The height of the cliff that a stone is dropped from the top, the splash it made as it hit the water below was heard 5.0 s later is 857.5 m.
Sound reflectionSound can bounce when it hits an object that has a hard surface and is not easily penetrated by waves. Examples of objects that can reflect sound are glass walls, iron, zinc, cave walls, and seabeds.
This reflected sound can be used to measure the depth of the sea or the distance of the walls of the cave passage. One of the uses of sound reflection is to know the depth of the sea or the distance of the cave walls.
The equation to determine the distance is:
s = ½ (v) (t)
s = distance or height
t = time
v = velocity of the sound wave (343 m/s)
We have,
Time = 5.0 s
Velocity of the sound wave = 343 m/s
So, the height of the cliff:
s = ½ (v) (t)
= ½ (343) (5)
= 857.5 m
The question is incomplete, it should be:
A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 5.0 s later. How high is the cliff?
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The four-wheeler has a weight of 335 lb and a center of gravity at G1, whereas the rider has a weight of 150 lb and a center of gravity at G2. If the engine can develop enough torque to cause the rear wheels to slip, determine the largest coefficient of static friction between the rear wheels and the ground so that the vehicle will accelerate without tipping over. What is this maximum acceleration? In order to increase the acceleration, should the rider crouch down or sit up straight from the position shown? Explain. The front wheels are free to roll. Neglect the mass of the wheels in the calculation.
To increase the acceleration rider should crouch down which decrease the height of CG from Ground & hence increase acceleration since both are Inversely proportional.
At each whee By condition of equilibrium we can find Load
Rear load Wr= 224lb=224*4.45=996.8N
Front load Wf= 261lb=261*4.45=1161.1N
(2.30, 1.62) here the CG of entire mass will be located
CG location Can be found by
x=sum(Wi*xi) /sum(Wi)
Load transfer due to acceleration is given by
DW=m*a*(h/L)
Where h= height of CG= 1.62ft=0.486m
L= wheel base=5ft=1.5m
m= total mass in kg=(335+150) *0.45=212.25kg
For front wheel to be lifted,
dW=Wf
solving we get acceleration value as
a= 16.884m/s² of acceleration will be max for tipping of vehicle.
We known that,
Tractive force = ma
Gives us tractive force as T=212.25*16.88=3582.78N
This will be achieved when friction force at rear end at max tractive condition Tractive force = static friction force.
T= Friction force= (coefficient of friction) *N
Where N= Wf+Wr for tipping condition
Coefficient of friction= 3582.78/(Wf+Wr) =1.72
For tipping of vehicle we need a coefficient of friction greater than & equal to 1.72.
In coefficient of Friction equation I have taken entire weight as normal since during tipping load will be shifted to rear wheel.
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g consider the woman doing push-ups in the figure. she has a mass of 44.2 kg, and the distance from her feet to her center of mass is 0.87 m, while the distance from her feet to her hands is 1.7 m.
The mass is 0.87 m and the energy exchange is 215.33J for the woman doing exercises.
The torque skilled via way of means of the frame with out the angular rotation however actions because of the rotational movement of some other item in touch with it's miles referred to as a static torque. While commencing a door, the door does not do beneathneath torque, however the connected to the door and a wall undergoes angular rotation.
Here we have
m= 56.6 kgg= 9.8m/sh=0.76maxis =1.55ma)thinking about torque approximately foot Fr = 271.ninety six N approx 272for every handFr' = 272 / 2= 136 NTorqueFt = 1435.fifty five Nusing verbal exchange of energy w= Fd = 1435× 0.15 = 215.33J p= w/ t = 215.35× 25/60 = 89.7 wRead more about mass;
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Find the unit vector in the direction of the vector 6i+8j
The unit vector in the direction of the vector 6i+8j is [tex]\frac{\sqrt{10}}{10} 3i-j[/tex]
Unit vectors are vectors whose magnitude is exactly 1 unit. These unit vectors are commonly used to indicate a direction, with a scalar coefficient providing the magnitude. It is also known as the Direction Vector.
A unit vector x in the direction of x is given by
[tex]x=\frac{x}{|x|}[/tex]
In this case, we have
[tex]|v|=\sqrt{6^{2}+2^{2} }[/tex]
=√40
=2√10
So a unit vector in the direction of v will be equal to as follows
[tex]v=\frac{1}{2\sqrt{10} } 6i-2j[/tex]
Taking 2 as a common from 6i - 2j, we get
[tex]v=\frac{1}{\sqrt{10} } 3i-j[/tex]
To rationalize the denominator, we will multiply √10 in the numerator as well as the denominator. After this, we will get
[tex]\frac{\sqrt{10}}{10} 3i-j[/tex]
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You have arranged that the magnetic field in a particular region of space is due North with a value of 0.0100 T. An electron enters the field traveling to the West with a speed of 3.0 percent of the speed of light. As a result, the electron experiences a magnetic force which is upwards. (a) What is the strength of this magnetic force? Give your answer in femtonewtons. (NOTE: The prefix "femto" represents 10-15., ก็เ (b) What is the amount of the resulting acceleration of the electron? Get the mass of an electron from the inside back cover of your textbook. Give your answer in Pm/s2 (NOTE: the prefix "P" is for "peta", which represents 10) Pm/s2 (c) What will be the result of the acceleration that you have calculated in part (b)? The speed of the moving electron will change but its direction of motion will remain constant.. Both the speed and direction of the moving electron will change The direction of the moving electron will change but its speed will remain constant.
(a) [tex]F = qv*B[/tex]
[tex]= 1.602*10^{-19} *9*10^{6} *10^{-2}[/tex]
[tex]= 1.4410*10^{-14}[/tex]
[tex]= 14.41 femtonewton[/tex]
Hence, the strength of the magnetic force is 14.41 femtonewtons
(b) [tex]a = \frac{F}{m}[/tex]
[tex]= \frac{1.4418*10^{-14} }{9.10*10^{-31} }[/tex]
[tex]= 1.584*10^{16} m/s[/tex]
The amount of resulting acceleration of the electron is [tex]1.584*10^{16} m/s[/tex]
(c) Since direction of acceleration is perpendicular to the direction of velocity. Both speed and direction will change.
A magnetic field is a vector field that represents the effect of magnetism on moving charges, currents, and magnetic bodies. A charge moving in a magnetic field experiences a force perpendicular to its velocity and magnetic field.
The magnetic field is generated by the movement of the intrinsic magnetic moment and charge of elementary particles associated with their fundamental quantum property of spin.
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The average global temperature and sea levels are rising, accelerating at an unprecedented rate. Suggest three ways in which human beings will be affected by these phenomena.
There are many potential ways in which human beings could be affected by rising global temperatures and sea levels. Some potential impacts include:
Extreme weather events: As global temperatures rise, the Earth's climate is becoming more unpredictable and extreme. This could lead to an increase in the frequency and severity of extreme weather events, such as heatwaves, droughts, floods, and hurricanes. These events can have devastating impacts on human communities, including loss of life, damage to infrastructure, and economic disruption.Loss of habitat and biodiversity: Rising sea levels could lead to the loss of coastal habitats, such as wetlands and coral reefs. This could have serious consequences for the many species that rely on these habitats for their survival. In addition, the loss of these habitats could also have negative impacts on human communities that depend on them for food, shelter, and other resources.Displacement of people: As sea levels rise, many coastal communities could become uninhabitable. This could force people to migrate inland, potentially leading to conflicts over resources and territory. In addition, the loss of coastal habitats could also disrupt the livelihoods of many people who rely on these habitats for their income, such as fishermen and tourism operators.Overall, the impacts of rising global temperatures and sea levels are likely to be complex and far-reaching, potentially affecting many aspects of human life.
A Doppler radar determines precipitation ___ by measuring changes in the ____ of the reflected radio wave.
A Doppler radar determines precipitation velocity by measuring changes in the frequency of the reflected radio wave.
Devices for Doppler microwave detection send a steady stream of low-energy microwave radiation toward a target location, then they examine the reflected signal. When the microwave source and the vehicle are moving relative to one another, the detector detects a change in the frequency of the waves that results.
The Doppler effect, also known as the Doppler shift, is a term used to explain changes in frequency caused by moving sources of sound or light with regard to an observer. As the source gets closer to the observer, the waves it is emitting become compressed and have a greater frequency.
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the drawing shows two transverse waves traveling on strings. the linear density of each string is 0.065 kg/m. the tension is provided by a 26-n block that is hanging from the string. find the speed of the wave in part (a) and in part (b) of the drawing.
The wave's speeds in parts (a) and (b) of the drawing are 13.7 m/s and 9.691 m/s, respectively.
(Tension / Mass per Unit Length) / sqrt = velocity
A) tension: m = 2.3 * 9.8 = 22.54 N; mass per unit length: 0.12 kg/m
velocity is equal to sqrt(22.54/0.12) = sqrt(187.83) = 13.7 m/s b. 2.3 * 9.8 / 2 = 11.27 N tension
velocity or speed is equal to sqrt(11.27/0.12) = sqrt(93.916) = 9.691 metres per second.
An oscillation of a wave that is perpendicular to the direction of the wave's progress is known as a transverse wave in physics. An oscillating wave moves in the opposite direction from a longitudinal wave, which moves in the same direction. Examples of transverse waves include water waves.
An easy illustration is provided by the waves that may be produced on a horizontal piece of string by anchoring one end and moving the other end up and down.
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How does a plastic stopper in a flask help reduce heat loss?
A plastic stopper in a flask can help reduce heat loss in several ways. Some of the main mechanisms by which a plastic stopper can help to reduce heat loss include:
Insulation: A plastic stopper can provide a layer of insulation between the hot liquid inside the flask and the surrounding air. This can help to slow the transfer of heat from the liquid to the air, and can keep the liquid hot for a longer period of time.Airtight seal: A plastic stopper can also create an airtight seal when it is inserted into the neck of the flask. This can help to prevent air from entering the flask, which can reduce heat loss through convection. When air is able to enter the flask, it can circulate around the hot liquid and carry heat away from it. By preventing air from entering the flask, a plastic stopper can help to keep the liquid hot for a longer period of time.Thermal conductivity: In addition, the specific material used for the plastic stopper can also affect its ability to reduce heat loss. Some plastics are better insulators than others, and have a lower thermal conductivity. This means that they can slow the transfer of heat more effectively, and can help to keep the liquid hot for a longer period of time.Overall, a plastic stopper in a flask can help to reduce heat loss by providing insulation, creating an airtight seal, and slowing the transfer of heat through its thermal conductivity. These mechanisms can help to keep the liquid hot for a longer period of time. ;)
two identical masses attached to identical springs with spring constant 33 n/m are coupled by a spring with spring constant l. the antisymmetric and symmetric normal modes have frequencies 15 hz and 12 hz.
If there is no friction and the mass's displacement out of its equilibrium position at x = 0 is "small," the motion of a mass linked to a spring is simple harmonic motion.
How does antisymmetric mode work?Mode of antisymmetry Four. .Both this and the modes mentioned above might have variants.
What are the usual frequency and mode?Free motion is described by its normal modes at specific frequencies.The term "natural" or "resonant frequencies" refers to these fixed frequencies of the a system's normal modes.Natural oscillations of a mass on a spring occur at the a single frequency.
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in the sun and seasons simulation in module 2, you observed that, at the equator, there are 12 hours of daylight year-round, because the angle between the apparent path of the sun across the sky and the horizon is always
There are always 12 hours of daylight at the equator because the apparent path of the sun across the sky and the horizon are always perpendicular to one another.
The "almost" equal lengths of day and night are caused by the sun's rays being bent or refracted, making it appear that the sun is above the horizon when it is actually below it. In addition, because the sun rises and sets later at higher latitudes (those farthest from the equator), the days become a little longer there. As a result, the length of the day will vary from around 12 hours to several days before and after the equinox and six and one-half minutes at the equator.
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A fully charged capacitor and a 0.22-H inductor are connected to form a complete circuit. If the circuit oscillates with a frequency of 2.9 ✕ 103 Hz, determine the capacitance of the capacitor. F
A complete circuit is created by connecting a completely charged capacitor to a 0.22-H inductor. 1.4 x 10⁻⁸ F is the capacitance of the capacitor if the circuit oscillates at a frequency of 2.9*10³ Hz.
C = capacitance
L = inductance = 0.22 H
f = frequency = 2900 Hz
using the formula
f = 1/(2[tex]\pi[/tex]√(LC))
2[tex]\pi[/tex] f = 1/√(LC)
inserting the values
2 x 3.14 x 2900 = 1/√(0.22 C)
C = 1.4 x 10⁻⁸ F
A capacitor is an electrical component with two terminals that has the ability to store energy in the form of an electric charge. Two electrical cables are used, and they are spaced apart. A vacuum or an insulating substance known as a dielectric can fill the gap between the conductors. The capacitor's capacitance is its capacity to hold charges.
By holding apart oppositely charged couples, capacitors store energy. A parallel plate capacitor has two metal plates with a space in between them and is the most basic type of capacitor design. But various capacitor types are made in a wide variety of shapes, sizes, lengths, girths, and materials.
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Star S1 is moving away from us at a speed of 0.8c. Star S2 is moving away from us in the opposite direction at a speed of 0.5c. The speed of S1 as measured by an observer on S2 is:
Star S₁ is moving away from us at a speed of 0.8c. Star S₂ is moving away from us in the opposite direction at a speed of 0.5c.The speed of S₁ as measured by an observer on S₂ is 0.5c.
We are given that,
Star S₂ is moving away in the opposite direction at a speed = v =+0.5c
Star S₁ is moving away from us at a speed = u =+0.8c
Speed of light = c = 3 × 10⁸ m/s
So that we can calculate the speed of S₁ as measured by an observer on S₂ by the formula , (which is multiple of c)
u/c = [(u/c)- (v/c)]/[(1 -uv)/c²]
Putting the values in above equation we get,
u/c = [(0.8 -0.5)]/[1 - (0.8)(0.5)]
u = 0.5c
Therefore, the speed of S₁ as measured by an observer on S₂ would be 0.5c in a direction away from earth .
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ieee standard for ethernet - amendment 5: physical layer specifications and management parameters for 10 mb/s operation and associated power delivery over a single balanced pair of conductors
Provides additions and appropriate modifications to IEEE Std 802.3 to 10 Mb/s physical layer specification and management parameters for operation.
On a single balanced copper twisted pair and associated optional power delivery Add. A single-pair Ethernet standard intended to support speeds up to 10 Mbps, extend Ethernet range up to 1,000 meters, and provide multidrop capability. An Ethernet physical layer standard ratified by the IEEE in late 2019 as part of the 802.3cg specification.
A standard for Ethernet network technology that uses a thicker version of coaxial cable. It is capable of transmitting data at speeds of 10 Mbit/s over distances of up to 500 meters using baseband transmission. 10Base5 is also known as Thicknet ThickWire Thick Ethernet, Thick Coaxial Ethernet.
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a 1271kg sample of water at 0c is cooled to -17c, and freezes in the process how much heat is liberated
The heat released in the whole process in which a sample of water is cooled and frozen is 4.64 * 10⁸ J.
The problem is to be divided into two steps. This process releases a certain quantity of heat,
Step 1) Solidification of the water at 0°C into ice.
Q = m* Lf
where,
m is the mass of the water
Lf stands for the latent heat of ice fusion
Substituting numbers, we find
Q₁ = (1271 kg)* (334 kJ/kg) = 424514 kJ = 4.2* 10⁸ J
step 2) The water became ice, now we need to cool it down to -36°C . This process releases a certain quantity of heat, which is
Q₂ = m* Cs* ΔT
where,
m is the mass of ice
Cs is the specific heat of ice
ΔT is the variation of temperature
By placing the numbers in the formula, we have,
Q₂ = (1271 kg)* (2050 J/kg K) (-17-0) = 44294350 J = - 0.44 * 10⁸ J
The negative sign means the heat is released by the system.
Thus, the overall process produces heat that is
Q = Q₁ + Q₂ = 4.2* 10⁸ J + 0.44 * 10⁸ J = 4.64 * 10⁸ J
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PLEASE SOMEBODY HELP A moon of mass m orbits a planet of mass M in uniform circular motion with a constant tangential velocity. The distance between the center of the moon and the center of the planet is R. The gravitational field strength at the center of the moon is Gmoon. A satellite of mass m/5 is halfway between the center of the moon and the center of the planet. What is the gravitational field strength at the satellite?
Evaluate using:
Gsatellite = Gmoon * M / (R^2 + R^2/4)^(3/2).
Explanation:
The gravitational field strength at the satellite is given by the formula Gsatellite = Gmoon * M / (R^2 + R^2/4)^(3/2)
To find the value of Gsatellite, we need to know the values of Gmoon, M, and R. We are given the mass of the moon (m) and the mass of the planet (M), but we are not given the value of Gmoon or R. However, we can use the information provided in the problem to find the value of Gmoon.
The gravitational field strength at the center of the moon (Gmoon) is equal to the product of the mass of the planet (M) and the tangential velocity of the moon (v) squared, divided by the distance between the center of the moon and the center of the planet (R).
Since the moon is in uniform circular motion with a constant tangential velocity, we can use the formula for centripetal acceleration (a = v^2/R) to find the value of Gmoon. We know that the mass of the moon (m) is accelerating toward the center of the planet, so the centripetal acceleration of the moon (a) is equal to the gravitational field strength at the center of the moon (Gmoon).
We can use the formula a = v^2/R to find the value of Gmoon by substituting the values for a, v, and R into the formula. We are given the value of v (the tangential velocity of the moon), and we can use the formula for centripetal acceleration (a = v^2/R) to find the value of a.
We can substitute the value of a into the formula Gmoon = M * a to find the value of Gmoon. Once we have the value of Gmoon, we can substitute it into the formula Gsatellite = Gmoon * M / (R^2 + R^2/4)^(3/2) to find the value of Gsatellite.
Therefore, to find the gravitational field strength at the satellite, we need to find the value of Gmoon by using the formula Gmoon = M * a and the values of M, a, and R provided in the problem. We can then use the value of Gmoon to find the value of Gsatellite using the formula Gsatellite = Gmoon * M / (R^2 + R^2/4)^(3/2).
a student of weight 671 n rides a steadily rotating ferris wheel (the student sits upright). at the highest point, the magnitude of the normal force f with arrown on the student from the seat is 543 n. (a) Does the student feel light or heavy there? (b) What is the magnitude of Fn the lowest point? If the wheels speed is doubled, what is the magnitude FN at the (c) highest and (d) lowest point?"
(a) The student feels heavy weighted.
(b) Magnitude of the force at lowest point = 799 N
(c) Highest magnitude = 1183 N
(d) Lowest magnitude = 159 N
The wheel rotates in the circular direction and while rotating the force tend to be at the center of the motion. This is called centrifugal force.
A
Centrifugal force pushing student upwards at the top Fc = (671 - 543)N = 128N
B
At the bottom, an equal centrifugal force Fb acts downwards on the student giving a total of :
Fc + 671 = 128 + 671 = 799N
C
The total force at the top when the centrifugal force is quadrupled = 671+ (4×128)
= (671 + 512 )N = 1183 N
D
The centrifugal force is quadrupled when speed is doubled ,therefore, the total force at the top = 671 - (4× 128)
= (671 - 512)N = 159N
Therefore, the above magnitudes can be found out by the use of centrifugal force.
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(a) The student feels heavy weighted.
(b) Magnitude of the force at lowest point = 799 N
(c) Highest magnitude = 1183 N
(d) Lowest magnitude = 159 N
The wheel rotates in the circular direction and while rotating the force tend to be at the center of the motion. This is called centrifugal force.
Centrifugal force pushing student upwards at the top Fc = (671 - 543)N = 128N
At the bottom, an equal centrifugal force Fb acts downwards on the student giving a total of :
Fc + 671 = 128 + 671 = 799N
The total force at the top when the centrifugal force is quadrupled = 671+ (4×128)
= (671 + 512 )N = 1183 N
The centrifugal force is quadrupled when speed is doubled ,therefore, the total force at the top = 671 - (4× 128)
= (671 - 512)N = 159N
Therefore, the above magnitudes can be found out by the use of centrifugal force.
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If a resistance of 500 £2 makes the ohmmeter pointer deflect to its scale, then the resistance which makes the pointer deflect to the scale of the ohmmeter is.
If a resistance of 500 ohms makes the ohmmeter pointer deflect to its scale, then the resistance that would cause the pointer to deflect to the scale of the ohmmeter would be equal to the range of the ohmmeter.
What is ohmmeter?Generally, The range of an ohmmeter refers to the range of resistances that it is designed to measure. If the ohmmeter has a range of 0 to 500 ohms, then a resistance of 500 ohms would cause the pointer to deflect to the scale of the ohmmeter.
If the ohmmeter has a range of 0 to 1,000 ohms, then a resistance of 500 ohms would not cause the pointer to deflect to the scale of the ohmmeter.
It is important to note that the range of an ohmmeter is typically marked on the front panel or in the user manual.
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Through a small window in a furnace, which contains a gas at a high temperature T, the spectral lines emitted by the gas molecules are observed. Because of molecular motions, each spectral line exhibits Doppler broadening. Show that the variation of the relative intensity I(2) with wavelength A in a line is given by me2(-Ao)2 2AT 10) ox exp where m is the molecular mass, c the speed of light, and Ag the mean wavelength of the line.
Variation of the relative intensity [tex]I(\lambda)[/tex] with the wavelength [tex]\lambda[/tex] in a line is given by [tex]$$I(\lambda) \propto e^{-\left(\frac{m c^2\left(\lambda-\lambda_0\right)^2}{2 k T \lambda_0^2}\right)}=\exp \left(-\left(\frac{m c^2\left(\lambda-\lambda_0\right)^2}{2 k T \lambda_0^2}\right)\right)$$[/tex]
As, for non-relativistic thermal velocities, the Doppler shift in frequency will be: [tex]$$f=f_0\left(1+\frac{v}{c}\right)$$[/tex]
where f is the observed frequency,
f_0 is the rest frequency,
v is the velocity of the emitter towards the observer
c is the speed of light.
As any volume part of the radiating body will have a distribution of speeds both toward and away from the observer, which will have the overall effect of widening the seen line.
If [tex]I(v)dv=P_v(v)dv[/tex] is the fraction of particles with velocity component v to (v+dv) along a line of sight
Then the value of the distribution of frequencies will be
[tex]$$I(f) \mathrm{d} f=P_f(f) \mathrm{d} f=P_v\left(v_f\right) \frac{\mathrm{d} v}{\mathrm{~d} f} \mathrm{~d} f$$[/tex]
where [tex]$v_f=c\left(\frac{f}{f_0}-1\right)$[/tex] is the velocity towards the observer corresponding to the shift of the rest frequency f0 to f.
So, we can write it as:
[tex]$$P_f(f) \mathrm{d} f=\frac{c}{f_0} P_v\left(c\left(\frac{f}{f_0}-1\right)\right) \mathrm{d} f$$[/tex]
As in the case of the thermal Doppler broadening, the velocity distribution is given by the Maxwell distribution:
[tex]$$P_v(v)=\sqrt{\frac{m}{2 \pi k T}} e^{-\frac{m v^2}{2 k T}} \mathrm{~d} v$$[/tex]
So,
[tex]$$P_f(f) \mathrm{d} f=\frac{c}{f_0} \sqrt{\frac{m}{2 \pi k T}} e^{-\left(\frac{m\left[c\left(\frac{f}{f_0}-1\right)\right]^2}{2 k T}\right)} \mathrm{d} f$$[/tex]
This can be simplified as
[tex]$$P_f(f) \mathrm{d} f=\sqrt{\frac{m c^2}{2 \pi k T f_0^2}} e^{-\left(\frac{m c^2\left(f-f_0\right)^2}{2 k T f_0^2}\right)} \mathrm{d} f$$[/tex]
While in the non-relativistic limit,
[tex]-\frac{f-f_0}{f_0} \approx \frac{\lambda-\lambda_0}{\lambda_0^2}.[/tex]
So, from above,
we will get:
[tex]$$I(\lambda) \propto e^{-\left(\frac{m c^2\left(\lambda-\lambda_0\right)^2}{2 k T \lambda_0^2}\right)}=\exp \left(-\left(\frac{m c^2\left(\lambda-\lambda_0\right)^2}{2 k T \lambda_0^2}\right)\right)$$[/tex]
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The correct questions may be like
Through a small window in a furnace, which contains a gas at a high temperature T, the spectral lines emitted by the gas molecules are observed. Because of molecular motions, each spectral line exhibits Doppler broadening. Show that the variation of the relative intensity [tex]I(\lambda)[/tex] with wavelength [tex]\lambda[/tex] in a line is given by [tex]$$I(\lambda) \propto \exp \left\{-\frac{m c^2\left(\lambda-\lambda_0\right)^2}{2 \lambda_0^2 k T}\right\}$$[/tex]
A new planet has been discovered ad given the name Planet X . The mass of Planet X estirated to be one-half that of Earth; and the radius of Planet X is estimated to be twice that of Earth The acceleration due to gravity on the surface of Planet X is gx: The acceleration due to gravity on the surface of Earth is 9E- Which of the following is correct expression? {9e 9x = #9e 9x = {91 9x = YE 29E
The mass of Planet X estirated to be one-half that of Earth; and the radius of Planet X is estimated to be twice that of Earth. The acceleration due to gravity on the surface of Planet X is gx: The correct formula is gx = (1/8)ge.
Let,
The Earth's mass is m.
The Earth's radius is r.
Gravitational acceleration on Earth's surface is equal to ge.
Then,
The answer to the query is
Planet X's mass is equal to (1/2)m.
Planet X's radius is equal to 2r.
gx is the acceleration brought on by gravity on the surface of planet X.
We are aware that the calculation for the gravitational acceleration is
g = GM/R^2
Where,
g = Gravitational acceleration
The gravitational constant is G, which is 6.674*10(11) Nm2/kg2.
M is the planet's mass.
r = Planetary Radius
Currently, for Earth, we have
gₑ = Gm/r^2 ............ (1) (1)
And we have planet X.
gₓ = Gm/2(2r)^2
gₓ = Gm/8r^2 ......... (2) (2)
Equations (1) and (2) are contrasted to obtain
gₑr^2 = 8gₓr^2
gₓ = (1/8)gₑ
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when the weight is moved downward, the pendulum slows down; when it is moved upward, the pendulum swings faster. if the rod has a mass of 1.23 kg and a length of 1.25 m and
The distance is 1.026 m, the mass be placed to give the pendulum a period of 2.00 seconds.
For complete question?You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of a rod with a movable weight on it. When the weight is moved downward, the pendulum slows down; when it is moved upward, the pendulum swings faster. If the rod has a mass of 1.23 kg and a length of 1.25 m and the weight has a mass of [10] kg, where should the mass be placed to give the pendulum a period of 2.00 seconds?
How to solve?mass of rod, M = 1.23 kg
Length, L = 1.25 m
mass, m = 10 kg
Time period, T = 2 s
Let the distance is d.
The formula of the time period is given by,
T=2π√1/3ML²+md²/(M+m)g
2x2=4π2x1/3×1.23×1.25×1.25+10d²/(1.23+10)×9.8
11.16=0.64+10d²
d=1.026m
How is the pendulum determined?You can use the equation L = (T/ 2)2*g to determine the length of a basic pendulum. where g is the acceleration caused by gravity and T is the simple pendulum's period of time.
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on two occasions during a football game, a defensive player collides with and stops a running back and a fullback. the fullback has more mass than the running back but they were both moving at the same speed prior to collision with the defensive player. during these two different collisions, the defensive player exerts the same force on each player for a prolonged time until the player is stopped. compared to the fullback, the less massive running back will experience momentum change, impulse, and collision time.
Compared to the fullback, the less massive running back will experience a greater momentum change, a greater impulse and a greater collision time.
The momentum change of an object can be calculated from knowledge of the object's mass and velocity change, represented by
P = m* Δv
When a force is exerted upon an object in a collision, the object is said to have experienced an impulse.
Impulse = momentum change
F * t = m* Δv
The essential difference between the fullback and the running back is their mass. Each player has same velocity change and encounters same force during the collision. The fact that two players have different mass means the more massive player encounters the greatest momentum change. Since, momentum change is equal to impulse, then the fullback must also encounter the greater impulse. This greater impulse is due to the fact that the same collision force must endure for a longer period of time.
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f the tolerance in the feature control frame applies to a feature of size and no material condition symbol is specified,
Form tolerances regulate the "form" of features and are frequently used as a size refinement, thus they do not need a datum reference.
Describe what you mean by a substance.The substance(s) from which a thing is created or composed: A long-lasting substance is stone. something that is utilized or created as crude or raw material: Paper is created using wood pulp as the main ingredient. any component component. a kind of cloth used to make dresses.
What kind of material is it?Metals, polymers, and ceramics have historically been the three main kinds of materials. Steel, fabric, and ceramics are a few examples. Typically, the sources, traits, and uses of these classes are very diverse.
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Egg #1 is thrown at a brick wall. Egg #2 is thrown at a pillow with the same speed as Egg #1. Both eggs end up at rest.
Which egg experiences a greater force?
Egg #1
Egg #2
O They experience the same force
Answer:
The answer is Egg #1. Egg #1 experiences a greater force.
air flows through a heating duct with a square cross-section with 6-inch sides at a speed of 5.9 ft/s. just before reaching an outlet in the floor of a room, the duct widens to assume a square cross-section with sides equal to 15 inches. compute the speed of the air flowing into the room (in ft/s), assuming that we can treat the air as an incompressible fluid
The speed of the air flowing into the room (in ft/s) is equal to 0.944 when air flows through a heating duct with a square cross-section with 6-inch sides at a speed of 5.9 ft/s.
By the equation of continuity
(Rate of volumetric flow rate is constant)
A1v1 = A2v2
(6×6)×5.9(inch2ft/sec) = (15×15)×v2
V2 = 4.5(81/169) ft/sec
V2 = 0.944 ft/sec
In contrast to velocity, which describes the speed and direction of an object's movement, speed is the rate of movement along a path. Alternatively, velocity is a vector while speed is a scalar quantity. For instance, the speed of a car on a road is shown by 50 km/h (31 mph), but the velocity is indicated by 50 km/h west.
Physicists employ the fundamental units of speed and velocity to express how things move through space, time, and in a particular direction.
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Consider four identical bulbs connected in series to a battery. A fifth bulb is then connected in parallel with the first four. What will this additional connection do? Please explain why.
a) Increase the illumination produced by bulbs connected in series
b) Decrease the illumination produced by bulbs in series
c) Tt will change the illumination shomehow, but we need to know the voltage on battery and resistances of bulbs to answer
d) Leave the illumination produced by bulbs in series unchanged
The addition of 5th bulb connected in parallel with the first four bulbs will leave the illumination produced by bulbs in series unchanged (option d)
The relation between resistance, current, and voltage is given by:
I = V / R
Where:
I = current
V = voltage
R = resistance
In the given problem, all bulbs have the same resistance.
- 4 bulbs connected in series.
Let Is be the current when the 4 bulbs are connected in series. Then:
Is = V/4R
- Added the 5th bulb in parallel with the first four
The equivalent resistance will be 4/5 R and the current drawn from the battery:
Ip = V / (4/5 R) = 5 x (V/4R) = 5 x Is
So the current drawn from the battery is 5 times the initial current, Is.
However, this current is divided into 2 branches, let:
Ia = current flows to 4 bulbs
Ib = current flows to 1 bulb
Using the Kirchoff law:
Ia x 4R = Ib x R
and
Ia + Ib = Ip = 5 Is
We get:
Ia = 1/5 x Ip = 1/5 x 5Is = Is
Ib = 4/5 x Ip = 4/5 x 5Is = 4Is.
Hence,
Ia = Is
Or in other word, the currents flow on the 4 bulbs, before and after the addition of the 5th bulb, are the same.
Therefore, the correct option is:
d) Leave the illumination produced by bulbs in series unchanged
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1. What is the largest wave on Earth? What is its wavelength, frequency, and period?
2. Is the wave a deep-water wave or a shallow-water? explain
1. The largest wave on Earth is a tsunami wave. It has a wavelength of up to 1000 kilometers, a frequency of 0.04 to 0.2 hertz, and a period of up to an hour.
2. Tsunami waves are deep-water waves, meaning that they originate in the deep ocean and move towards the shoreline. The wave energy is transmitted across the entire water column, and it is refracted as it approaches the coast. This refraction causes the wave to slow down and build in height, resulting in the massive waves we often see in news coverage.
WILL MARK BRAINLIEST:
Which of these ALL require direction?
Displacement, Velocity, Force
Force, Time, Acceleration
Speed, Velocity, Distance
Mass, time, speed
Answer:
Force, Time, Acceleration
Explanation:
Displacement, velocity, and force do not necessarily require direction, but acceleration, which is the rate of change of velocity, does require direction. Speed, velocity, and distance do not require direction. Mass is a scalar quantity and does not have a direction. Time is also a scalar quantity and does not have a direction.
eudora ran from her home to her secret laboratory at an average speed of 12\text{ km/h}12 km/h12, start text, space, k, m, slash, h, end text. she then took one of her jetpacks and flew to her school at an average speed of 76\text{ km/h}76 km/h76, start text, space, k, m, slash, h, end text. eudora traveled a total distance of 120120120 kilometers, and the entire trip took 222 hours.
Eudora spent 0.5 hours from home to library and 1.5 hrs from library to home.
Given,
Home to library, Average speed S₁ = 12 km/hr
Average speed S₂ = 76 km/hr
Total distance = 120 km
Total Time = 2 hrs
Let us take the time taken to go from home to library is x. So, from library to school will be 2-x.
S₁ = 12 km/hr
T₁ = x
S₂ = 76 km/hr
T₂ = 2-x
We know that, distance = speed * time
D₁ = 12x
D₂ = 76(2-x) = 152 - 76x
D₁ + D₂ = 120
12x + 152 - 76x = 120
152 - 64x = 120
64x = 32
x = 0.5 hrs
T₂ = 2-x = 2- 0.5 = 1.5 hrs
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Calculate the total Av required to transfer between two coplanar circular Earth orbits of radii r 1 parameters p = 13,457 km and e = 0.76. Ans.: 7.086 km/s 12,750 km and r2 = 31,890 km, respectively, using a transfer ellipse having parameters p=13.475 km and e=0.76
The total velocity required (Av) to transfer from one circular orbit to another is 7.086 km/s
The total velocity required (Av) to transfer from one circular orbit to another is given by Av = √(μ/p)(2/r1 - 1/r2), where μ is the standard gravitational parameter of the central body and p is the semi-latus rectum of the transfer ellipse.
Given the parameters for the two circular orbits, r1 = 12,750 km and r2 = 31,890 km, and for the transfer ellipse, p = 13,475 km and e = 0.76, we can calculate the total velocity required to transfer between the two orbits as follows:
Av = √(μ/p)(2/r1 - 1/r2)
Av = √(3.986E+13/13,475)(2/12,750 - 1/31,890)
Av = √(2.946E+10)(0.0690 - 0.0314)
Av = √(2.946E+10)(0.0375)
Av = 7.086 km/s
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A long straight wire is positioned in the plane of a conducting square loop of side A and resistance R. The wire is parallel to the closest side of the square loop and at a distance A from it. What is the average power dissipated in the loop when the current in the wire is I = I0sin(ωt).
A. μ0/2π. Il㏒e. a+I/a. the average power dissipated in the loop is when the current in the wire is I = I0sin(ωt).
How to solve ?As the field of current wire passing the loop is same in direction but not uniform in magnitude.
we use integration method for finding the flux.
The same flux ,a thin rectangular strip of length l and with dx, is given by
dϕB=Bx.dS=B(x)dScos180∘.
Magnetic field due to a long straight wire carrying current I is given by B=μ0I/2πx, and area, dS=l×dx.
∴ ϕB=∫dϕB=−∫μ0/2πIlx.dx=−μ0/2πIl[logex]x=a+l/x=a
=−μ0/2π.Illoge.a+I/a
How can I determine the average power lost by a resistor?Any equation linking power to current, voltage, and resistance may be used to calculate the power wasted by each resistor because all three variables are known. Since each resistor receives its full voltage, let's use P=V2R P = V 2 R.
Power consumption and power dissipation are they equal?Power consumption refers to the device's overall power usage. The portion of power used by objects unrelated to the targeted tasks is known as power dissipation.
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