Moon Illusion is an Optical Phenomenon. This occurs because when the moon is near the horizon, Earthly things that resemble the moon on your retina generate a misunderstanding or inaccurate perception of distance and size.
The brain interprets the low moon as being bigger since it is lower in the sky than the higher moon.
The mechanism behind size-distance perception in everyday life, which transforms planar pictures that fall on the retina into a perception of hard objects moving in space, is related to the moon illusion.
What happens when the mechanism acts in an uncommon circumstance is what causes the moon illusion. In typical perception, the angular size of the light image stimulating our eyes expands or shrinks as rigid objects move in depth (distance).
The sense of hard objects with shifting depth perception is produced naturally by the brain in response to these changing stimuli. The ground and horizon make the moon appear comparatively close when it is close to the horizon.
Due to the moon's shifting apparent position in depth while the light stimulus stays constant, the moon appears extraordinarily huge due to the brain's size-distance mechanism.
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which of the following describes your ability to identify the sound of a musical instrument you have heard before
psychology
Answer: Recognition.
Explanation:
I'm pretty sure this is right !
The gauge pressure in a helium gas cylinder is initially 27atm . After many balloons have been blown up, the gauge pressure has decreased to 5atm .
One atm less than the absolute pressure is present in the gauge pressure. It is the variance between the absolute pressure inside the tank and the pressure outside. Therefor, the ratio is 0.185.
It is positive for pressures above atmospheric pressure and negative for pressures below atmospheric pressure; gauge pressure is the pressure as compared to atmospheric pressure. Any fluid that is not contained has its pressure increased by the atmospheric pressure. The pressure that exists inside the Earth's atmosphere is known as atmospheric pressure, also spelled barometric pressure (after the barometer). 101,325 Pa is the definition of the standard atmosphere, which is represented by the sign atm (1,013.25 hPa).
5/27 = 0.185 atm is the pressure ratio.
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If 1 cm on the globe represents 555 km in the real world, how many kilometers would you have traveled in 16.8 cm? (round)
9324 km
On travelling 16.8 cm, the distance travelled in kilometres is 9324 kilometres.
As per the known fact, the ratio of map distance will be equal. So, we will write the two ratio and equate it. Let us assume the distance travelled in 16.8 cm be x. Representing the information of equation form -
1 : 555 = 16.8 : x
x = 16.8 × 555
Performing multiplication on Right Hand Side of the equation to find the value of x (Denominator of 1 can be ignored)
x = 9324 kilometres
Thus, the distance travelled in 9324 kilometres.
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Why do we think that Venus has so much more atmospheric gas than Earth?
Most of the gases that have been released from volcanoes on Earth later returned to the surface.
Venus has gained much more gas through outgassing than has Earth.
Earth has lost much more gas to thermal escape than has Venus.
Because of its lack of magnetic field, Venus has been able to gain gas through the impacts of solar wind particles, while Earth has not gained gas in this way.
We think Venus has more atmospheric gas than Earth because of its lack of magnetic field, Venus has been able to gain gas through the impacts of solar wind particles, while Earth has not gained gas in this way.
Venus's densest atmosphere of the four terrestrial planets consists of 96% carbon dioxide. Venus's surface atmospheric pressure is 92 times greater than Earth's. With an average surface temperature of 735 K (462 °C; 863 °F), Venus is the hottest planet in the Solar System. The planet has no carbon cycle that captures carbon in rocks and surface features, and no organic life that can sequester carbon in the form of biomass.
Venus may have had oceans, but those oceans are evaporating due to increasing temperatures caused by the continuous greenhouse effect. Much of the water may have photodissociated, and the solar wind has sent free hydrogen rage into space as a result of the lack of an internal magnetic field on Venus. The surface of Venus itself is deserted, dry, and punctuated with rock that is periodically renewed by volcanic activity.
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Linearity is an extremely useful property of equations in physics. If an equation is linear, it allows you to easily add together simple contributions from individual elements to get a solution for a more complex situation. One example of a simple linear equation in classical mechanics is F = ma. If you find the individual accelerations of a particular body due to various individual forces, you can then add those accelerations to find the acceleration due to all of the forces acting together. If Newton's second law were not linear, introductory physics would be a much more difficult and confusing subject!
The schrodinger's equation for [tex](A\phi_1+AU\Psi_1)+(B\phi_2+BU\Psi_2)[/tex] for the particular U and E can be expressed as [tex](A\phi_1+AU\Psi_1)+(B\phi_2+BU\Psi_2)=E(A\Psi_1+B\Psi_2)[/tex].
The solution to the equation is based on Fourier's Eigen Values method. This is the expression for any mathematical function as the sum of an infinite series of other periodic functions. The trick is to find the right functions with the right amplitudes so that when they are superimposed, they give the desired solution.
As a result, the solution to Schrondinger's equation, the system's wave function, was replaced by the wave functions of the individual series, natural harmonics of each other, resulting in an infinite series. Shrodinger discovered that replacement waves described individual states of the quantum system, and their amplitudes indicated the relative importance of that state to the entire system.
The schoringer 's equation is
[tex]-\frac{h^2}{2m}\frac{d\psi^2}{dx^2}+U(x) \psi=E\psi[/tex]
here, the E is the energy of the particle.
since the wave function [tex]\psi_1[/tex] is the solution of the schrodinger's equation,
[tex]-\frac{h^2}{2m}\frac{d\psi_1^2}{dx^2}+U(x) \psi=E\psi_1[/tex]
Since, the wave function [tex]\psi_2[/tex] is the solution of the schrodinger's equation,
[tex]-\frac{h^2}{2m}\frac{d\psi_2^2}{dx^2}+U(x) \psi=E\psi_2[/tex]
Then,
[tex](A\phi_1+AU\Psi_1)+(B\phi_2+BU\Psi_2)=(A(-\frac{h^2}{2m}\frac{d\Psi_1^2}{dx^2})+AU\Psi_1)+(B(-\frac{h^2}{2m}\frac{d\Psi^2}{dx^2})+BU\Psi_2)\\\\=A(-\frac{h^2}{2m}\frac{d\Psi_1^2}{dx^2}+U\Psi_1)+B(-\frac{h^2}{2m}\frac{d\Psi_1^2}{dx^2}+U\psi_2)\\\\=A(-\frac{h^2}{2m}\frac{d\Psi_1^2}{dx^2}+U(x)\Psi_1)+B(-\frac{h^2}{2m}\frac{d\Psi_1^2}{dx^2}+U(x)\psi_2)\\\\=A(E\Psi_1)+B(E\Psi_2)\\\\=E(A\Psi_1+B\Psi_2)[/tex]
Thus, the schrodinger's equation for particular U and E can be expressed as [tex]E(A\Psi_1+B\Psi_2)[/tex].
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Your question is incomplete, here is the complete question in the figure,
I have got the answers to these question but I need to find the working out pls if you are able to answer at least one please do
calm :), thanks mods
A uniform, rigid, thin board that is 2.4m long and weighs 200N is attached to a post by a pivot at point P and hangs over the edge of a building. A crate weighing 400N is attached to the right end of the board. The left end of the board, .8m from the pivot, is attached to a light vertical wire anchored at the other end to keep the board from rotating.
1. Draw and label vectors on a horizontal board to represent forces acting on the board (not components). Show each force vector originating at its point of application.
2. Calculate the magnitudes of the forces exerted on the board by the post and by the wire. If you need to draw anything other than what you have shown in part 1 to assist in your solution, use the space below. Do not add anything to the figure in part 1
3. The rotational inertia of the board alone about its center is (1/12)Ml2 , where M is the mass of the board and L is its length. Calculate the rotational inertia of the combined board-crate system about point P
4. Suppose that the wire breaks and the board begins to pivot about point P. Calculate each of the following
(a) The magnitude of the initial angular acceleration of the board-crate system.
(b) The magnitude of the initial linear acceleration of the left end of the board.
1. Diagram (Attached above)
2. By equilibrium of forces of vertical direction,
T + Wa +Wc = Rp
T + 200 + 400 = Rp
Rp - P = 600N ------------------(i)
Taking moment of forces about P (Clockwise positive),
Wb * 0.4m + Wc * (0.4 + 1.2m) – T * 0.8m = 0
200 * 0.4 + 400 * 1.6 – 0.8T = 0
0.8T = 80 + 640
T = 900N
Putting T in equation (i),
Rp - 900= 600
Rp = 1500N
So, Rp= 1500N, T = 900N, Wb = 200N, Wc = 400N
3. Moment of inertia about C.O.M. ‘O’ is Icom= 1/12 ML2
Moment of inertia about ‘P’ by parallel axis theorem
Ip = Icom + M * (OP)2
Ip= 1/12 ML2+ M * (0.4)2
For net MOI about P, mass of crate Mc at distance PC from P will also be taken in account.
Mc = 400 / 9.8 = 40.8Kg, PC = 1.6m
Mass of board M = 200/9.8 = 20.4 Kg
Length of board L = 2.4m
So, Net M.O.I. about P,
Ip net = Icom + M*(0.4)2 + Mc * (PC)2
Ip net= 1/12 * 20.4*(2.4)2 + 20.4*(0.4)2+ 40.8*(1.6)2
Ip net = 117.5 Kg m2
4.
a. As, the moment of inertia about point P, Ip net =117.5 Kg m2
Torque experienced when wire is cut
τ = Wb * 0.4 + Wb * (0.4 + 1.2)
τ = 200 * 0.4 + 400 *1.6
τ = 720 N-m
As τ = I α, where α is angular acceleration.
720 = 117.5 * α
α = 6.127 rad / s2
So, magnitude of initial angular acceleration is α = 6.127 rad / s2
b. Initial Acceleration (linear) of left most point A
aA = α * AP
aA= 6.127 * 0.8
aA= 4.9 m/s2
So, linear acceleration of point A is 4.9 m/s2
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An astronaut's pack weighs 19.0 N when she is on earth but only 3.84 N when she is at the surface of moon. Part A What is the acceleration due to gravity on this moon? Express your answer with the appropriate units.
The weight of an astronaut's pack is 19.0 N on earth and 3.84 N on moon. The acceleration due to gravity on the moon is 1.98 m/s²
According to the Newton's second law of motion, the relation between force acted on an object with its acceleration is given by:
F = ma
Where:
F = force acted on the object
m = mass
a = acceleration
Weight is another name for the force due to gravity. If an object is brought from the earth to the moon, its weight will be different because the gravitational accelerations are different. However, its mass remains the same.
m = F / a
In case a = g = acceleration due to gravity:
m = w/g
Let
g = acceleration due to gravity on the earth = 9.8 m/s²
g₂ = acceleration due to gravity on the moon.
Hence,
19/g = 3.84/ g₂
g₂ = 3.84/19 x 9.8 = 1.98 m/s²
Hence, the acceleration due to gravity on the moon is 1.98 m/s²
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A wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly couple to the same left.
What fraction of the original rotational kinetic energy is lost?
The fraction of the original rotational kinetic energy lost due to the sudden coupling of the second wheel with twice the rotational inertia of the first is 100%.
When the two wheels are coupled, their rotational energies are combined and shared. Since the second wheel is initially at rest, its rotational energy is 0 and thus all the original rotational kinetic energy is lost.
The rotational kinetic energy of the first wheel is given by KE = 1/2Iω2, where I is the rotational inertia and ω is the angular velocity. Since the second wheel has twice the rotational inertia, the combined rotational kinetic energy is 1/2(I1 + 2I2)ω2. Since I2 is initially 0, the combined rotational kinetic energy is 1/2I1ω2, which is half of the original rotational kinetic energy. Therefore, 100% of the original rotational kinetic energy is lost.
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Two pipes of identical diameter and material are connected in parallel. The length of pipe A is three times the length of pipe B. Assuming the flow is fully turbulent in both pipes and thus the friction factor is independent of the Reynolds number and disregarding minor losses, determine the ratio of the flow rates in the two pipes.
According to the question the ratio of the flow rates in the two pipes is [tex]\sqrt{3}[/tex] /1
What in physics is flow?Fluid Flow, a branch of fluid dynamics, is concerned with fluids. It involves the movement of a fluid under the influence of uneven forces. As soon as unbalanced pressures are applied, this motion will persist.
Briefing:Lets take
Length of pipe B = L
Length of pipe A = 3 L
Discharge in pipe A = Q₁
Discharge in pipe B = Q₂
[tex]h_{f}[/tex] = (FLQ²)/12.1.[tex]d^{5}[/tex]
F=Friction factor, Q=Discharge,L=length
d=Diameter of pipe
here all only Q and L is varying and all other quantity is constant
So we can say that
LQ²= Constant
L₁Q₁²=L₂Q²₂
By putting the values
3LQ₁²=LQ²₂
Therefore
[tex]Q_{1}[/tex]/ [tex]Q_{2}[/tex] =[tex]\sqrt{3}[/tex]/ 1
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Unpolarized light of Intensity Io is incident on a pair of (ideal) polarizers, as shown below. The direction of polarization of the first polarizer is vertical. The second polarizer has unknown orientation.For each part below, determine whether it is possible for the light that reaches the observer to have the given value of intensity.If so: Determine the direction of polarization of the second polarizer. Show all work.If not: Explain by not.a. No light reaches the observerb. Light of intensity (2*Io)/3 reaches the observer.c. Light of intesity Io/2 reaches the observer.d. Light of intensity Io/4 reaches the observer.
The un polarization of light at each state was explained and second polarizer is I2=I1=Io/2 and first polarizer is I2=Io/4=Io/2 cos^(q).
What is polarization?
Polarized light waves are defined as those light waves that only travel in one plane. The polarization of light is the process by which unpolarized light waves are converted to polarized light waves.
What is un polarization ?
Unpolarized light is a form of electromagnetic radiation that vibrates in more than one plane. Unpolarized light is the kind that comes from the sun, a desk lamp, or a candle flame.
a) When the pass axis of the two polarizers are perpendicular to each other.
b) Not possible, because after first polarizer intensity becomes Io/2.
from malus law after second polarizer I2=I1 cos^2(q)
I2=2Io/3, cos^2(q)=4/3, q=cos^-1(1.15) Which does not exist.
c) When the pass axis of the second polarizer is parallel to first i.e q=0
I2=I1=Io/2.
d) I2=Io/4=Io/2 cos^(q)
cos^(q)=1/2, q=45 degree w.r.t pass axis of the first polarizer.
Therefore, the un polarization of light at each state was explained and second polarizer is I2=I1=Io/2 and first polarizer is I2=Io/4=Io/2 cos^(q).
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in the figure, a conducting metal bar of mass 0.5 kg sliding along conducting rails is connected to a battery of emf 18 v and resistor r
The bar's maximum magnitude of the velocity under these circumstances is 240 m/s.
What does a definition of mass mean?In physics, mass is a statistical expression of inertia, a basic characteristic of all matter. It essentially refers to a body of matter's resistance to changing its speed or location in response to the use of a force.
Briefing:The metal bar is experiencing the force.
F = iLB = ELB/R
The bar will increase in speed until the battery emf equals the induced emf.
Eind = vLB
current, and the force is eliminated.
Then the maximum velocity v = E/LB
v = 18/0.30*0.25
v = 240 m/s
Hence Option (5) is correct.
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The complete question is-
In the figure, a conducting metal bar of mass [tex]$0.5 \mathrm{~kg}$[/tex] sliding along conducting rails is connected to a battery of emf [tex]$18 \mathrm{~V}$[/tex] and resistor [tex]$R=200 \Omega$[/tex] to form a circuit. The rails are [tex]$30 \mathrm{~cm}$[/tex] apart and the entire arrangement is embedded in a uniform magnetic field [tex]$0.25 \mathrm{~T}$[/tex] directed into the page, as shown. Find the maximum velocity [tex]$v$[/tex] the bar attains under these conditions.
(1) [tex]$345 \mathrm{~m} / \mathrm{s}$[/tex]
(2) [tex]$68 \mathrm{~m} / \mathrm{s}$[/tex]
(3) [tex]$98 \mathrm{~m} / \mathrm{s}$[/tex]
(4) [tex]$135 \mathrm{~m} / \mathrm{s}$[/tex]
(5) [tex]$240 \mathrm{m} / \mathrm{s}$[/tex]
The pressure of water vapor over ice is 3.88 mm Hg at – 2C and 4.58 mm Hg at 0 C. Estimate in J mol^(-1) the heat of vaporization of ice at -1 C.
The heat of vaporization of ice at -1°C is equal to 1.02 ×10⁵J/mol
What is the heat of vaporization?The Heat of Vaporization can be defined as the amount of heat that requires to be absorbed to vaporize a certain amount of liquid at a given temperature.
We know that the Clausius Clapeyron equation can be written as:
[tex]\displaystyle lln\frac{P_2}{P_1} =\frac{-\triangle H}{R}(\frac{T_1-T_2}{T_1T_2} )[/tex]
Given the initial pressure of water, P₁ = 3.88 mm Hg
The final pressure of the water, P₂ = 4.58 mm Hg
The initial temperature, T₁= - 1°C = 272 K
The final pressure of water, T₂ = 0°C = 273 K
[tex]\displaystyle ln\frac{4.58}{3.88} =\frac{-\triangle H}{8.314}(\frac{272-273}{272\times 273} )[/tex]
ΔH = 1.02 × 10⁵J/mol
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dentica bcams of light arc incident on thrcc diffcrent pairs of (idcal) polarizers. Thc double arrow drawn on cach polarizer represents its direction of polarization. Case Unpolarized light Case B Unpolarized light Case C Unpolarized light Suppose that the incident light in each case were unpolarized. Rank the three cases (A-C) according to the intensity of the light transmitted past the sccond polarizer; from largest to smallest: If, for any casc, no light is transmitted past the second polarizer; state 50 explicitly. Explain your easoning Now suppose that Ihe incident light in each case were polarized in the vertical direction. Rank the three cases A-C) according to the intensity of the light transmitted past the second polarizer; from largest to smallest If; for any case; no light is transmitted past the second polarizer; state explicitly. Explain your reasoning:
Unpolarized light has light waves that are oriented in all directions while polarized light has all the light waves oscillating in one direction. An example for unpolarized is the Sun, or any other simple type of light.
Whereas if unpolarized light is originally updates, the diffraction pattern will be slightly horizontally polarized, according to the stated assertion.
The light that strikes the object as incident light from the source but also reflects off it is alluded to as refract. It makes no difference what the flashlight is or what you are photographing. When light strikes anything, it will always be transformed and reflected.
A portion of the light is dispersed around a different wavelength; this is called Raman scattering.
Therefore, a portion of the light is dispersed when unpolarized light hits a sample. Rayleigh scattering is the process wherein most of the dispersed light shares the same frequency as the incoming light.
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10 rooster head questions
Answer:
How does a rooster's head differ from that of a hen?What is the function of a rooster's crest on its head?Can a rooster's head and neck change color during mating season?How does a rooster use its head and beak during courtship displays?How does a rooster use its head to establish dominance in a flock?Can a rooster's head and beak be used as a weapon against predators?How does a rooster's head and beak help it to find food?How does a rooster's head and beak help it to eat and digest food?Can a rooster's head and beak be used for vocalizations, such as crowing?What are some common health issues that can affect a rooster's head and beak?Here are ten questions about the story entitled "cabeza de gallo":
Who are the main characters in the play?What is the central theme of the story?What kind of narrator tells the story?What was the rooster game about?Where is the place where most of the story takes place?Were the protagonist's predictions correct?What holiday was being celebrated?Which were knocked down as creatures struck in the abdomen?Who is the author of the work?What lesson or message does the story convey?¿What is a tale?A story is a narration, either oral or written, about a story that may have happened in real life, or is the product of the author's imagination.
The stories have the characteristic of having few characters, and are shorter than any common text.
¡Hope this helped!
When a country burns its forests, neighboring countries may suffer from air pollution as well. Why are air pollutants produced by forest fires in one country able to travel to neighboring countries?
Answer:
Air pollutants produced by forest fires in one country are able to travel to neighboring countries because of the way air moves and circulates around the globe. The Earth's atmosphere is made up of a mixture of gases, including oxygen and carbon dioxide, which are capable of moving and circulating through the air. When pollutants, such as those produced by forest fires, are released into the atmosphere, they can be carried by wind and other atmospheric phenomena, allowing them to travel over long distances and affect other regions and countries. This is why air pollution from forest fires in one country can affect neighboring countries.
An object attached to one end of a string moves in a circle at constant speed. Which of the following is correct?
the object is accelerating as it moves
the object's velocity is the same as its speed
the object does not require a force to keep its state of circular motion
if the string breaks, the object will keep its circular motion
If the string breaks, the object will move Radially away from the center of the cicl
The object is accelerating as it moves in a circle at constant speed. This is because acceleration is a measure of how an object's velocity changes over time, and the velocity of an object moving in a circle is constantly changing direction.
The object's velocity is not the same as its speed. Speed is a scalar quantity that refers to the distance an object travels per unit of time, while velocity is a vector quantity that refers to the speed and direction of an object's motion. The velocity of an object moving in a circle will vary as the object moves around the circle, even if its speed remains constant.
An object moving in a circle does require a force to keep its state of circular motion. This force is known as the centripetal force, and it is directed towards the center of the circle. The centripetal force is necessary to keep the object moving in a circular path, rather than following a straight-line path.
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In shape and component stars, elliptical galaxies are most like our own
A. galactic center
B. spiral arms
C. nuclear bulge
D. halo
E. companions, the Magellanic Clouds
d
The correct option is D : Halo , In comparison to spiral galaxies, elliptical galaxies seem smooth, ellipsoidal, or spherical and have far less structure. Galaxy clusters are generally home to elliptical galaxies.
There are stars, planets, interstellar dust, and gases in elliptical, spiral, and irregular galaxies alike. The same gravitational forces that create other galaxies also create these three different sorts of galaxies.
Old stars, a tiny amount of gas, and dust can be found in elliptical galaxies. In contrast to open clusters, stars in these galaxies coalesce into globular clusters. Individual stars, nebula, and clusters are visible in some irregular galaxies but are invisible in other irregular galaxies.
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a toroid has a 50.8 cm inner diameter and a 53.7 cm outer diameter. it carries a 24.7 a current in its 604 coils.(figure 1) (figure 2). Determine the range of values for B inside the toroid. There should be two values, B(min) and B(max), in mT.
The range of values for B inside the toroid is: B(min) = 1.37 mT and B(max) = 3.37 mT
The range of values for B inside the toroid can be determined using the formula B = μ0nI/l, where μ0 is the permeability of free space, n is the number of turns, I is the current, and l is the average length of the toroid. The average length of the toroid is equal to the average diameter, which is (50.8 cm + 53.7 cm)/2 = 52.25 cm. Therefore, B = μ0nI/l = (4π×10^-7 H/m)(604 turns)(24.7 A)/(52.25 cm) = 0.0022 H/m. Since 1 mT = 10^-3 T, the range of values for B inside the toroid is 1.37 mT (B(min)) and 3.37 mT (B(max)).
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the luminosity of star a is 26 time bigger than luminosity of star b. star a is also known to be 3 times closer than star b
The luminosity of star a is 78 times bigger than star b.
The luminosity of star a is 78 times bigger than star b because the luminosity of star a is 26 times bigger than the luminosity of star b but star a is also 3 times closer than star b, so the difference in luminosity is amplified by a factor of 3. This means the luminosity of star a is 26 times 3, or 78 times bigger than star b.
The luminosity of a star is directly related to its distance from us. The closer a star is, the brighter it will appear. This is because the amount of light we receive from a star is inversely proportional to the square of its distance from us.
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A rectangular swimming pool has dimensions of 15 m x 30 m and is 3.5 m deep. When the pool is filled with fresh water, what is the force on the side of the pool due to the water?
The total force on the side of the pool due to the water is 2275 N/m^2 x (15 m x 3.5 m) = 1523125 N.
To find the force on the side of the pool due to the water, we first need to calculate the volume of the water in the pool. Since the pool has dimensions of 15 m x 30 m and is 3.5 m deep, its volume is 15 m x 30 m x 3.5 m = 1575 cubic meters. Next, we need to calculate the weight of the water in the pool. The weight of an object is equal to its mass multiplied by the acceleration due to gravity. The density of fresh water is about 1000 kg/m^3, so the mass of the water in the pool is 1575 m^3 x 1000 kg/m^3 = 1575000 kg. Since the acceleration due to gravity is approximately 9.8 m/s^2, the weight of the water in the pool is 1575000 kg x 9.8 m/s^2 = 15442500 N. Finally, we can calculate the force on the side of the pool due to the water. Since the force is distributed evenly over the surface of the side of the pool, the force on each square meter of the side of the pool is 15442500 N / (15 m x 3.5 m) = 2275 N/m^2. Therefore, the total force on the side of the pool due to the water is 2275 N/m^2 x (15 m x 3.5 m) = 1523125 N.
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(Figure 1) shows a thin liquid film bounded on the right side by a sliding wire that is attached to a spring with spring constant 0.50 N/m. The spring is stretched by 1.4 cm. Figure 1 of 1 < > 0.50 N/m 6.0 cm Mai W Part A What is the liquid's surface tension in mN/m? Express your answer in millinewtons per meter. || ΑΣφ ? 7= mN/m
The liquid's surface tension in mN/m is 58.33.
Surface tension is the tendency of the surface of a liquid at rest to contract into the smallest possible surface area. Surface tension allows objects denser than water, such as razor blades and insects, to float on the surface without being partially submerged.
Given:
Spring constant of the spring = k = 0.5 N/m
Distance by which the string is stretched = d = 1.4 cm = 0.014 m
Force of the spring on the wire = Fs
Fs = kd
Surface tension of the liquid = γ
Length of the sliding wire = L = 6 cm = 0.06 m
Surface tension force of the liquid on the wire = Ft
A liquid film has 2 surfaces therefore,
Ft = 2γL
The force of the spring on the wire and the surface tension force on the wire will be equal.
Fs = Ft
kd = 2γL
(0.5)(0.014) = 2γ(0.06)
0.007 = 0.12γ
γ = 0.05833 N/m
Converting the surface tension from Newton per meter to milli-Newton per meter, (1 N = 103 mN)
γ = (0.05833) x (103) mN/m
γ = 58.33 mN/m
Surface tension of the liquid = 58.33 mN/m
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which of the following is not a strong argument for the theory that some large elliptical galaxies formed as the result of galaxy collisions?
The majority of galaxies in the universe are elliptical galaxies, and galaxy collisions are frequent.
Mergers produce large elliptical galaxies. There is proof that galaxies develop through mergers and collisions. Star formation is triggered by galaxic collisions. All of the gas is transformed into stars before a disc can develop because the higher gas density produces stars more quickly. High redshift elliptical galaxies lack young, blue stars. Old red stars with erratic orbits in several planes are abundant in elliptical galaxies, which have a small amount of gas and dust. The largest galaxies that we are currently aware of are enormous elliptical galaxies. The majority of galaxies found now are small elliptical galaxies. Elliptical galaxies often have very little cold gas.
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complete question: Which of the following is not a strong argument for the theory that some large elliptical galaxies formed as the result of galaxy collisions?
A) Elliptical galaxies dominate the population in dense galaxy clusters.
B) Some ellipticals have stars and gas that rotate opposite to the rest of the galaxy.
C) Some elliptical galaxies are surrounded by shells of stars.
D) Computer simulations predict that the product of a galaxy collision is generally an elliptical galaxy.
E) Galaxy collisions are common and most galaxies in the universe are elliptical.
At point A, 4.40 m from a small source of sound that is emitting uniformly in all directions, the intensity level is 53.0 dBPart AWhat is the intensity of the sound at A?Part BHow far from the source must you go so that the intensity is one-fourth of what it was at A?Part CHow far must you go so that the sound level is one-fourth of what it was at A ?Part DDoes intensity obey the inverse square law?Part EDoes intensity level obey the inverse square law?
No, intensity level does not follow inverse square law.
The Inverse Square Law Formula: What Does It Mean?The square of the distance between the source and the observer determines how intense the light is for the observer. This demonstrates that the intensity of light is proportional to a value multiplied by 1/d² as the distance to a light source increases.
Here,
intensity level = 53 dB
distance, d = 4.4 m
part A)
let the intensity is I
Intensity level = 10×log(l/1×10⁻¹²)
53 = 10 ×log(1/1×10⁻¹²)
solving for I
I = 1.995 ×10⁻⁷ W/m²
the intensity of sound at A is 1.995 ×10⁻⁷W/m²
part B)
for the sound intensity to be one fourth.
as sound intensity = 1/(4×π×d²)
hence, the distance must be made two times,
new distance = 4.4 ×2
new distance = 8.8 m
the distance from the source is 8.8 m
part C)
distance to move from point A = 8.8 - 4.4
distance to move from point A = 4.4 m
part D)
yes, the intensity follow the inverse square law
part E)
for the intensity level,
as I = 10 × log(P/(4×π×d²×10))
hence, intensity level does not follow inverse square law.
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the gravitational constant g was first measured accurately by henry cavendish in 1798. he used an exquisitely sensitive balance to measure the force between two lead spheres whose centers were 0.19 m apart. one of the spheres had a mass of 138 kg, while the mass of the other sphere was 0.93 kg.
The answer is
3.5*10^-8:1
Explanation:
F(gravitation) = G*M1M2/r^2
G = gravitational constant = 6.67*10^-11 Nm^2/kg^2
M1 = mass of the heavier sphere,
M2 = mass of the lighter sphere
r = their distance apart
Fg = 6.67*10^-11*188*0.93/(0.19)^2
Fg = 3.23*10^-7N
The ratio of the force of gravitation between them to the weight of the lighter sphere = 3.23*10^-7: (0.93*9.81) = 3.5*10^-8. : 1 by dividing both side by (0.93*9.81)
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in fig. 12-30, trying to get his car out of mud, a man ties one end of a rope around the front bumper and the other end tightly around a utility pole 18 m away. he then pushes sideways on the rope at its midpoint with a force of 550 n, displacing the center of the rope 0.30 m, but the car barely moves.what is the magnitude of the force on the car from the rope? (the rope stretches somewhat.)
The force on the car from the rope is 27496N.
The sum of forces in y direction is zero.
F = Tsinθ - Tsinθ =0
F - 2T SINθ = 0
T = F/2sinθ
Tan θ = .28/18 = 0.015
θ = [tex]tan^{-1}[/tex] 0.015 = 0.8593
T = 550 / 2sin (0.8593)
T = 27500 N
Fcar = Tcosθ
Fcar = Tcos[0.8593]
Fcar = 27496N
All the planets are maintained in their orbits around the sun by the force of gravity. Many of the large-scale structures in the Universe are caused by gravity because the gravitational attraction between the original gaseous stuff in the Universe allowed it to combine and form stars, which then condensed into galaxies. Although the reach of gravity is limitless, its effects diminish with increasing distance.
The general theory of relativity, which Albert Einstein proposed in 1915, is the theory that most accurately describes gravity. According to this theory, gravity is not a force but rather the curvature of spacetime brought on by an imbalance in the distribution of mass, which causes masses to move along geodesic arcs.
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a 345 pf capacitor is charged to 145 v and then quickly connected to a 165 mh inductor. frequency of oscillation
A 165 mh inductor was attached right after that. The frequency of the oscillation is f=7957.74 Hz.
What is motion-induced oscillation?Periodic or oscillatory motion is defined as a motion that repeats itself. A restoring force or torque causes the object in this motion to oscillate about its equilibrium position.
For Class 8, what is an oscillation?Oscillation is a revolving motion between two states or locations. The side-to-side swing of a pendulum or the up-and-down motion of a spring with a weight are two examples of periodic motions that repeat themselves in a regular cycle and are considered oscillations.
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difference between constant and
variable work
Work done on an object by a constant force is known as Constant Work, whereas the work done on an object by variable force is known as Variable Work.
In constant work or work done by constant force, the magnitude and direction of the force are constant or they do not change. Therefore, constant work is simply calculated by force acting on the object multiplied by displacement of the object.
But in case of variable work, things are not so easy. In this scenario, the force's magnitude and direction may alter at any point while the job is being done. The majority of the work we do on a daily basis is an instance of variable force work. The same calculation involves integration and is fairly difficult.
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two wires cross at a 400 without electrical contact. find the magnitude and direction of the magnetic field at point p. point p is 10 cm from the wire intersection and equally distant from both wires.
Both wire carrying current in same direction with the same magnitude A current-carrying wire produces a magnetic field and the Bio-Savart law enables us to calculate the magnitude and direction of this magnetic field. at any point where the magnetic field due to the segment of current-carrying wire is given by equation
[tex]B=\frac{\mu_0}{2 \pi} \frac{l}{r}[/tex]
where r is the distance between the wire and the point and I is the current of the wire. At point , the distance between it and the intersection is . So, its distance from the horizontal wire is B
[tex]r_1=(4 \mathrm{~cm}) \sin 75^{\circ}=3.86 \mathrm{~cm}[/tex]
This the same distance from the second wire as we are given
[tex]r_1=r_2=3.86 \mathrm{~cm}[/tex]
if we apply the right-hand rule, we find that the magnetic field at point of each wire has direction out of the magnetic field at point is the summation of both magnetic fields
[tex]B_1=\frac{\mu_0}{2 \pi} \frac{1}{r_2}+\frac{\mu_0}{2 \pi} \frac{1}{r_2}=2\left(\frac{\mu_0}{2 \pi} \frac{1}{r_1}\right) \text { (2) }[/tex]
Now, we can use the values for to get the magnetic field at point by
[tex]B_1 & =2\left(\frac{\mu_0}{2 \pi} \frac{1}{r_1}\right) \\& =2\left(\frac{\left(4 \pi \times 10^{-7} T \cdot m / A\right)(5 \mathrm{~A})}{2 \pi\left(3.86 \times 10^{-2} \mathrm{~m}\right)}\right) \\& =5.2 \times 10^{-5} \mathrm{~T}\end{aligned}[/tex]
At point , the horizontal while applies a magnetic field with direction out of the page while the other wire applies a magnetic feild with direction into . As both wires exert the same magnetic field at point is zero
At point , the horizontal while applies a magnetic field with direction out of the page while the other wire applies a magnetic feild with direction into the page. As both wires exert the same magnetic field at point is zero.
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Student 1 brings a negatively charged rod with an insulating handle near to but not touching an uncharged metal sphere and finds that the objects attract. She wants the sphere to be repelled by the rod. Student 2, who is uncharged, suggests that this can be accomplished if he touches the sphere. Which of the following indicates whether this will work, and if so, gives a procedure that will cause the repulsion? (A) Yes, Student 2 could touch the sphere while the rod is held near it, because that would make the sphere negatively charged. (B) Yes, Student 2 could touch the sphere while the rod is kept far away, because that would make the sphere negatively charged. (C) Yes, Student 2 could touch the sphere while the rod is kept far away, because that would make the sphere positively charged. (D) No, there is no way that having Student 2 touch the sphere will work.
No, there is no way that having Student 2 touch the sphere will work. There will be no repulsion by touching the sphere whether the rod is kept near or away. No way cause the sphere negatively charged.
Charging an uncharged body by bringing near to it a charged body is called charging by induction. In the given case negatively charged rod is brought near to uncharged sphere, then all positive charges on it realigns towards the end of the rod by attraction.
When it is touched while keeping the rod closer, all negative charges are grounded passing through body. Thus sphere has now only positively charges. When rod is removed, the positive charges distributes uniformly ands sphere becomes positively charged.
In no way sphere will be negatively charged for repulsion to occur.
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