Final answer:
The direction of the resultant force acting on an electron moving upward on the page through a magnetic field directed into the page is to the left, as per the left-hand rule.
Explanation:
The question deals with the Lorentz force experienced by a charged particle, in this case an electron, moving through a magnetic field. When an electron with initial velocity directed from the bottom edge to the top edge of the page moves through a uniform magnetic field directed into the page, the force acting on it can be determined using the left-hand rule, since electrons have a negative charge. Holding your left thumb up (direction of the electron's velocity) and the fingers into the page (direction of the magnetic field), the resultant force will be directed to the left of the page. Therefore, the correct answer to the direction of the resultant force acting on the electron is c. to the left.
1. The workpart in a turning operation is 88 min in diameter and 400 mm long. A feed of 0.25 mm/rev is used in this operation. If cutting speed is 3.5 m/s, the too should be changed in every 3 workparts, but if the cutting speed is 2.5 m/sec, the tool can be used to produce 20 pieces between the tool changes. Determine the cutting speed that will allow the tool to be used for 50 parts between tool changes.
Find the given attachments
A plane electromagnetic wave traveling in the positive direction of an x axis in vacuum has components Ex = Ey = 0 and Ez = (5.2 V/m) cos[(π × 1015 s-1)(t - x/c)].(a) What is the amplitude of the magnetic field component? (b) Parallel to which axis does the magnetic field oscillate? (c) When the electric field component is in the positive direction of the z axis at a certain point P, what is the direction of the magnetic field component there? Assume that the speed of light is 2.998*108 m/s.
Answer:
Explanation:
Ez = (5.2 V/m)
Magnitude of electric field = 5.2 V/m
Magnitude of magnetic field be B then
Magnitude of electric field / Magnitude of magnetic field = c , c is velocity of light
5.2 / B = 2.998 x 10⁸
B = 5.2 / 2.998 x 10⁻⁸
= 1.73 x 10⁻⁸ T.
b ) E X B = direction of velocity of light , E is electric vector , B is magnetic V.
E is along z - axis , velocity is along x -axis , then magnetic field must be in - Y direction.
c ) E and B are in phase so when E is in positive z axis , directio of magnetic field must be in - y direction.
Find the charge q(t) on the capacitor and the current i(t) in the given LRC-series circuit. L = 1 h, R = 100 Ω, C = 0.0004 f, E(t) = 20 V, q(0) = 0 C, i(0) = 3 A q(t) = C i(t) = A
Find the maximum charge on the capacitor. (Round your answer to four decimal places.)
Answer:
The maximum charge around the capacitor is 0.03170189C
Explanation:
See attached file
In an RLC series circuit with given values for resistance, inductance, and capacitance, we can calculate the charge on the capacitor five cycles later and fifty cycles later using the equation q(t) = q(0) * e^(-(R/L)t)
Explanation:In an RLC series circuit with a resistance of 7.092 ohms, an inductance of 10 mH, and a capacitance of 3.0 µF, the charge on the capacitor can be calculated using the equation:
q(t) = q(0) * e^(-(R/L)t)
Given that the initial charge on the capacitor is 8.0 µC, we can calculate the charge five cycles later and fifty cycles later using this equation.
(a) To find the charge five cycles later, we need to find the time period of one cycle. The time period can be calculated using the formula:
T = 2π√(LC)
Once we have the time period, we can calculate the time it takes for five cycles to pass and substitute it in the equation to find the charge.
(b) To find the charge fifty cycles later, we can use the same approach as in part (a), but substituting the time it takes for fifty cycles to pass.
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1. A 500-g block is placed on a level, frictionless surface and attached to an ideal spring. At t = 0 the block moves through the equilibrium position with speed vo in the –x direction, as shown below. At t = sec, the block reaches its maximum displacement of 40 cm to the left of equilibrium. a. Determine the value of each of the following quantities. Show all work. • period: • spring constant: b. Using x(t) = A cos(t + o) as the solution to the differential equation of motion: i. Determine the form of the function v(t) that represents the velocity of the block. ii. Evaluate all constant parameters (A, , and o) so as to completely describe both the position and velocity of the block as functions of time. c. Consider the following statement about the situation described above. "It takes the first seconds for the block to travel 40 cm, so the initial speed vo can be found by dividing 40 cm by seconds." Do you agree or disagree with this statement? If so, explain why you agree. If not, explain why you disagree and calculate the initial speed vo of the block.
a. The period (T) of the motion is [tex]\(2\pi\)[/tex] seconds, and the spring constant (k) is [tex]\(m\omega^2\) where \(m\) is the mass, and \(\omega\)[/tex] is the angular frequency. For the given problem, [tex]\(T = 2\pi\) seconds and \(k = \frac{m}{\pi^2}\).[/tex]
b. i. The velocity function [tex]\(v(t)\) is given by \(v(t) = -A\omega\sin(\omega t + \phi_0)\).[/tex]
ii. Evaluating the constant parameters:
[tex]\(A = 0.4\ m\) (maximum displacement),[/tex]
[tex]\(\omega = \frac{2\pi}{T} = 1\ \text{rad/s}\),[/tex]
[tex]\(\phi_0 = -\frac{\pi}{2}\) (phase angle).[/tex]
c. Disagree. The block doesn't travel 40 cm in the first [tex]\(\pi\)[/tex] seconds. The correct initial speed [tex]\(v_0\) is found using \(v_0 = A\omega\), so \(v_0 = 0.4\ m \times 1\ \text{rad/s} = 0.4\ \text{m/s}\).[/tex]
Explanation:a. The period of the motion [tex](\(T\))[/tex] is the time taken for one complete oscillation. For simple harmonic motion, [tex]\(T = 2\pi/\omega\), where \(\omega\)[/tex] is the angular frequency. In this case, [tex]\(T = 2\pi\)[/tex] seconds. The spring constant [tex](\(k\))[/tex] is related to the angular frequency by [tex]\(k = m\omega^2\), where \(m\)[/tex] is the mass. Substituting [tex]\(T\) into the formula for \(\omega\), we find \(k = \frac{m}{\pi^2}\).[/tex]
b. i. The velocity function [tex]\(v(t)\)[/tex] is the derivative of the displacement function [tex]\(x(t)\). For \(x(t) = A\cos(\omega t + \phi_0)\), \(v(t) = -A\omega\sin(\omega t + \phi_0)\).[/tex]
ii. To fully describe the motion, we find [tex]\(A\), \(\omega\), and \(\phi_0\). \(A\)[/tex] is the amplitude, given as 0.4 m. [tex]\(\omega\)[/tex] is calculated from the period, resulting in 1 rad/s. [tex]\(\phi_0\)[/tex] is the phase angle, set to [tex]\(-\frac{\pi}{2}\)[/tex] to match the initial condition.
c. The statement is incorrect. The block does not travel 40 cm in the first [tex]\(\pi\)[/tex] seconds. The initial speed [tex](\(v_0\))[/tex] can be found using [tex]\(v_0 = A\omega\), which yields \(0.4\ m \times 1\ \text{rad/s} = 0.4\ \text{m/s}\).[/tex]
PROBLEM-10 GRAVITATIONAL POTINTIAL ENERGY In a typical professional downhill ski race, athletes start 820 m above where they cross the finish line. Assume that they start at rest, friction is negligible, and their mass is 80 kg. What is the magnitude of the change in gravitational potential energy of the Earth-athlete system over the course of the race? (in J )
Answer:
The magnitude of the change in gravitational potential energy of the earth-athlete system over the course of the race is 643,536Joules
Explanation
Potential energy is one of the form of mechanical energy and it is defined as the energy possessed by a body due to virtue of its position. When the body is under gravity, it possesses an energy called the gravitational potential energy.
Gravitational potential energy is expressed as shown:
GPE = mass × acceleration due to gravity × height
Given mass of athlete = 80kg
height covered = 820m
acceleration due to gravity = 9.81m/s
GPE = 80×9.81×820
GPE = 643,536Joules
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius RRR which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2cm2. How large does the radius RRR of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/mmV/m at the receiver? For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius RRR refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal
Complete Question
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.
How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?
For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.
Give your answer in centimeters, to two significant figures.
Answer:
The radius of the dish is [tex]R = 18cm[/tex]
Explanation:
From the question we are told that
The radius of the orbit is = [tex]R = 35,000km = 35,000 *10^3 m[/tex]
The power output of the power is [tex]P = 1 kW = 1000W[/tex]
The electric vector amplitude is given as [tex]E = 0.1 mV/m = 0.1 *10^{-3}V/m[/tex]
The area of thereciever is [tex]A_R = 5cm^2[/tex]
Generally the intensity of the dish is mathematically represented as
[tex]I = \frac{P}{A}[/tex]
Where A is the area orbit which is a sphere so this is obtained as
[tex]A = 4 \pi r^2[/tex]
[tex]= (4 * 3.142 * (35,000 *10^3)^2)[/tex]
[tex]=1.5395*10^{16} m^2[/tex]
Then substituting into the equation for intensity
[tex]I_s = \frac{1000}{1.5395*10^{16}}[/tex]
[tex]= 6.5*10^ {-14}W/m2[/tex]
Now the intensity received by the dish can be mathematically evaluated as
[tex]I_d = \frac{1}{2} * c \epsilon_o E_D ^2[/tex]
Where c is thesped of light with a constant value [tex]c = 3.0*10^8 m/s[/tex]
[tex]\epsilon_o[/tex] is the permitivity of free space with a value [tex]8.85*10^{-12} N/m[/tex]
[tex]E_D[/tex] is the electric filed on the dish
So since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish
Now making the eletric field intensity the subject of the formula
[tex]E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }[/tex]
substituting values
[tex]E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }[/tex]
[tex]= 7*10^{-6} V/m[/tex]
The incident power on the dish is what is been reflected to the receiver
[tex]P_D = P_R[/tex]
Where [tex]P_D[/tex] is the power incident on the dish which is mathematically represented as
[tex]P_D = I_d A_d[/tex]
[tex]= \frac{1}{2} c \epsilon_o E_D^2 (\pi R^2)[/tex]
And [tex]P_R[/tex] is the power incident on the dish which is mathematically represented as
[tex]P_R = I_R A_R[/tex]
[tex]= \frac{1}{2} c \epsilon_o E_R^2 A_R[/tex]
Now equating the two
[tex]\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R[/tex]
Making R the subject we have
[tex]R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }[/tex]
Substituting values
[tex]R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }[/tex]
[tex]R = 18cm[/tex]
An inexperienced catcher catches a 126 km/h fastball of mass 160 g within 1.36 ms, whereas an experienced catcher slightly retracts his hand during the catch, extending the stopping time to 13.6 ms. What is the average force imparted to the two gloved hands during the catches of the inexperienced catcher
Answer:
4117.65 N
Explanation:
Speed of ball, u = 126 km/h = 35 m/s
Mass of ball, m = 160 g = 0.16 kg
Time interval, t = 1.36 ms = 0.00136 s
We can calculate the force as a measure of the momentum of the ball:
F = P/t
Momentum, P, is given as:
P = mv
Therefore:
F = (mv) / t
F = (0.16 * 35) / (0.00136)
F = 4117.65 N
The force imparted to the two glove d hands of the inexperienced catcher is 4117.65 N.
Consider two inductors, with the first inductor experiencing a current which is rising at twice the rate as that through the second inductor. Suppose the first inductor has half the EMF across it as the second inductor. How does the inductance of the first inductor compare to that of the second inductor?
Answer:
Inductance of first is one-4th that of the second inductor
Explanation:
See attached file
An infinite slab of charge of thickness 2z0 lies in the xy-plane between z=−z0 and z=+z0. The volume charge density rho(C/m3) is a constant. Part A Use Gauss's law to find an expression for the electric field strength inside the slab (−z0≤z≤z0). Express your answer in terms of the variables rho, z, z0, and constant ϵ0.
Final answer:
The electric field inside an infinite slab with a constant volume charge density is found using Gauss's law, resulting in an expression where the electric field is directly proportional to both the distance from the midplane and the charge density, and inversely proportional to the permittivity of free space.
Explanation:
Finding the Electric Field Inside an Infinite Slab Using Gauss's Law
To find the electric field strength inside an infinite slab of charge lying in the xy-plane and having a uniform volume charge density (ρ), we employ Gauss's law. Gauss's law states that the net electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). For an infinite slab of thickness 2z0, positioned between z = -z0 and z = +z0, we choose a Gaussian surface in the form of a rectangular box (pillbox) that extends symmetrically above and below the xy-plane. This choice is motivated by the symmetry of the slab and the desire to have an electric field that is either parallel or perpendicular to the faces of the box.
The charge enclosed by the Gaussian surface is given by Q = ρA(2z), where A is the cross-sectional area of the box parallel to the xy-plane and 2z is the extent of the box in the z-direction within the slab. Applying Gauss's law, the electric flux through the box is Ψ = EA = Q/ε0, where E is the magnitude of the electric field inside the slab and is directed along the z-axis. Simplifying, we find E = ρz/ε0. It follows that the electric field strength inside the slab, for -z0 ≤ z ≤ z0, is directly proportional to the distance z from the midplane and to the volume charge density ρ, and inversely proportional to the permittivity of free space ε0.
Using Gauss's law, the electric field strength inside the infinite slab of charge is found to be E = (ρ * z) / ε0.
To find the electric field strength inside an infinite slab of charge with uniform volume charge density ρ, we will use Gauss's law.
The slab extends from z = -z₀ to z = z₀ in the z-direction.
Step-by-Step Solution:
Consider a Gaussian surface in the form of a pillbox centered at a point within the slab at height z, with thickness 2z and cross-sectional area A.By symmetry, the electric field E points in the z direction and is perpendicular to the faces of the pillbox.The charge enclosed within this Gaussian surface is ρ * A * (z + z₀).Gauss's law states that the net electric flux through this surface is Φ = ∮E・dA = q_enclosed / ε₀.The electric flux through the pillbox is Φ = 2 * E * A, as the field is perpendicular to two faces of area A.Using Gauss's law: 2E * A = (ρ * A * (z + z₀)) / ε₀Simplifying, E = (ρ * (z + z₀)) / (2ε₀).The electric field strength inside the slab, for -z0 ≤ z ≤ z0, is given by:
E = (ρ * z) / ε₀
The human circulatory system is closed-that is, the blood pumped out of the left ventricle of the heart into the arteries is constrained to a series of continuous, branching vessels as it passes through the capillaries and then into the veins as it returns to the heart. The blood in each of the heart’s four chambers comes briefly to rest before it is ejected by contraction of the heart muscle. If the aorta (diameter da) branches into two equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches?
(a) √da (b) da/√2 (c) 2da
(d) da/2
Answer:
(b) da/√2
Explanation:
Detailed explanation and calculation is shown in the image below
A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base of the hill. (a) At what rate was it rotating at the base of the hill? (b) What was the rotational kinetic energy then? (Ignore rolling friction and assume total mechanical energy is conserved). [Hint: Soccer ball is a thin walled h
Answer with Explanation:
We are given that
Diameter=d=22.6 cm
Mass,m=426 g=[tex]426\times 10^{-3} kg[/tex]
1 kg=1000 g
Radius,r=[tex]\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m[/tex]
1m=100 cm
Height,h=5m
[tex]I=\frac{2}{2}mr^2[/tex]
a.By law of conservation of energy
[tex]\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh[/tex]
[tex]\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh[/tex]
[tex]v=\omega r[/tex]
[tex]gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2[/tex]
[tex]\omega^2=\frac{6}{5r^2}gh[/tex]
[tex]\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s[/tex]
Where [tex]g=9.8m/s^2[/tex]
b.Rotational kinetic energy=[tex]\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J[/tex]
Rotational kinetic energy=8.35 J
A typical stellar spectrum (a plot of intensity versus wavelength) includes a number of deep indentations in which the intensity abruptly falls and then rises. These deep indentations are called ____ lines.
Answer:
They are called absorption lines
Explanation:
Absorption lines are defined as dark lines or lines having reduced intensity, on an ongoing spectrum. A typical example is noticed in the spectra of stars, where gas existing in the outer layers of the star absorbs some of the light from the underlying thermal blackbody spectrum.
A typical stellar spectrum includes dark lines called absorption lines.
Explanation:A typical stellar spectrum includes a number of deep indentations in which the intensity abruptly falls and then rises. These deep indentations are called absorption lines. Absorption lines are dark lines in a spectrum that correspond to specific wavelengths of light that have been absorbed by elements in the star's outer layers.
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The shock absorbers in an old car with mass 1160 kg are completely worn out. When a 970 N person climbs slowly into the car, it deforms 3.0 cm. The car is now towed down the road (with the person inside). The car hits a bump, and starts oscillating up and down with an amplitude of 6.4 cm.
Model the car and person as a single body on a spring and find the period and frequency of oscillations.
Answer:
[tex]f = 0.806\,hz[/tex], [tex]T = 1.241\,s[/tex]
Explanation:
The problem can be modelled as a vertical mass-spring system exhibiting a simple harmonic motion. The spring constant is:
[tex]k = \frac{970\,N}{0.03\,m}[/tex]
[tex]k = 32333.333\,\frac{N}{m}[/tex]
The angular frequency is:
[tex]\omega = \sqrt{\frac{32333.333\,\frac{N}{m} }{1258.879\,kg} }[/tex]
[tex]\omega = 5.068\,\frac{rad}{s}[/tex]
The frequency and period of oscillations are, respectively:
[tex]f = \frac{5.068\,\frac{rad}{s} }{2\pi}[/tex]
[tex]f = 0.806\,hz[/tex]
[tex]T = \frac{1}{0.806\,hz}[/tex]
[tex]T = 1.241\,s[/tex]
What attraction and repulsion behavior do the north and south poles of a magnet exhibit? Explain the left-hand rule for the flux direction in conductors. What is the most common source of power for most HVACR systems? What is the purpose for transformers? What is a solenoid valve?
Answer:
1.attraction and repulsion behavior do the north and south poles of a magnet exhibit?A.Like poles repel each other, and opposite poles attract.
2.Explain the left-hand rule for flux direction in conductors. Ans.The left-hand rule is used to determine the direction of rotation of the flux around the conductor
.3.What is the most common source of power for most HVACR systems?Ans.Alternating current is the most common source of power used for most HVACR systems.
4.What is the purpose of transformers?Ans..Transformers are used to increase or decrease incoming voltages to meet the requirements of the load.
5.What is a solenoid valve?Ans.Solenoid valves are electrically operated valves that open or close automatically by energizing or de-energizing a coil of wire.6.Is there an electrical connection between the primary and secondary windings of a transformer?A.There is no electrical connection between the primary and secondary windings.
Explanation:
The poles of a magnet attract and repel based on their polarity. The left-hand rule explains how magnetic fields interact with electric currents. Most HVACR systems use electricity, and transformers convert voltage levels to transmit power efficiently. A solenoid valve controls fluid or gas flow through electromagnetism.
Explanation:The north and south poles of a magnet exhibit a behavior where like poles repel and opposite poles attract. This means that north pole to north pole or south pole to south pole will repel, while north pole to south pole will attract.
The left-hand rule for flux direction, often used in electromagnetism, states that if you point your thumb in the direction of the current, the direction in which your fingers curl represents the direction of the magnetic field lines. This is significant in understanding the interaction of electric current with magnetic fields in conductors.
The most common source of power for most HVACR (Heating, Ventilation, Air Conditioning, and Refrigeration) systems is electricity. This power is used to run the system's various components including compressors, fans, and controls.
Transformers are used in electric power systems to convert voltages from one level to another. This is mainly used to transmit electricity over long distances with low energy losses and is essential in power grids.
A solenoid valve is an electromechanical device used to control the flow of a fluid or gas. When an electric current is passed through the coil, a magnetic field is created causing the valve to open or close.
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In a thunder and lightning storm there is a rule of thumb that many people follow. After seeing the lightning, count seconds to yourself. If it takes 5 seconds for the sound of the thunder to reach you, then the lightning bolt was 1 mile away from you. Sound travels at a speed of 331 meters/second. How accurate is the rule of thumb? Express your answer as a percent error.
Answer:
2.837% less than actual value.
Explanation:
Based on given information let's calculate our value.
S = Vxt = 331m/s x 5s = 1655m, that is the total distance that sound would travel in 5 seconds.
1mile = 1609.34meters.
percentage error is.
[tex]\frac{actual-calculated}{actual} *100 = \frac{1609.34-1655}{1609.34} *100 = -2.83%[/tex]
negative indicates less than actual value.
The rule of thumb that states the lightning bolt was one mile away for every five seconds between seeing the flash and hearing the thunder is not very accurate. The actual distance the lightning bolt would be approximately 3.19 miles away, which results in a percent error of approximately 219%. The rule tends to underestimate the distance to the lightning bolt.
Explanation:The "rule of thumb that many people follow" during a thunder and lightning storm is based on the fact that light travels much faster than sound. When you see a flash of lightning, the sound wave created by the thunder associated with the lightning bolt takes more time to reach the observer than the light from the flash.
According to the rule of thumb, we estimate one mile per five seconds because sound travels at a speed of approximately 331 meters per second. However, to calculate the actual distance in miles, you would multiply the time (in seconds) by the speed of sound and convert to miles (1 meter = 0.00062137 miles). This gives an actual distance of about 0.00063741 miles/second. Therefore, for every second delay between the lightning and the thunder, the lightning bolt would be about 0.00063741 miles away.
So, if we see a flash of lightning and hear the thunder 5 seconds later, according to the accurate calculation, the lightning bolt was just over 3.19 miles away (5 seconds * 0.00063741 miles/second). That's a sizeable difference from the one-mile estimate given by the rule of thumb. To find the percent error, we subtract the accepted value from the experimental value, divide by the accepted value, and multiply by 100. That gives us a percent error of approximately 219%. So the rule of thumb is not particularly accurate.
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A -turn rectangular coil with length and width is in a region with its axis initially aligned to a horizontally directed uniform magnetic field of mT and set to rotate about a vertical axis with an angular speed of rad/s. What is the maximum induced emf in the rotating coil? What is the induced emf in the rotating coil at ? What is the maximum rate of change of the magnetic flux through the rotating coil?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The maximum emf is [tex]\epsilon_{max}= 26.8 V[/tex]
The emf induced at t = 1.00 s is [tex]\epsilon = 24.1V[/tex]
The maximum rate of change of magnetic flux is [tex]\frac{d \o}{dt}|_{max} =26.8V[/tex]
Explanation:
From the question we are told that
The number of turns is N = 44 turns
The length of the coil is [tex]l = 15.0 cm = \frac{15}{100} = 0.15m[/tex]
The width of the coil is [tex]w = 8.50 cm =\frac{8.50}{100} =0.085 m[/tex]
The magnetic field is [tex]B = 745 \ mT[/tex]
The angular speed is [tex]w = 64.0 rad/s[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = \epsilon_{max} sin (wt)[/tex]
Where [tex]\epsilon_{max}[/tex] is the maximum induced emf and this is mathematically represented as
[tex]\epsilon_{max} = N\ B\ A\ w[/tex]
Where [tex]\o[/tex] is the magnetic flux
N is the number of turns
A is the area of the coil which is mathematically evaluated as
[tex]A = l *w[/tex]
Substituting values
[tex]A = 0.15 * 0.085[/tex]
[tex]= 0.01275m^2[/tex]
substituting values into the equation for maximum induced emf
[tex]\epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0[/tex]
[tex]\epsilon_{max}= 26.8 V[/tex]
given that the time t = 1.0sec
substituting values into the equation for induced emf [tex]\epsilon = \epsilon_{max} sin (wt)[/tex]
[tex]\epsilon = 26.8 sin (64 * 1)[/tex]
[tex]\epsilon = 24.1V[/tex]
The maximum induced emf can also be represented mathematically as
[tex]\epsilon_{max} = \frac{d \o}{dt}|_{max}[/tex]
Where [tex]\o[/tex] is the magnetic flux and [tex]\frac{d \o}{dt}|_{max}[/tex] is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is
[tex]\frac{d \o}{dt}|_{max} =26.8V[/tex]
Suppose the maximum safe intensity of microwaves for human exposure is taken to be 1.39~\mathrm{W/m^2}1.39 W/m 2 . If a radar unit leaks 10.0~\text{W}10.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Assume that the power spreads uniformly over the area of a sphere with no complications from absorption or reflection. (Note that early radar units leaked more than modern ones do. This caused identifiable health problems, such as cataracts, for people who worked near them.)
Answer:
0.763 m
Explanation:
Intensity I = power P ÷ area A of exposure (spherical area of propagation)
I = P/A
A = P/I
Power = 10.0 W
Intensity = 1.39 W/m^2
A = 10/1.39 = 7.19 m^2
Area A = 4¶r^2
7.19 = 4 x 3.142 x r^2
7.19 = 12.568r^2
r^2 = 7.19/12.568 = 0.57
r = 0.753 m
Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would require quantum mechanics to describe, that is, would show both particle and wave properties? Which objects would act like everyday objects, and be adequately described by classical mechanics? object quantum or classical? A virus with a mass of 9.4 x 10-17 g, 280. nm wide, moving at 0.50 µm/s. classical quantum A buckyball with a mass of 1.2 x 10-21 g, 0.7 nm wide, moving at 37. m/s. classical quantum A mosquito with a mass of 1.0 mg, 6.3 mm long, moving at 1.1 m/s. classical quantum A turtle with a mass of 710. g, 22. cm long, moving at 2.8 cm/s. classical quantum
All objects are described using classical mechanics, with the exception of the buckyball, which requires a quantum mechanical description due to its small mass and high speed that result in a significant de Brogli wavelength.
Explanation:Whether an object shows particle-wave duality, requiring quantum mechanics to describe, or behaves like everyday objects, described by classical mechanics, is determined by the de Broglie wavelength of the object. Calculating the de Broglie wavelength (λ) can be accomplished using the equation λ = h/mv, where h is Planck's constant and m and v are the object's mass and velocity, respectively.
For the virus, buckyball, mosquito, and turtle, the calculated de Broglie wavelengths are so small compared to their physical dimensions that quantum mechanical effects would be negligible. Hence, all these objects can be described using classical mechanics. The buckyball is the only exception. Owing to its tiny mass and high speed, its de Broglie wavelength becomes significant in relation to its size, requiring quantum mechanical description.
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Objects that are atomic or subatomic in size require quantum mechanics for their description, while larger objects can be adequately explained by classical mechanics. In the provided cases, a virus and a buckyball would require quantum mechanics, while a mosquito and a turtle would be described by classical mechanics.
Explanation:To determine whether an object would be adequately described by classical mechanics or require quantum mechanics, we need to consider its size, mass, and speed. Under the domain of quantum mechanics are typically objects that are atomic or subatomic in size, meaning they exhibit wave-particle duality due to their small size and significant motion.
1. A virus: Being atomic in size, a virus would require quantum mechanics for its description. Specifically, it would demonstrate both particle and wave properties due to its small size.
2. A buckyball: Given its subatomic mass, size, and significant speed, a buckyball falls into the quantum domain, displaying both particle and wave properties.
3. A mosquito and 4. A turtle : These are larger, macroscopic objects. Their motion and behavior can be adequately described using classical mechanics as they primarily exhibit particle properties, their wave-like characteristics being effectively obscured due to their larger mass and size. Their behaviors conform to the laws defined under classical mechanics and do not exhibit the quantum nature that smaller objects do.
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A steady stream of projectiles, with identical linear momenta, collides with a fixed target. The average force on the target is proportional to *
Answer:
The average force on the target is proportional to:
- The number of projectiles hitting the target.
- The mass of the projectiles.
- If the time increases or decreases
Explanation:
This is a problem that applies momentum and the amount of movement. Where this principle can be explained by the following equation:
m*v1 + Imp1_2 = m*v2
where:
∑F *Δt = Imp1_2 = impulse. [N*s]
m*v1 = mass by velocity before the impact [kg*m/s]
m*v2 = mass by velocity after the impact [kg*m/s]
When a problem includes two or more particles, each particle it can be considered separately and equation is written for every particle.
We clear the expression of force in the equation:
∑F *Δt = m*v2 - m*v1
In this equation if we have a different number of particles, given by the value n, we see that the equation is proportional to the number of particles
∑F *Δt = n*m*v2 - n*m*v1
Therefore the average force on the target is proportional to the number of projectiles hitting the target.
The Force F is also increased or decreased if the mass of the projectiles is changed. Therefore it is also proportional to the mass of the projectiles.
The Force F also changes if the time increases or decreases.
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibration amplitude of 0.05mm at 2 Hz (a typical resonant frequency of a building). a) If there is no damping, how much transmitted vibration (amplitude of displacement) does the microscope experience
Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is = [tex]5.676*10^{-3} \ mm[/tex]
Explanation:
The motion of the ceiling is y = Y sinωt
y = 0.05 sin (2 π × 2) t
y = 0.05 sin 4 π t
K = 25 lb/ft × 4 sorings
K = 100 lb/ft
Amplitude of the microscope [tex]\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}][/tex]
where;
[tex]\epsilon = 0[/tex]
[tex]W_n = \sqrt { \frac{k}{m}}[/tex]
= [tex]\sqrt { \frac{100*32.2}{200}}[/tex]
= 4.0124
replacing them into the above equation and making X the subject of the formula:
[tex]X =[/tex] [tex]Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}[/tex]
[tex]X =[/tex] [tex]0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}[/tex]
[tex]X =[/tex] [tex]5.676*10^{-3} \ mm[/tex]
Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = [tex]5.676*10^{-3} \ mm[/tex]
The centripetal acceleration might better be expressed as −ω2r⃗ (t)−ω2r→(t) because it is a vector. The magnitude of the centripetal acceleration is v2radial/Rvradial2/R. The magnitude of the centripetal acceleration is v2tangential/Rvtangential2/R. A particle that is going along a path with local radius of curvature RRR at speed sss experiences a centripetal acceleration −s2/R−s2/R. If you are in a car turning left, the force you feel pushing you to the right is the force that causes the centripetal acceleration. In these statements vradialvradialv_radial refers to the component of the velocity of an object in the direction toward or away from the origin of the coordinate system or the rotation axis. Conversely, vtangentialvtangentialv_tangential refers to the component of the velocity perpendicular to vradialvradialv_radial. Identify the statement or statements that are false.
Answer:
false b) a = v²(radial) / r and e)
Explanation:
Let's review given statement separately
a) centripetal acceleration
a = v² / r
linear and angular velocity are related
v = w r
we substitute
a = w²r
this acceleration is directed to the center of the circle, so the vector must be negative
a = - w r2
the bold are vector
True this statement
b) the magnitude is the scalar value
a = v² / r
where v is the tangential velocity, not the radial velocity, so this statement is
False
c) This is true
a = v²/ r
this speed is tangential
d) Newton's second law is
F = m a
if the acceleration is centripetal
F = m (- v² / r)
we substitute
F = m (- s² / R)
the statement is true
e) when the car turns to the left, the objects have to follow in a straight line, which is why
you need a force toward the center of the circle to take the curve.
Consequently there is no outward force
This statement is false
What is energy?
O
A. A form of sound
O
B. The ability to do work
O
c. The number of atoms in an object
O
D. The size of an object
In a ballistics test, a 26 g bullet traveling horizontally at 1000 m/s goes through a 35-cm-thick 400 kg stationary target and emerges with a speed of 950 m/s . The target is free to slide on a smooth horizontal surface.
How long is the bullet in the target?
Answer: 3.60*10^-4 s
Explanation:
Given
Mass of bullet, m = 26 g = 0.026 kg
Initial speed of bullet, u = 1000 m/s
Length of target, s = 35 cm = 0.35 m
Mass of target, M = 400 kg
Final speed of bullet, v = 950 m/s
Using equation of motion
v² = u² + 2as, making a subject of formula, we have
a = (v² - u²) / (2*s)
a = 950² - 1000² / 2 * 0.35
a = 902500 - 1000000 / 0.7
a = -97500 / 0.7
The acceleration = - 1.39*10^5 m/s² ( it is worthy of note that the acceleration is negative)
now, using another equation of motion, we have
v = u + a*t
we know our a, so we make t subject of formula
time t = (v-u) / a
t = (950 - 1000) / -1.39*10^5
t = -50 / -1.39*10^5
t = 3.60*10^-4 s
A uniform-density wheel of mass 9 kg and radius 0.40 m rotates on a low-friction axle. Starting from rest, a string wrapped around the edge exerts a constant force of 13 N for 0.72 s. (a) What is the final angular speed? Entry field with correct answer 5.2 radians/s (b) What was the average angular speed? Entry field with incorrect answer 1.872 radians/s (c) Through how big an angle did the wheel turn? Entry field with incorrect answer 1.3478 radians/s (d) How much string came off the wheel? Entry field with incorrect answer 0.8469 m
Answer:
Explanation:
Moment of inertia of wheel = 1/2 x mR² , m is mass and R is radius of wheel
= .5 x 9 x .4²
= .72 kg m²
Torque created on wheel by string = T x r , T is tension and r is radius of wheel .
13 x .4 = 5.2 N m
angular acceleration α = torque / moment of inertia
= 5.2 / .72
= 7.222 rad /s²
a ) final angular speed = α x t , α is angular acceleration , t is time.
= 7.222 x .72
= 5.2 rad /s
b )
θ = 1/2 α t² , θ is angle turned , t is time
= .5 x 7.222 x .72²
= 1.872 rad
average angular speed = θ / t
= 1.872 / .72
= 2.6 rad /s
c )
angle turned = 1.872 rad ( discussed above )
d )
length of string coming off
= angle rotated x radius of wheel
= 1.872 x .4
= .7488 m .
74.88 cm
The final angular speed of the wheel is 5.2 radians/s. The average angular speed is 0.52 radians/s. The wheel turned through an angle of 0.3478 radians and 0.1391 m of string came off the wheel.
To find the final angular speed, we can use the equation ω = φt + 0.5αt^2, where ω is the final angular speed, φ is the initial angular speed (which is 0 since the wheel starts from rest), α is the angular acceleration, and t is the time. We are given that the force applied to the wheel is 13 N for 0.72 s.
Using the equation F = mα, we can rearrange it as α = F/m, where F is the force and m is the mass of the wheel. Plugging in the values, we get α = 13/9. Now we can substitute the values into the first equation and solve for ω: ω = (0)(0.72) + 0.5(13/9)(0.72)^2 = 5.2 radians/s.
To find the average angular speed, we use the formula ωavg = φ + (1/2)αt, where ωavg is the average angular speed. Since the initial angular speed φ is 0, we can simplify the equation to ωavg = (1/2)αt. Plugging in the values, we get ωavg = (1/2)(13/9)(0.72) = 0.52 radians/s.
To find the angle through which the wheel turned, we can use the equation θ = φt + 0.5αt^2. Since the initial angular speed φ is 0, the equation simplifies to θ = 0.5αt^2. Plugging in the values, we get θ = 0.5(13/9)(0.72)^2 = 0.3478 radians.
To find how much string came off the wheel, we can use the formula s = rθ, where s is the length of string and r is the radius of the wheel. Plugging in the values, we get s = (0.4)(0.3478) = 0.1391 m.
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The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The answer is
[tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]
[tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]
Explanation:
From the question we are told that
The first string has a frequency of [tex]f_1 = 230 Hz[/tex]
The period of the beat is [tex]t_{beat} = 0.99s[/tex]
Generally the frequency of the beat is
[tex]f_{beat} = \frac{1}{t_{beat}}[/tex]
Substituting values
[tex]f_{beat} = \frac{1}{0.99}[/tex]
[tex]= 1.01 Hz[/tex]
From the question
[tex]f_2 - f_1 = f_{beat}[/tex] for [tex]f_2[/tex] having a higher tension
So
[tex]f_2 - 230 = 1.01[/tex]
[tex]f_2 = 231.01Hz[/tex]
From the question
[tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]
[tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]
Substituting values
[tex]\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}[/tex]
[tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]
For [tex]f_2[/tex] having a lower tension
[tex]f_1 - f_2 = f_{beat}[/tex]
So
[tex]230 - f_2 = 1.01[/tex]
[tex]f_2 = 230 -1.01[/tex]
[tex]= 228.99[/tex]
From the question
[tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]
[tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]
Substituting values
[tex]\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}[/tex]
[tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]
Between this and the next assignment, we want to get a better under- standing of how light interacts with the eye. Here are two questions to get us started, focused on diffraction (i.e., the spreading of light when it passes through a narrow opening). A. To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright
Answer:
θ₁ = 3.35 10⁻⁴ rad , θ₂ = 8.39 10⁻⁵ rad
Explanation:
This is a diffraction problem for a slit that is described by the expression
sin θ = m λ
the resolution is obtained from the angle between the central maximum and the first minimum corresponding to m = 1
sin θ = λ / a
as in these experiments the angle is very small we can approximate the sine to its angle
θ = λ / a
In this case, the circular openings are explicit, so the system must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ / D
where D is the diameter of the opening
let's apply this expression to our case
indicates that the wavelength is λ = 550 nm = 550 10⁻⁹ m
the case of a lot of light D = 2 mm = 2 10⁻³ m
θ₁ = 1.22 550 10-9 / 2 10⁻³
θ₁ = 3.35 10⁻⁴ rad
For the low light case D = 8 mm = 8 10⁻³
θ₂ = 1.22 550 10-9 / 8 10⁻³
θ₂ = 8.39 10⁻⁵ rad
A heat conducting rod, 0.90 m long, is made of an aluminum section, 0.20 m long, and a copper section, 0.70 m long. Both sections have a cross-sectional area of 0.0004 m^2. The aluminum end and the copper end are maintained at temperatures of 30*C and 230*C respectively. The thermal conductivities of aluminum and copper are 205 and 385 W/m ? K, respectively. The temperature of the aluminum-copper junction in the rod, in *C, is closest to:
Answer:
is closest to 100*C temperature at the aluminum-copper junction
Explanation:
The expression for calculating the resistance of each rod is given by
[tex]R =\frac{ L}{ kA}[/tex]
Now; for Aluminium
[tex]R_{al} =\frac{ 0.20 }{ 205*0.0004}[/tex]
[tex]R_{al}[/tex] = 2.439
For Copper
[tex]R_{Cu}=\frac{0.70}{385*0.0004}[/tex]
[tex]R_{Cu} = 4.545[/tex]
Total Resistance [tex]R = R_{al} + R_{Cu}[/tex]
= 2.439 + 4.545
= 6.9845
Total temperature difference = 230*C + 30*C
= 200 *C
The Total rate of heat flow is then determined which is = [tex]\frac{ total \ temp \ difference}{total \ resistance }[/tex]
=[tex]\frac{200}{ 6.9845 }[/tex]
= 28.635 Watts
However. the temperature difference across the aluminium = Heat flow × Resistance of aluminium
= 28.635 × 2.349
= 69.84 *C
Finally. for as much as one end of the aluminium is = 30 *C , the other end is;
=30*C + 69.84*C
= 99.84 *C
which is closest to 100*C temperature at the aluminum-copper junction
Projectile motion is a combination of which two types of motion?
Answer:horizontal motion and vertical motion
Explanation:projectile motion is a combination of both vertical and horizontal motion
Answer:
C. Horizontal and Vertical
Explanation:
edge
For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that is initially at 4.3°C is dropped into boiling water at 100°C. The heat transfer coefficient at the surface of the egg is estimated to be 800 W/m2⋅K. If the egg is considered cooked when its center temperature reaches 74°C, determine how long the egg should be kept in the boiling water. Solve this problem using the analytical one-term approximation method. The thermal conductivity and diffusivity of the eggs can be approximated by those of water at room temperature to be k = 0.607 W/m·°C and α = 0.146×10−6 m2/s.
Using the analytical one-term approximation method, we can determine how long an egg should be kept in boiling water to reach a specific temperature at its center.
Explanation:In this problem, we are given the initial and final temperatures of an egg, as well as its diameter, the heat transfer coefficient, and the thermal conductivity and diffusivity of water. We are asked to find how long the egg should be kept in boiling water in order for its center temperature to reach a certain value.
To solve this problem, we can use the analytical one-term approximation method, which involves calculating the Biot number and the dimensionless temperature profile. By equating the temperature at the center of the egg to the desired final temperature, we can solve for the time required.
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A 1500 kg car carrying four 90 kg people travels over a "washboard" dirt road with corrugations 3.7 m apart. The car bounces with maximum amplitude when its speed is 20 km/h. When the car stops, and the people get out, by how much does the car body rise on its suspension?
Answer:
Car body rise on its suspension by 0.0309 m
Explanation:
We have given mass of the car m = 1500 kg
Mass of each person = 90 kg
Speed of the car [tex]v=20km/hr=20\times \frac{5}{18}=5.555m/sec[/tex]
Distance traveled by car d = 3.7 m
So time period [tex]T=\frac{distance}{speed}=\frac{4}{5.55}=0.72sec[/tex]
Frequency [tex]f=\frac{1}{T}=\frac{1}{0.72}=1.388Hz[/tex]
Angular frequency is [tex]\omega =2\pi f=2\times 3.14\times 1.388=8.722rad/sec[/tex]
Angular frequency is equal to [tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]8.722 =\sqrt{\frac{k}{1500}}[/tex]
k = 114109.92 N/m
Now weight of total persons will be equal to spring force
[tex]4mg=kx[/tex]
[tex]4\times 90\times 9.8=114109.92\times x[/tex]
x = 0.0309 m