a. The energy density of the magnetic field ([tex]\(u_B\)[/tex]) at the surface of the wire is [tex]\(3.08875 \times 10^{-7} \, \text{J/m}^3\)[/tex]
b. The energy density of the electric field ([tex]\(u_E\)[/tex]) at the surface of the wire is [tex]\(3.74883 \times 10^{-3} \, \text{J/m}^3\).[/tex]
To calculate the energy density of the magnetic field ([tex]\(u_B\)[/tex]) and the electric field ([tex]\(u_E\)[/tex]) at the surface of the wire, we first need to find the magnetic field strength (B) and the electric field strength (E) at that point.
Given:
- Current (I) = 11 A
- Wire diameter (d) = 2.8 mm = 2.8 x 10^-3 m
- Resistance per unit length (R) = 3.7 Ω/km = 3.7 x 10^3 Ω/m
We can calculate the radius (r) of the wire:
[tex]\[ r = \frac{d}{2} = \frac{2.8 \times 10^{-3}}{2} = 1.4 \times 10^{-3} \, \text{m} \][/tex]
(a) Magnetic Field Energy Density ([tex]\(u_B\)[/tex]):
The magnetic field strength at the surface of the wire (B) is given by Ampère's Law:
[tex]\[ B = \frac{\mu_0 I}{2\pi r} \][/tex]
[tex]\[ B = \frac{4\pi \times 10^{-7} \times 11}{2\pi \times 1.4 \times 10^{-3}} \][/tex]
[tex]\[ B = \frac{4 \times 11 \times 10^{-7}}{2 \times 1.4 \times 10^{-3}} \][/tex]
[tex]\[ B = \frac{4 \times 11}{2 \times 1.4} \times 10^{-4} \][/tex]
[tex]\[ B = \frac{44}{2.8} \times 10^{-4} \][/tex]
[tex]\[ B = 15.714 \times 10^{-4} \][/tex]
[tex]\[ B = 1.571 \times 10^{-3} \, \text{T} \][/tex]
The energy density of the magnetic field ([tex]\(u_B\)[/tex]) is given by:
[tex]\[ u_B = \frac{B^2}{2\mu_0} \][/tex]
[tex]\[ u_B = \frac{(1.571 \times 10^{-3})^2}{2 \times 4\pi \times 10^{-7}} \][/tex]
[tex]\[ u_B = \frac{2.471 \times 10^{-6}}{8\pi \times 10^{-7}} \][/tex]
[tex]\[ u_B = \frac{2.471}{8} \times 10^{-6} \][/tex]
[tex]\[ u_B = 0.308875 \times 10^{-6} \][/tex]
[tex]\[ u_B = 3.08875 \times 10^{-7} \, \text{J/m}^3 \][/tex]
(b) Electric Field Energy Density ([tex]\(u_E\)[/tex]):
The electric field strength (E) at the surface of the wire is given by Ohm's Law:
[tex]\[ E = \frac{V}{r} \][/tex]
[tex]\[ E = \frac{IR}{r} \][/tex]
[tex]\[ E = \frac{11 \times 3.7 \times 10^3}{1.4 \times 10^{-3}} \][/tex]
[tex]\[ E = \frac{40.7 \times 10^3}{1.4 \times 10^{-3}} \][/tex]
[tex]\[ E = \frac{40.7}{1.4} \times 10^3 \][/tex]
[tex]\[ E = 29.071 \times 10^3 \][/tex]
[tex]\[ E = 29.071 \times 10^3 \, \text{V/m} \][/tex]
The energy density of the electric field ([tex]\(u_E\)[/tex]) is given by:
[tex]\[ u_E = \frac{\varepsilon_0 E^2}{2} \][/tex]
[tex]\[ u_E = \frac{8.85 \times 10^{-12} \times (29.071 \times 10^3)^2}{2} \][/tex]
[tex]\[ u_E = \frac{8.85 \times 10^{-12} \times 845.96 \times 10^6}{2} \][/tex]
[tex]\[ u_E = \frac{7.49766 \times 10^{-3}}{2} \][/tex]
[tex]\[ u_E = 3.74883 \times 10^{-3} \][/tex]
So, the energy density of the magnetic field ([tex]\(u_B\)[/tex]) at the surface of the wire is [tex]\(3.08875 \times 10^{-7} \, \text{J/m}^3\)[/tex] and the energy density of the electric field ([tex]\(u_E\)[/tex]) at the surface of the wire is [tex]\(3.74883 \times 10^{-3} \, \text{J/m}^3\).[/tex]
When is an electron in an atom likely to move from one energy level to another
Explanation:
When a electron is collided with a photon with exactly the same energy it would require to get to any of the farther orbits,electron transition takes place to an orbit depending on the energy of the photon.
When electrons emit exactly the same amount energy that is difference between the current energy level and the new level,then the electron makes a transition to the new level.
An electron in an atom is likely to move from one energy level to another when it gains or loses energy.
Explanation:
The change in an electron’s position with respect to energy levels is termed as Atomic electron transition. In spite of having similar charge and mass, the energy level of any electron in an atom, surrounding the nucleus, differs depending on its orbital position from the nucleus.
Electrons positioned nearest to the atomic nucleus carry least energy. To move an electron from its original ground state energy level to a higher level; energize the atom and it will excite the electrons thus making it move from its lower energy stable state to an unstable state with higher energy level. Releasing energy of an atom decreases the energy level of its excited electrons, de-energizing it and thus stabilizing the atom.
Two speakers face each other, and they each emit a sound of wavelength (lambda). One speaker is 180 (degrees)
%u2218out of phase with respect to the other. If we separate the speakers by a distance1.5 (lambda)
,how far from the left-most speaker should we place a microphone in order to pick up the loudest sound? Ignore reflections from nearby surfaces. Select all that apply.
A. 3/4 lambda
B. 0 lambda
C. 1/2 lambda
D. 1/4 lambda
E. 1 lambda
Answer:
D. 1/4 lambda
Explanation:
Given:
Initially out of phase speakers by 180°.
Then on being separated by 1.5 λ.
The initial case of speakers being out of phase by 180°
In such a case we obtain destructive interference because the vectors of the waveform are equal and opposite in direction so they cancel out each other.On separating the speakers by a distance of 1.5 λ.
We obtain that the waves-forms are in-phase and form constructive interference leading to the highest value at the crest of the wave pattern, i.e. we obtain the loudest sound when we place a microphone at a distance of one-fourth of the wavelength.A small bolt with a mass of 33.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical direction with a frequency of 3.05 Hz. What is the maximum amplitude that the piston can oscillate without the bolt losing contact with the piston's surface?
Answer:
0.027m
Explanation:
the bolt loses contact with the piston only when acceleration due to gravity equals acceleration of piston
ω² * A = g where ω is angular velocity, A amplitude, g acceleration due to gravity
ω is given by 2πf, ω² is 4π²f²
A= g/4π²f² depending on the value of g used either 10m/s² or 9.8m/s²,
i used 10m/s² in this answer
The specific heat capacity of copper is three times the specific heat capacity of lead. Equal masses of copper and lead are heated from room temperature to the temperature of boiling water. To achieve, this the amount of heat added to the copper is______times the heat added to lead.
Answer:
Three times
Explanation:
The amount of heat added to a substance when its temperature is increased from T_1 to T_2 is given by
[tex]Q=mc(T_2-T_1)[/tex]
c= specific heat capacity
m= mass of the substance
⇒Q∝c
that is if c is increased three times the amount of heat added is also increased three times. Therefore, the amount of heat added to the copper is three times the heat added to lead.
Why is it important to develop alternative energy sources
Answer:
The alternative energy sources are defined as those resources that are used in place of the natural and non-renewable resources. This resources plays an important role in the conservation of natural resources.
The fossil fuels are the resources on which the people are directly dependent. Burning up of these fossils leads to the emission of carbon, which has a direct impact on earth. A small increase in the amount of carbon dioxide can lead to the increase in the surface temperature of earth.
In addition to this, these fossil fuels such as coal, petroleum, oil and natural gases are found to be present in a limited proportion, and it is a very expensive process to obtain these resources, so sustainable development method must be adopted in order to save this natural resources for the future generation.
Some of the examples of alternative resources that are widely used in place of fossil fuels are wind energy, solar energy, tidal energy, biomass energy and bio-fuels.
Thus, it is very important to develop and use alternative resources.
Final answer:
Developing alternative energy sources is vital for mitigating climate change, reducing reliance on finite fossil fuels, and enhancing economic opportunity and security. Renewable energy such as solar, wind, and hydropower are crucial for the sustainable transition.
Explanation:
Developing alternative energy sources is crucial for a number of reasons. First, as renewable energy sources like solar and wind become more technically superior and cost-effective compared to fossil fuels, market forces will naturally encourage a transition away from non-renewable resources. This helps prevent dependency on finite resources that will eventually deplete.
Second, climate change concerns necessitate the development of alternative energy sources. Political measures may impose financial penalties on fossil fuel use to mitigate harmful ecological impacts, prompting a migration towards more sustainable energy choices.
Lastly, alternative energy sources like solar, wind, and hydropower can provide increased economic opportunity, job creation, and energy security. Especially in light of climate-related disasters that reveal the vulnerabilities of fossil-fuel-dependent energy systems, diversification is key. It is also critical to plan now for a transition to sustainable energy to avoid the potential "energy trap" where too much energy is consumed in building new infrastructure, leaving insufficient resources for society's needs.
An object of mass m is dropped from a height h above the surface of a planet of mass M and radius R. Find the speed of the object when it hits the planet surface. Enter your answer as an expression in terms of m, M, R, h and the universal constant of gravity G.
Final answer:
The speed of an object of mass m when it hits the planet's surface after being dropped from height h is found by using the conservation of energy, yielding the formula v = √[2GM(1/R - 1/(R+h))] where G is the gravitational constant, M the planet's mass, and R its radius.
Explanation:
To find the speed v of an object of mass m when it hits the surface of a planet, we will use the principles of energy conservation in the context of gravitational fields. The total mechanical energy (potential plus kinetic) at the beginning and at the end must be equal since we are ignoring air resistance and any other non-conservative forces.
The initial potential energy when the object is at height h above the planet is given by the gravitational potential energy formula U = -G(Mm)/(R+h) and the initial kinetic energy is zero as the object is initially at rest. When the object reaches the surface of the planet, its potential energy is U = -G(Mm)/R and its kinetic energy is K = (1/2)m*v^2. Conservation of energy dictates that the initial total energy equals the final total energy:
-G(Mm)/(R+h) = -G(Mm)/R + (1/2)m*v^2
Solving for v, the speed of the object at the planetary surface, we get:
v = √[2GM(1/R - 1/(R+h))]
This expression shows that the object's speed increases with a greater initial height h, a more massive planet M, and decreases with a larger planetary radius R.
Are wind-driven currents found near the surface of the oceans or in the deep waters of the ocean?
Wind-driven currents are found NEAR THE SURFACE of the oceans
Explanation:
Surface currents are at the interphase between the hydrosphere and atmosphere. Therefore the feel the greatest effect of drag by wind currents, especially prevailing winds (that blow predominantly in one direction like westerlies and easterlies) within the lower atmosphere. The deep currents, on the other hand, are more influenced by Coriolis effect of the earth’s rotation. It is these differences in influences of surface and deeper currents that cause Ekman transport in oceans.
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To get off a frozen lake, a 70 kg person removes his shoe of mass 0.175 kg and throws it horizontally away from the shore at a velocity of 3.2 m/s. If the person is 5.15 m from the shore, how long do they take to reach the shore?
Answer:
the person will be in the shore at 10.73 minutes after launch the shoe.
Explanation:
For this we will use the law of the lineal momentum.
[tex]L_i = L_f[/tex]
Also,
L = MV
where M is de mass and V the velocity.
replacing,
[tex]M_i V_i = M_{fp}V_{fp} + M_{fz}V_{fz}[/tex]
wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.
so, we will take the direction where be launched the shoe as negative. then:
(70)(0) = (70-0.175)([tex]V_fp[/tex]) + (0.175)(-3.2m/s)
solving for [tex]V_fp[/tex],
[tex]V_fp[/tex] = [tex]\frac{(3.2)(0.175)}{69.825}[/tex]
[tex]V_fp[/tex] = 0.008m/s
for know when the person will be in the shore we will use the rule of three as:
1 second -------------- 0.008m
t seconds-------------- 5.15m
solving for t,
t = 5.15m/0.008m
t = 643.75 seconds = 10.73 minutes
The time to reach the shore is 643.75 seconds.
The problem presented involves conservation of momentum on a frictionless surface, which is a physics concept. To solve for the time it takes the person to reach the shore, we can use the principle that the momentum before throwing the shoe is equal to the momentum after throwing the shoe, as there are no external forces acting on the system (since friction is ignored).
The initial momentum of the system is zero because the person is not moving. After throwing the shoe, the momentum of the shoe can be calculated using the formula: p = m × v, where p is momentum, m is mass, and v is velocity. The momentum of the shoe is 0.175 kg × 3.2 m/s, which must be equal and opposite to the momentum of the person. Thus, the velocity of the person, v_p, can be found through the equation m_shoe × v_shoe = m_person × v_person.
Once the velocity of the person is calculated, the time, t, taken to cover the distance to the shore can be found using the formula: t = d / v_p, where d is the distance to the shore.
Carrying out the calculations gives us:
p_shoe = 0.175 kg × 3.2 m/s = 0.56 kg m/s
v_person = p_shoe / m_person = 0.56 kg m/s / 70 kg = 0.008 m/s
t = d / v_person = 5.15 m / 0.008 m/s = 643.75 seconds
The person will take approximately 643.75 seconds to reach the shore.
Lenses are described as convergent or divergent depending on how they refract light. What is the difference between these two types of lenses?
Answer:
Explanation:
Convergent lens is the lens which converges the rays of light falling on it. It is thicker at the middle and thinner at the edges. It is also known as convex lens.
Divergent lens is the lens which diverge the rays of light falling on it. It is thinner at the middle and thicker at the edges. It is also called as concave lens.
Convergent lenses, or convex lenses, thicken in the middle and focus light rays to a point, forming real or virtual images; divergent lenses, or concave lenses, are thinner in the middle, causing the light rays to spread out and only form virtual images. Snell’s law describes the refraction process in both lens types as being influenced by differences in the index of refraction.
Explanation:Difference Between Convergent and Divergent Lenses
Lenses are often categorized as either convergent (convex) or divergent (concave) depending on how they refract light. A converging lens is thicker at the middle and causes parallel light rays entering the lens to converge, or come together, at a single point known as the focal point. This lens can create both real and virtual images, the nature of which depends on the position of the object relative to the lens. In contrast, a divergent lens is thinner at the middle and causes parallel light rays to spread apart or diverge after passing through the lens. Divergent lenses only form virtual images.
Snell's law helps explain the refraction in both types of lenses, considering the difference in the index of refraction between the lens material and the surrounding air. The lens power is determined by its ability to bend light, described by the lens's focal length, with more powerful lenses having shorter focal lengths.
A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30 with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the mand is:A) –200 JB) 61 JC) 140 JD) 200 JE) 260 J
Answer:
The work done by the man is 260 J.
(E) is correct option.
Explanation:
Given that,
Force = 80 N
Distance = 5.0 m
Angle = 30°
Acceleration = 1.5 m/s²
Let F be the man's force
We need to calculate the force
Using balance equation
[tex]F-mg\sin\theta=ma[/tex]
[tex]F-80\sin30=\dfrac{80}{9.8}\times1.5[/tex]
[tex]F=\dfrac{80}{9.8}\times1.5+80\sin30[/tex]
[tex]F=52\ N[/tex]
We need to calculate the work done by the man
Using formula of work done
[tex]W=F\dotc d[/tex]
[tex]W=52\times 5.0[/tex]
[tex]W=260\ J[/tex]
Hence, The work done by the man is 260 J.
If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the man is E) 260 J
Explanation:Work is the quantity of energy transferred by displacement. The following is formula of work done
[tex]\Delta KE = F*d[/tex]
Whereas the following is formula for balance equation:
[tex]F-mg\sin\theta=ma[/tex]
The equation above is come from Newton's Second Law. It applies to a wide range of physical phenomena, but it is not a fundamental principle like the Conservation Laws and also it applies only if the force is the net external force.
Force = 80 N, thereforeMass = Force/g = 80 N / 10 = 8 kgDistance = 5.0 mAngle = 30°Acceleration = 1.5 m/s²For the Case of fictionless slope, the speed at the bottom of a frictionless incline does not depend upon the angle of the incline
Then the work done by the man is...
[tex]w = mgd* sin\theta + \Delta KE\\w = 80 N *5.0 m *sin 30 + (F*d)\\w = 80 N *5.0 m *\frac{1}{2} + (F*d)\\w = 40 N *5.0 m+ (F*d)\\w = 40 N *5.0 m+ (m*a*d)\\w = 200 Nm+ (8*1.5*5)\\w = 200 Nm+ 60\\w = 260 J\\[/tex]
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A 1.0-m long wire is carrying a certain amount of current. The wire is placed perpendicular to a magnetic field of strength.20 T. If the wire experiences a force of 0.60 N, what is themagnitude of the current moving through the wire?
A) 1A
B) 5A
C) 4A
D) 2A
E) 3A
Answer:
Current, I = 3 A
Explanation:
It is given that,
Length of the wire, l = 1 m
Magnetic field acting on the wire, B = 0.2 T
The magnetic force acting on the wire, F = 0.6 N
Let the current flowing through the wire is given by I. The magnetic force acting on an object in the uniform magnetic field is given by :
[tex]F=ILB\ sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=ILB[/tex]
[tex]I=\dfrac{F}{LB}[/tex]
[tex]I=\dfrac{0.6\ N}{1\ m\times 0.2\ T}[/tex]
I = 3 A
So, the current flowing through the rod 3 A.
A satellite has a mass of 5850 kg and is in a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The period of the orbit is two hours. The radius of the planet is 4.15 x 10 to the 6th power m. What is the true weight of the satellite when itis at rest on the planet's surface?
Answer:
W = 24.28 kN
Explanation:
given,
Mass of satellite = 5850 Kg
height , h = 4.1 x 10⁵ m
Radius of planet = 4.15 x 10⁶ m
Time period = 2 h
= 2 x 3600 = 7200 s
Time period of satellite
[tex]T = \dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}[/tex]
R is the radius of planet
h is the height of satellite
[tex]T^2 = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{g}}[/tex]
now calculation of acceleration due to gravity
[tex]g = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{T^2}}[/tex]
[tex]g = \dfrac{4\pi^2}{(4.15\times 10^6)^2}\ {\dfrac{(4.15\times 10^6+4.1\times 10^5)^3}{(7200)^2}}[/tex]
g = 4.15 m/s²
True weight of satellite
W = m g
W = 5850 x 4.15
W = 24277.5 N
W = 24.28 kN
True weight of the satellite is W = 24.28 kN
The true weight of the satellite, when the satellite is at rest on the surface of the planet, is 24.28 kN.
What is the time period of satellites?Time period of satellites is the total time taken by a satellite to complete a full orbit around a body. It can be given as,
[tex]T=\dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}[/tex]
Here, (R) is the radius of the body, and (g) is the gravitational acceleration force.
In a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The period of the orbit is 2 hours and the radius of the planet is 4.15 x 10 to the 6th power m.
To find the weight of the satellite, first find the value of gravitation acceleration using the time period formula as,
[tex]2=\dfrac{4\pi^2}{4.15\times10^6}\sqrt{\dfrac{(4.15\times10^6+4.1\times10^5)^3}{g}}\\g=4.15\rm m/s^2[/tex]
The weight of the body is mass time gravity. As the satellite has a mass of 5850 kg and value of g is 4.15 m/s². Thus, the weight of it is,
[tex]W=5850\times4.15\\W=24277.5\rm N\\W=24.28\rm \; kN[/tex]
Thus, the true weight of the satellite, when the satellite is at rest on the surface of the planet, is 24.28 kN.
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A coin with a diameter of 2.11 cm is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 19.0 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular deceleration of 1.26 rad/s2 , how far does the coin roll before coming to rest?
Answer:
9.49596 m
Explanation:
[tex]\omega_f[/tex] = Final angular velocity = 0
[tex]\omega_i[/tex] = Initial angular velocity = 19 rad/s
[tex]\alpha[/tex] = Angular acceleration = -1.26 rad/s²
[tex]\theta[/tex] = Angle of rotation
Equation of rotational motion
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-19^2}{2\times -1.26}\\\Rightarrow \theta=143.25396\ rad[/tex]
Converting to m
[tex]143.25396\times \pi d=143.25396\times \pi\times 0.0211=9.49596\ m[/tex]
The distance the coin rolls before it stops is 9.49596 m
Circle the letter of the sentence that tells how Bohr' model of the atom differed from Rutherford's model A. Bohr's model focused on the nucleus B. Bohr's model focused on the protons C. Bohr's model focused on the neutrons D. Bohr's model focused on the electrons
Answer:
A. Bohr's model focused on the nucleus
What is the difference between an sbb galaxy and an sb galaxy?
Answer:
One of the most important difference between the two galaxies" is that the sbb galaxy has a lot of material and bar structures passing thourgh the center and the sb galaxy does not have this".
Explanation:
Basically we have 3 types of galaxies (ellipticals, spirals, and irregulars)
Sb galaxies
On this type of galaxies the "intermediate type of spiral typically has a medium-sized nucleus. They can contain stars, star clouds, and interstellar gas and dust. Usually show wide dispersions in terms of their shape. Other members of this subclass have arms that begin tangent to a bright, while still others reveal a small, bright spiral pattern inset into the nuclear bulge. In any of these cases, the spiral arms may be set at different pitch angles. is important to mention that a pitch angle is defined as the angle between an arm and a circle centred on the nucleus and intersecting the arm".
SB galaxies
"For these types of galaxies the luminosities, dimensions, spectra, and distributions of the barred spirals tend to be indistinguishable from those of normal spirals".
One special case occuers at the SBb systems, that "have a smooth bar as well as relatively smooth and continuous arms. In some galaxies of this type, the arms start at or near the ends of the bar, with conspicuous dust lanes along the inside of the bar that can be traced right up to the nucleus"
One of the most important difference between the two galaxies" is that the sbb galaxy has a lot of material and bar structures passing thourgh the center and the sb galaxy does not have this".
Early on in the story the narrator mentions the sound of wind through the wires strung overhead: "The wind in the wires took up the story . . ."Taking the plot's rising action at this point into account, which of the following literary devices is the author most likely using?
A.creative license
B.irony
C.extended metaphor
D.personification
Answer:
option D.
Explanation:
The correct answer is option D.
The irony is the figure of speech which represents the contradiction between what is stated and what actually the scenario is.
the statement by the poet does not have any contradictory statements.
The extended metaphor is a part of speech which is used when there is a comparison between two unlike things in the paragraph.
in the given statement of the poem, there is no comparison.
Personification is the part of speech where human quality are given to non-living things.
in the give statement wind in the wire took up the story can be taken as the human quality.
so, the statement part of speech is personification.
A small car of mass 643 kg is parked behind a small truck of mass 1415 kg on a level road. The brakes of both the car and the truck are off so that they are free to roll with negligible friction. A 43 kg woman sitting on the tailgate of the truck shoves the car away by exerting a constant force on the car with her feet. The car accelerates at 2 m/s². What is the acceleration of the truck? Answer in units of m/s².
Answer:
0.882 m/s²
Explanation:
m = Mass of car = 643 kg
a = Acceleration of the car = 2 m/s²
[tex]a_2[/tex] = Acceleration of the truck
Force exerted on the car
[tex]F=ma\\\Rightarrow F=643\times 2\\\Rightarrow F=1286\ N[/tex]
Mass of the truck with the woman = [tex]M=1415+43=1458\ kg[/tex]
From Newton's second law
[tex]F=Ma_2\\\Rightarrow a_2=\frac{1286}{1458}\\\Rightarrow a=0.882\ m/s^2[/tex]
The acceleration of the truck is 0.882 m/s²
Final answer:
The acceleration of the truck can be found by using Newton's second and third laws. Equal and opposite forces act on the car and the truck, and with the mass of each known, calculations can provide the truck's acceleration.
Explanation:
To find the acceleration of the truck when the woman pushes the car, we use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. When the woman pushes the car with a certain force, the car pushes back on the woman (and hence the truck she is sitting on) with an equal and opposite force. This means the force causing the car to accelerate is the same force causing the truck to accelerate, but in the opposite direction.
Let's denote F as the force with which the woman pushes the car. Given the mass of the car (mcar = 643 kg) and its acceleration (acar = 2 m/s²), we can find the force using Newton's second law of motion: F = mcar × acar.
The same force is responsible for the acceleration of the truck in the opposite direction. Therefore, for the truck with mass mtruck = 1415 kg and acceleration atruck, Newton's second law gives us the relationship: F = mtruck × atruck. By substituting the force F from the previous calculation, we can solve for atruck, which is the acceleration of the truck.
As a system increases in volume, it absorbs 51.0 J of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.545 atm. The final volume of the system is 55.6 L. What was the initial volume of the system if the internal energy of the system decreased by 106.0 J?
Answer:52.72 L
Explanation:
Given
Energy absorb by system [tex]Q=51 J[/tex]
Piston is working against a pressure of [tex]P=0.545 atm[/tex]
Final volume of system [tex]V_f=55.6 L[/tex]
change in internal Energy [tex]\Delta U=-106 J[/tex]
Work done by system [tex]W=P(V_f-V_i)[/tex]
According to first law of thermodynamics
[tex]Q=dU+PdV[/tex]
[tex]51=-106+W[/tex]
[tex]W=157 J[/tex]
[tex]157=0.545\times 10^5\times 10^{-3}(55.6-V)[/tex]
[tex]157=54.5\times (55.6-V)[/tex]
[tex]55.6-V=2.88[/tex]
[tex]V=52.72 L[/tex]
Two pop cans are at rest on a stand. A firecracker is placed between the cans and lit. The firecracker explodes and exerts equal and opposite forces on the two cans. Assuming the system of two cans to be isolated, the post-explosion momentum of the system ____.
a. is dependent upon the mass and velocities of the two cans
b. is dependent upon the velocities of the two cans (but not their mass)
c. is typically a very large value
d. can be a positive, negative or zero value
e. is definitely zero
Answer:
e. is definitely zero
Explanation:
Given that
At initial condition the speed of the pop cans is zero.
We know that linear momentum
P = Mass x velocity
P = m v
At initial condition v = 0
P= 0
If there is no any external force then the linear momentum of the system will be conserve.And given that ,consider the system isolated.
Therefore the answer is e.
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) the amplitude, (b) the frequency, (c) the total energy, and (d) the kinetic energy and potential energy when x = 0.360 m.
Answer:
(a) A = 0.650 m
(b) f = 1.3368 Hz
(c) E = 17.1416 J
(d) K = 11.8835 J
U = 5.2581 J
Explanation:
Given
m = 1.15 kg
x = 0.650 cos (8.40t)
(a) the amplitude,
A = 0.650 m
(b) the frequency,
if we know that
ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)
⇒ f = 1.3368 Hz
(c) the total energy,
we use the formula
E = m*ω²*A² / 2
⇒ E = (1.15)(8.40)²(0.650)² / 2
⇒ E = 17.1416 J
(d) the kinetic energy and potential energy when x = 0.360 m.
We use the formulas
K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)
and
U = (1/2)*m*ω²*x² (the potential energy)
then
K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)
⇒ K = 11.8835 J
U = (1/2)*(1.15)*(8.40)²*(0.360)²
⇒ U = 5.2581 J
The amplitude (A) of the oscillation is 0.650 m, and the frequency (f) is approximately 1.34 Hz. To determine the total energy and the kinetic and potential energy at the point when x = 0.360 m, the spring constant is needed.
Explanation:In order to answer your question about a 1.15-kg mass oscillating according to the equation x = 0.650 cos(8.40t), we can use the facts about simple harmonic motion.
(a) The amplitude is the coefficient of the cosine function, which is 0.650 m.
(b) The frequency (f) can be determined from the coefficient of the t inside the cosine function. In simple harmonic motion, the angular frequency (ω) equals 2πf. Therefore f = ω / 2π = 8.40 / 2π = 1.34 Hz.
(c) The total energy (E) of an oscillator is given by the formula E = 1/2 kA². However, without the spring constant (k) given in the question, this cannot be determined.
(d) When x = 0.360 m, the potential energy (PE) is given by PE = 1/2 kx², and the kinetic energy (KE) can be calculated by subtracting the potential energy from the total energy. But again, without the spring constant, these cannot be determined.
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Technician A says that maintenance-free batteries use a large amount of water. Technician B says that maintenance-free batteries use materials that reduce the amount of water used by the battery. Which technician is correct
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B
Answer:
Option B
Explanation:
Maintenance-free battery uses Recombination Electrolyte in order to limit the oxygen and hydrogen formation during charging.
Each plate of the battery uses a glass micro-fiber separator that absorbs the whole liquid electrolyte in its pore thus resulting in no free acid.
As the battery charges to its fullest, the oxygen produced at the positive plate moves via the pores of the separator to the negative plate.
Initially it reacts and forms [tex]PbSO_{4}[/tex], i.e., Lead Sulphate, then on further charging it gets converted to Pb, i.e., lead.
Due to this, the negative plate is not able to reach the correct potential in order to liberate hydrogen, and hence no water formation takes place.
Thus technician B is correct.
A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is dropped onto the spring.
(a) How far does the object compress the spring?
m
(b) Repeat part (a), but this time assume a constant air-resistance force of 0.750 N acts on the object during its motion.
m
(c) How far does the object compress the spring if the same experiment is performed on the Moon, where g = 1.63 m/s2 and air resistance is neglected?
m
Answer:
(A) l = 0.39 m
(B) l =0.38 m
(C) l = 0.14 m
Explanation:
Answer:
Explanation:
Answer:
Explanation:
from the question we are given the following values:
mass (m) = 1.6 kg
height (h) = 1.05 m
compression of spring (l) = ?
spring constant (k) = 330 N/m
acceleration due to gravity (g) = 9.8 m/s^{2}
(A) initial potential energy of the object = final potential energy of the spring
potential energy of the object = mg(1.05 + l)
potential energy of the spring = 0.5 x k x l^{2} (k= spring constant)
therefore we now have
mg(1.05 + l) = 0.5 x k x l^{2}
1.6 x 9.8 x (1.05 + l) = 0.5 x 300 x l^{2}
15.68 (1.05 + l) = 150 x l^{2}
16.5 + 15.68l = 150l^{2}
l = 0.39 m
(B) with constant air resistance the equation applied in part A above becomes
initial P.E of the object - air resistance = final P.E of the spring
mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}
1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l) = 0.5 x 300 x l^{2}
(16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}
16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}
15.71 + 14.93l = 150^{2}
l =0.38 m
(C) where g = 1.63 m/s^{2} and neglecting air resistance
the equation mg(1.05 + l) = 0.5 x k x l^{2} now becomes
1.6 x 1.63 x (1.05 + l) = 0.5 x 300 x l^{2}
2.608 (1.05 +l) = 0.5 x 300 x l^{2}
2.74 + 2.608l = 150 x l^{2}
l = 0.14 m
The compression of the spring when it is dropped from 1.05 m is 0.37 m.
The compression of the spring when air resistance is considered 0.36 m.
The compression of the spring when air resistance is neglected and gravity is 1.63 is 0.14 m.
The given parameters;
mass of the object, m = 1.6 kginitial position of the object, h₀ = 1.05 mspring constant, k = 330 N/mThe compression of the spring is determined by applying the principle of conservation of energy;
[tex]\frac{1}{2} kx^2 = mgh\\\\\frac{1}{2} kx^2 = mg(1.05 + x)\\\\kx^2 = 2mg(1.05 + x)\\\\330x^2 = 2\times 1.6 \times 9.8(1.05 + x)\\\\330x^2 = 32.93 + 31.36x\\\\330x^2 - 31.36x - 32.93 = 0\\\\a = 330, \ b = -31.36, \ c = -32.93\\\\x = \frac{-b \ \ +/- \ \sqrt{b^2 -4ac} }{2a} \\\\x = \frac{-(-31.36) \ \ +/- \ \sqrt{(-31.36)^2 -4(330\times -32.93)} }{2(330)}\\\\x = 0.37 \ m[/tex]
Considering air resistance, the compression of the spring is calculated as follows;
[tex]\frac{1}{2} kx^2 = mg(1.05+ x) - F(1.05+ x)\\\\\frac{1}{2} \times 330 x^2 = 1.6\times 9.8(1.05 + x) - 0.75(1.05 + x)\\\\165x^2 = 16.46 + 15.68x - 0.79 - 0.75x \\\\165x^2 -14.93x - 15.67 = 0\\\\a = 165, \ \ b = -14.93 \ \ c = -15.67 = 0\\\\x = \frac{-b \ \ +/- \ \sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{-(-14.93) \ \ +/- \ \sqrt{(-14.93)^2 - 4(15.67)} }{2(165)}\\\\x = 0.36 \ m[/tex]
The compression of the spring when air resistance is neglected and gravity is 1.63;
[tex]\frac{1}{2} kx^2 = mg(1.05 + x)\\\\\frac{1}{2} \times 330 x^2 = 1.6 \times 1.63(1.05 + x)\\\\165 x^2 = 2.74 + 2.61x \\\\165 x^2 - 2.61x- 2.74 = 0\\\\a = 165, \ \ b = -2.61, \ c = \ -2.74\\\\x = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{-(-2.61) \ \ +/- \ \ \sqrt{(-2.61) ^2 - 4(165\times -2.74)} }{2(165)}\\\\x = 0.14 \ m[/tex]
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The work function of titanium metal is 6.93 × 10⁻¹⁹ J. Calculate the kinetic energy of the ejected electrons if light of frequency 1.216 × 10¹⁵ s⁻¹ is used to irradiate the metal.
We can calculate the kinetic energy of the ejected electrons in the photoelectric effect by applying the equation KE = hf - BE, where 'KE' is kinetic energy, 'h' is Planck's constant, 'f' is the frequency of the light, and 'BE' is the binding energy or work function of a metal (in this case, titanium).
Explanation:The photoelectric effect is the process where light of sufficient energy (in this case, frequency) shone on a metal surface can release electrons (photoelectrons). Kinetic energy of these ejected electrons can be calculated according to the equation: KE = hf - BE. Here, 'KE' is the kinetic energy we're striving to find, 'h' is Planck's constant, 'f' is the frequency of the incident light, and 'BE' is the binding energy or work function of the electron.
For titanium metal, you've provided that the work function (BE) is 6.93 × 10⁻¹⁹ J. The frequency ('f') of the light used is 1.216 × 10¹⁵ s⁻¹. And, Planck's constant ('h') approximately equals 6.63 × 10⁻³⁴ Js.
To execute the calculation: KE = (6.63 × 10⁻³⁴ Js x 1.216 × 10¹⁵ s⁻¹) - 6.93 × 10⁻¹⁹ J. The solution will provide your kinetic energy in Joules
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A tennis player swings her 1000 g racket with a speed of 15.0 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 18.0 m/s. The ball rebounds at 40.0 m/s. How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
If the tennis ball and racket are in contact for 7.00 , what is the average force that the racket exerts on the ball?
Answer:
[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.
[tex]F=18.86\ N[/tex]
Explanation:
Given:
mass of the racket, [tex]m_r=1000\ g[/tex]mass of ball, [tex]m_b=60\ g[/tex]initial speed of racket, [tex]u_r=15\ m.s^{-1}[/tex]initial speed of ball, [tex]u_b=18\ m.s^{-1}[/tex]final speed of ball, [tex]v_b=40\ m.s^{-1}[/tex]time of contact of racket with the ball, [tex]t=0.07\ s[/tex]By the law of conservation of momentum:
[tex]m_r.u_r+m_b.u_b=m_r.v_r+m_b.v_b[/tex]
where: [tex]v_r=[/tex] final velocity of the racket
[tex]1000\times 15+60\times 18=1000\times v_r+60\times 40[/tex]
[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.
By the Newton's second law of motion:
[tex]F=\frac{dp}{dt}[/tex] ............................(1)
where:
dp = change in momentum
dt = change in time
Change in momentum of ball:
[tex]\Delta p_b=m_b.v_b-m_b.u_b[/tex]
[tex]\Delta p_b=60\times 10^{-3}\times (40-18)[/tex]
[tex]\Delta p_b=1.32\ kg.m.s^{-1}[/tex]
Now, using eq.(1):
[tex]F=\frac{1.32}{0.07}[/tex]
[tex]F=18.86\ N[/tex]
The tennis racket will be moving at 11.52 m/s immediately after the impact. The average force exerted on the ball by the racket, during the brief 7 ms contact time, is approximately 497 N.
The problem can be solved using the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision if no external forces act on the system. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity.
The momentum before the collision (pi) for both the racket and the ball is:
pi,racket = mracket \\times vi,racket = 1.0 kg \\times 15.0 m/s = 15 kg\\cdot m/s
pi,ball = mball \\times vi,ball = 0.060 kg \\times (-18.0 m/s) = -1.08 kg\\cdot m/s
The negative sign indicats that the ball is moving in the opposite direction to the racket. Total initial momentum Pi,total = pi,racket + pi,ball.
The momentum after the collision (pf) for the ball is:
pf,ball = mball \\times vf,ball = 0.060 kg \\times 40.0 m/s = 2.4 kg\\cdot m/s
Using the conservation of momentum, Pi,total = Pf,total, i.e.,
15 kg\\cdot m/s - 1.08 kg\\cdot m/s = pf,racket + 2.4 kg\\cdot m/s
Solving for pf,racket, the momentum of the racket after collision:
pf,racket = 15 - 1.08 - 2.4 = 11.52 kg\\cdot m/s
The velocity of the racket after the collision (vf,racket) can be found by dividing pf,racket by the mass of the racket:
vf,racket = pf,racket / mracket = 11.52 kg\\cdot m/s / 1.0 kg = 11.52 m/s
To calculate the average force exerted on the ball by the racket, we can use the change in momentum (impulse) and divide it by the time of contact. The change in momentum of the ball (\\Delta p) is the momentum after minus the momentum before:
\\Delta p = pf,ball - pi,ball = 2.4 kg\\cdot m/s - (-1.08 kg\\cdot m/s) = 3.48 kg\\cdot m/s
The average force (F) exerted on the ball is:
F = \\Delta p / \\Delta t = 3.48 kg\\cdot m/s / 0.007 s = 497.14 N
The racket will be moving at 11.52 m/s immediately after the impact, and the average force that the racket exerts on the ball is approximately 497 N.
A light horizontal spring has a spring constant of 138 N/m. A 3.85 kg block is pressed against one end of the spring, compressing the spring 0.103 m. After the block is released, the block moves 0.290 m to the right before coming to rest. The acceleration of gravity is 9.81 m/s 2 . What is the coefficient of kinetic friction between the horizontal surface and the block?
Answer:
0.06683
Explanation:
m = Mass of block = 3.85 kg
g = Acceleration due to gravity = 9.81 m/s²
[tex]\mu[/tex] = Coefficient of kinetic friction
x = Compression of spring = 0.103 m
k = Spring constant = 138 N/m
Work done against friction is given by
[tex]W=m\mu gs\\\Rightarrow W=\mu 3.85\times 9.81\times 0.29\\\Rightarrow W=10.952865\mu[/tex]
The potential energy of the spring is given by
[tex]P=\frac{1}{2}kx^2\\\Rightarrow P=\frac{1}{2}\times 138\times 0.103^2\\\Rightarrow P=0.732021\ J[/tex]
The potential energy and the work done against friction will balance the system
[tex]0.732021=10.952865\mu\\\Rightarrow \mu=\frac{0.732021}{10.952865}\\\Rightarrow \mu=0.06683[/tex]
The coefficient of kinetic friction between the horizontal surface and the block is 0.06683
A force F~ = (1.8 N)ˆı + (3 N) ˆ is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis. The force is applied at the point R~ = (4.1 m)ˆı + (5.4 m) ˆ. Find the z-component of the net torque. Answer in units of N m.
Answer:
The z-component of the net torque is: 2.58 (N*m)
Explanation:
We need to apply the torque equation ([tex]T=RXF[/tex]), where T is the torque, R is the distance at the point where the force is applied and F is the force, and remembering the cross product ([tex]AXB=abSin(\beta)(n)[/tex]), where A and B are the vectors and ab are the magnitudes, (beta) the angle between A and B in the plane containing them, and (n) is the direction of the vector given by the right-hand rule. Knowing those things, we can get:[tex]T=RXF=4.1(i)+5.4(j)X1.8(i)+3(j)=0+4.1*3(k)+5.4*1.8(-k)+0=2.58(k)(N*m)[/tex] that is the z-component of the torque.
The torque on the object is -2.58Nm
Data;
F = (1.8i + 3j)Nr = (4.1i + 5.4j)mNet TorqueThe net torque is calculated as the product of the force and distance
[tex]t = r * F[/tex]
Let's substitute the values and solve
Since this is a dot product, we can easily multiply through
[tex]\tau = r * F\\\tau = (1.8*5.4) - (3 * 4.1) = -2.58Nm[/tex]
Using 2 by 2 matrix method, the torque on the object is -2.58Nm
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How is the air found in soils different from the air in the atmosphere?
Answer:
Air found in the soil is trapped in it and is not exposed to the air currents, poor in oxygen and rich in moisture.
Explanation:
Air found in the soil is trapped in it and is not exposed to the air currents.
It contains more moisture than the atmospheric air.
It is rich in carbon dioxide and poor in oxygen.
The air found in the soil forms a part of the lithosphere and not the atmosphere.
Nitrogen trapped in the soil is used by the plants to make protein, carbon dioxide is used for the photosynthesis and oxygen is used for the respiration of roots and microorganisms.
A ray of light strikes the midpoint of one face of an equiangular crown glass prism (n=1.52) at an angle of incidence of 30.0.
(a.) Find the angle of reflection at the first surface of the prism.
(b.) Find the angle of refraction
Answer:
Angle of reflection at plane surface of prism = 30°
Explanation:
Angle of reflection = angle of incidence at any surface plane or curved .
Here angle of incidence at plane surface of prism = 30°
Angle of reflection at plane surface of prism = 30°
If i be the angle of incidence and r be angle of refraction
sin i / sin r = refractive index
= 1.52
sin r = sin i / 1.52
= sin 30 / 1.52
= .5 / 1.52
= .3289
r = 19.2 °
As your hand moves back and forth to generate longitudinal pulses in a spiral spring, your hand completes 2.91 back-and-forth cycles every 3.67 s. The velocity of the pulse in the spring is 0.925 cm/s. What is the wavelength? Answer in units of m.
Answer:
Wavelength, [tex]\lambda=0.011\ m[/tex]
Explanation:
Given that,
Number of cycles in a spiral spring is 2.91 in every 3.67 s
The velocity of the pulse in the spring is 0.925 cm/s, v = 0.00925 m/s
To find,
Wavelength
Solution,
Number of cycles per unit time is called frequency of a wave. The frequency of the longitudinal pulse is,
[tex]f=\dfrac{2.91}{3.67}=0.79\ Hz[/tex]
The wavelength of a wave is given by :
[tex]\lambda=\dfrac{v}{f}[/tex]
[tex]\lambda=\dfrac{0.00925\ m/s}{0.79\ Hz}[/tex]
[tex]\lambda=0.011\ m[/tex]
So, the wavelength of the longitudinal pulse is 0.011 meters. Hence, this is the required solution.
To find the wavelength, the frequency is calculated by dividing the number of cycles by the total time. The velocity is then converted from cm/s to m/s. The wavelength is calculated using these values in the equation for wave speed to get approximately 0.012 meters.
Explanation:In the scenario described in your question, your hand completes 2.91 back-and-forth cycles every 3.67 seconds, resulting in longitudinal pulses in a spiral spring. Here, we have frequency and wave velocity, and we need to find the wavelength. We can determine the frequency by dividing the number of cycles by the total time and the wavelength using the formula for wave speed: v = fλ, where v is the velocity, f is the frequency, and λ is the wavelength.
Frequency (f) = number of cycles / total time = 2.91 cycles / 3.67 s = 0.793 Hz
Velocity (v) = 0.925 cm/s = 0.00925 m/s (since we need the answer in meters)
To find the wavelength, we can use the formula v = fλ, to rearrange this to solve for λ, we get λ = v/f. So, λ = 0.00925 m/s / 0.793 Hz = 0.01166 meters or approximately 0.012 meters.
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Light of wavelength λ travels through a medium with an index of refraction n1before striking a thin film with an index of refraction n2 at an angle of incidence of 0. Some of the light is reflected off of and some is transmitted through the n2 thin film. The transmitted light travels a distance t through the thin film before encountering the n1 medium again. Some of the light reflects off of the n1 medium.What is the effective path length difference between the light that reflected off of the n2 medium and the light that reflected off the n1 medium, given that n1>n2?
A. t+λ/n2
B. 2t+λ/n2
C. 2t+λ/(2n2)
D. t+λ/(2n2)
Answer:
option C
Explanation:
The correct answer is option C
A light that transmits through n₂ travels t distance before reflection off the n₁ medium and again travels distance t before reaching the point from where it entered n₂ medium. Hence it travels 2 t distance more than the light that is reflected off n₂.
It( light entering n₂) also travels an additional distance equal to, half of the wavelength, when reflected off n₁ ( as n₁ is greater than n₂).
Wavelength in n₂ is = [tex]\dfrac{\lambda}{n_2}[/tex]
Hence, path length difference = [tex]2t +\dfrac{\lambda}{2 n_2}[/tex]
The effective path length difference between the light that reflected off of the n2 medium and the light that reflected off the n1 medium is
Path length difference = 2t+λ/(2n2)
What is wavelength?Generally, Wavelength is simply defined as the distance between points in the adjacent cycles of a waveform signal channeled through a space.
In conclusion, A light that travels through n2, travels in a time t, a distance before reflection from the n1 channel, travels a distance d before reaching the point from where it entered n2 channel.
Path length difference = 2t+λ/(2n2)
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