The angular speed after 10.0 seconds is approximately 2733 rad/s. it takes approximately 24.5 seconds for the rotor to come to rest.
In this problem, we are given the initial angular velocity of a rotor (1940 rad/s) and its angular acceleration (79.3 rad/s^2). We are asked to find the angular velocity after 10.0 seconds and the time it takes for the rotor to come to rest. To find the angular velocity after 10.0 seconds, we can use the formula omega_f = omega_i + alpha*t, where omega_f is the final angular velocity, omega_i is the initial angular velocity, alpha is the angular acceleration, and t is the time elapsed.
Plugging in the given values, we find that the final angular velocity is approximately 2733 rad/s. To find the time it takes for the rotor to come to rest, we set the final angular velocity to 0 and solve for t. We can use the same formula as before: omega_f = omega_i + alpha*t. Plugging in the values for omega_i and alpha, we find that it takes approximately 24.5 seconds for the rotor to come to rest.
In summary, the angular velocity of the rotor after 10.0 seconds is approximately 2733 rad/s, and it takes approximately 24.5 seconds for the rotor to come to rest.
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through a refinery, fuel ethanol is flowing in a pipe at a velocity of 1 m/s and a pressure of 101300 pa. the refinery needs the ethanol to be at a pressure of 2 atm (202600 pa) on a lower level. how far must the pipe drop in height in order to achieve this pressure? assume the velocity does not change. (hint: use the bernoulli equation. the density of ethanol is 789 kg/m3 and gravity g is 9.8 m/s2. pay attention to units!)
Δh = -13.101 m Far must the pipe drop in height in order to achieve this pressure.
13,101 m must be removed from the height.
Bernoulli went on to say that at whatever point along the fluid flow, the total amount of pressure, kinetic energy, and potential energy per volume is the same.
P+ 1/2 ρv² + ρgh = constant
Given
v = 1 m/s ⇒ constant
P₁ = 101300 Pa (1 atm)
P₂ = 202600 Pa (2 atm)
ρ ethanol = 789 kg/m3
g = 9.8 m/s2
Required
The height(lower level)
Solution
The Bernoulli equation becomes when v₁=v₂ constant
ρgh₁+P₁=ρgh₂+P₂
Solve for the value (height difference):
P₁-P₂=ρgh₂-ρgh₁
P₁-P₂=ρg(h₂-h₁)
(P₁-P₂)/(ρg) = Δh
- 101300 kg/ms² / 789 kg/m³ × 9.8 m/s² = Δh
Δh = -13.101 m
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A train is approaching you at very high speed as you stand next to the tracks. Just as an observer on the train passes you, you both begin to play the same recorded version of a Beethoven symphony on identical MP3 players. (a) According to you, whose MP3 player finishes the symphony first? (b) What If? According to the observer on the train, whose MP3 player finishes the symphony first?
(c) Whose MP3 player actually finishes the symphony first?
(a) According to you, your player finishes the symphony first.
(b) According to the observer on the train, the observer's player finishes the symphony first.
Because of the variables in the given problem move relative to one another. The person standing next to the track is travelling at a much slower relative speed than the one seated on the train since the train is moving at a much higher speed. In this scenario, the clock that is located in the train will be operating at a slower rate in comparison to the frame of reference that is considered to be stationary.
So, according to you, your player finishes the symphony first. Ans according to the observer on the train, the observer's player finishes the symphony first.
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the reason that buoyant force acts upward on a submerged object is that question 12 options: if it acted downward, nothing would float. it acts in a direction to oppose gravity. the weight of fluid displaced reacts with an upward force. upward pressure against the bottom is greater than downward pressure against the topof the submerged object.
Upward pressure against the bottom is greater than downward pressure against the top of the submerged object.
Buoyant Force:
The upward force applied to an object that is fully or partially submerged in a fluid is known as the buoyant force. Upthrust is another name for this upward thrust. A body submerged partially or completely in a fluid appears to shed weight or to be lighter, due to the buoyant force.
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shows a particle of mass m at distance x along the axis of a very thin ring of mass m and radius r. calculate the gravitational potential energy of these two masses. Use what you know about the relationship between force and potential energy to find the magnitude of the gravitational force on when it is at position .
When a very thin ring is at position x>0 and at distance x along its axis, the gravitational force acting on it is 4GMm/4x2-L2.
With a gravitational pull that diminishes with distance as 1/r2, every body in the cosmos is attracted to every other body. The gravitational force of attraction a body experiences must be proportionate to its mass if two bodies of different masses accelerate at the same rate (i.e., a body twice as heavy experiences twice the gravitational force).
The gravitational force experienced by a minuscule unit mass placed at any point P in space is known as the gravitational field at that point.
Using the equation gravitational potential = GMm/r and further integrating on a short dx length at x from the rod's origin, we may determine this relationship.
U equals GMm/Lln2x-2/2x+2.
separating now with regard to dx, we obtain
F equals 4GMm/4x2L2.
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In the situation below, ring is stationary in space and initially sits perpendicular to large magnetic field B1 0.59 T. After mT The area of the ring is A 0,61 m2: (a) What the Initial flux, Di when the ring sits In; (b) What the flnal flux, Bi When the ring slt; (c) What is the change flux between the Initial and fina scenarios
When the ring is stationary, the initial flux, Di, is 0.3713. Final flux, Bi, 2.107*10-3Tm If the ring is there and the difference in flux between the initial and final scenarios is 0.3692 Tm2,
The total number of magnetic field lines passing through a coil or other given area is referred to as initial flux. It is the area's typical magnetic field component that travels through. Flux is modelled mathematically as the surface integral of a vector field, where F is the vector field and d A is the vector area of surface A, directed as the surface normal.
The initial flow is 0.97*0.49, which is 0.371 Tm2.
9.3*10-3T*0.49m2 = 2.107*10-3Tm2 is the final flux.
The flux change is calculated as (0.3713 - 2.107*10-3Tm2 = 0.3692Tm2).
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(II) A ski gondola is connected to the top of a hill by a steel
cable of length 800 m and diameter 1.7 cm As the gondola
comes to the end of its run, it bumps into the terminal and
sends a wave pulse along the cable. It is observed that it
took 17 s for the pulse to return. (a) What is the speed of the
pulse? (b) What is the tension in the cable?
FT=8278.857 N In the arteries of a man, the pulse wave moves at a rate of 4 to 10 meters per second.Its velocity is somewhat influenced by the blood flow through the artery under consideration
What is the pulse?Your pulse is how you can feel your heart rate. Every time your heart beats, it squeezes and propels blood through the network of arteries in your body. Your pulse is the pressure in your arteries going up briefly as your heart pushes out more blood to keep circulation going.
Does a pulse's speed alter?The amplitude, length, & pulse width of a pulse have no effect on its speed.The breadth or length of a pulse changing has no effect on the pulse speed.
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At an instant in time, the electric and magnetic fields of an electromagnetic wave are given by E = -9.13 times 10^3k V/m and B = -3.04 times 10^-11i T. Find the Poynting vector for this wave. (Express your answer in vector form.) S = W/m^2
the pointing vector for the given wave is [tex]219.26*10^{-9}j[/tex]
Electromagnetic waves, or EM waves, are waves produced by oscillations between electric and magnetic fields. In other words, EM waves consist of oscillating magnetic and electric fields.
The electric and magnetic fields of electromagnetic waves are perpendicular (right angles) to each other. They are also perpendicular to the EM wave direction.
S = (1/μ[tex]_{o}[/tex] )E*B ,
where
μ[tex]_{o}[/tex] is the vacuum permeability;
E is the electric field;
B is the magnetic flux density.
And,
1/μ[tex]_{o}[/tex] [tex]= 7.9*10^{5} units[/tex]
so,
[tex]S = 7.9*10^{5} (-9.13*10^{-3} ) *(-3.04*10^{-11} )[/tex] ∵ [tex](kxi = j)[/tex] [tex]S = 219.26*10^{-9} j[/tex]
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A 20-N crate starting at rest slides down a rough 3 m long ramp, inclined at 30 degree with the horizontal. The force of friction between crate and ramp is 6 N. What is the kinetic energy of the crate at the bottom of the ramp? Zero 2 J 8 J 12 J 32 J
The velocity of the crate is 12 m/s and kinetic energy of the crate at the bottom of the ramp is 144.06 J.
Given that,
Weight of the crate = 20 N
Length of the ramp = 3 m
Inclination of the ramp with horizontal = 30°
The friction between the ramp and the crate = 6 N
Potential energy of the crate at the top = 20 * 3 sin 30° = 30 J
Energy lost due to friction = 6 N* 3 m = 18 J
Let the speed of the crate at the bottom be v.
Applying the law of conservation of energy,
30 - 18 = 1/2* (20/9.8)* v²
(12* 24)/2.04 = v²
v² = 141.18
v = 11.88 m/s ≈ 12 m/s
The kinetic energy of the crate at the bottom of the ramp = 1/2* m* v²
⇒ 1/2* (20/9.8)* 141.18 = 144.06 J
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A football player pushes a 670-N blocking sled. The coefficient of static
friction
between sled and grass is 0.73, and the coefficient of kinetic friction
between sled and grass is 0.68.
a. How much force must the player exert to start the sled in motion?
b. How much force is required to keep the sled in motion?
c. Answer the same two questions with a 100-kg coach standing on the back
of the sled.
The force must the player exert to start the sled in motion and keep it in motion is 489.1 Newtons and 455.6 Newtons;
Given values,
Weight of blocking sled = 670 N
Coefficient of static friction between sled and grass = 0.73
Coefficient of kinetic friction between sled and grass = 0.68
a) Force required to start the sled in motion = Static friction acting on the sled
= coefficient of static friction * weight
= 0.73 * 670
= 489.1 N
b) Force to keep the sled in motion = kinetic friction acting on the sled
= coefficient of kinetic friction * weight
= 0.68 * 670
=455.6N
c) Now,
Total weight = weight of the sled + weight of the coach
where the weight of the coach = mass of the coach * acceleration due to gravity;
= 670 + 100*9.8
=1650
Force required to start the sled and the coach = Static friction acting on the sled and the coach
= 0.73 * 1650
= 1204.5N
Force req to keep the sled and the coach moving = Kinetic friction acting on the sled and the coach
= 0.68 * 1650
= 1122N
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compare the frequency of an electromagnetic wave with the frequency of the vubrating charge that produces the wave
An electromagnetic wave's frequency is the same as the frequency of the vibrating charge that causes it. The amount of vibrations, or back and forth motions, that occur in a charge in a second, is known as its frequency.
What is meant by electromagnetic waves?Electromagnetic waves, often known as EM waves, are waves that are produced when an electric field and a magnetic field vibrate together. In other words, the oscillation of the magnetic and electric fields results in electromagnetic (EM) waves. In contrast to mechanical waves, electromagnetic waves can move through voids. This suggests that electromagnetic waves can travel into empty space in addition to through solid materials like air and rock.
What is the significance of electromagnetic?Food-borne microorganisms can be eliminated by the widespread use of electromagnetic radiation in food preparation. Recent research on the microwave or radio wave portion of the electromagnetic spectrum has found promising results for the inactivation of germs in food preparation.
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a block of mass 5kg starts from rest at point a and is accelerated by a force of 5 n over a total distance from point a to c as shown. the distance measured from point a to b is exactly 10 meters and is frictionless. the distance from b to c is rough and its length is unknown. the block finally comes to rest at point c. if the coefficient of kinetic friction between the rough surface and the mass is 0.5, determine the total distance the mass slides (from a to c).
The total distance when the mass slides of 5kg starts from rest at point a and is accelerated by a force of 5N is 50 m.
F=μ.d
ma=μ.d
d=ma/μ
d=5×5/0.5
d=50 m
Distance, which is independent of direction, is the length between two points or things. Distance only considers the total magnitude and disregards the start and finish positions because it is a scalar property. Distance can only have a positive or zero value because it is a scalar attribute; it cannot be negative. The meter (m) is the most widely used unit of measurement for distance, though kilometers (km) can also be used to express longer distances and centimeters (cm) or millimeters can be used to express shorter distances (mm). The distance traveled is frequently represented by the letter D when calculating distance.
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A 570 N physics student stands on a bathroom scale in an 900 kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 400 N.Find the magnitude of the acceleration of the elevator.
Part B
Find the direction of the acceleration of the elevator.
up or down
Part C
What is the acceleration if the scale reads 610 ?
Part D
If the scale reads zero, should the student worry? Explain.
Essay answers are limited to about 500 words (3800 characters maximum, including spaces).
part E
What is the tension in the cable in part A?
Part F
What is the tension in the cable in part D?
The acceleration of the elevator when the scale reads 400 N is 0.18m/[tex]s^{2}[/tex] downwards.
The acceleration of the elevator when the scale reads 610 N is 0.04m/[tex]s^{2}[/tex] upwards.
Yes, the student should worry because the elevator is in free fall, that is, there are no supporting cables on the elevator.
The tension in the cable for part A is 450 N.
The tension on the cable for part D is 0.
The given parameters;
weight of the student, W = 570 Nmass of the student and the elevator, m = 900 kgWhen the scale of the elevator reads 400 N, is moving downwards since the apparent weight is less than the real weight
R = mg + ma
400 = 570 + ma
400 - 570 = ma
- 170 = ma
a = [tex]\frac{-170}{m}[/tex] = [tex]\frac{-170}{900}[/tex] = 0.18m/[tex]s^{2}[/tex] , downwards
so, The acceleration of the elevator when the scale reads 400 N is 0.18m/[tex]s^{2}[/tex] downwards.
The scale of the elevator will read more when it is moving upwards;
610 = mg + ma
610 = 570 + ma
610 - 570 = ma
ma =40
a = [tex]\frac{40}{m}[/tex] = [tex]\frac{40}{900}[/tex] = 0.04m/[tex]s^{2}[/tex] , upwards
So, the acceleration of the elevator when the scale reads 610 N is 0.04m/[tex]s^{2}[/tex] upwards.
When the scale reads, the upward acceleration is equal to the acceleration due to gravity and the student will be in free fall.
R = m(g - a)
g = a
R = m(0)
R = 0
Thus, the student should worry because the elevator is in free fall, that is, there are no supporting cables on the elevator.
The tension in the cable for part A is 400 N.
The tension in the cable for part D is 0.
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which statement about saturn's rings is not true? which statement about saturn's rings is not true? the rings must look much the same today as they did shortly after saturn formed. some features of the rings are shaped by small moons that actually orbit within the ring system. the rings are so thin that they essentially disappear from view when seen edge-on. the large gap known as the cassini division is shaped by an orbital resonance with the moon mimas, which orbits well outside the rings.
The rings must look much the same today as they did shortly after Saturn formed.
The Rings of Saturn:
The Saturnian ring system is the largest ring system in the Solar System. They are made up of innumerable tiny particles that circle Saturn and range in size from micrometers to meters. With a very small amount of stony material, the ring particles are virtually exclusively formed of water ice. Regarding how they form, there is still no agreement. Newer evidence from Cassini indicates that the rings developed relatively late, despite theoretical models suggesting they were likely to have formed early in the Solar System's history.
It is believed that the fragments of comets, asteroids, or shattered moons that broke apart before they reached Saturn and were blasted apart by the planet's strong gravity make up Saturn's rings. They are composed of countless quantities of tiny pieces of rock and ice that have been dusted with other substances.
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you are standing at the top of a cliff that has a stairstep configuration. there is a vertical drop of 6 m at your feet, then a horizontal shelf of 6 m , then another drop of 4 m to the bottom of the canyon, which has a horizontal floor. you kick a 0.16 kg rock, giving it an initial horizontal velocity that barely clears the shelf below.1. What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? The acceleration of gravity is 9.8 m/s2 . Consider air friction to be negligible. Answer in units of m/s.
2. How far from the bottom of the second cliff will the projectile land? Answer in units of m.
1. The initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.
2. The projectile will land at 9.54 m of the second cliff d = 9.54 m
1. To find the horizontal velocity of the rock we need to use the following equation:
d = v * t
Where:
d: is the distance traveled by the rock
t: is the time
The time can be calculated as follows:
t = √2d/g
g: is gravity = 9.8 m/s²
t = √2d/g = 1.20s
Now, the horizontal velocity of the rock is:
v = d/t = 6.67m/s
Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.
To calculate the distance at which the projectile will land, first, we need to find the time:
t = √2d/g = 1.43s
the distance is:
d = v * t = 9.54m
Therefore, the projectile will land at 9.54 m of the second cliff.
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two possible geometric shapes for the clf3 molecule are trigonal planar and t-shaped. the student does some research and learns that the molecule has a dipole moment. which of the shapes is consistent with the fact that the clf3 molecule has a dipole moment? justify in terms of bond polarity and molecular structure.
The trigonal planar shape is consistent with the fact that the CLF3 molecule has a dipole moment.
This shape is formed by three identical polar bonds that are arranged in a flat plane. Each of the three bonds has a slightly negative charge on the chlorine atom and a slightly positive charge on the fluorine atoms. This arrangement of charge creates a permanent dipole moment across the molecule, which is a result of the arrangement of the polar bonds and the molecular structure.
The t-shaped structure, on the other hand, has a single polar bond and two non-polar bonds, which would lead to a net dipole moment of zero, since the charges would cancel out. Therefore, the trigonal planar shape is the most consistent with the fact that the CLF3 molecule has a dipole moment.
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one study gave several suggestions for how much imagery an athlete should practice. which of these was not one of the suggestions?
The one that is not a suggestion is the number of repetitions should remain the same regardless of the sport's skill. The correct option is c.
Who is an athlete?An athlete is someone who practices for and participates in sporting activities, either professionally or just for enjoyment. Examples of athletes include those who bowl for a team, run in neighborhood 5K races or take home an Olympic gold medal.
The biggest advantages of imagery are seen in tasks with mostly cognitive components for increasing the learning and performance of motor abilities.
Therefore, the correct option is c, The number of repetitions should remain the same regardless of the sport skill.
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The question is incomplete. Your most probably complete question is given below:
There should be three to four sessions per week.
Sessions should be around 15 minutes in length.
The number of repetitions should remain the same regardless of the sport skill.
If concentration wanes, break imagery sessions into shorter blocks.
Imagery practice should be systematic.
A uniform30 cmruler of mass50 gis attached to a friction-less vertical wall by a nail through one end. A horizontal force F=0.545 Nis applied to the end of the rod. If the rod is tilted at 40∘ from the vertical, what is the angular acceleration of the rod (around the nail) at this instant? Please enter a numerical answer below. Accepted formats are numbers or "e" based scientific notation e.g.0.23,−2,1e6, 5.23e-8
The value of the angular acceleration of the rod is approximately equal to 0.0055 rad/s².
To find the angular acceleration of the rod, we need to first determine the torque acting on the rod around the nail. The torque is given by the formula:
torque = force * distance * sin(angle)
Where: 'force' is the applied force F, 'distance' is the distance between the point where the force is applied and the point around which the torque is calculated (in this case, the nail) and 'angle' is the angle between the force and the distance vector
Plugging in the values, we get:
torque = 0.545 N * 30 cm * sin(40°) = 4.14 N cm
We can then use the formula for torque to calculate the angular acceleration:
torque = moment of inertia * angular acceleration
The moment of inertia of a rod about an end is given by:
moment of inertia = mass * length² / 12
Plugging in the values, we get:
moment of inertia = 50 g * (30 cm)² / 12 = 750 g*cm^2
Solving for angular acceleration, we get:
angular acceleration = torque / moment of inertia = 4.14 N cm / 750 gcm² = 0.0055 rad/s².
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A newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its surface, in terms of the free-fall acceleration g at the surface of the earth?
A. 2/9 g
B. 2/3 g
C. 3/4 g
D. 4/3 g
A newly discovered planet has twice the mass and three times the radius of the earth. The free-fall acceleration at its surface, in terms of the free-fall acceleration g at the surface of the earth is 2/9 g
What is Gravitational Acceleration?The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s² is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration or free-fall acceleration . At sea level, the standard value of g on earth's surface is 9.8 m/s².
The free fall aceleration can be calculate as follows:
The acceleration due to gravity on Earth is
g= GM
R²
The acceleration due to gravity on the new planet is
gₙ= G2M
(3 R)²
gₙ= G2M
(3 R)²
Dividing the two equations we get
G2M
gₙ = 9 R² = 2 g
g GM 9
R²
Therefore, the free-fall acceleration at its surface, in terms of the free-fall acceleration g at the surface of the earth is 2/9 g.
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an object is intiially hanging in equilibrium froma vertical spring. the object is pulled down 15 cm
The item is lowered 15 cm from its equilibrium point shown above before being let go at time t = 0. The item subsequently begins to oscillate with a 2.0 s period. Assuming that x = 0 represents equilibrium, upward motion is the positive direction.
What does a "object response" mean?A sentence's object is the portion on which the verb performs its action. In conventional English, a sentence is usually written with the object coming first. The verb, or what performs the action, is the subject.
What functions do things have?Things may be done to objects as well as done to them. A function or method object, for instance, may be coded to change the contents of a data structure or variable.
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if an object is at a pole, it blank accelerating.target 1 of 6 we can truly write blank, such that the apparent weight is exactly mg .target 2 of 6 at the blank, the object is moving in a very large circular path as the earth rotates.target 3 of 6 this means that technically blank.target 4 of 6 that is, the weight force toward the center of the circular path must be slightly blank so it can provide a centripetal force.target 5 of 6 thus n
The answer related to Equator of following are : 1) is not, 2) n – mg = may = 0, 3) equator, 4) n – mg = may < 0, 5) larger, 6) equator
What is Equator?
The equator, an invisible circle that spans the centre of the planet and runs parallel to it at its widest point, is located halfway between the North and South Poles. The areas of Earth where the equator crosses are among the warmest because they are closest to the sun.
The equator exists in other places besides just Earth. There is an equator on each planet and star in the sky. An invisible line that lies evenly between the two poles of each rotating planet or star, including the sun in our solar system, is referred to as the equator.
The majority of the equator travels across the ocean, as seen in the image below.
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A special type of electro-
magnetic wave that has a
frequency which we can see:
T
iuf the angle between the horizintaon lad tje direction of the 5.0 meters per second vecpol;tiy decreases from 30 degrees to 30 degrees, the horitzontal distance the ball travels will
The horizontal distance the ball travels is 2.209 meters ,if it has velocity of 4 m/s
Given,
velocity (v) = 5 m/s
angle (Θ) = 30+30 = 60
gravitational constant (g) = 9.8 m/s
We know that the distanced travelled by a ball can be find using the below formulae
distance (D) = v² sin2Θ / g
D = 5²×sin 120 / 9.8
D = 2.209 meters
Aerial distance is the distance between two points or locations that is calculated in the air, as opposed to the ground or a road, which is the route that an aeroplane takes to get from one location to another.
Distance in the air must be measured when flying since it is crucial information for the aviation industry. The nautical mile is a measurement that is widely used to compute distances in space.
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Will a piece of paper produce the same kind of reflection as the mirror? Explain your answer.
Answer:
A piece of paper will not produce the same kind of reflection as a mirror. A mirror is a highly polished, smooth surface that is able to reflect light in a way that produces a clear and detailed image. A piece of paper, on the other hand, is a rough and porous surface that is not able to reflect light in the same way. When light hits a piece of paper, it is absorbed by the fibers and absorbed, scattered, and absorbed again in multiple directions, which produces a diffuse and distorted reflection. Therefore, a piece of paper will not produce the same kind of reflection as a mirror.
water flows through the pipe contraction shown. for the given 0.2-m difference in the manometer level, determine the flowrate as a function of the diameter of the small pipe, d.
The flow rate as a function of diameter is Q^2 = 2.24 × 10^-4×D^4 / 10^-4 - D^4 when 0.2 m difference in the manometer level.
Given: The pipe's diameter at point 1 isD1=0.1 m D1=0.1 m
D2=D at point 2 dia. of pipe
And the pipe's flow rate is Q.
Velocity at point 1, where V1=Q/D12/4.
V1= Q/π×(0.1) (0.1)
^2/ 4
Speed at point 2; V2=Q/D2
^2/ 4
As atmospheric pressure causes pipes 1 and 2 to both open, both are 0; 1 and Bernoulli equation application at point 2
P1/pg+v12/2g+ h1= P2/pg+ v22/2g+ H2, where v12 / 2g+ h1= v2 / 2g+ H2.
Put the velocity and height values in the equation above.
Q^ 2/( π/4)^2×(1/D^4-l/(0.1)^4)= 2×g×(h1-h2) (h1-h2)
Q^2 = 2.24 × 10^-4×D^4 / 10^-4 - D^4
This represents the flow rate as a function of diameter.
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At which latitude on the Earth is the speed of Earth's rotation the slowest?
Answer:
North and South poles
Explanation:
a full moon appears larger on the horizon than it does overhead not because of the refraction of the atmosphere but because of an optical illusion. objects in the foreground influence the perceived size of objects in the background, and when the moon is on the horizon amid trees and houses the brain interprets it as being larger than when it appears overhead in the absence of those objects.
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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To understand the basic principles underlying interference.
One of the most important properties of waves is the principle of superposition. The principle of superposition for waves states that when two waves occupy the same point, their effect on the medium adds algebraically. So, if two waves would individually have the effect "+1" on a specific point in the medium, then when they are both at that point the effect on the medium is "+2." If a third wave with effect "-2" happens also to be at that point, then the total effect on the medium is zero. This idea of waves adding their effects, or canceling each other's effects, is the source of interference.
Part C
In talking about interference, particularly with light, you will most likely speak in terms of phase differences, as well as wavelength differences. In the mathematical description of a sine wave, the phase corresponds to the argument of the sine function. For example, in the function y=Asin(kx), the value of kx at a particular point is the phase of the wave at that point. Recall that in radians a full cycle (or a full circle) corresponds to 2? radians. How many radians would the shift of half a wavelength from the previous part correspond to?
Express your answer in terms of ?.
Shift of half a wavelength from the previous part correspond to π radian.
What is the principle of interference?Interference is a phenomenon in which two waves combine by adding displacement together at every single point in the space and time, to form a resultant wave of greater, lower, or same amplitude.
The superposition principle explains that when two or more waves overlap in space, then the resultant disturbance is equal to the algebraic sum of individual disturbances
Given, Wavelength= λ
We have to find the phase difference when wavelength is half of the initial wavelength.
So, the path difference= Δx = λ/2
As, phase difference= 2π/λ * Δx
So, phase difference= 2π/λ * λ/2
Thus, phase difference= π radian
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A student wants to increase the temperature of an ideal gas in a cylinder that is initially at room temperature. The cylinder has a movable piston with negligible friction. Which of the following correctly indicates an action the student could take to increase the temperature of the gas and why the temperature increases? Pull the poston out and Then wit expand violently and increase in temperature Pull the piston out slowly. The as will have time to absorb encesy from the surroundings by heatinst as it expands. Push the piston in rapidly. The temperature of the gas will increase because no enery will be transferred to the surroundinis bw coolinu Push the piston in slowtv: The gas will have time to absorb energy from the surroundings by heating as it is compressed
In order to increase the temperature of an ideal gas in a cylinder that is initially at room temperature, the statement that correctly indicates the appropriate action is to push the piston in rapidly. The temperature of the gas will increase because no energy will be transferred to the surrounding by cooling. (Option C)
An ideal gas refers to a theoretical gas which is composed of many randomly moving point particles that are not subject to interparticle interactions. The concept is useful as it obeys the ideal gas law, which is a simplified equation of state, and is amenable to analysis under statistical mechanics. For an ideal gas, as the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container. The force exerted by the particles per unit of area on the container is the pressure, so as the temperature increases the pressure must also increase. Hence, when the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly.
Note: The question is incomplete. The complete question probably is: A student wants to increase the temperature of an ideal gas in a cylinder that is initially at room temperature. The cylinder has a movable piston with negligible friction. Which of the following correctly indicates an action the student could take to increase the temperature of the gas and why does the temperature increases? A) Pull the piston out rapidly. The gas will expand violently and increase in temperature B) Pull the piston out slowly. The gas will have time to absorb energy from the surroundings by heating as it expands. C) Push the piston in rapidly. The temperature of the gas will increase because no energy will be transferred to the surrounding by cooling. D) Push the piston in slowly. The gas will have time to absorb energy from the surroundings by heating as it is compressed
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A sphere of mass mis dropped from the top of a building and reaches the ground before achieving terminal velocity. The force of air resistance that acts on the sphere as it falls is given by F, where is a positive constant and is the velocity of the sphere. What happens to the magnitude of the sphere's velocity and acceleration, and to the distance it falls during each second, as the sphere approaches the ground? Magnitude of Yelocity Magnitude of Acecloration Distance of all Each Second (A) Increases Increases Increases Increases Decreases Increases (C) Increases Decreases Decreases Decreases Increases Decreases (1) Decreases Decreases Increases
The magnitude of the sphere's velocity, acceleration, and to the distance it falls during each second is A)Increases Increases Increases .So, correct option is A.
We know that when an object is coming from a large height towards the surface of earth, due to air resistance it feels huge amount of opposition force which basically tries to decrease its velocity. The equation of force will be given by
F=mg + air resistance force
Since acceleration due to gravity is also responsible for acting of force, therefore, F will increase as when object comes closer to the surface air resistance is very high and due to that value of F is very large.
Same because of the high resistance force, velocity and distance will be also increases.
Hence, correct option is A.
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Part A In addition to their remarkable top speeds, cheetahs have impressive cornering abilities. In one study, the sideways force on the feet of a 45 kg cheetah, with its claws dug into the ground, was found to be 960 N What was the cheetah's centripetal acceleration? Express your answer with the appropriate units. HÃ… ? a = Value Units Submit Request Answer Part B What was the radius of the turn if the cheetah was running at 20 m/s? Express your answer with the appropriate units. HA ?
Centripetal acceleration of cheetah is 21.3 m/s^2. the radius of the turn was approximately 18.9 meters.
To find the centripetal acceleration, we can use the formula a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the turn.
In this case, we are given the mass of the cheetah (45 kg) and the sideways force on its feet (960 N), so we can use the formula F = ma to find the acceleration.
F = ma 960 N = (45 kg)(a) a = 960 N / 45 kg a = 21.3 m/s^2
Now that we have the acceleration, we can use the formula a = v^2/r to find the radius of the turn if the cheetah was running at 20 m/s.
a = v^2/r 21.3 m/s^2 = (20 m/s)^2/r r = (20 m/s)^2 / 21.3 m/s^2 r = 18.9 m
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