A town interested in installing wind turbines to generate electricity measures the speed of the wind in the town over the course of a year. They find that most of the time, the wind speed is pretty slow, while only rarely does the wind blow very fast during a storm. What would be the best measure of the central wind speed they should report to the mayor?
Answer:
A town interested in installing wind turbines to generate electricity measures the speed of the wind in the town over the course of a year. They find that most of the time, the wind speed is pretty slow, while only rarely does the wind blow very fast during a storm. What would be the best measure of the central wind speed they should report to the mayor?
The median is the best measure of the central wind speed they should report to the mayor because the distribution is not symmetric. The distribution of wind speeds is skewed.
Explanation:
Based on the scenario described in the question, the median is the best measure of the central wind speed they should report to the mayor because the distribution is not symmetric. The distribution of wind speeds is skewed.
The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.081.08 m. She sets the pendulum swinging, and her collaborators carefully count 101101 complete cycles of oscillation during 2.00×1022.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2
Complete Question
The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2
Answer:
The acceleration due to gravity is [tex]g = 167.2 \ m/s^2[/tex]
Explanation:
From the question we are told that
The length of the simple pendulum is [tex]L = 1.081.08 \ m[/tex]
The number of cycles is [tex]N = 101[/tex]
The time take is [tex]t = 2.00 *10^{2 \ }s[/tex]
Generally the period of this oscillation is mathematically evaluated as
[tex]T = \frac{N}{t }[/tex]
substituting values
[tex]T = \frac{101}{2.0*10^2 }[/tex]
[tex]T = 0.505 \ s[/tex]
The period of this oscillation is mathematically represented as
[tex]T = 2 \pi \sqrt{\frac{l}{g} }[/tex]
making g the subject of the formula we have
[tex]g = \frac{L}{[\frac{T}{2 \pi } ]^2 }[/tex]
[tex]g = \frac{4 \pi ^2 L }{T^2 }[/tex]
Substituting values
[tex]g = \frac{4 * 3.142 ^2 * 1.08 }{505.505^2 }[/tex]
[tex]g = \frac{4 * 3.142 ^2 * 1.08 }{0.505^2 }[/tex]
[tex]g = 167.2 \ m/s^2[/tex]
g A fiber optic is made by covering a thin fiber core material with a cladding material. An optical fiber works on the principle of total internal reflection, from this we can conclude that: A) The refractive index of the core material is less than the refractive index of the cladding material. B) The refractive index of the cladding material is less than the refractive index of the core material. C) They must have the same refractive index. D) It does not matter what the refractive index is. E) The refractive index of the cladding must be exactly twice that of the core.
Answer:(b)
Explanation:
Optical fiber works on the principle of total internal reflection . Total internal reflection occurs when the following two conditions are met
(1) When the angle of incidence of light is greater than the critical angle and,
(2) When a ray of light travels from a optically denser medium to optically rarer medium .
We know that light bend away from normal when it travels from denser to rarer medium. If we we choose such an angle that the refraction angle is [tex]90^{\circ}[/tex] then the incident angle is called critical angle.
So, the refractive index of cladding must be must be less than the refractive index of core material.
What best describes an impulse acting on an object
An impulse is the product of force and the time interval during which that force is applied, resulting in a change of an object's momentum. It can be described both mathematically (J = F⋅Δt) and visually (area under the force-time curve). An impulse leads to an object's acceleration or deceleration and affects both speed and direction.
Explanation:An impulse acting on an object is a concept in physics that describes the effect of a force applied over a period of time. It is the product of the average force and the time duration during which the force acts, resulting in a change in the object's momentum. The impulse experienced by an object can result in acceleration or deceleration, dependent on the direction of the force. Moreover, impulse is not just about the magnitude of force, but also the duration over which it is applied. A key point is that an impulse can be delivered either by a large force over a short period or a smaller force over a longer period, tailored to the specifics of a situation.
Impulse is measured as the change in momentum, which is the mass of the object multiplied by its velocity (mv). The formula for impulse is typically represented as J = F⋅Δt, where J represents impulse, F the force, and Δt the change in time. If you graph force versus time, the area under the curve represents the impulse, visually demonstrating the relationship between force, time, and momentum change.
It is essential to understand that impulse not only influences the speed of an object but also its direction of motion. For example, when a tennis player hits a ball, the racket imparts an impulse to the ball changing its momentum. The total impulse given by multiple forces is considered the net impulse, which is the sum of all individual impulses over a specified time.
Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin film of oil with n2 =1.75. On the other side of the thin film is glass with n3 =1.5. What is the minimum non-zero value of the film thickness d that will cause the reflections from both sides of the film to interfere constructively?
Answer:
Explanation:
On both sides of the film , the mediums have lower refractive index.
for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is
2μt = ( 2n +1 ) λ / 2
μ is refractive index of film ,t is thickness of film λ is wavelength of light
n is order of fringe
for minimum thickness
n = 0
2μt = λ / 2
t = λ / 4μ
= 670 / 1.75 x 4
= 95.71 nm .
A. When light of wavelength 250 nm is incident on a metal surface, the maximum speed of the photoelectrons is 4.0 × 105 m/s, what is the work function of the metal in electron volts? [10 marks] B. Assume that a 100-W light bulb gives off 2.50% of its energy as visible light of wavelength 500 nm. How many photons of visible light are given off in 1.0 min? Page 7 of 7 C. A small blood vessel near the skin surface has a radius of 10 μm, a length of 1 μm and the pressure drop along the blood vessel is 2.50 Pa. The viscosity of blood is 0.0027 Pa.s. i. What is the volume flow rate of blood through this blood vessel? What is the velocity of blood flow? [6 marks] ii. Vasodilation causes the radius of this blood vessel to increase to 12 μm, while leaving the pressure drop along the vessel unchanged? What is the volume flow rate through this blood vessel now? What is the velocity of blood flow? D. Many radioisotopes have important industrial, medical, and research applications. One of these is Cobalt-60, which has a half-life of 5.20 years and decays by the emission of a beta particle (energy 0.31 MeV) and two gamma photons (energies 1.17 MeV and 1.33 MeV). A scientist wishes to prepare a Cobalt-60 sealed source that will have an activity of at least 10 Ci after 30 months of use. i. What minimum initial mass of Cobalt-60 is required? [6 marks] ii. At what rate will the source emit energy after 30 months? [4 marks] E. The radioactive isotope Gold-198 has a half-life of 64.80 hrs. A sample containing this isotope has an initial activity of 40.0 μCi. Calculate the number of nuclei that will decay in the time interval from 10 hrs to 12 hrs. [10 marks]
Answer:
Explanation:
Part A
- Work function ( ∅ ) is the minimum energy required by the photon to knock an electron out of the metal surface. That is the portion of energy of a photon transferred to an electron so that it can escape the metal.
- We can mathematically express it:
∅ = Ep - Ek
= [tex]\frac{h*c}{lamba} - 0.5*m_e*v^2_e\\[/tex]
Where,
Planck's constant ( h ) = 6.6261*10^-34
Speed of light ( c ) = 3*10^8 m/s
mass of an electron ( m_e ) = 9.1094*10^-31 kg
Given:-
Incident light's wavelength ( λ ) = 250*10^-9 m
The maximum speed o electron ( v_e ) = 4*10^5 m/s
Solution:-
- Plug the values into the expression derived before:
∅ = [tex]\frac{(6.6261*10^-^3^4)*(3*10^8)}{250*10^-^9} \\\\[/tex]
∅ = [tex]7.22257*10^-^1^9 J * \frac{6.242*10^1^8 eV}{J}[/tex]
∅ = 4.508 eV ... Answer
Part B
- 2.5% of the energy emitted by a 100-W light bulb was visible light with wavelength ( λ ) = 500*10^-9 m. In 1.0 min the amount of energy harbored by a stream of photons are:
[tex]E_p = n_p*\frac{h*c}{lambda} = P*t*e[/tex]
Where,
Planck's constant ( h ) = 6.6261*10^-34
Speed of light ( c ) = 3*10^8 m/s
Given:-
visible light's wavelength ( λ ) = 500*10^-9 m
Power of light bulb ( P ) = 100 W
Time taken ( t ) = 1.0 min = 60 s
Portion of energy as light ( e ) = 0.025
Solution:-
- Plug the values into the expression derived before:
[tex]n_p = \frac{(lambda)*(P)*(t)*(e)}{h*c} \\\\n_p = \frac{(500*10^-^9)*(100)*(60)*(0.025)}{(6.6261*10^-^3^4)*(3*10^8)} \\[/tex]
n_p = 3.773 * 10^20 ... Answer
Part C
- A blood vessel of radius ( r ) with length ( L ) carries blood with viscosity ( μ ). The pressure drop ( ΔP ) in the blood vessel was witnessed.
- Pressure loss ( ΔP ) in a cylindrical blood vessel is given by the Darcy's equation given below:
[tex]\frac{dP}{p} = f*\frac{L}{D}*\frac{v^2}{2}[/tex]
Where,
ρ: Density of blood
f: Friction factor
D: Diameter of vessel
v: Average velocity
- The friction factor is a function of Reynolds number and relative roughness of blood vessel. We will assume the blood vessel to be smooth, round and the flow to be laminar ( later verified ).
- The flow rate ( Q ) in a smooth blood vessel subjected to laminar flow conditions is given by the Poiseuille's Law. The law states:
[tex]Q = \frac{\pi*dP*r^4 }{8*u*L}[/tex]
- The velocity ( v ) in a circular tube is given by the following relation:
[tex]v = \frac{Q}{\pi*r^2 }[/tex]
Given:-
dP ( Pressure loss ) = 2.5 Pa
radius of vessel ( r ) = 10μm = 10*10^-6 m
viscosity of blood ( μ ) = 0.0027 Pa.s
Length of vessel ( L ) = 1μm = 10^-6 m
Solution:-
- Use the Poiseuille's Law to determine the flow rate ( Q ) of the blood in the vessel:
[tex]Q = \frac{\pi*(2.5)*(10*10^-^6)^4 }{8*(0.0027)*(10^-^6)}[/tex]
Q = 3.6361*10^-12 m^3 / s
- The corresponding velocity ( v ) of the blood flow would be:
[tex]v = \frac{3.6361*10^-^1^2}{\pi*(10*10^-^6)^2 }[/tex]
v = 0.01157 m/s
- Use the Poiseuille's Law to determine the flow rate ( Q ) of the blood blood in the enlarged vessel ( r = 12 μm = 10*10^-6 m ) :
[tex]Q = \frac{\pi*(2.5)*(12*10^-^6)^4 }{8*(0.0027)*(10^-^6)}[/tex]
Q = 7.54*10^-12 m^3 / s
- The corresponding velocity ( v ) of the blood flow would be:
[tex]v = \frac{7.53982*10^-^1^2}{\pi*(12*10^-^6)^2 }[/tex]
v = 0.01666 m/s
Part D
- A radioactive isotope of Cobalt ( Co - 60 ) undegoes Beta decay ( 0.31 MeV ) and emits two gamma rays of energy ( 1.17 & 1.33 ) MeV.
- The radioactive decay for the ( Cobalt - 60 ) can be expressed in form of an equation:
[tex]_2_7Co ( 60 ) ---> _2_8 Ni ( 60 ) + e^- + v_e^- + gamma[/tex]
- The half life ( T_1/2 ) of the Co-60 can be used to determine the decay constant ( λ ):
λ = [tex]lambda = \frac{Ln(2)}{T_1/2}\\[/tex]
Where,
T_1/2 = 5.2 yrs = 1.68*10^8 s
Hence, the decay constant is
λ = [tex]\frac{Ln ( 2 ) }{1.68*10^8}[/tex] = 4.22*10^-9 s^-1
- The activity ( A ) of any radioactive isotope is function of time ( t ) defined by negative exponential distribution:
[tex]A = A_o*e^(^-^l^a^m^b^d^a^*^t^)[/tex]
Where,
A_o: The initial activity ( Bq )
- The activity of the radioactive isotope Co-60 was A = 10 Ci after t = 30 months. The initial activity ( A_o ) can be determined:
[tex]A_o= \frac{A}{e^(^-^l^a^m^b^d^a^*^t^)} \\\\A_o= \frac{(10Ci)*(3.7*10^1^0 Bq/Ci)}{e^(^-^4^.^2^2^*^1^0^-^9*^7^.^8^8^*^1^0^7^)} \\[/tex]
A_o = 5.016 * 10^11 Bq
- The initial number of nuclei in the sample ( N_o ) is given by:
[tex]N_o = (A_o) / (lambda)\\\\N_o = \frac{5.016*10^1^1}{4.22*10^-^9} = 1.22*10^2^2[/tex]
- The initial mass of Co-60 used as a sample can be determined:
[tex]m_o = M_r*N_o\\\\m_o = (59.933822u)*(1.66*10^-^2^7)*(1.22*10^2^2)\\[/tex]
m_o = 12.2 * 10^-6 kg ... Answer
- The total energy ( E ) released from the beta decay transformation:
E = E(β) + E(γ1) + E(γ2) = 0.31 + 1.17 + 1.33 = 2.81 MeV
- The rate at which the source emits energy after 30 months:
P = E*A = [tex]( 2.81 MeV * \frac{1.6*10^-^1^3 J}{MeV} ) * ( 10 Ci * \frac{3.7*10^1^0 Bq}{Ci} )[/tex]
P = 0.166 W .. Answer
In an ultrahigh vacuum system, the pressure is measured to be 8.4 × 10−11 torr (where 1 torr = 133 Pa). The gas molecules have a molecular diameter of 2.2 × 10−10 m and the temperature is 310 K. Avogadro's number is 6.02214×1023 1/mol. Find the number of molecules in a volume of 0.87 m3 . Answer in units of molecules.
Answer:
The number of molecules in the volume is [tex]N_v = 2.27109* 10^{12}[/tex] molecules
Explanation:
From the question we are told that
The pressure of the ultrahigh vacuum is [tex]P = 8.4*10^{-11} torr = 8.4*10^{-11} * 133 = 1.1172 *10^{-8}Pa[/tex]
The molecular diameter of the gas molecules [tex]d = 2.2*10^{-10} m[/tex]
The temperature is [tex]T = 310 \ K[/tex]
Avogadro's number is [tex]N = 6.02214 *10^{23}\ l/mol[/tex]
The volume of the gas is [tex]V = 0.87 m^3[/tex]
From the ideal gas law[[tex]PV = nRT[/tex]] that the number of mole is mathematically represented as
[tex]n = \frac{PV}{RT}[/tex]
Where R is the gas constant with a value [tex]R = 8.314\ J/mol[/tex]
Substituting values
[tex]n = \frac{1.1172 *10^{-8} * 0.87}{8.314 * 310}[/tex]
[tex]n = 3.771*10^{-12} \ mole[/tex]
The number of molecules is mathematically represented as
[tex]N_v = n * N[/tex]
Substituting values
[tex]N_v = 3.771*10^{-12} * 6.02214 *10^{23}[/tex]
[tex]N_v = 2.27109* 10^{12}[/tex] molecules
A person uses 25.0 J of kinetic energy to push an object for 11.0 How are work and power affected if the person uses the same amount of kinetic energy to push the object in less time ?
Complete question:
A person uses 25.0 J of kinetic energy to push an object for 11.0 s How are work and power affected if the person uses the same amount of kinetic energy to push the object in less time ?
Answer:
The power will increase, and the amount of work will remain the same
Explanation:
Given;
Kinetic energy, K = 25.0 J
time of work, t = 11.0 s
Power = work / time = Energy / time
This equation shows that power is inversely proportional to time
Also, Energy is directly proportional to work (both are measured in Joules)
Since the person will use the same amount of kinetic energy to push the object in less time.
It means that energy will be constant (work done will not change) and the time will be reduced.
Power and time are inversely proportional, decrease in time means increase in power.
Thus, the power will increase, and the amount of work will remain the same
A cylindrical bar of metal having a diameter of 21.0 mm and a length of 210 mm is deformed elastically in tension with a force of 46800 N. Given that the elastic modulus and Poisson's ratio of the metal are 60.9 GPa and 0.34, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number. (a) 0.443 mm (b) mm
Answer:
The the elongated length is [tex]\Delta L = 0.4 \ mm[/tex]
The change in diameter is [tex]\Delta d = - 0.0136\ mm[/tex]
Explanation:
From the question we are told that
The diameter of the cylindrical bar is [tex]d = 21.0 \ mm = \frac{21}{1000} = 0.021 \ m[/tex]
The length of the cylindrical bar is [tex]L= 210 \ mm = 0.21 \ m[/tex]
The force that deformed it is [tex]F = 46800 \ N[/tex]
Elastic modulus is [tex]E = 60.9 \ GPa = 60.9 *10^{9}Pa[/tex]
The Poisson's ratio is [tex]\mu = 0.34[/tex]
Generally elastic modulus is mathematically represented as
[tex]E = \frac{\sigma }{\epsilon}[/tex]
Where
[tex]\epsilon[/tex] is the strain which is mathematically represented as
[tex]\epsilon = \frac{L}{\Delta L}[/tex]
Where [tex]\Delta L[/tex] is the elongation length
[tex]\sigma[/tex] is the stress on the cylinder which is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
Where F is the force and
A is the area which is calculated as
[tex]A = \frac{\pi} {4} d^2[/tex]
Substituting values
[tex]A = \frac{\pi}{4} * (0.021)[/tex]
[tex]A = 0.000346 \ m^2[/tex]
So the stress is
[tex]\sigma = \frac{46800}{0.000346}[/tex]
[tex]\sigma = 1.35 *10^{8} \ N \cdot m^2[/tex]
Thus the elastic modulus is
[tex]E = \frac{1.35 *10 ^{8}}{\frac{\Delta L}{L} }[/tex]
making [tex]\Delta L[/tex] the subject
[tex]\Delta L = \frac{EL}{1.35 *10^{8}}[/tex]
Substituting values
[tex]\Delta L = \frac{1.35 *10^{8} * 0.21}{1.35 *10^{8}}[/tex]
[tex]\Delta L = \frac{1.35 *10^{8} * 0.21}{60.9*10^{9}}[/tex]
[tex]\Delta L = 0.0004 \ m[/tex]
Converting to mm
[tex]\Delta L = 0.0004 * 1000[/tex]
[tex]\Delta L = 0.4 \ mm[/tex]
Generally the poisson ratio is mathematically represented as
[tex]\mu = - \frac{\frac{\Delta d }{d} }{\frac{\Delta L }{L} }[/tex]
The negative sign indicate a decrease in diameter as a result of the force
making [tex]\Delta d[/tex] the subject
[tex]\Delta d = - \mu * \frac{\Delta L }{L } * d[/tex]
Substituting values
[tex]\Delta d = - 0.34 * \frac{0.0004 }{0.210 } * 0.021[/tex]
[tex]\Delta d = - 1.36 *10^{-5} \ m[/tex]
Converting to mm
[tex]\Delta d = - 0.0136\ mm[/tex]
White light, with a uniform intensity across the visible wavelength 400 – 690 nm, is perpendicularly incident on a water film, having index of refraction n2 = 1.33 and a thickness of 320 nm, that is suspended in air. At what wavelength is the light reflected by the film brightest to an observer? Leave your final answer in nanometers.
Answer:
Wavelength at which the light reflected by the film is brightest = 567.5 nm
Explanation:
We are given;
index of refraction n2 = 1.33
Thickness;(t) = 320 nm
Now the wavelength at which the light reflected by the film is brightest is gotten from the formula for path difference in critical interference as;
Path difference = (m + ½)(λ/n)
Where;
path difference = 2 x thickness = 2(320) = 640 nm
λ = Wavelength at which the light reflected by the film is brightest
n is Refractive index
m is an integer = 0,1,2,3...
Thus; at m = 0;
We have;
640 = (0 + ½)(λ/1.33)
640 = (λ/2.66)
λ = 640 x 2.66
λ = 1702.4 nm
at m = 1;
We have;
640 = (1 + ½)(λ/1.33)
640 = (3/2)(λ/1.33)
λ = 640 x 1.33 x 2/3
λ = 567.5 nm
at m = 2;
We have;
640 = (2 + ½)(λ/1.33)
640 = (5/2)(λ/1.33)
λ = 640 x 2 x 1.33/5
λ = 340.5 nm
Since we are told that the wavelength is between 400 – 690 nm.
Thus, the wavelength at which the light reflected by the film is brightest is the higher value gotten that is between 400nm and 690nm.
Thus, Wavelength at which the light reflected by the film is brightest = 567.5 nm
The light reflected by the water film will appear brightest to an observer at a wavelength of approximately 854.4 nm.
Explanation:The wavelength of light in a medium is given by λn = λ/ʼn, where λ is the wavelength in vacuum and ʼn is the medium's index of refraction. In water, which has an index of refraction of n = 1.33, the range of visible wavelengths is 285 to 570 nm. When light is incident on the water film, it will reflect predominantly at the wavelengths where the film's thickness is an integer multiple of half the wavelength. The brightest reflection will occur when the film's thickness is such that the reflected wavelength is at its maximum in the range of visible light.
To find the wavelength of the brightest reflection, we can use the equation λn = 2nt, where λn is the wavelength in vacuum, n is the refractive index of the film, and t is the thickness of the film. Given the refractive index n2 = 1.33 and the thickness t = 320 nm, we can solve for λn.
Plugging in the values:
λn = 2nt = 2(1.33)(320 nm) ≈ 854.4 nm
Therefore, the wavelength of the light reflected by the film that appears brightest to an observer is approximately 854.4 nm.
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What is one way to increase the amplitude if a wave in a medium?
A. By applying a vibration at the natural frequency of the medium
B. By increasing the velocity of the wave source
C. By moving the source of the wave away from the receptor or observer
D. By increasing the frequency of the wave source
Answer: c
Explanation:
How has God designed sound waves so that a person is able to hear all the different instruments in a musical piece at the same time?
Answer: when the wave encounters something, it can bounce (reflection) or be bent (refraction). In fact, you can "trap" waves by making them bounce back and forth between two or more surfaces. Musical instruments take advantage of this; they produce pitches by trapping sound waves.
Explanation: Any bunch of sound waves will produce some sort of noise. But to be a tone - a sound with a particular pitch - a group of sound waves has to be very regular, all exactly the same distance apart. That's why we can talk about the frequency and wavelength of tones.
Final answer:
The human ear hears different instruments in music at once due to the superposition of sound waves and the ear's ability to process overlapping sounds, determining their pitch, loudness, and direction.
Explanation:
The human ear can hear all the different instruments in a musical piece simultaneously because of the principle of superposition of sound waves. Sound waves combine without affecting each other, so multiple sounds can reach our ears at the same time. Additionally, our ears and brain are capable of processing and interpreting these overlapping waves, enabling us to differentiate between the various sources of sound.
This is partly due to the design of the ear, which is sensitive to a wide range of frequencies, and the ear's ability to discern pitch, loudness, and direction of sound. The cochlea inside the ear uses temporal and place theories to help us perceive pitch, which allows us to enjoy the complexities of music and speech.
An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 491 Hz when the train is approaching him and 472 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find? (Assume the speed of sound in air is 343 m/s.)
The speed of the train can be calculated using the formula for the Doppler effect. Using the given frequencies and the speed of sound in air, the student can calculate the speed of the train to be approximately 8.7 m/s.
Explanation:To calculate the speed of the train, we can use the formula for the Doppler effect. The formula is given by:
Δf/f = v/c
Where Δf is the change in frequency, f is the frequency observed when the train is at rest, v is the speed of the train, and c is the speed of sound in air.
Using the given frequencies of 491 Hz and 472 Hz, and the speed of sound in air of 343 m/s, we can calculate the speed of the train:
v = (Δf/f) * c = (491 - 472) / 491 * 343 = 8.7 m/s
Therefore, the student finds that the speed of the train is approximately 8.7 m/s.
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Given the observed frequencies of 491 Hz when approaching and 472 Hz when receding, the calculated speed of the train is approximately 6.66 m/s. This calculation assumes the speed of sound in air is 343 m/s.
The question involves using the Doppler Effect to calculate the speed of the train. The Doppler Effect formula for a source moving towards a stationary observer is:
[tex]f'_{approach} = f \times (v + v_o)/(v - v_s)[/tex]
and when the source is moving away:
[tex]f'_{recede} = f \times (v - v_o)/(v + v_s)[/tex]
where:
f' = observed frequencyf = source frequencyv = speed of sound in air (343 m/s)v₀ = speed of observer (0 m/s, since student is stationary)[tex]v_s[/tex] = speed of source (train)Given:
[tex]f'_{approach}[/tex] = 491 Hz[tex]f'_{recede}[/tex] = 472 HzFirst, solve for the source frequency (f) using both equations.
For the approaching train:
[tex]491 = f \times (343) / (343 - v_s)[/tex]
For the receding train:
[tex]472 = f \times (343) / (343 + v_s)[/tex]
Divide the two frequency equations to eliminate f:
(491 / 472) = (343 + [tex]v_s[/tex]) / (343 - [tex]v_s[/tex])
Cross-multiplying and solving for [tex]v_s[/tex]:
491 × (343 - [tex]v_s[/tex]) = 472 × (343 + [tex]v_s[/tex])
491 × 343 - 491 × [tex]v_s[/tex] = 472 × 343 + 472 × [tex]v_s[/tex]
168313 - 491 × [tex]v_s[/tex] = 161896 + 472 × [tex]v_s[/tex]
168313 - 161896 = 491 × [tex]v_s[/tex] + 472 × [tex]v_s[/tex]
6417 = 963 × [tex]v_s[/tex]
[tex]v_s[/tex] = 6417 / 963
[tex]v_s[/tex] ≈ 6.66 m/s
Therefore, the student calculates the speed of the train to be approximately 6.66 m/s.
In an oscillating LC circuit, L = 4.24 mH and C = 3.02 μF. At t = 0 the charge on the capacitor is zero and the current is 2.38 A. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time t > 0 is the rate at which energy is stored in the capacitor greatest, and (c) what is that greatest rate?
Answer:
a) 2.693*10^-4 C
b) 8.875*10^-5 s
c) 2.96 W
Explanation:
Given that
Inductance of the circuit, L = 4.24 mH
Capacitance of the circuit, C = 3.02 μF
Current in the circuit, I = 2.38 A
See attachment for calculations
Answer:
a) 0.269 mC
b) 0.355 ms
c) 1.39W
Explanation:
a) To find the charge off the capacitor you start by using the following expression for the charge in the capacitor:
[tex]q=Qsin(\omega t)[/tex]
next, you calculate the current I by using the derivative of q:
[tex]I=\frac{dq}{dt}=Q\omega cos(\omega t)\\\\for \ t= 0:\\\\I=Q\omega\\\\Q=\frac{I}{\omega}\\\\\omega=\frac{1}{\sqrt{LC}}\\\\Q=I\sqrt{LC}[/tex] ( 1 )
L: inductance = 4.24*10^{-3}H
C: capacitance = 3.02*10^{-6}F
I: current = 2.38 A
you replace the values of the parameters in (1):
[tex]Q=(2.38A)(\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}F)})=2.69*10^{-4}C=0.269mC[/tex]
b) to find the time t you use the following formula for the energy of the capacitor:
[tex]u_c=\frac{q^2}{2C}=\frac{Q^2sin^2(\omega t)}{2C}[/tex]
the maximum storage energy in the capacitor is obtained by derivating the energy:
[tex]\frac{du_c}{dt}=\frac{2\omega Q^2sin(\omega t)cos(\omega t)}{2C}=0\\\\\frac{du_c}{dt}=\frac{\omega Q^2 sin(2\omega t)}{2C}=0\\\\sin(2\omega t)=0\\\\2\omega t= 2\pi\\\\t=\frac{\pi}{\omega}=\pi\sqrt{LC}=\pi\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}F)}=3.55*10^{-4}s=0.355\ ms[/tex]
hence, the time is 0.355 ms
c) The greatest rate is obtained for duc/dt evaluated in t=0.355 ms:
[tex]\frac{du_c}{dt}=\frac{2Q^2sin(2\frac{t}{\sqrt{LC}})}{\sqrt{LC}}[/tex]
[tex]\frac{du_c}{dt}=\frac{(2.69*10^{-4}C)^2sin(2\frac{3.55*10^{-4}s}{\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}C)}})}{2(3.02*10^{-6}C)\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}C)}}=-1.39W[/tex]
The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:
garage door openers: 40.0 MHz
standard cordless phones: 40.0 to 50.0 MHz
baby monitors: 49.0 MHz
FM radio stations: 88.0 to 108 MHz
cell phones: 800 to 900 MHz
Global Positioning System: 1227 to 1575 MHz
microwave ovens: 2450 MHz
wireless internet technology: 2.4 to 2.6 GHz
Which of the following statements correctly describe the various applications listed above? Check all that apply.
a.) All these technologies use radio waves, including low-frequency microwaves.
b.) All these technologies use radio waves, including high-frequency microwaves.
c.) All these technologies use a combination of infrared waves and high-frequency microwaves.
d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.
g.) All the technologies emit waves with a wavelength in the range of 0.01 to 10.0 km.
The correct statements are: (b) All these technologies use radio waves, including high-frequency microwaves. (d) Microwave ovens emit in the same frequency band as some wireless Internet devices.
Explanation:The correct statements that describe the various applications listed above are:
b.) All these technologies use radio waves, including high-frequency microwaves.d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.Statement a.) is incorrect because not all technologies listed use low-frequency microwaves. Statement c.) is incorrect because not all technologies listed use a combination of infrared waves and high-frequency microwaves. Statement e.) is incorrect because wireless Internet devices do not have the shortest wavelength among the technologies listed. Statement f.) is incorrect because the wavelengths of the technologies listed vary. Statement g.) is incorrect because the wavelengths of the technologies listed also vary.
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The correct answers are option (a) and option (d). All these technologies use radio waves, including low-frequency microwaves and Microwave ovens emit in the same frequency band as some wireless Internet devices.
The electromagnetic (EM) spectrum includes a wide range of frequencies, which are used in various modern technologies. Here are analyses correlating with the provided frequency ranges:
All these technologies use radio waves, including low-frequency microwaves: This statement is correct because the frequency ranges provided fall within the radio wave section of the EM spectrum (300 kHz to 300 GHz).All these technologies use radio waves, including high-frequency microwaves: This statement is partially correct because they indeed use radio waves, but not all fall under high-frequency microwaves.All these technologies use a combination of infrared waves and high-frequency microwaves: This statement is incorrect as none of the mentioned applications utilize infrared waves.Microwave ovens emit in the same frequency band as some wireless Internet devices: This is correct. Microwave ovens operate at 2450 MHz, which overlaps with wireless internet technologies operating at 2.4 GHz (or 2400 MHz).The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above: This is incorrect. Different technologies listed operate within varying ranges, some of which have shorter wavelengths.All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m: This is incorrect because not all of the frequencies provided correspond to this specific wavelength range.All the technologies emit waves with a wavelength in the range of 0.01 to 10.0 km: This is also incorrect since the provided frequencies exceed these wavelength ranges.Migratory birds are able to use the earth's magnetic field to navigate even when clouds and darkness prevent them from having visual references for their flight. The range of sensitivity of such birds extends to magnetic fields as small as about a third of the earth's natural field. If such a bird is flying past a power line which carries 105 amps, and if we assume that the minimum field detectable at 60 Hz is the same as the minimum field detected at DC, at what distance could the bird detect the presence of the power line? Why will the real answer be much smaller? (think about how power lines are arranged)
Answer:
The receptors that sense the Earth's magnetic field are probably located in the birds' eyes. Now, researchers at Lund University have studied different proteins in the eyes of zebra finches and discovered that one of them differs from the others: only the Cry4 protein maintains a constant level throughout the day and in different lighting conditions
Explanation:
Final answer:
The bird will be able to detect the power line at a much smaller distance than calculated due to the concentration of the magnetic field close to the wire.
Explanation:
The distance at which a migratory bird could detect the presence of a power line can be calculated using the equation for the magnetic field produced by a long straight wire. However, since the bird's sensitivity to magnetic fields is much smaller than the field produced by the power line, the real answer will be much smaller than the calculated distance. This is because power lines are typically arranged in such a way that the magnetic field they produce is concentrated close to the wire, diminishing rapidly with distance.
. Block m1 slides along a frictionless surface at speed v1 = 4 m/s. Then it undergoes a onedimensional elastic collision with stationary block m2 = 2m1. Next, block m2 undergoes a one-dimensional elastic collision with stationary block m3 = 2m2. (a) What is the speed of block m3? Are the (b) speed, (c) kinetic energy, and (d) momentum of block m3 greater than, less than, or the same as the initial values for m1?
Answer:
a) v3 = 1 m/s
c) K3 < K1
d) p3 = p1
Explanation:
a) To solve this problem you use the conservation of the linear momentum in elastic collision.
In the first case you have:
[tex]p_i=p_f\\\\m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex]
but the second block is at rest, then v2i = 0m/s:
[tex]m_1v_{1i}=m_1v_{1f}+m_2v_{2f}[/tex]
Furthermore, you can assume that the first object stops just after the collision with the second one. From this last expression you obtain the value of the second object:
[tex]v_{2f}=\frac{m_1v_{1i}}{m_2}\\\\m_2=2m_1\\\\v_{2f}=\frac{m_1(4m/s)}{2m_1}=2\ m/s[/tex]
Then, you use the conservation of momentum for the second case, in which the second objects impact the third one:
[tex]m_2v'_{2i}+m_3v_{3i}=m_2v'_{2f}+m_3v_{3f}\\\\v_{3i}=0\\\\m_2v'_{2i}=m_2v'_{2f}+m_3v_{3f}\\\\v_{2f}=0\\\\m_2v'_{2i}=m_3v_{3f}\\\\v_{3f}=\frac{m_2v'_{2i}}{m_3}[/tex]
where again it has assumed that the second object stops, just after the impact with the third object. v'_2i = v_2f (in order to distinguish). BY using the fact m3 = 2m2 you obtain:
[tex]v_{3f}=\frac{m_2(2m/s)}{2m_2}=1\ m/s[/tex]
Then, you obtain that v3 < v2 < v1
c) The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
you compute for all the three objects:
[tex]K_1=\frac{1}{2}m_1(4m/s)^2=8m_1\ m^2/s^2\\\\K_2=\frac{1}{2}m_2(2m/s)^2=\frac{1}{2}(2m_1)(4m^2/s^2)=4m_1\ m^2/s^2\\\\K_3=\frac{1}{2}m_3=(1m/s)^2=\frac{1}{2}(2m_2)(1\ m^2/s^2)=\frac{1}{2}(2(2m_1))(1 m^2/s^2)=2m_1\ m^2/s^2[/tex]
then, k3 < k2 < k1
d) For the momentum you have:
[tex]p_1=4m_1\ m/s\\\\p_2=m_2(2m/s)=(2m_1)(2m/s)=4m_1\ m/s\\\\p_3=m_3(1m/s)=(2m_2)(1m/s)=(2(2m_1))(1m/s)=4m_1\ m/s[/tex]
p1 = p2 = p3
1. Car Down Incline w Friction An automobile weighing 4250 lb starts from rest at point A on a 6o incline and coasts through a distance of 500 ft to point B. The brakes are then applied, causing the automobile to come to a stop at point C, 70 ft from B. Knowing that slipping is impending during the braking period and neglecting air resistance and rolling resistance, determine (a) the speed of the automobile at point B, (b) the coefficient of static friction between the tires and the road.
Answer:
Explanation:
Let θ be the inclination
downward acceleration on an inclined plane
= g sinθ
= 32 x sin6
a = 3.345 ft /s
a ) for knowing the speed at point B
v² = u² + 2 a s , v is final velocity , u is initial velocity , a is acceleration and s is distance travelled .
v² = 0 + 2 x 3.345 x 500
= 3345
v = 57.8 ft /s
from point B to C , the car decelerates so we shall find deceleration
v² = u² + 2 a s
0 = 3345 + 2 x a x 70 ( v becomes u here )
a = - 23.9 m /s²
net force on car during deceleration
= μmgcosθ - mg sinθ where μ is coefficient of static friction ,
= mg ( μcosθ - sinθ )
deceleration = g ( μcosθ - sinθ )
g ( μcosθ - sinθ ) = 23.9
( μcosθ - sinθ ) = .74
μcosθ = .74 + .104
= .8445
μ = .8445 / .9945
= .85 .
Final answer:
To calculate the maximum deceleration of a car heading down a 6° slope under different road conditions, we can use the coefficient of static friction. On dry concrete, the deceleration is approximately 2.12 m/s². On wet concrete, the deceleration is approximately 1.62 m/s². On ice, the deceleration is approximately 2.04 m/s².
Explanation:
To calculate the maximum deceleration of a car heading down a 6° slope, we need to consider the road conditions. Assuming the weight of the car is evenly distributed on all four tires, and that the tires are not allowed to slip during the deceleration, we can calculate the deceleration for different road conditions.
(a) On dry concrete, the coefficient of static friction can be calculated using the equation μs = tan(θ), where θ is the angle of the slope. In this case, the coefficient of static friction is approximately 0.105 and the maximum deceleration is approximately 2.12 m/s².
(b) On wet concrete, the coefficient of static friction is typically lower than on dry concrete. Let's assume a coefficient of 0.08. In this case, the maximum deceleration is approximately 1.62 m/s².
(c) On ice, assuming a coefficient of static friction of 0.100, the same as for shoes on ice, the maximum deceleration is approximately 2.04 m/s².
A cyclist is moving up a slope that is at an angle of 19 to the horizontal. The mass of the cyclist and the bicycle is 85 kg. What is the component of the weight of the cyclist and bicycle parallel to the slope and what is the normal reaction force on the bicycle from the slope?
Answer:
Explanation:
Given
Slope of inclination [tex]\theta =19^{\circ}[/tex]
Mass of cyclist and bicycle is [tex]m=85\ kg[/tex]
When cyclist is going up then there is two components of its weight , one is parallel to inclination and other is perpendicular to inclination
Weight of person is mg
it can be resolved in [tex]mg\cos \theta[/tex] and [tex]mg\sin \theta [/tex]
[tex]mg\cos \theta[/tex] is perpendicular to the inclination
and [tex]mg\sin \theta[/tex] is parallel to inclination as shown in diagram
Parallel component [tex]=mg\sin \theta=85\times 9.8\times \sin 19=271.2\ N[/tex]
Normal reaction [tex]N=mg\cos \theta =85\times 9.8\times \cos 19=787.61\ N[/tex]
The normal reaction on the bicycle by the inclined slope is 787.62 N.
The parallel on the bicycle along the inclined slope is 271.2 N.
The given parameters;
angle of the slope, = 19mass of the bicycle, m = 85 kgThe normal reaction on the bicycle by the inclined slope is calculated as; follows;
[tex]F_n = W cos \theta\\\\F_n = mg \ cos \theta \\\\F_n = 85 \times 9.8 \times cos \ (19)\\\\F_n = 787.62 \ N[/tex]
The parallel on the bicycle along the inclined slope is calculated as
[tex]F_x = Wsin\theta\\\\F_x = mg \ sin \ \theta \\\\F_x = 85 \times 9.8 \times sin(19) \\\\F_ x = 271 .2 \ N[/tex]
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how much force is required to accelerate an 1800 kg truck 3 m/s
Answer:
5,400 Newtons.
Explanation:
Force = 1,800 kg * 3 m/s^2 = 5,400 kg*m/s^2 = 5,400 N
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
USE THE ABOVE INFORMATION TO ANSWER QUESTION 3 TO 6
The focal length of a convex mirror is 10 cm and the object distance is 20
Determine the image height if the peg place at the object distance of 5 cm from the concave mirror
4. Determine the image height if the peg place at the object distance of 15 cm from the concave mirror
5. Determine the image height if the peg place at the object distance of 20 cm from the concave mirror
6. Determine the image height if the peg place at the object distance of 30 cm from the concave mirror
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
Please help me with this question :
A soap bubble has the index of refraction n = 1.33. There is air both inside and outside
the bubble.
a/ What wavelength (in air) of visible light is most strungly reflected from a point on a soap bubble where
its wall is 290 nm thick? To what color does this correspond?
b/ Repeat part a/ for a wall thickness of 340 nm. What is your observation about the two answers a/ and b/?
Answer:
514.27 ( wavelength )the color is green
602.93 nm ( orange color )the observation is that there is a change of visible color
Explanation:
A) wavelength of visible light that is most strongly reflected from a point on a soap
refraction n = 1.33
wall thickness (t) = 290 nm
2nt = (2m +1 ) ∝/2 -----equation 1
note when m = 0
therefore ∝ = 4nt/ 1 = 4 * 1.33 * 290 = 1542.8nm we will discard this
when m = 1
equation 1 becomes
∝ = 4nt/3 =( 4 * 1.33 * 290) / 3 = 1542.8 / 3 = 514.27 ( wavelength )
the color is green
B) the wavelength when the wall thickness is 340 nm
∝ = 4nt / 2m +1
where m = 1
∝ = (4 * 1.33 * 340 ) / 3 = 1808.8 / 3 = 602.93 nm ( orange color )
the observation is that there is a change of visible color
Final answer:
The wavelength of visible light most strongly reflected for a soap bubble wall thickness of 290 nm and index of refraction of 1.33 is 770 nm, corresponding to the red color. For a thickness of 340 nm, the wavelength is 904 nm, which falls outside the visible spectrum, in the infrared range. As the thickness increases, the reflected wavelength shifts toward longer wavelengths.
Explanation:
The phenomenon described in the question is known as thin film interference, which is a physical effect that occurs when light waves reflected off the top and bottom surfaces of a thin film interfere with each other. To find the wavelength (λ) of visible light that is most strongly reflected, we use the formula for constructive interference in thin films:
2nt = mλ,
where n is the index of refraction of the film, t is the thickness of the film, and m is the order of the fringe (which is an integer).
For part (a), we have a soap bubble thickness of 290 nm and an index of refraction n = 1.33. Assuming the light is perpendicular to the surface (m = 1 for the first order of constructive interference), we calculate the wavelength using the given formula as follows:
λ = 2nt/m = 2 * 1.33 * 290 nm / 1 = 770 nm.
The wavelength of 770 nm corresponds to red light in the visible spectrum.
For part (b), with a wall thickness of 340 nm, we calculate the wavelength in a similar fashion:
λ = 2 * 1.33 * 340 nm / 1 = 904 nm.
However, a wavelength of 904 nm falls outside the visible spectrum and cannot be seen as a color. It is in the infrared range.
Comparing answers from (a) and (b), we observe that as the thickness of the soap bubble increases, the wavelength of light most strongly reflected shifts towards the longer wavelengths, moving out of the visible range into the infrared.
A one-dimensional particle-in-a-box may be used to illustrate the import kinetic energy quantization in covalent bond formation. For example, the electronic energy change associated with the reaction H+H H2 may be modeled by treating each reactant H atom as an electron in a one-dimensional box of length LH 5a0 (the 99% electron density diameter of hydrogen), and treating he diatomic H2 as a one-dimensional box of length LH2 RB+5ao (where ao is the Bohr radius of hydrogen and Re 0.74 Å is the experimental bond length of H2). (a) Use the above particle-in-a-box model to model ance of predict the bond formation energy of H2, and compare your result with the experi- mental value of -436 kJ/mol. (b) What interactions have been neglected in the above calculation and what does your result imply with regard to the importance of kinetic energy quantization in covalent bond formation?
Answer:
a) 423.64 KJ / mole
Explanation:
The pictures below explains it all in the calculation and i hope it helps you
As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a channel in a microprocessor. If we know that an electron is somewhere along the 50 nm length of the channel, what is ∆vx? If we treat the elec- tron as a classical particle moving at a speed at the outer edge of the uncertainty range, how long would it take to traverse the channel?
Answer:
a) ∆x∆v = 5.78*10^-5
∆v = 1157.08 m/s
b) 4.32*10^{-11}
Explanation:
To solve this problem you use the Heisenberg's uncertainty principle, that is given by:
[tex]\Delta x\Delta p \geq \frac{\hbar}{2}[/tex]
where h is the Planck's constant (6.62*10^-34 J s).
If you assume that the mass of the electron is constant you have:
[tex]\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}[/tex]
you use the value of the mass of an electron (9.61*10^-31 kg), and the uncertainty in the position of the electron (50nm), in order to calculate ∆x∆v and ∆v:
[tex]\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s[/tex]
[tex]\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}[/tex]
If the electron is a classical particle, the time it takes to traverse the channel is (by using the edge of the uncertainty in the velocity):
[tex]t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s[/tex]
A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 240 V to a primary coil of 280 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages? (b) If the maximum input current is 5.00 A, what are the maximum output currents (each used alone)?
Answer:
A) i) at V_s = 5.6V
N_s = 6.53 turns
ii) at V_s = 12V, N_s = 14 turns
iii) at V_s = 480V, N_s = 560 turns
B) i) at V_s = 5.6V
I_s = 214.29A
ii) at V_s = 12V, I_s = 100A
iii) at V_s = 480V, N_s = 2.5A
Explanation:
A) The formula for calculating number of turns in the secondary coil is gotten from;
V_p/V_s = N_p/N_s
Making N_s the subject we have;
N_s =( V_s*N_p)/V_p
Where;
V_p is input voltage
V_s is output voltage
N_p is number of turns in primary coil
N_s is number of turns in secondary coil
We are given V_p = 240V and N_p = 280 turns
Thus;
i) at V_s = 5.6V,
N_s = (5.6*280)/240
N_s = 6.53 turns
ii) at V_s = 12V,
N_s = (12*280)/240
N_s = 14 turns
iii) at V_s = 480V,
N_s = (480*280)/240
N_s = 560 turns
B) The formula for calculating maximum output current in the secondary coil is gotten from;
I_s = (V_p*I_p)/V_s
Where;
I_s is maximum output current
V_p is input voltage
I_p is maximum input current
V_s is output voltage
We are given I_p = 5A
Thus;
i) at V_s = 5.6V,
I_s = (240*5)/5.6
I_s = 214.29 A
ii) at V_s = 12V
I_s = (240*5)/12
I_s = 100 A
iii) at V_s = 480V
I_s = (240*5)/480
I_s = 2.5 A
Final answer:
The number of turns in the secondary coil of the transformer to produce outputs of 5.60, 12.0, and 480 V are approximately 6.5, 14, and 560 turns respectively. The associated maximum output currents for each voltage, assuming 100% efficiency, are approximately 214.29 A for 5.60 V, 100.00 A for 12.0 V, and 2.50 A for 480 V.
Explanation:
The question involves a multipurpose transformer where the primary coil receives an input voltage of 240 V and has 280 turns, and we need to calculate the number of turns in the secondary to produce various output voltages of 5.60, 12.0, and 480 V. Secondly, we have to determine the maximum output currents for a maximum input current of 5.00 A using the transformer's power conservation principle.
(a) To find the number of turns in each part of the secondary coil, we can use the transformer equation:
VP / VS = NP / NS
For each output voltage, we can solve for NS (number of turns in the secondary):
For 12.0 V output: NS = (12.0 V / 240 V) × 280 turns = 14 turns
For 480 V output: NS = (480 V / 240 V) × 280 turns = 560 turns
(b) Power conservation in transformers (assuming 100% efficiency) indicates that PP = PS (power in primary equals power in secondary), so:
For 480 V: IS = (240 V × 5.00 A) / 480 V = 2.50 A
5. A 30.0-g rifle bullet traveling 185 m/s embeds itself in a 3.15-kg pendulum hanging on a 2.85-m-long string, which makes the pendulum swing upward in an arc. Determine: a. The vertical component of the pendulum’s maximum displacement. (10pts) b. The horizontal component of the pendulum’s maximum displacement. (10pts) c. The angle of the pendulum’s maximum displacement with the vertical. (10pts)
Answer:
Explanation:
We shall apply law of conservation of momentum to calculate the common velocity of pendulum and bullet after collision
m v = ( m + M ) V
m is mass of bullet M is mass of pendulum , v is velocity of bullet and V is their final velocity after collision.
V = m v / ( m + M )
= .030 x 185 / 3.18
= 1.745 m /s
The kinetic energy of bullet+pendulum will be converted into potential energy.
1/2 (m+M) V² = ( m + M ) g h
h is height by which pendulum and bullet rises after collision
Putting the values
.5 x 3.18 x 1.745² = 3.18 x 9.8 x h
h = .155 m
If θ be the angle that the pendulum is making at deflected position
l - lcosθ = h , l is length of pendulum
2.85 ( 1 - cosθ) = .155
1 - cosθ = .054
cosθ = .946
θ = 19 degree
Horizontal displacement = 2.85 sin 19
= .9278 m
92.78 cm
Using the principle of conservation of momentum and kinetic energy transformation, we calculate the vertical displacement of the pendulum after the bullet embeds. This is then used to compute the horizontal displacement using the Pythagorean theorem. Finally, the angle of displacement is determined with the trigonometric principle of arcsine.
Explanation:To solve this question, we will use the principle of conservation of momentum to the bullet and pendulum system. After the bullet embeds into the pendulum, they move together, and due to the conservation of momentum, the joint velocity will be lower than the initial velocity of the bullet. We can calculate this joint velocity using the formula: (Mass of bullet x bullet velocity + Mass of pendulum x pendulum velocity) / (Mass of bullet + Mass of pendulum).
Next, we calculate the kinetic energy of the bullet-pendulum system after the bullet embeds itself using the formula: 1/2 x Mass of system x Velocity of system². This kinetic energy is then transformed into potential energy as the pendulum swings upward, which can be calculated using the formula: Mass of system x gravitational acceleration (9.81 m/s²) x vertical displacement.
We can solve for vertical displacement by equating the kinetic and potential energies. The horizontal displacement can then be calculated by using the Pythagorean theorem, where the hypotenuse is the length of the pendulum string, and one side is the length of the string less the vertical displacement (calculated earlier). Using these, the horizontal displacement is calculated as the square root of (length of string² - vertical displacement²).
The angle of maximum displacement with the vertical can then be calculated using the principle of trigonometry, namely the Arcsine formula, where arcsin (vertical displacement/length of string).
Learn more about Physics of a bullet embedding into a pendulum here:https://brainly.com/question/34643408
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A 17.0 m long, thin, uniform steel beam slides south at a speed of 28.0 m/s. The length of the beam maintains an east-west orientation while sliding. The vertical component of the Earth's magnetic field at this location has a magnitude of 30.0 µT. What is the magnitude of the induced emf between the ends of the beam (in mV)?
Answer:14.2 mV
Explanation:
Given
Length of steel beam [tex]L=17\ m[/tex]
Magnetic field [tex]B=30\ \muT[/tex]
speed of beam [tex]v=28\ m/s[/tex]
Now consider as beam slides it encloses an area of rectangle of
width [tex]w=17\ m and length L'=v\times t [/tex]
Area [tex]A=17v\cdot t[/tex]
and Induced EMF is [tex]V=-\frac{d(BA)}{dt}[/tex]
[tex]V=-B\frac{d(17vt)}{dt}[/tex]
[tex]V=-17Bv[/tex]
[tex]V=-17\times 30\times 10^{-6}\times 28[/tex]
[tex]V=-14.2\ mV[/tex]
Magnitude of EMF[tex]=14.2\ mV[/tex]
You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work on water resources project. The department will be overseeing the construction of a dam to create a large fresh water lake that will be approximately 17 meters deep. A horizontal pipe 1.2 meters long and 3 cm in diameter will pass through the dam at a depth of 7 meters to allow for release of the water in emergencies and for sampling. In normal situations, a plug will secure the pipe opening. 1)What will be the total force on the left side of the plug
Answer:
Check the explanation
Explanation:
1) Pressure acting on the plug = Patm + P
Pressure = Patm + rho*g*h (Here h = D2)
Pressure = 101325 + 1000*9.8*7
Pressure = 169925 Pa
so, Force = PA
Force = 169925*pi*0.0152
Force = 120.1 N
A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the distance of point charge on the line joining the charge B. Where the resultant electric field is zero
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
How much current will pass through a 12.5 ohm resistor when it is connected to ta 115 volt source of electrical potential?
Answer:
9.2 amperes
Explanation:
Ohm's law states that the voltage V across a conductor of resistance R is given by [tex]V = R I[/tex]
Here, voltage V is proportional to the current I.
For voltage, unit is volts (V)
For current, unit is amperes (A)
For resistance, unit is Ohms (Ω)
Put R = 12.5 and V = 115 in V=RI
[tex]115=12.5I\\I=\frac{115}{12.5}\\ =9.2\,\,amperes[/tex]