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!
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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,
Related Questions
an apple weighs 1.20 n. when you hang it from the end of a long spring of force constant 1.44 n/m and negligible mass, it bounces up and down in shm. if you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (because the angle is small, the back and forth swings do not cause any appreciable change in the length of the spring.)
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The unstretched length of the spring is 3.333 m
Calculation :
k = Spring constant = 1.5 N/m
g = Acceleration due to gravity = 9.81 m/s²
l = Unstretched length
Frequency of SHM motion is given by
[tex]f_{s} = \frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex]
Frequency of pendulum is given by
[tex]f_{p} = \frac{1}{2\pi }\sqrt{\frac{g}{l} }[/tex]
Given in the question
[tex]f_{s} = \frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex] = [tex]f_{s} = \frac{1}{2\pi }\sqrt{\frac{g}{l} }[/tex]
[tex]\frac{1}{2\pi }\sqrt{\frac{k}{m} }= \frac{1}{2}\frac{1}{2\pi }\sqrt{\frac{k}{m} }[/tex]
l = [tex]\frac{4gm}{k}[/tex]
l = [tex]\frac{4*9.8*\frac{1.2}{9.8} }{1.44}[/tex]
l = 3.333 m
The unstretched length of the spring is 3.333 m
The word "force" has a precise meaning. At this level it is appropriate to describe the force as pushing or pulling. A force does not contain or "have within" an object. A force is applied from one object to another. The concept of force is not limited to living things and inanimate things
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Figure 7-40 shows a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.6 m, so the cart slides from x1 = 5.0 m to x2 = 1.0 m. During the move, the tension in the cord is a constant 28.0 N. What is the change in the kinetic energy of the cart during the move?
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The cord's tension remains constant during the movement at 28.0 N. 94.08 J is the change in the cart's kinetic energy during motion.
the force exerted on the cart is the tension T in the chord itself. so,
F=T
Calculate the displacement of the cart.
d=[tex]\sqrt{x_{1}^{2}+h^{2} } -\sqrt{x_{2}^{2}+h^{2} }[/tex]
subsitute 5.0m for x₁,1.0m for x₂,and 1.60m for h.
d=[tex]\sqrt{5.0^{2}+1.60^{2} } -\sqrt{1.0^{2}+1.60^{2} }[/tex]
d=3.36m
calculate the change in kinetic energy of the cart.
ΔK=Fd
substitute 28.0N for F and 3.36m for d.
ΔK=28.0×3.36
ΔK=94.08 J
Tension is a force that runs the length of a medium, particularly one that is conveyed by a flexible medium like a rope or cable.
An action-reaction pair of forces acting at either end of the aforementioned elements is what is referred to as tension. Consider a rope; aside from the endpoints, every section of the rope experiences tension force in both directions. The ends feel strained on one side and the force of the attached weight. The tension fluctuates in several places along the string.
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rosa a block of mass 3 kg slides down an inclined plane at an angle of ??p with a massless tether attached to a pulley with mass 1 kg and radius 0.5 m at the top of the incline (see the following figure). the pulley can be approximated as a disk. the coefficient of kinetic friction on the plane is 0.4. what is the acceleration of the block? os 10
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If the coefficient of kinetic friction on the plane is 0.4, the acceleration of the block is 0.3 m/s² and tension is T = 1.82N.
A property of a body that is a measure of inertia is usually viewed as a measure of the amount of matter contained in the body that causes it to have weight in a gravitational field. Mass is a dimensionless quantity that describes the amount of matter in a particle or object.
Calculation:-
Tension force = I/R² × a
= 0.5 × 3.564
T = 1.782 N.
acceleration = 1.782/3
= 0.3 m/s²
The mass of an object is a measure of its inertial properties or the amount of matter it contains. The weight of an object is a measure of the force exerted on it by gravity, or the force required to support it. Mass can best be understood as the amount of matter present in an object or body. Everything around us has mass.
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process command line args and get the simulation parameters (e.g., alg, quantum, inputfile) create/initialize the necessary data structures (ready q and io q (double linked lists of pcb),
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In the operating system's command-line shell, parameters are provided after the program's name. Typically, we define main() with two arguments: a list of command-line arguments and a number representing the number of command-line arguments.
In C#, how do you acquire command-line arguments for a process?We'll talk about utilizing the Environment class to retrieve and show command-line arguments in this article. Therefore, we use the Environment Class's CommandLine property to do this task. To locate the command line for the active process, utilize this attribute. Message Type: This property returns a string as its return type.
Which of these methods accepts parameters from the command line?Only the main() method can accept parameters from the command line.
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BONUS: Make a sketch and label all the forces. A 2.00 kg box slides down an inclined plane
(ramp) at an angle of 20.0 degrees at a constant velocity. If the horizontal length of the
inclined plane is 2.00 m (you need to determine the actual slope length or hypotenuse):
(a) Calculate the work done by the force of gravity.
(b) Calculate the work done by the force of kinetic friction.
By signing my name to the test below, I pledge that the work that I have turned in for this
test is entirely my own and I have neither given nor received help from anyone else.
Answers
The amount of labour that gravity does is 28.58J. And the work produced by kinetic friction is -28.58J. A force that acts between moving surfaces is referred to as kinetic friction.
We may determine the slope length or hypotenuse using elementary trigonometry: cos(20) = 2.00. (slope length)
slope length: 2.00 m/cos(20) = 2.13 m
The box covers this distance while moving at a constant speed.
In terms of the forces, the total force parallel to the inclined plane acting on the box is given by the equation: F = W - f = mgsin(20) - f, where f represents friction and W is weight.
Work = Wd = (+13.2N)(2.13m) = 28.58J, which is the amount of work done by the force of kinetic friction.
The force acts in the same direction as the displacement, resulting in work.
The kinetic friction force exerts the following work:
Work= -fd =(-13.42N)(2.13m) = -28.58J
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in which of the following cases will there be a current in the ammeter? (1) the switch has just been closed (2) the switch has been closed a long time (3) the switch has just been opened (4) the switch has been opened a long time
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The presence of a current in an ammeter depends on the state of an external electrical switch. If the switch is closed, there will be a current in the ammeter; if the switch is opened, there will be no current in the ammeter. The length of time the switch has been in a particular state does not affect the presence of a current in the ammeter.
The presence of a current in an ammeter depends on the state of an external electrical switch. A switch is a device used to break or make an electrical connection, and can be used to control the flow of electricity in a circuit. In the following cases, the presence of a current in the ammeter can be determined:
(1) If the switch has just been closed, there will be a current in the ammeter. When the switch is closed, it creates an electrical connection, allowing electricity to flow through the circuit, and thus a current to be registered on the ammeter.
(2) If the switch has been closed for a long time, there will still be a current in the ammeter. This is because the switch is still in the closed position, allowing electricity to flow through the circuit, and thus a current to be registered on the ammeter.
(3) If the switch has just been opened, there will be no current in the ammeter. When the switch is opened, it breaks the electrical connection, preventing electricity from flowing through the circuit, and thus no current will be registered on the ammeter.
(4) If the switch has been opened for a long time, there will still be no current in the ammeter. This is because the switch is still in the open position, preventing electricity from flowing through the circuit, and thus no current will be registered on the ammeter.
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evaporation is a cooling process and condensation isquestion 16 options:a cooling process also.a warming process.neither a warming nor cooling process.can not tell with the information given.
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Evaporation is a cooling process and condensation is a warming process.
Condensation is the transition of a region of matter from a gaseous state to a liquid state and the opposite of vaporization. This word most often refers to the water cycle. Condensation is the process by which water vapor becomes a liquid. This is the opposite of evaporation, which turns liquid water into vapor. Condensation occurs in one of two ways. When the air cools to its dew point or becomes saturated with water vapor, it can no longer hold water.
Evaporation is always a cooling process. For evaporation to occur, energy is needed by the liquid molecules that are changing phase into gas. This energy is taken from the surroundings in the form of heat energy. So when evaporation occurs, the temperature of the surroundings will decrease, making evaporation a cooling process.
Condensation occurs when warmer, moist air comes into contact with cooler surfaces such as frames, windows, and other superstructures, or cooler areas inside an insulating shell (where moisture has penetrated the vapor barrier).
Therefore, evaporation is a cooling process and condensation is a warming process.
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The drawing shows box 1 resting on a table, with box 2 resting on top of box 1. A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2, and the other end is connected to box 3. The weights of the three boxes are W1 = 51.5 N, W2 = 39.0 N, and W3 = 25.3 N. Determine the magnitude of the normal force that the table exerts on box 1.
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The force that the table exerts on the box1 is 65.2N
As given in the question we are assuming box 2 and 3 aren't touching.
The forces on box 2 are:
gravitational force
Force of tension
Normal force
The box isn't accelerating so
T + N2 - Fg2 = 0
The forces acting on the box 3 are:
gravitational force
Tension
Since box 3 isn't accelerating
T - Fg3 = 0
The tension force will be same in both cases so:
Fg3 + N2 - Fg2 = 0
So
N2 = Fg2 -Fg3
N2 = Fg2 - Fg3
= 39-25.3 = 13.7 N
Remember forces are equal and opposite.
Now,
write the equation for the forces which are acting on Block 1
There is the force of gravity
normal force acting from the table and
A force which is equal and is opposite to the normal force on block 2
N1 - F(g1) - N2 = 0
N1 = Fg1 +N2
N1 = 51.5 + 13.7 = 65.2 N
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The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. Suppose that a horizontal wind blows in streamline flow with a speed of 10.5 m/s outside a large pane of plate glass with dimensions 6.00 m Times 1.10 m. Assume the density of air to be constant at 1.20 kg/m^3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is 21 m/s, twice as high.
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Given,
Speed of wind v = 10.5 m/s
Dimension of plate = 6 m times 1.10 m
The density of air d = 1.20 kg/m³
Therefore,
F = (1/2)dv²A
A = 6(1.10)m² [From dimension]
F = (1/2)(1.20)(10.5)²(6)(1.10)
F = 436.59 N
The force on the window pane is 436.59 N.
(a) if the wind speed is 21 m/s
F = (1/2)(1.20)(21)²(6)(1.10)
F = 1746.36 N
The force on the window pane is 1746.36 N.
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An engineer wanted to put a speed limit at the enterance to atrain at 500maway as it comes to rest to a plant form his plan is give exactly 3/2 minutes for the plate to decelerate uniformaly within the 500m distance what must bethe maximum speed at the entrance?
Answers
Answer:
50
Explanation:
100×50÷1000= 50
this is easy question
Porosity decreases Choose one: A. with decreasing compaction of sediments or rock. B. when rocks develop joints or faults. C. when fluids passing through a rock dissolve parts of the rock. D. with the cementing of sediments by mineral grains from groundwater. The water table Choose one: O A. usually lies within a few meters of the surface in deserts. O B. rises to a higher elevation during the dry season. O C. mimics the topography of the land it underlies. O D. is always located beneath the ground surface.
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A decrease in porosity occurs when rocks form joints or faults.
Option B is correct
What impact does porosity have on groundwater?Porosity is the entire available area for groundwater to live in. A rock's or sediment's porosity impacts how much water it can hold. The porosity of sediments and sedimentary rocks is influenced by the size, shape, and degree of sorting of the grains as well as the degree of cementation.
What impact do permeability and porosity have on groundwater?A good permeability must coexist with good porosity for groundwater to enter a rock. The pore gaps between the rock grains must be linked for a rock to be permeable and allow water to pass through it.
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A laser is used in eye surgery to weld a detached retina back into place. The wavelength of the laser beam is 581 nm, and the power is 1.5 W. During surgery, the laser beam is turned on for 0.050 s. During this time, how many photons are emitted by the laser?
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The laser beam emits 1.939 x 10¹⁷ photons after being activated for 0.050 seconds.
Explain what a laser is.A laser emits an extremely focused beam of energy that may be used in a variety of equipment and technologies. Laser Amplification by Stimulation of Radiation Emission is what the letters in the term laser stand for. Light Amplifier by Stimulated Radiation Emission is what the elements in the term laser stand for.
Briefing:As we are aware,
This pace of energy is equivalent to power in numbers.
As things evolve throughout time,
using mathematics,
power : P = E /t
Total Energy : E = P t
= (1.5)(0.050)
= 0.075 J
The laser beam's wavelength is = λ = 514 nm
One photon's energy is given by E ' = h c /λ
= (6.626068 × 10⁻³⁴ )(3 x 10⁸)/(514 x10⁻⁹)
= 0.03867 x 10⁻¹⁷ J
The amount of photons the laser beam emits:
= (total energy) / (energy per photon)
= 0.075 J / 0.03867 x 10⁻¹⁷ J
= 1.939 x10¹⁷ photons
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Consider the 52.0 kg mountain climber in the figure. 15° (a) Find the tension in the rope in N) and the force that the mountain climber must exert with her feet in N) on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms. tension in rope force on feet ON (b) What is the minimum coefficient of friction between her shoes and the cliff?
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a. the force that the mountain climber must exert with her feet in N) on the vertical rock face to remain stationary is 273 N.
b. the minimum coefficient of friction between her shoes and the cliff is 0.268.
What is the term of friction?
The force that prohibits one solid object from traveling across another is known as friction. Static friction, sliding friction, rolling friction, and fluid friction are the four main categories of friction.
a).
Balancing the forces in x-direction,
F cos(15) = T sin(31)
F = T sin(31)/cos(15) ...... (1)
balancing the forces in y-direction,
T cos(31) + F sin(15) = 52 x 9.8 = 509.6
T cos(31) + (T sin(31)/cos(15)) sin(15) = 509.6
T = (509.6 cos(15))/(cos(31) cos(15) + sin(31) sin(15))
T = 512 N
Force, F = 512 sin(31)/cos(15)
= 273 N
b)
Coefficient of friction,
u = sin(15)/cos(15)
= 0.268
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Five different wind shear profiles are shown below. Please place a triangle around the letter beneath the wind shear profile that is ideal for a supercell. Please place a circle around the letter beneath the wind shear profile that is ideal for a squall line 200 mb 250 mb 300 mb 400 mb 500 mb 600 mb 700 mb 850 mb 000 mb A. C. Wind shear is calculated by finding the difference between the wind speed values at two different heights. Using the wind profile you identified as the ideal environment for a supercell, next you will calculate the 0-1 km and 0-5 km wind shear values. This means we will find the difference between the surface wind (lowest wind barb on the sounding) and the speed of the wind at lkm and 5 km. The atmospheric pressure at 1km above sea level is typically very close to 850 mb. The pressure at 5 km above sea level is very close to 500 mb. Please calculate the 0-1 km and 0-5 km wind shear values in knots (kts). 0-1km shear value 0-5km shear value
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At the surface, the wind speed is between 13 to 17 knots. At 1 km or 850 MB level, the wind speed is between 48 to 52 knots. Thus, the 0-1 km wind shear is (48 -13) or (52-17) = 35 knots.
At 5 km or 500 MB level, the wind speed is between 68 to 72 knots. Thus, the 0-5 km wind shear is (68-13) or (72-17) = 55 knots.
Wind speed describes how fast air moves past a particular point. This can be averaged over specific time units. B. Miles per hour or instantaneous speed is reported as peak wind speed gust or squall. The wind is the natural movement of air or other gases against the surface of a planet.
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When a 40-N force, parallel to the incline and directed up theincline, is applied to a crate on
a frictionless incline that is 30? above the horizontal, theacceleration of the crate is 2.0m/s2,
up the incline. The mass of the crate is:
A. 3.8kg
B. 4.1kg
C. 5.8kg
D. 6.2kg
E. 10 kg
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The mass of the crate when it is directed up the incline will be equal to 5.8 kg that is option C is correct.
This problem can be solved using the concept of Newton's second law of motion. The mathematical expression of this will be
F = ma
where F is the force, m is mass and a is the acceleration.
Now, net force will be expressed as
Fₙ = 40N - mg (sin 30°) = m(2 m/s²)
where 40 N is the force parallel to incline and mg (sin 30°) is the vertical component of force.
From this we solve for m as
m = 40/6.9
m = 5.8 kg
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A clean nickel surface (work function 5.1eV), is exposed to light of wavelength 206nm. What is the maximum speed of the photoelectrons emitted from this surface?
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A pristine nickel surface with a work function of 5.1 eV is exposed to 206 nm light. The photoelectrons released from this surface have a maximum speed of 521.8 x 10^3 m/s.
An electron of the metal can receive photon energy from incident light, which is equal to the planck's constant times the light's frequency.
f = c/∧
assuming that all of the photon energy has been utilized
First in removing electrons from metals (work function)
Second, the leftover is used to furnish the photoelectron with kinetic energy:
V(m) = √2/M(hc/∧)-5.1eV.
where: M = 9.11 x10^3 kg; is the mass of nickel
h = 6.63 x 10 -34 Js; is plancks constant
c = 3 x 10^8 m/s; is the speed of light
∧ = wavelength of light = 206 nm = 206x10-9m
v(m) = (2 x 6.63 x 10-34 x 3 x10^8)/206 x 10-9 x 9.11x10^3 - 5.1 eV
v(m) = 521.8 x 10^3 m/s.
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A car driving at 20 m/s accelerates 2.5 m/s2 for a distance of 350 meters. What is the car's final velocity?
Answers
The time of the travel with 20m/s is 17.5 seconds. Thus, the final velocity of the car will be 63 m/s.
What is velocity?Velocity of an object is the measure of the distance travelled per unit time. The rate of change in velocity is called the acceleration.
Given that, initial velocity u = 20 m/s
distance = 350 meters.
time of travel = distance/ velocity
= 350 m/20 m/s = 17.5 s.
Acceleration = 2.5 m/s²
the final velocity v = u + at.
v = 20 + (2.5 × 17.5 ) = 63 m/s.
Therefore, the final velocity of the car is 63 m/s.
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60 m/s while piece c moves in the positive x-direction at 70 m/s. we want to find the speed and direction of piece b.
Answers
The speed and direction of piece b is 11.33m/s positive x direction.
Total momentum = 0 (at rest)
Applying conservation of linear momentum,
Momentum after explosion = momentum before explosion
MaVa + MbVb + McVc = 0
100(-60) + 150Vb + 70(86) = 0
Vb = 11.33m/s
Speed of Vb = 11.33m/s
In Newtonian mechanics, momentum is the result of an object's mass and velocity, more precisely linear momentum or translational momentum. This mechanical quantity is typical. It has. a direction. and a magnitude., making. it a vector. quantity. The momentum of an item is expressed as: [tex]displaystyle mathbf p =m mathbf v.mathbfp = m mathbf v[/tex] if m is the object's mass and v is its velocity (also a vector quantity).
The kilogram metre per second (kg m/s), which is a unit of measurement for momentum in the International System of Units (SI), is a unit of measurement for a newton-second.
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A 2.20-m-long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. Part A What will be the speed of the upper end of the pole just before it hits the ground?
Answers
the speed of the upper end of the pole just before it hits the groundv is = 7.27 m / s
Explanation:
At both the greatest and lowest points, we shall practise energy conservation.
first. The highest point
Em₀ = U
Em₀ = m g y
final. smallest point
At this point, we adjust the reference system so that the potential energy is 0
= K
= ½ m v²
how to conserve energy
Em₀ =
mg y = ½ m v²
v = √2 gy
v = √ (2 9.8 2.7) (2 9.8 2.7)
v = 7.27 m / s
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multiple choice question the ratio that measures the reasonableness of accounts receivable outstanding, and can be used to estimate the average collection period of accounts receivable is the ratio.
Answers
The ratio that measures the reasonableness of accounts receivable outstanding, and can be used to estimate the average collection period of accounts receivable is the accounts receivable turnover ratio.
The amount owed to a business for goods or services that have been delivered or utilized, but for which consumers have not yet made payment is known as accounts receivable, or AR. A current asset is defined as accounts receivable on the balance sheet. Any sum that clients owe for purchases they made with credit is considered AR. Account receivable is the term used to describe any unpaid invoices or cash that a business is owed by customers. Accounts that a company is entitled to get as a result of delivering a good or service are referred to by the phrase. Receivables, also known as accounts receivable, are a type of line of credit that a business extends to customers. Typically, the terms of receivables stipulate that payments must be made within a reasonable amount of time.
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What device can be used to spot the differences between AC and DC?
A. Transformer
B. Generator
C. Motor
D. Oscilloscope
Answers
Answer:
Check explaination
Explanation:
A - Transformer is the right option 100%
A member ABC has a mass of 2.4kg and is attached to pin support at B. An 800 gram sphere D strikes the end of the member ABC with a vertical downward velocity v1 = 3.0 m/s. knowing that L=750 mm and that the coefficient of restitution between the sphere and the Plank is 0.5. Determine immediately after impact:
a) The angular velocity of member ABC.
b)The velocity of the sphere.
Answers
(a)The angular velocity v(d)' is 0.936 m/s
(b)The velocity of the sphere is 3 rad/s
Let us assume that
angular velocity of member ABC is v(d)'
velocity of the sphere = ω'
Write the moment of inertia of the member ABC
I(G) = 1/12 ML²
I(G) = 1/12*(2.4)(0.75m)²
I(G) = 0.112 kgm²
After the impact
ω = ω'(ccw)
ν'(G) = (L/4)ω' (upwards)
ν'(A) = (L/4) ω' (downwards)
Take the moments about B
m(d)v(d)(L/4) + 0 = m(d)v'(d)(L/4) + I(G)ω' + Mv'(G)(L/4)
m(d)v(d)(L/4) = m(d)v'(d)(L/4) + I(G)ω' + M(L/4)²ω'
m(d)v(d)(L/4) = m(d)v'(d)(L/4) + ] I(G) + M(L/4)² ]ω'
(0.8)(3)(0.75/4) = (0.8)v'(d)(0.75/4) + [ 0.112 + (2.4)(0.75/4)²
0.45 = 0.15v(d)' + 0.1963ω' -------------------- (eq1)
From the coefficient of restitution,
v(d)' - v(A)' = -e [ v(d) - v(A) ]
v(d)' - (L/4)ω' = -(0.5) [3-0]
v(d)' - (0.75/4)ω' = - 1.5
v(d)' - 0.1875ω' = - 1.5 ---------------------- (eq 2)
From eq 1 and eq 2
v(d)' = 0.936 m/s
ω' = 3 rad/s
Therefore, the angular velocity is 0.936 m/s and the velocity of the sphere is 3 rad/s.
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a permanent dipole and charged particle lie on the x-axis and are separated by a distance as indicated in the figure. the dipole consists of positive and negative charge separated by a distance where . the solo particle has a negative charge and is located at .
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The vector quantity with a magnitude equal to the product of the two equal charges with opposing polarity separated by a distance, d, is defined as the electric dipole moment associated with it.
The distance between the charges and a direction running along the line between the charges from the negative to the positive charge.
Two charges totaling 10.0 C and spaced 2.00 cm apart make up the dipole. The positive charge is further away from the line of charge than the negative charge, and the dipole's axis forms a 35.0° angle with the x-axis.
The dipole will be subject to a force that is Charges 2.0108C separated by a distance of 2.0103m make up an electric dipole. As seen in the picture, it is positioned so that the negative charge is 2.0 cm away from a long line charge with a linear charge density of 4.0104Cm1. The pressure on the Axis.
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15. A 5.0 g coin sliding to the right at 25.0 cm/s makes an elastic head-on collision with a 15.0 g coin that is initially at rest. After the collision, the 5.0 g coin moves to the left at 12.5 cm/s. Find the final velocity of the other coin. (12.5 cm/s, Right)
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Answer: Finally, we can use this value to find the final velocity of the 15.0 g coin. Since velocity is defined as the change in position over time, and the 15.0 g coin is initially at rest, its final velocity must be equal to its final momentum divided by its mass. Therefore, the final velocity of the 15.0 g coin is 187.5 g cm/s / 15.0 g = 12.5 cm/s. Since the 5.0 g coin is moving to the right and the 15.0 g coin is moving
Explanation: To solve this problem, we can use the conservation of momentum. Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. In an elastic collision, momentum is conserved, meaning that the total momentum of the system before the collision is equal to the total momentum after the collision.
We can use this principle to find the final velocity of the 15.0 g coin. First, we need to calculate the initial momentum of the system. The 5.0 g coin has an initial momentum of 5.0 g * 25.0 cm/s = 125 g cm/s. Since the 15.0 g coin is initially at rest, its initial momentum is 0 g cm/s. The total initial momentum of the system is therefore 125 g cm/s.
After the collision, the 5.0 g coin moves to the left at 12.5 cm/s, so its final momentum is 5.0 g * (-12.5 cm/s) = -62.5 g cm/s. Since momentum is conserved, the final momentum of the system must be equal to the initial momentum of the system, so the final momentum of the 15.0 g coin must be 125 g cm/s - (-62.5 g cm/s) = 187.5 g cm/s.
Finally, we can use this value to find the final velocity of the 15.0 g coin. Since velocity is defined as the change in position over time, and the 15.0 g coin is initially at rest, its final velocity must be equal to its final momentum divided by its mass. Therefore, the final velocity of the 15.0 g coin is 187.5 g cm/s / 15.0 g = 12.5 cm/s. Since the 5.0 g coin is moving to the right and the 15.0 g coin is moving
The nebular theory proposes the sun and planets formed from a rotating cloud of interstellar gases and dust called a
celestial sphere
protoplanet
solar nebula
planetesimal
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The nebular theory proposes the sun and planets formed from a rotating cloud of interstellar gases and dust called a Solar nebula
What is nebular theory?
The creation of solar systems is explained by the nebular theory. According to the explanation, stars are produced from clouds of interstellar gas and dust, and the term "nebula" is Latin for "cloud." It takes around 100 million years for an undifferentiated cloud to develop into a star system with planets and moons.
The nebular theory proposes the sun and planets formed from a rotating cloud of interstellar gases and dust called a solar nebula
According to the nebular theory, the Sun and all of the plants in the solar system descended from the solar nebula, an interstellar cloud of gases and dust. The term "nebula" itself is a Latin word for cloud. The nebula began to collapse over time, and the centre of the collapse (the central mass) eventually turned into a protostar.
It should be noted that this protostar was distinct from our current star in that it lacked the continuous nuclear fusion process that our stars do now. When nuclear fusion begins in the centre of a protostar, it transforms into a star.
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A ship sends sonar and receives an echo after 2 seconds. If the speed of sound waves in water is 1500 m/s, then the depth of the sea is
A 1500 m
B 500 m
C 1000 m
D 750 m
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After 2seconds of sending sonar, a ship hears an echo. The depth of the sea is 1500 meters if the sound wave speed in water is 1500 m/s.
Let sea depth equal d.
Considering that the sound wave's water speed (v) is 1500 m/s
The sound must travel to the ocean's bottom before being reflected back to the ship, resulting in a total distance of two dimensions for the sound.
Time (t) = 2s for receiving the echo
Total distance traveled times the speed of sound
1500 m/s is equal to 2d/2s = 1500 m.
Consequently, the water is 1500 meters deep.
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a block with a mass of m slides at a constant velocty v0 on a horizontal frictionless surface. if the spring constant iis k , what ii the maximm compresion in the spring
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If the block's mass is m, block's initial velocity is v and the spring constant is k, the maximum compression in the spring is sqrt(mv² /k)
Applying the law of conservation of energy, since it is a frictionless surface, all the kinetic energy of the block will be converted to the potential energy of the spring.
The kinetic energy of the block:
Ek = 1/2 . mv²
Where:
m = mass of block
v = block's initial velocity
The potential energy of the spring:
Ep = 1/2 . kx²
Where:
k = spring constant
x = maximum compression in the spring
It holds:
Ek = Ep
1/2 . mv² = 1/2 . kx²
x = sqrt(mv² /k)
Hence, the maximum compression in the spring is sqrt(mv² /k)
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a motorist drives south at 25m/s for 3.00 min, then turns west and travels at 20m/s for 2.60 min, and finally travels northwest at 28m/s for 1.00 min. For this 6.00min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive x axis point east.
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a) D = -4273(x)-2327(y)
b) average speed v = 23.33m/s
c) average velocity V = -13.51m/s south of west.
In order to find the displacement we have to divide the motion on its X and Y components:
dx1 = 25*2*60min = 3000
dx2 = 30*1*60s/min(cos45) = 1273 m
The total X displacement is the algebraic sum of the movement:
dx = -3000 - 1273 = -4273
We have to do the same on Y:
dy1 = 20*3*60 = 3600
dy2 = 30*1*60(sin45) = 1273
dy = -3600 + 1273 =-2327m
D=-4273(x)-2327(y)
The angle is given by:
α = arctg(dy/dx)
α = 28.57 degrees
That is 28.57 degrees south of west.
the total displacement magnitude is given by:
D = √{(4273) - (2327)}
D= 4865
The average speed is given by:
v = (20*3)+(25*2)(1*60)/6*60
v =23.33 m/s
and the Average velocity is given by:
|v| = 4865/6*60
|v| =13.51 m/s
Therefore, the average velocity is V = -1351 m/s with an angle of 28.57 degrees south of west.
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a) D = -4273(x)-2327(y)
b) average speed v = 23.33m/s
c) average velocity V = -13.51m/s south of west.
In order to find the displacement we have to divide the motion on its X and Y components:
dx1 = 25*2*60min = 3000
dx2 = 30*1*60s/min(cos45) = 1273 m
The total X displacement is the algebraic sum of the movement:
dx = -3000 - 1273 = -4273
We have to do the same on Y:
dy1 = 20*3*60 = 3600
dy2 = 30*1*60(sin45) = 1273
dy = -3600 + 1273 =-2327m
D=-4273(x)-2327(y)
The angle is given by:
α = arctg(dy/dx)
α = 28.57 degrees
That is 28.57 degrees south of west.
the total displacement magnitude is given by:
D = √{(4273) - (2327)}
D= 4865
The average speed is given by:
v = (20*3)+(25*2)(1*60)/6*60
v =23.33 m/s
and the Average velocity is given by:
|v| = 4865/6*60
|v| =13.51 m/s
Therefore, the average velocity is V = -1351 m/s with an angle of 28.57 degrees south of west.
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Why is the following situation impossible? Figure shows Superman attempting to drink cold water through a straw of length â„“=12.0 m. The walls of the tabular straw are very strong and do not collapse. With his great strength, he achieves maximum possible suction and enjoys drinking the cold water.
Answers
(a) The maximum height to which Superman can lift the water is 10.32874 m
(b) On the Moon there is no atmosphere so no atmospheric pressure which means when the straw is placed in water water will not rise in the tube.
Pₐ = Atmospheric pressure = 101325 Pa
g = Acceleration due to gravity = 9.81 m/s²
h = Height of water
ρ = Density of water = 1000 kg/m³
If the walls of the tube do not collapse that means that maximum pressure inside will be the atmospheric pressure
Atmospheric pressure is given by
Pₐ = ρgh
height = Pₐ /ρg
h = 101325 / 1000 * 9.81
h = 10.328 m
The maximum height is 10.32874 m.
(b) On the Moon there is no atmosphere so no atmospheric pressure which means when the straw is placed in water water will not rise in the tube.
Therefore, the maximum height to which Superman can lift the water is 10.32874 m.
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[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: Superman attempts to drink water through a very long vertical straw. With his great strength, he achieves maximum possible suction. The walls of the tubular straw don't collapse.
(a) Find the maximum height through which he can lift the water.
(b) Still thirsty, the Man of Steel repeats his attempt on the Moon, which has no atmosphere. Find the difference between the water levels inside and outside the straw.]
The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. Suppose that a horizontal wind blows in streamline flow with a speed of 10.5 m/s outside a large pane of plate glass with dimensions 6.00 m Times 1.10 m. Assume the density of air to be constant at 1.20 kg/m^3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is 21 m/s, twice as high.
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Force is 436.59 N and the speed is 1.746 kN.
A) Force F = P*A = 0.5*rho*v^2*A = 0.5*1.2*10.5^2*6*1.1 = 436.59 N
B) If v is doubled then F increases by 4 times since F is proportional to the square of the speed.
then F2 = 4*436.59 = 1746.35 N = 1.746 kN
A body can change its condition of rest or motion through the application of force, which is an external agent. It has. both a magnitude. and. a direction. The application of force is the point at which force is applied, and the direction in which the force is applied is referred to as the direction of the force. If motion is conceivable, gravitational forces can separate phases of various densities. One option is widespread melting, but high temperatures are not anticipated for the early Earth (Chapter 2). Sorting as a result of deeply buried ices outgassing is a more likely explanation. Such processes would have altered the moment of inertia, necessitating a modification of angular velocity.
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