The efficiency of the inclined plane is 40%.
The efficiency of the inclined plane can be calculated by dividing the output work by the input work and multiplying by 100% to get a percentage.
Efficiency = (Output work / Input work) x 100%
In this case, the input work is 25 J and the output work is 10 J.
Efficiency = (10 J / 25 J) x 100%
Efficiency = 0.4 x 100%
Efficiency = 40%
Therefore, the efficiency of the inclined plane is 40%.
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An astronaut weighs 8.00 × 102
newtons on the
surface of Earth. What is the weight of the astronaut
6.37 × 106
meters above the surface of Earth?
The weight of the astronaut 6.37 × 106 meters above the surface of the Earth would be 160 N.
Weight of an astronautThe weight of an object depends on its mass and the gravitational field it is in. Near the surface of the Earth, the acceleration due to gravity is approximately 9.81 m/s².
We can use the formula F = mg to calculate the weight of the astronaut at the surface of the Earth:
The gravitational field weakens as the distance from the center of the Earth increases, thus, we can use the formula for gravitational acceleration at a distance r from the center of the Earth:
g' = g (R / (R + h))²g' = 9.81 m/s² x (6.37 × 106 m / (6.37 × 106 m + 6.37 × 106 m))²g' = 1.96 m/s²Now we can calculate the weight of the astronaut at this height:
F' = mg'F' = mass x (1.96 m/s²)We don't know the mass of the astronaut, but we can use the weight at the surface of the Earth to find it:
F = mgm = F / g= 8.00 × 102 N / 9.81 m/s²
= 81.63 kg
F' = (81.63 kg) x (1.96 m/s²)
F' = 160 N
Therefore, the weight of the astronaut 6.37 × 106 meters above the surface of the Earth is approximately 160 N.
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A crate (60 kg) is in an elevator traveling upward and slowing down at 6 m/s2. find the normal force exerted on the crate by the elevator. assume g
The normal force exerted on the crate by the elevator is 294 N. The normal force is the force exerted by a surface perpendicular to an object in contact with it.
In this case, the crate is in contact with the floor of the elevator. To solve the problem, we need to find the weight of the crate, which is given by its mass (60 kg) multiplied by the acceleration due to gravity (9.8 m/s2).
So the weight of the crate is 588 N. The force exerted on the crate by the elevator is the normal force.
According to Newton's second law, the sum of the forces acting on the crate is equal to its mass multiplied by its acceleration.
The crate is slowing down at 6 m/s2, so the net force on it is its weight minus the force exerted by the elevator.
Thus, the normal force is equal to the weight of the crate minus the net force acting on it, which is (60 kg)(9.8 m/s2) - (60 kg)(6 m/s2) = 294 N. Therefore, the normal force exerted on the crate by the elevator is 294 N.
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Two electrons (-1.6 x 10-1°c) in an atom are separated by 3.4 x 10-11 m. what is the electrostatic force
between them? is it attractive or repulsive?
The electrostatic force between the electrons is approximately 2.31 x 10⁻²⁸ N, acting in a repulsive manner.
To find the electrostatic force between the two electrons, we will use Coulomb's Law, which states that the electrostatic force (F) between two charged particles is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. Mathematically, this is represented as:
F = (k * q1 * q2) / r²
Where k is Coulomb's constant, approximately equal to 8.99 x 10⁹ Nm²/C². In this problem, both electrons have a charge of -1.6 x 10⁻¹⁹ C, and they are separated by a distance of 3.4 x 10⁻¹¹ m. Plugging these values into the equation, we get:
F = (8.99 x 10⁹ Nm²/C² * (-1.6 x 10⁻¹⁹ C) * (-1.6 x 10⁻¹⁹ C)) / (3.4 x 10⁻¹¹ m)²
Calculating the force:
F ≈ 2.31 x 10⁻²⁸ N
Since both electrons have negative charges, the electrostatic force between them is repulsive. This is because like charges repel each other, while opposite charges attract. In this case, the two electrons have the same negative charge, which causes them to repel one another.
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When a wind-up toy is set in motion, elastic potential energy that was stored in a compressed spring is converted into the __________ of the toy’s moving parts
Answer:When a wind-up toy is set in motion, elastic potential energy that was stored in a compressed spring is converted into the kinetic energy of the toy's moving parts.
Explanation:
a 55 kg man runs at a speed of 4 m/s. find his kinetic energy
Answer:
EK = 440 Joule
Explanation:
Known:
m = 55 kg
v = 4 m/s
Ek = ?
Equation to solve this is:
Ek = 1/2 m [tex]v^{2}[/tex]
Ek = 1/2 . (55) . [tex](4)^{2}[/tex]
Ek = 440 J
Which class of fires consists of flammable liquids including.
Answer:
class b
Explanation:
flamabal liquids , gas alcohol or oil
A solenoid had 1040 turns and is 4.4 cm long. If it carries a current of 31 A, what is the magnetic field it creates?
Answer:
.92 T
Explanation:
This is just a plug-and-chug question.
Here is the formula: B = uni
u = vacuum permeability = 4pi * 10^-7 this is a given constant
n = turns per meter = 1040/ (4.4*10^-2)
i = current = 31 A also given by the problem
so B = .92 T
The unit of the magnetic field is Tesla ("T")
If a galaxy is receeding from the earth at 560,000 mi/hr, how far away is it from earth?
The distance of the galaxy from Earth is approximately 34 Mpc or 111 million light-years away.
The distance of the galaxy from Earth can be calculated using Hubble's law, which states that the recessional velocity of a galaxy is proportional to its distance from Earth.
The proportionality constant is known as the Hubble constant, denoted by H. The current estimated value of the Hubble constant is approximately 73.3 km/s/Mpc.
To convert the given velocity from miles per hour to kilometers per second, we need to divide it by 2.237 × 10^5. Thus, the recessional velocity of the galaxy in km/s is approximately 2510 km/s.
Using Hubble's law, we can calculate the distance of the galaxy from Earth by dividing its velocity by the Hubble constant.
Therefore, the distance of the galaxy from Earth is approximately 34 Mpc or 111 million light-years away.
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A motorcyclist traveling due north at 50km/h. the wind appears to come from north west at 60km/h . what is the true velocity of the wind
The true velocity of the wind is approximately 43.10 km/h.
To find the true velocity of the wind when a motorcyclist is traveling due north at 50 km/h, and the wind appears to come from the northwest at 60 km/h, we can use vector addition.
Step 1: Break the wind's apparent velocity into its north and west components. Since the wind is coming from the northwest, the north and west components will be equal.
Using the Pythagorean theorem (a² + b² = c²) to find the components:
North component:
a = 60 * cos(45°)
= 60 * 0.707
= 42.43 km/h
West component:
b = 60 * sin(45°)
= 60 * 0.707
= 42.43 km/h
Step 2: Subtract the motorcyclist's northward velocity from the north component of the wind's apparent velocity:
True north component of the wind:
42.43 - 50 = -7.57 km/h (southward)
Step 3: Combine the true north and west components of the wind's velocity using the Pythagorean theorem:
True wind velocity = √((-7.57)² + (42.43)²)
= √(57.36 + 1800.06)
= √1857.42
≈ 43.10 km/h
The true velocity of the wind is approximately 43.10 km/h.
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ightning is an electrostatic discharge between two electrically charged regions that allows electrons in a negatively charged region to flow back to the positive region. how did these regions in thunderstorms get oppositely charged to begin with?
The process of charge separation, driven by updrafts and downdrafts in a thunderstorm, causes regions within the storm cloud to become oppositely charged, leading to the Electrostatic discharge known as lightning.
Lightning occurs due to electrostatic discharge between two electrically charged regions within a thunderstorm. These regions become oppositely charged through a process called charge separation.
Charge separation begins when updrafts and downdrafts within a thunderstorm cause ice particles, hail, and water droplets to collide. During these collisions, electrons are transferred between particles, resulting in some particles becoming positively charged while others become negatively charged.
The lighter, positively charged ice particles are carried upward by the updrafts, accumulating at the top of the storm cloud. Conversely, the heavier, negatively charged particles, such as hail, are carried downward by gravity and downdrafts, accumulating at the base of the cloud.
This separation of charges creates an electric field between the top and bottom regions of the cloud. When the electric field becomes strong enough, it overcomes the air's insulating properties, allowing electrons to flow from the negatively charged region to the positively charged region. This flow of electrons results in a lightning discharge.
In summary, the process of charge separation, driven by updrafts and downdrafts in a thunderstorm, causes regions within the storm cloud to become oppositely charged, leading to the electrostatic discharge known as lightning.
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A 1. 5\,\text {kg}1. 5kg1, point, 5, start text, k, g, end text mass attached to an ideal spring oscillates horizontally with an amplitude of 0. 50\,\text m0. 50m0, point, 50, start text, m, end text. the spring constant is 85\,\dfrac{\text n}{\text m}85 m n 85, start fraction, start text, n, end text, divided by, start text, m, end text, end fraction
This scenario represents a simple harmonic motion with a specific period, frequency, maximum velocity and acceleration.
The given scenario describes a mass oscillating horizontally attached to an ideal spring with a spring constant of 85 N/m and an amplitude of 0.50 m. The ideal spring is assumed to have no mass, damping or friction, and therefore it is a simple harmonic oscillator.
The period of oscillation is calculated as T = 2π√(m/k)
where m is the mass and k is the spring constant.
Substituting the given values, we get T = 2π√(1.5/85) = 0.449 s.
The frequency of oscillation is f = 1/T = 2.23 Hz.
The maximum velocity of the mass can be found using the equation vmax = Aω,
where A is the amplitude and ω is the angular frequency.
Substituting the given values, we get vmax = 0.50 × √(k/m) = 0.50 × √(85/1.5) = 5.06 m/s.
The maximum acceleration of the mass can be found using the equation amax = Aω^2 = 0.50 × (2πf)^2 = 7.96 m/s^2.
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Two new materials have been discovered. One is shiny
and has a metallic look, while the other is dull and
has a non-metallic look. Although you think that one
is a conductor and the other an insulator, you want
to be certain. Describe a test you could do to test the
conductivity of these two materials.
Answer:
placing an object between two free ends of a wire in a circuit. the circuit must have a bulb, wires, crocodile clips and a switch (optional) .
Explanation:
get a material between the clips . if the bulb lights up, the object is a conductor and is it doesn't its an insulator.
A landscaper uses 15. 00 newtons of force to push a lawn mower. How much work, in joules, does the landscaper use to move the lawn mower?
The landscaper uses 75.00 joules of work to move the lawn mower.
Work is the product of force and displacement, in the direction of the force.
Given that the landscaper uses a force of 15.00 N to push a lawn mower, the amount of work done depends on the distance the mower is pushed.
If we assume that the mower is pushed a distance of 5 meters, the work done can be calculated as follows:
Work = force x distance x cos(theta)
where theta is the angle between the force and the direction of displacement, which we assume to be zero degrees in this case. Therefore, the work done can be calculated as:
Work = 15.00 N x 5 m x cos(0) = 75.00 J
Therefore, the landscaper uses 75.00 joules of work to move the lawn mower.
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Estimate how long dr.mann went without human contact while on the ice planet:
the lazarus mission was ___ years ago. it takes ____ years to get to saturn from earth. copper and dr. brand were on miller’s planet for approximately _____ years. this means that the total time dr. mann went without human contact is: _____ years.
The Lazarus mission was mentioned to be around 10 years ago in the movie Interstellar. It takes approximately 6 years to travel from Earth to Saturn. Copper and Dr. Brand were on Miller's planet for approximately 23 years. This means that the total time Dr. Mann went without human contact is estimated to be around 33 years.
It is stated in the movie that it takes approximately 6 years to travel from Earth to Saturn, where the wormhole that leads to other galaxies was located. This indicates that the Lazarus mission likely took 6 years to reach Saturn from Earth.
During the Lazarus mission, two of its members, Dr. Mann and Dr. Brand, were sent to explore separate planets that were potentially suitable for human colonization. Dr. Mann was sent to a planet named Mann's planet, while Dr. Brand was sent to Miller's planet.
On Miller's planet, time was significantly dilated due to its proximity to a massive black hole, known as Gargantua. For every hour that passed on the planet, approximately 7 years passed outside the planet's gravitational influence.
This means that the time Dr. Brand and the other crew members spent on Miller's planet felt like 23 years had passed for the rest of the universe.
Considering the 6-year journey to Saturn, the time spent on Miller's planet, and the 10 years that had already passed since the Lazarus mission, the total time that Dr. Mann went without human contact can be estimated to be around 33 years.
This is derived from adding the 6-year journey, the 23 years on Miller's planet, and the 10 years prior to the Lazarus mission.
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on page 1 you had to draw the direction the earth moved without gravity in position 1 and 2. what was different about the path the earth took when there was not any gravity? * 5 points the earth moved in a straight line and did not move around the sun the orbit of the earth was faster the earth moved around the sun but travelled in a straight line the orbit of the earth stayed the same when you increased the mass of the sun, what happened to the gravity force? * 5 points it increased it stayed the same it decreased when you increased the distances, what happened to the gravity force? * 5 points it decreased it increased it stayed the same when you decreased the distances, what happened to the gravity force? * 5 points it increased it decreased it stayed the same when you increased the gravity, what happened to orbital speed? * 5 points it stayed the same it decreased it increased what are the 2 factors that affect the force of gravity? * 5 points the speed and size (mass) of an object the shape of the orbit and the distance between the objects nothing can change gravity because it is a natural force the size (mass) and distance between the objects
Changes in mass, distance, and gravity affect the force of gravity and orbital speed; the force of gravity is directly related to the size (mass) and distance between objects.
Changes in mass and distance affect the force of gravity and orbital speed: increasing mass or decreasing distance increases the force of gravity and orbital speed, while decreasing mass or increasing distance decreases them. The force of gravity is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. The two factors that affect the force of gravity are the size (mass) and distance between the objects.
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--The complete question is, What are the effects of changes in mass, distance, and gravity on the force of gravity and orbital speed? What is the relationship between the force of gravity and the size and distance of the objects?--
A 5.10 kg cast-iron skillet is heated on the stove from 295 k to 450 k. how much heat had to be transferred to the iron (specific heat of iron is 450j/kg k)?
The amount of heat transferred to the cast-iron skillet is approximately 351,450 J.
To calculate the amount of heat transferred to the cast-iron skillet, we can use the formula:
Q = m * c * ΔT
where:
Q is the heat transferred,
m is the mass of the skillet,
c is the specific heat capacity of iron, and
ΔT is the change in temperature.
Given:
m = 5.10 kg (mass of the skillet)
c = 450 J/(kg*K) (specific heat capacity of iron)
ΔT = 450 K - 295 K (change in temperature)
Let's calculate the heat transferred:
Q = (5.10 kg) * (450 J/(kg*K)) * (450 K - 295 K)
Q = 5.10 kg * 450 J/(kg*K) * 155 K
Q ≈ 351,450 J
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At t = 0 what are the two smallest positive values of x for which the probability function |ψ(x,t)|2 is a maximum?
The two smallest positive values of x for which [tex]\left|\Psi(x,t)\right|^2[/tex] is a maximum at t=0 are π/2k and 3π/2k.
To find the values of x for which the probability function [tex]\left|\Psi(x,t)\right|^2[/tex] is maximum at t=0, we need to calculate [tex]\left|\Psi(x,t)\right|^2[/tex] and find its maximum values.
The probability density [tex]\left|\Psi(x,t)\right|^2[/tex] gives the probability of finding the particle at position x at time t. In this case, the wave function is given by:
[tex]\Psi(x,t) = A\left[e^{i(kx-\omega t)}-e^{i(2kx-4\omega t)}\right][/tex]
So, the probability density is:
[tex]\left|\Psi(x,t)\right|^2 &= A^2 \left[e^{i(kx-\omega t)} - e^{-i(kx+\omega t)}\right]\left[e^{-i(kx+\omega t)} - e^{i(kx-\omega t)}\right]\&= A^2 \left[2 - 2\cos(2kx-4\omega t)\right][/tex]
Now, at t=0, the probability density reduces to:
[tex]\left|\Psi(x,0)\right|^2 = A^2 \left[2 - 2\cos(2kx)\right][/tex]
We want to find the two smallest positive values of x for which [tex]\left|\Psi(x,0)\right|^2[/tex] is the maximum. Since cos(2kx) varies between -1 and 1, [tex]\left|\Psi(x,0)\right|^2[/tex] varies between 0 and [tex]4A^2[/tex].
To find the maximum values of [tex]\left|\Psi(x,0)\right|^2[/tex], we need to find the values of x where cos(2kx) takes its minimum values. The minimum value of cos(2kx) is -1, which occurs when 2kx = (2n+1)π, where n is an integer.
Thus, the two smallest positive values of x for which [tex]\left|\Psi(x,0)\right|^2[/tex] is maximum are given by:
2kx = π and 2kx = 3π
So, the values of x are:
x = π/2k and x = 3π/2k
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Ferris wheel has a diameter of 76 m and completed one revolution every 20 min.
a)Calculate the tangential speed the car
b) Calculate the magnitude to the centripetal acceleration of one of the car
The tangential speed of a point on the Ferris wheel is approximately 2.01 m/s. the magnitude of the centripetal acceleration of a point on the Ferris wheel is approximately 0.106 m/s².
The tangential speed of a point on the Ferris wheel is given by the formula:
v = (2πr) / T
where v is the tangential speed, r is the radius of the Ferris wheel (half the diameter), and T is the time taken to complete one revolution.
In this case, the diameter of the Ferris wheel is 76 m, so its radius is 38 m. It completes one revolution every 20 min, so the time taken is T = 20 min = 1200 s. Substituting these values in the formula, we get:
v = (2π × 38 m) / 1200 s
≈ 2.01 m/s
The centripetal acceleration of a point on the Ferris wheel is given by the formula:
a = v² / r
where a is the magnitude of the centripetal acceleration, v is the tangential speed, and r is the radius of the Ferris wheel.
In this case, we have already calculated the tangential speed to be approximately 2.01 m/s, and the radius of the Ferris wheel is 38 m. Substituting these values in the formula, we get:
a = (2.01 m/s)² / 38 m
≈ 0.106 m/s²
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Please hurry, being timed. And no links !!!
A ball weighing 10 kg rolls down a frictionless incline with a 50 degree angle to the horizontal. If the balls initial velocity was 0 m/s, how much does the mechanical energy of the system change by the time the ball reaches its destination.
A. Increases by 12%
B. Increases by 58%
C. Decreases by 12%
D. Does not change
The mechanical energy of the system decreases by 12% (152.70 J/751.98 J x 100%) when the ball reaches its destination. Therefore, the correct answer is C. Decreases by 12%.
What is Energy?
Energy is a fundamental physical quantity that describes the ability of a system to do work. In other words, energy is the capacity of a system to produce changes in itself or in its environment. It is a scalar quantity, which means that it is characterized only by its magnitude and not by its direction.
When the ball reaches the bottom of the incline, it will have a velocity v given by v = [tex]\sqrt{2gh}[/tex], where h is the height of the incline. Substituting the values, we get v = [tex]\sqrt{29.817.66}[/tex] = 10.99 m/s.
The final kinetic energy of the ball is given by (1/2)m[tex]v^{2}[/tex] = (1/2)[tex]1010.99^{2}[/tex] = 599.28 J.
The final potential energy of the ball is 0, since it is at ground level. Therefore, the total mechanical energy of the system at the bottom of the incline is the sum of the final kinetic and potential energies, which is 599.28 J.
The initial mechanical energy of the system is the potential energy of the ball at the top of the incline, which is 751.98 J.
Therefore, the change in mechanical energy is:
Delta E = 599.28 - 751.98 = -152.70 J
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What does it mean to you to be healthy? In your answer, give three attributes that you believe healthy people have.
If all of the gravitation potential energy of the apple on the tree is transferred to the spring when it is compressed. What is the spring constant of this spring?
Answer:
360 N/m
Explanation:
you pull a sled with a package on it across a snow covered flat lawn. if u apply a force of 46.8 N to the sled, it accelerates at 0.75 m/s. what is the combined mass of the package and sled
The combined mass of the package and the sled can be found to be 62.4 kg.
How to find the combined mass ?We can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F net = ma), to solve this problem.
F net = ma
46.8 N = (m sled + m package) × 0.75 m/s²
To find the combined mass of the sled and package, we need to add their individual masses together. Therefore, we can rewrite the equation as:
46.8 N = (m sled + m package) × 0.75 m/s²
46.8 N / 0.75 m/s² = m sled + m package
62.4 kg = m sled + m package
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A baseball is hit straight up at an initial velocity of 30m/s if the ball has a negative acceleration of about 10m/s2 how long does the ball take to reach the too of its path
The ball takes about 3.06 seconds to reach the top of its path.
When a ball is thrown or hit straight up, it will reach its maximum height at the point where its vertical velocity becomes zero.
At this point, the ball's acceleration will be equal to the acceleration due to gravity, which is -9.8 m/s².
Using the equation of motion for an object with constant acceleration, we can find the time it takes for the ball to reach the top of its path:
vf = vi + at
where vf is the final velocity (which is zero when the ball reaches its maximum height), vi is the initial velocity (30 m/s), a is the acceleration (-9.8 m/s²), and t is the time we're looking for.
Rearranging the equation, we get:
t = (vf - vi) / a
Since the final velocity is zero, we have:
t = -vi / a
Substituting the values, we get:
t = -30 m/s / (-9.8 m/s²)
t = 3.06 seconds
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(a)(i) A galaxy moves away from the Earth at a speed of 3.9 × 104 km/s.
The speed of light is 3.0 × 105 km/s.
Light from the galaxy is emitted with a wavelength of 6.2 × 10−7 m.
Calculate the change in the wavelength of the light that is received by an observer on the Earth.
(ii) Calculate the wavelength of the light that is received by the observer on the Earth.
(b)One of the pieces of evidence for the Big Bang theory is the red-shift of galaxies. Explain how the red-shift of galaxies supports the Big Bang theory.
Wavelength of the light that is received by the observer on the Earth is 5.4 x 10⁻⁷m.
a) Speed of the galaxy, v = 3.9 x 10⁴ m/s
Speed of light, c = 3 x 10⁵ m/s
Wavelength of the light emitted from the galaxy, λ = 6.2 x 10⁻⁷m
(i) The expression for the change in wavelength of the light that is received by the observer on the Earth is given by,
Δλ = (v/c)λ
Δλ = (3.9 x 10⁴/3 x 10⁵) 6.2 x 10⁻⁷
Δλ = 8.06 x 10⁻⁸m
(ii) Wavelength of the light that is received by the observer on the Earth,
λ' = λ - Δλ
λ' = 5.4 x 10⁻⁷m
b) Redshifts have been observed for almost all galaxies. When light from far-off galaxies travels from the galaxy to our telescopes, it is stretched (made redder) by the universe's expansion, which is known as "cosmological redshift".
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If the force between two charges is initially 1800 N then what will it be if one of the charges is moved 3x farther away?
When one of the charges is moved 3 times farther away, the force between the two charges will be 200 N.
The force between two charges is described by Coulomb's Law, which states that the force (F) is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them:
F = k × (q1 × q2) / r²
Here, k is Coulomb's constant.
Initially, the force between the two charges is 1800 N. Let's assume the initial distance between the charges is r. Now, one of the charges is moved 3 times farther away, making the new distance between the charges 3r.
To find the new force, we can apply Coulomb's Law again:
F_new = k × (q1 × q2) / (3r)²
Notice that k × (q1 × q2) / r² = 1800 N (initial force). To make calculations easier, we can replace the expression with 1800 N:
F_new = 1800 N / 3²
F_new = 1800 N / 9
F_new = 200 N
So, when one of the charges is moved 3 times farther away, the force between the two charges will be 200 N. This demonstrates the inverse-square relationship between the force and the distance in Coulomb's Law.
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A carnival ride initially rotates counterclockwise at rad 2. 0 but comes to rest with a constant acceleration S over an angular displacement of 6. 0 rad. What is the angular acceleration? Answer using a coordinate system where counterclockwise is positive. Round answer to two significant digits.
The angular acceleration of the carnival ride is approximately -0.33 rad/s² (rounded to two significant digits).
Angular acceleration is defined as the rate of change of angular velocity with respect to time. It is measured in radians per second squared. In this problem, the carnival ride initially rotates counterclockwise at a rate of 2.0 radians per second and comes to rest over an angular displacement of 6.0 radians with a constant acceleration.
To find the angular acceleration of the carnival ride, we can use the following equation:
ω² = ω₀² + 2αθ
where ω is the final angular velocity (0 rad/s since the ride comes to rest), ω₀ is the initial angular velocity (2.0 rad/s, counterclockwise), α is the angular acceleration, and θ is the angular displacement (6.0 rad, counterclockwise).
Since counterclockwise rotation is considered positive in the given coordinate system, we have:
0² = (2.0 rad/s)² + 2α(6.0 rad)
Rearranging to solve for α:
α = - (2.0 rad/s)² / (2 × 6.0 rad)
α = - 4.0 / 12.0 = -0.33 rad/s²
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Calculate and compare the gravitational force and the electrical force between two protons that are separated by 4. 25x 10 -15 m (G = 6. 67 x 10 -11 Nm 2 /kg 2 , e = 1. 60 x 10 -19 C, m p = 1. 67 x 10 -27 kg)
The electrical force is approximately [tex]10^{43}[/tex] times larger than the gravitational force. This is because the electrical force is much stronger than the gravitational force, due to the large difference in the strength of the fundamental forces involved.
To calculate the gravitational force between two protons, we use the equation:
[tex]F_g = G * (m_p)^2 / r^2[/tex]
where
G is the gravitational constant,
[tex]m_p[/tex] is the mass of a proton, and
r is the distance between the centers of the two protons.
Plugging in the values given, we get:
[tex]F_g = 6.67 * 10^-11 Nm^2/kg^2 * (1.67 * 10^-27 kg)^2 / (4.25 * 10^-15 m)^2[/tex]
[tex]= 1.72 x 10^{-51} N[/tex]
To calculate the electrical force between the protons, we use the equation:
[tex]F_e = k * (q_p)^2 / r^2[/tex]
where
k is the Coulomb constant,
[tex]q_p[/tex] is the charge of a proton, and
r is the distance between the centers of the two protons.
Plugging in the values given, we get:
[tex]F_e = 9 *10^9 Nm^2/C^2 * (1.6 * 10^-19 C)^2 / (4.25 * 10^-15 m)^2[/tex]
= 2.32 x [tex]10^{-8}[/tex] N
Comparing the two forces, we see that the electrical force is much larger than the gravitational force. The electrical force is approximately [tex]10^{43[/tex]times larger than the gravitational force.
This is because the electrical force is much stronger than the gravitational force, due to the large difference in the strength of the fundamental forces involved.
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A certain vibrating string on a piano has a length of 74 cm and forms a standing wave having two antinodes. (a) Which harmonic does this wave represent?
(b) Determine the wavelength of this wave
(c) how many nodes are there if 20.0 Newton find the fundamental frequency I'm the next three frequencies that could cause standing wave patterns on the street
A standing wave with two antinodes on a vibrating string represents the 1st overtone, which is also known as the 2nd harmonic. The wavelength is 148 cm. The 2nd harmonic already has two antinodes, so for the 3rd, 4th, and 5th harmonics, there will be 3, 4, and 5 antinodes, respectively.
(a) A standing wave with two antinodes on a vibrating string represents the 1st overtone, which is also known as the 2nd harmonic.
(b) To determine the wavelength of this wave, first, recall that the length of the string is half of the wavelength for the 2nd harmonic. So, we can use the following formula:
Length of the string = Wavelength / 2
Now, plug in the given values:
74 cm = Wavelength / 2
To find the wavelength, multiply both sides by 2:
Wavelength = 74 cm × 2 = 148 cm
(c) If the tension in the string is 20.0 N, first, we need to find the fundamental frequency. In a standing wave pattern with 1 antinode (1st harmonic), the length of the string is equal to half of the wavelength. So, the wavelength of the fundamental frequency is:
Wavelength (1st harmonic) = 2 × Length of the string = 2 × 74 cm = 148 cm
To find the next three frequencies that could cause standing wave patterns on the string, we will look at the 3rd, 4th, and 5th harmonics. For each harmonic, the number of nodes increases by 1. The 2nd harmonic already has two antinodes, so for the 3rd, 4th, and 5th harmonics, there will be 3, 4, and 5 antinodes, respectively.
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Both objects are released from rest and the pulley turns without slipping the coefficient of kinetic friction between the 2kg object and the surface is 0. 40. Calculate the angular acceleration of the pulley.
a. 34. 25 rad/s^2
b. 36. 17 rad/s^2
c. 39. 22 rad/s^2
d. 46. 57 rad/s^2
The angular acceleration of the pulley is approximately [tex]39.22 rad/s^2[/tex].
What does the term "angular acceleration" mean?
The angular acceleration, which is frequently denoted by the symbol and stated in radians per second per second, is the rate at which the angular velocity changes over time.
Here is the calculation:
The net force acting on the 2 kg object is the difference between the tension in the string and the frictional force. Using Newton's second law, we can write:
[tex]F_{net} = ma\\T - f_k = ma[/tex]
The moment of inertia of the pulley can be calculated using the formula for the moment of inertia of a disk:
[tex]I = (1/2)mr^2[/tex]
The torque due to the tension can be calculated as:
[tex]\tau_T = T*(r/2)[/tex]
The torque due to the frictional force can be calculated as:
[tex]\tau_f = f_k*(r/2)[/tex]
The net torque can be calculated as the difference between the torque due to the tension and the torque due to the frictional force:
[tex]\tau_{net} = \tau_T - \tau_f[/tex]
Finally, the angular acceleration can be calculated using Newton's second law for rotational motion:
[tex]\tau_{net} = I*\alpha[/tex]
Substituting the values and solving for α, we get:
[tex]\alpha = (T - f_k)/(1/2mr^2) = (2/3)g(\mu_k - sin\theta)[/tex]
where g is the acceleration due to gravity, [tex]\mu_k[/tex] is the coefficient of kinetic friction, and θ is the angle of the incline.
Using the given values, we get:
[tex]\alpha = (2/3)9.81(0.40 - sin(30)) = 39.22 rad/s^2[/tex]
Therefore, the angular acceleration of the pulley is approximately [tex]39.22 rad/s^2[/tex].
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The farthest bright galaxies that modern telescopes are capable of seeing are up to:.
The farthest bright galaxies that modern telescopes are currently capable of seeing are up to several billions of light-years away. The exact distance depends on various factors such as the sensitivity and resolution of the telescope, observational techniques, and the brightness of the galaxy itself.
Modern telescopes, such as the Hubble Space Telescope and large ground-based observatories equipped with advanced instruments, have greatly advanced our ability to observe and study distant galaxies. These telescopes can detect and capture the light from galaxies that existed when the universe was relatively young.
Through deep field observations and gravitational lensing techniques, astronomers have been able to observe galaxies that are more than 13 billion light-years away. These observations provide valuable insights into the early universe and its evolution.
It's important to note that the term "bright" is relative and can vary depending on the context and specific criteria used for brightness. Additionally, ongoing advancements in telescope technology continue to push the limits of observation, and future telescopes and space missions are expected to enable us to see even farther into the universe.
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