The locus of possible positions for the third charge Q2, given Q1 = 21 C and Q2 = -2.2 C, is represented by two separate curves on a graph, determined by the equation r2 = sqrt((2.2 * r1^2) / 21).
Given two point charges of magnitude Q at specific positions, the task is to determine the locus (possible positions) of a third charge Q2, such that the total electric field at a specific point is zero.
This locus represents the positions where the net electric field due to the two charges cancels out. The specific scenario is when the original charges are 21 and -2,2.
To find the locus of the possible positions of the third charge, we need to consider the electric field due to the two original charges. The electric field at any point due to a point charge is given by Coulomb's Law: E = k * (Q / r^2), where E is the electric field, k is the electrostatic constant, Q is the charge, and r is the distance from the charge.
For the total electric field to be zero at the point (0,1,0), the electric field vectors due to the two charges must have equal magnitudes but opposite directions. By setting up the equations for the electric fields due to each charge and considering their magnitudes and directions, we can determine the locus of possible positions for the third charge Q2.
Specifically, if the original charges are 21 and -2,2, the locus of possible positions for the third charge Q2 can be found by solving the equations derived from Coulomb's Law with the given charge magnitudes and positions. By solving these equations, we can determine the specific coordinates that satisfy the condition of zero net electric field at the point (0,1,0).
It is important to note that the complete mathematical derivation and calculation of the locus would require solving the equations explicitly using the given charge values and positions.
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b) Given three 2-inputs AND gates, draw how you would produce a 4-inputs AND gate. (3 marks)
To create a 4-input AND gate using three 2-input AND gates, you can use the following configuration: (The picture is given below)
In this configuration, the inputs A1 and B1 are connected to the first 2-input AND gate, inputs A2 and B2 are connected to the second 2-input AND gate, and inputs A3 and B3 are connected to the third 2-input AND gate. The outputs Y1 and Y2 from the first two AND gates are then connected to the inputs of the third AND gate.
The outputs Y1, Y2, and Y of the three AND gates are connected together, resulting in a 4-input AND gate with inputs A1, B1, A2, B2, A3, B3, A4, and B4, and output Y.
By appropriately connecting the inputs and outputs of the three 2-input AND gates, we can achieve the desired functionality of a 4-input AND gate.
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A standing wave on a string has 2 loops ( 2 antinodes). If the string is 2.00 m long, what is the wavelength of the standing wave? 1.00 m 4.00 m 0.500 m 2.00 m A simple pendulum is made of a 3.6 m long light string and a bob of mass 45.0 grams. If the bob is pulled a small angle and released, what will the period of oscillation be? 1.21 s 2.315 4.12 s 3.81 s A block is attached to a vertical spring attached to a ceiling. The block is pulled down and released. The block oscillates up and down in simple harmonic motion and has a period . What would be true of the new period of oscillation if a heavier block were attached to the same spring and pulled down the same distance and released? The new period would be less than T The new period would be greater than T The new period would still be T The heavier block would not oscillate on the same spring
1. the wavelength of the standing wave is 4.00 m. 2. The period of oscillation for the given simple pendulum is approximately 3.81 seconds. 3. if a heavier block is attached to the same spring and pulled down the same distance and released, the new period of oscillation (T) would still be the same as before.
1. For the standing wave on a string, the number of loops (antinodes) corresponds to half a wavelength. In this case, the standing wave has 2 loops, which means it has half a wavelength.
Given the length of the string is 2.00 m, we can determine the wavelength of the standing wave by multiplying the length by 2 (since half a wavelength corresponds to one loop):
Wavelength = 2 × Length = 2 × 2.00 m = 4.00 m
Therefore, the wavelength of the standing wave is 4.00 m.
2. Regarding the second question about the simple pendulum, the period of oscillation for a simple pendulum can be calculated using the formula:
Period (T) = 2π√(L/g)
where L is the length of the pendulum and g is the acceleration due to gravity.
Given:
Length (L) = 3.6 m
Mass (m) = 45.0 grams = 0.045 kg
Acceleration due to gravity (g) ≈ 9.8 m/s²
Using the formula, we can calculate the period:
T = 2π√(L/g)
= 2π√(3.6/9.8)
≈ 2π√(0.367)
Calculating the approximate value:
T ≈ 2π(0.606)
≈ 3.81 s
Therefore, the period of oscillation for the given simple pendulum is approximately 3.81 seconds.
3. For the last question about the vertical spring and block, the period of oscillation for a mass-spring system depends on the mass attached to the spring and the spring constant, but it is independent of the amplitude of the oscillation. Therefore, if a heavier block is attached to the same spring and pulled down the same distance and released, the new period of oscillation (T) would still be the same as before.
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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 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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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.
Angular velocity refers to the rate at which an object rotates around a fixed axis. It is a vector quantity, expressed in radians per second, and represents the object's rotational speed and direction.Learn more about angular velocity here;
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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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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.
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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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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A uniformly charged conducting spherical shell of radius Ro and surface charge density o, is spinning with constant angular velocity o. Calculate the magnetic field B and vector potential à in (20 marks) all space.
To calculate the magnetic field (B) and vector potential (Ã) in all space due to a uniformly charged conducting spherical shell spinning with constant angular velocity.
The current density can be expressed as
J = σv,
The Biot-Savart law as well:
à = (μ₀/4π) * ∫(J / r) * dV.
As a result, the magnetic field and vector potential inside the shell will be zero.
Therefore, the expressions for B and à in all space due to uniformly charged conducting spherical shell spinning with constant angular velocity will be zero inside the shell and calculated using appropriate integrals outside shell.
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The Sun appears at an angle of 55.8° above the horizontal as viewed by a dolphin swimming underwater. What angle does the sunlight striking the water actually make with the horizon? (Assume nwater = 1.333. Enter an answer between 0° and 90°.)
__________________°
The Sun appears at an angle of 55.8° above the horizontal as viewed by a dolphin swimming underwater. The angle at which sunlight strikes the water, relative to the horizon, is approximately 49.3°.
To find the angle at which sunlight strikes the water, we can use Snell's law, which relates the angles of incidence and refraction when light passes through a boundary between two media.
The Snell's law equation is:
n₁ × sin(θ₁) = n₂ × sin(θ₂)
Given:
Angle of incidence (θ₁) = 55.8°
Index of refraction of water (n₂) = 1.333 (approximate value for water)
We want to find the angle of refraction (θ₂) when light passes from air (n₁ = 1) into water (n₂ = 1.333).
Rearranging the equation, we have:
sin(θ₂) = (n₁ / n₂) × sin(θ₁)
Plugging in the values:
sin(θ₂) = (1 / 1.333) × sin(55.8°)
Calculating:
sin(θ₂) ≈ 0.7479
To find the angle θ₂, we can take the inverse sine (arcsine) of the calculated value:
θ₂ ≈ arcsin(0.7479)
Calculating:
θ₂ ≈ 49.3°
Therefore, the angle at which sunlight strikes the water, relative to the horizon, is approximately 49.3°.
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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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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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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(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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Drag the tiles to the correct boxes to complete the pairs. Identify the type of chemical reaction that is described.
Answer:
Synthesis= the one about leaves
Neutralization= the vinegar one
Combustion= the one where the food burns
decomposition- the one about water breaking down
Explanation:
sorry if I'm wrong with any of these. decomposition and synthesis may be the other way round i wasn't sure
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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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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Two pistons of a hydraulic lift have radii of 2.67 cm and 20.0 cm. A mass of 2.00×10 3
kg is placed on the larger piston. Calculate the minimum downward force needed to be exerted on the smaller piston to hold the larger piston level with the smaller piston. N
The minimum downward force required to exert more force for the smaller piston to hold a larger piston is 266.52 N
Radii of pistons = 2.67 cm and 20.0 cm
Mass of pistons = [tex]2.00*10^{3}[/tex]
Pressure = Force / Area
The areas of the pistons:
Area1 = π *[tex]r1^2[/tex]
Area2 = π * [tex]r2^2[/tex]
We need to equate both pistons, then we get:
Pressure1 = Pressure2
F1 / Area1 = F2 / Area2
F1 / (π * [tex]r1^2[/tex] ) = F2 / (π * [tex]r2^2[/tex] )
The weight can be calculated as:
Weight = mass * gravity
Weight = [tex]2.00 * 10^3 kg * 9.8 m/s^2[/tex]
F1 = (F2 * Area1) / Area2
F1 = [tex]((2.00 * 10^3 kg * 9.8 m/s^2)[/tex] * (π * [tex]r1^2[/tex] ) * (π * [tex]r2^2[/tex] )
F1 = [tex](2.00 * 10^3 kg * 9.8 m/s^2 * r1^2) / r2^2[/tex]
F1 = [tex](2.00 * 10^3 kg * 9.8 m/s^2 * (2.67 cm)^2) / (20.0 cm)^2[/tex]
F1 = 266.52 N
Therefore, we can conclude that the minimum downward force needed is 266.52 N.
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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 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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An infinitely long solid insulating cylinder of radius a = 3 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density p = 22 HC/m³. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 19 cm, and outer radius c = 22 cm. The conducting shell has a linear charge density λ = -0.47μC/m. R(0,d) P 2 P(d,d) 5) The charge density of the insulating cylinder is now changed to a new value, p' and it is found that the electric field at point P is now zero. What is the value of p'? HC/m³ Submit
The new charge density [tex]\(p'\)[/tex] of the insulating cylinder, the electric field at point P is set to zero by considering the electric fields due to both the insulating cylinder and the conducting shell. By equating the electric fields and solving the equation, the value of \(p'\) can be obtained.
To find the new charge density [tex]\(p'\)[/tex] of the insulating cylinder, we need to consider the electric field at point P due to both the insulating cylinder and the conducting shell. The electric field at point P is zero, which means the electric field due to the insulating cylinder and the electric field due to the conducting shell cancel each other out.
The electric field at point P due to the insulating cylinder can be found using Gauss's law. Since the cylinder is symmetric and has a uniform charge density, the electric field inside the cylinder is given by [tex]\(E = \frac{p}{2\epsilon_0}\)[/tex], where [tex]\(\epsilon_0\)[/tex] is the permittivity of free space
The electric field at point P due to the conducting shell is given by [tex]\(E = \frac{\lambda}{2\pi\epsilon_0}\left(\frac{1}{d}-\frac{1}{\sqrt{d^2+(b+c)^2}}\right)\), where \(d\)[/tex] is the distance from the center of the cylinder.
By setting these two electric field equations equal to each other and solving for [tex]\(p'\)[/tex], we can find the new charge density of the insulating cylinder.
Note: The values of [tex]\(d\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are not provided in the question, so the specific numerical value of [tex]\(p'\)[/tex] cannot be determined without that information.
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. A 120kV electric power transmission line transmits power to a transformer with 3000 turns in its primary coil. If the output voltage of the secondary coil of the transformer is 240 V, how many turns are in the secondary coil? A. 6000 B. 6 C. 60 D. 600
The number of turns in the secondary coil is 1500. The correct option is not given in the options.
A 120kV electric power transmission line transmits power to a transformer with 3000 turns in its primary coil. If the output voltage of the secondary coil of the transformer is 240 V, then we have to find the number of turns in the secondary coil.
Let's calculate the number of turns in the secondary coil of the transformer.By the formula of a transformer, the primary voltage (Vp) times the primary turns (Np) equals the secondary voltage (Vs) times the secondary turns (Ns).
Hence,Vp * Np = Vs * NsVp = 120 kVVs = 240 V Np = 3000 Ns.Now, substitute the given values in the above equation.120 kV × 3000 = 240 V × Ns360000 = 240 NsNs = 1500 turns.
Therefore, the number of turns in the secondary coil is 1500. So, the correct option is not given in the options.
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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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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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A beam of radiation is propagating inside a dielectric of refractive index n= 1.5 and is incident on a dielectric/free space interface. If the angle of incidence is 80° and the radiation has a wavelength of 500 nm in free space, calculate the distance outside the medium at which the electric field amplitude has dropped to 10% of its value at the surface. (2 marks) Explain the meaning of the term frustrated total internal reflection, and describe any advantages or disadvantages arising from this phenomenon. (2 marks)
The angle of incidence, refractive index, and wavelength are used to determine the critical angle and the angle of refraction at the interface. From there, the distance can be calculated using trigonometry and the decay equation.
To calculate the distance outside the dielectric at which the electric field amplitude drops to 10% of its value at the surface, we need to consider the decay of the electric field in the dielectric material. The angle of incidence (80°) and the refractive index (n = 1.5) are used to determine the critical angle and the angle of refraction at the interface between the dielectric and free space. With these angles, we can calculate the distance at which the electric field amplitude drops to 10% of its value.
Frustrated total internal reflection refers to the phenomenon where total internal reflection does not occur at the interface between two mediums, such as from a higher refractive index medium to a lower refractive index medium. This can happen when the angle of incidence exceeds the critical angle, but instead of all the light being reflected, a small portion of it is transmitted into the second medium. Frustrated total internal reflection can be advantageous in applications like optical fibers and waveguides, where it allows controlled transmission of light. However, it can also be disadvantageous when trying to achieve complete reflection, such as in certain optical devices or when designing systems that rely on total internal reflection for efficient light confinement.
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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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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 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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