. The disc begins to tum at a constant rate of 6.32 radio2. How many revolutions does the discoth 7000
To determine the number of revolutions the disc completes in 7000 seconds, to convert the angular velocity from radians per second to revolutions per second and then multiply it by the time duration.
The angular velocity of the DVD is given as 6.32 rad/s. One revolution is equal to 2π radians, so we can convert the angular velocity from rad/s to revolutions per second by dividing it by 2π. Thus, the angular velocity in revolutions per second is 6.32 rad/s / (2π rad/rev) ≈ 1.003 rev/s.
To find the number of revolutions the disc completes in 7000 seconds, we multiply the angular velocity in revolutions per second by the time duration. Therefore, the number of revolutions is 1.003 rev/s * 7000 s ≈ 7010 revolutions.
The DVD rotating at a constant rate of 6.32 rad/s will complete approximately 7010 revolutions in a time span of 7000 seconds.
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A coil is wrapped with 191 turns of wire around the perimeter of a circular frame (radius = 9 cm). Each turn has the same area, equal to that of the circular frame. A uniform magnetic field perpendicular to the plane of the coil is activated. This field changes at a constant rate of 20 to 80 mT in a time of 2 ms. What is the magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT? Give your answer to two decimal places.
The magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT is approximately 7.64 V.
Number of turns = 191
Radius = 9 cm = 0.09 m
Initial magnetic field = 20 mT
Final magnetic field = 80 mT
Time = 2 ms = 2 x 10^-3 s
The induced emf in the coil is given by Faraday's law:
ε = -N∆B/∆t
where ε is the induced emf, N is the number of turns, ∆B is the change in magnetic field, and ∆t is the time interval.
Substituting the given values, we get:
ε = -191 × (80 - 20) mT / 2 x 10^-3 s
ε = -7640 mT/s
The magnitude of the induced emf is 7640 mV. Rounding to two decimal places, we get:
ε = 7640.0 mV = 7.64 V
Therefore, the magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT is 7.64 V.
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