Helpppppppppp!!!!!!!!!!...​

Given data:

Acceleration due to gravity on Mars, a = 3.77 m/s²Mass of rock, m = 43 g = 0.043 kg

Time is taken, t = 9.5 s

Let h be the height from which the rock falls.

Using the kinematic equation of motion, h = ut + 1/2at²Where,u = Initial velocity = 0m/sa = Acceleration = 3.77 m/s²t = Time taken = 9.5 s

Substitute the given values,

h = 0 + 1/2 × 3.77 × (9.5)²h = 1689.5 m

The 43 g rock will fall from rest to a distance of 1689.5 m in 9.5 s if the only force acting on it was the gravitational force due to Mars.

According to the second law of motion, when a force acts on an object, it produces acceleration in that object. And the formula to calculate acceleration is given below,

F = m × a

Where,

F = acting on the object

m = mass of the object

a = acceleration produced by the force

Given data:

Mass of the rock, m = 43 g = 0.043 kg

Acceleration due to gravity on Mars, a = 3.77 m/s²

We know that gravitational force acting on an object of mass m due to a planet of mass M is given by the formula:

F = G (M m)/r²

Where,

G = Gravitational constant

M = Mass of the planet

m = Mass of the object

r = Distance between the object and the planet

Given data:

Mars is the planet acting on the rock, so its mass is M = 6.39 × 10²³ kg

The rock falls from rest, so the initial velocity of the rock is u = 0

Distance traveled by the rock from rest in time t is given by the formula,

h = ut + 1/2at²Substitute the given values,

h = 0 + 1/2 × 3.77 × (9.5)²h = 1689.5 m

Hence, the 43 g rock will fall from rest to a distance of 1689.5 m in 9.5 s if the only force acting on it was the gravitational force due to Mars.

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Answer: (D) an egg colliding with a brick wall is not the best example of an elastic collision.

An elastic collision is defined as a collision in which both kinetic energy and momentum are conserved. Kinetic energy is transferred from one object to another during a collision in which they both collide. The energy lost in a collision is known as inelastic energy.

The best example of an elastic collision is:

A) a collision between two billiard balls and

C) a basketball bouncing off the floor.

A collision between two billiard balls A collision between two billiard balls is the best example of an elastic collision. When two billiard balls collide, kinetic energy is transferred from one ball to the other. However, because the billiard balls are made of elastic materials, the energy is not lost during the collision. As a result, both the momentum and the kinetic energy of the billiard balls remain unchanged after the collision.A basketball bouncing off the floorWhen a basketball bounces off the floor, it is also an example of an elastic collision. When the basketball hits the floor, kinetic energy is transferred from the basketball to the floor.

However, because the basketball is made of an elastic material, the energy is not lost. As a result, the basketball bounces back up off the floor with the same kinetic energy that it had before it hit the floor.

Therefore, a basketball bouncing off the floor is also an example of an elastic collision. An egg colliding with a brick wall is an example of an inelastic collision. In an inelastic collision, some of the kinetic energy is lost during the collision. As a result, both the momentum and the kinetic energy of the objects are not conserved.

Therefore, D) an egg colliding with a brick wall is not the best example of an elastic collision.

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The correct statement that explains the observation using Newton's laws is objects at rest tend to remain in their current state of motion unless acted upon by an unbalanced force, but objects in motion require a continual application of force to stay in motion. Here option A is correct.

According to Newton's first law of motion, an object will continue moving at a constant velocity in a straight line unless acted upon by an external force. In this case, when the cannonball is shot horizontally from the cannon, it initially possesses a forward velocity due to the force applied by the cannon. However, once the cannonball is in motion, the only forces acting on it are gravity and friction.

Gravity acts vertically downward, causing the cannonball to accelerate downward. Friction acts horizontally in the opposite direction to the motion of the cannonball. As the cannonball moves forward, friction opposes its motion and gradually slows it down.

Since there is no force continuously propelling the cannonball forward, and the forces of friction and gravity act on it, the cannonball eventually comes to a stop and falls to the ground. Hence option A is correct.

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The first diffraction maximum fringe will be located at a distance of approximately 0.022 meters from the central maximum when light of wavelength 550 nm falls on a slit that is 3.40 × 10-3 mm wide, and the screen is positioned 12.0 meters away.

When light passes through a narrow slit, it undergoes diffraction, resulting in a pattern of bright and dark fringes on a screen. The position of these fringes can be calculated using the formula for the angular position of the mth order fringe: θ = mλ/d, where θ is the angular position, λ is the wavelength of light, m is the order of the fringe, and d is the width of the slit.

In this case, we are interested in the first order fringe (m = 1). Plugging in the given values, we can find the angular position of the first diffraction maximum: θ = (1)(550 × 10^-9 m) / (3.40 × 10^-3 mm). Note that we convert the width of the slit to meters for consistency.

To determine the position of the fringe on the screen, we can use the small-angle approximation: x ≈ rθ, where x is the distance from the central maximum, r is the distance between the slit and the screen, and θ is the angular position in radians.

Given that r = 12.0 m and θ is calculated as above, we can find the position of the first diffraction maximum fringe: x ≈ (12.0 m)(1)(550 × 10^-9 m) / (3.40 × 10^-3 mm).

Evaluating this expression gives us x ≈ 0.022 meters, which is approximately the distance from the central maximum to the first diffraction maximum fringe.

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Answer:

Gravitational energy

Thermal energy

Mechanical energy

The workers have approximately 1.30 seconds to reach the piano before it reaches the bottom of the ramp. This can be determined by analyzing the motion of the piano as it descends the inclined ramp.

To determine the time the workers have to reach the piano, we need to consider the motion of the piano as it moves down the inclined ramp. The time can be calculated using the equations of motion and trigonometry.

Given that the back of the truck is 1.5 m above the ground and the ramp is inclined at 30 degrees, we can use trigonometry to find the height of the ramp.

The height of the ramp can be calculated as the vertical distance traveled by the piano along the ramp, which is equal to the vertical displacement from the back of the truck to the ground. By applying trigonometry, the height of the ramp is found to be 0.75 m.

Next, we can use the equation of motion for vertical motion with constant acceleration to determine the time it takes for the piano to descend from the top to the bottom of the ramp. The equation is:

[tex]h = (1/2) * g * t^2[/tex]

where h is the height of the ramp, g is the acceleration due to gravity, and t is the time. Substituting the known values, we get:

[tex]0.75 = (1/2) * 9.8 * t^2[/tex]

Simplifying and solving for t, we find that t is approximately 1.30 seconds. Therefore, the workers have approximately 1.30 seconds to reach the piano before it reaches the bottom of the ramp.

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The length of the ramp is first calculated using trigonometry given the height and the angle (30°). Upon acquiring the length, the time it would take for the piano to slide down this length (ramp) under the influence of gravity is found using equations of motion, giving a value of 0.78 seconds.

This question requires us to use the principles from kinematics and basic trigonometry. We can begin by calculating the length of the inclined ramp using trigonometry. The sine of the angle of inclination equals the vertical height divided by the length of the ramp, thus length (L) of the ramp is equal to height (h) divided by sin(angle), which is L = 1.5 m / sin(30°) which equals 3 m.

Once we know the length, we can then calculate the time for an object to slide down the ramp under the influence of gravity (g = 9.8 m/s²). This can be done using the equation of motion, which is s = ut + 0.5gt² (where s is distance, u is initial velocity, t is time, and g is gravitational acceleration). Assuming no initial velocity we get t = √(2s/g), which gives time (t) as √(2×3/9.8) = 0.78 seconds.

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Answer: is c

Explanation:

Answer:

0 Newton

Explanation:

We have the formula F=m*a. Given the mass = 12kg and a = 0 ( I assume that the rock is rolling down at a constant velocity of 4.7m/s which its acceleration is 0)

So we have F = 12*0 = 0 N.

Hope that is what you are looking for.

When we jump on a concrete surface ,the feet are more seriously hurt than while jumping on sand because if we fall on  concrete surface there will very less time to make momentum zero , force will hit with greater intensity and if we fall on any soft surface ,enough time will be their to make momentum zero and force will decrease

It states that the time rate of change of momentum of a body is equal to force imposed on it The body whose mass is constant , newton's 2nd law of motion can be written as F = ma

When we jump on a concrete ,the feet are more seriously hurt than while jumping on sand because if we fall on  concrete  there will very less time to make momentum zero , due to which force doesn't decrease and hit with greater intensity

but if we fall on any soft surface  then it will take long time for us to stop and  enough time will be their to make momentum zero , in that time span the  intensity of force will decrease as rate ( by newtons 2nd law) has increased and feet won't hurt that much

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The electric flux through the square sheet is [tex]4.4 * 10^4 Nm^2/C[/tex], when a uniform electric field of magnitude [tex]1.1 * 10^4 N/C[/tex] is perpendicular to a square sheet with sides 2.0 m long.

The electric flux through a closed surface is given by the formula:

[tex]\[ \Phi = \mathbf{E} \cdot \mathbf{A} \][/tex]

where [tex]\(\Phi\)[/tex] is the electric flux, [tex]\(\mathbf{E}\)[/tex] is the electric field, and [tex]\(\mathbf{A}\)[/tex] is the area vector of the surface. In this case, the electric field [tex]\(\mathbf{E}\)[/tex] is perpendicular to the square sheet, and the magnitude of the electric field is given as [tex]1.1 * 10^4 N/C[/tex].

The area of the square sheet is [tex]\(A = (2.0 \, \text{m})^2 = 4.0 \, \text{m}^2[/tex]). Since the electric field is perpendicular to the surface, the angle between the electric field and the area vector is 0 degrees.

Substituting the values into the formula, we have:

[tex]\[ \Phi = (1.1 \times 10^4 \, \text{N/C}) \cdot (4.0 \, \text{m}^2) = 4.4 \times 10^4 \, \text{N} \cdot \text{m}^2/\text{C} \][/tex]

Therefore, the electric flux through the square sheet is [tex]4.4 * 10^4 Nm^2/C[/tex].

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Answer:

The force exerted by one source object on another target object always creates another force at the target object that pushes back on the source object with the same ... or His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.

The leak will take 120/11 hours to empty the full cistern. When both pipes are open, then the rate of flow of water = (R₁ + R₂) = C/10 hours

Let the capacity of the tank be C. Let the rate of the first pipe be R₁ and the rate of the second pipe be R₂. When both pipes are open, then the rate of flow of water = (R₁ + R₂) = C/10 hours (i)When the first pipe is open for 3 hours, then it fills = 3R₁ volume of water.

Let x be the volume of the tank leaked in 3 hours. Now, the effective rate of the first pipe = R₁ - x/3.

The effective rate of flow of water when both the pipes are open = (R₁ - x/3 + R₂) liters per hour. This effective rate fills the tank in 10 + 3 = 13 hours.

Using the formula of the flow of water, we get: C = (R₁ - x/3 + R₂) * 13 C/10

= (R₁ - x/3 + R₂)13/10

= R1 + R₂ - x/3x/3

= R₁ + R₂ - 13/10C

Now, we know that when the first pipe is open for 3 hours, then it fills 3R₁ volume of water.

Therefore, 3R₁ = C - x... (ii)

From equations (i) and (ii), we get:

C/10 = R₁ + R₂C/15

= R₁ + R₂ - 13/10C + x/3

Adding both equations, we get:

C/10 + C/15

= 2(R₁ + R₂) - 13/10C + x/3C

= 1/6 [(20R₁ + 20R₂) - 13C + 2x]

But we know that 3R₁ = C - x

Therefore, x = C - 3R₁

Therefore, C = 1/6 [(20R₁ + 20R₂) - 13C + 2(C - 3R₁)]

Solving the above equation, we get: C = 120R₁/11

Therefore, the leak will take 120/11 hours to empty the full cistern.

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Answer:

6000kg

Explanation:

Given parameters:

Initial velocity  = 10m/s

Final velocity  = 15m/s

Time = 10s

Force  = 3000N

Unknown:

Mass of the bus = ?

Solution:

Force is the mass multiplied by acceleration.

        Force = mass x acceleration

 Acceleration = [tex]\frac{final velocity - initial velocity }{time}[/tex]

  Force  = mass x  [tex]\frac{final velocity - initial velocity }{time}[/tex]

Insert the parameters and solve;

           3000 = mass x [tex]\frac{15 - 10}{10}[/tex]

            3000 = [tex]\frac{mass}{2}[/tex]

  Mass  = 6000kg

ρ=2.5⋅g⋅cm3............

Explanation:

ρ=Mass/Volume=5⋅g2⋅cm3,

and thus typically it has the units g⋅mL−1 or g⋅cm−3.

Here, mass=5⋅g, and volume=2⋅m3.

Their quotient is the density as shown. And this could be any solid material (generally solids are denser than liquids). You will have to look up tables of densities for common materials.

The magnitude of the force that the charge distribution q exerts on q is given by (k * |q * q'|) / a^2, where k is the Coulomb's constant, q and q' are the magnitudes of the charges, and a is the distance between them.

To calculate the magnitude of the force that the charge distribution q exerts on q, we can use Coulomb's Law. Coulomb's Law states that the force between two charged objects is given by the equation:

F = (k * |q1 * q2|) / r^2

where:

F is the magnitude of the force

k is the Coulomb's constant (approximately 9 x 10^9 N m^2/C^2)

q1 and q2 are the magnitudes of the charges

r is the distance between the charges

Let's use the variables q, q', a, and r to represent the given charge distribution and distance:

q1 = q

q2 = q'

r = a

Now we can substitute these values into the formula:

F = (k * |q1 * q2|) / r^2

F = (k * |q * q'|) / a^2

Therefore, the magnitude of the force that the charge distribution q exerts on q is given by (k * |q * q'|) / a^2, where k is the Coulomb's constant, q and q' are the magnitudes of the charges, and a is the distance between them.

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Answer:

True

Explanation:

Mitochondria is the power-house of almost every cell due to the fact that the cytoplasm stores energy, then sends it to the mitochondria. (at the needed times, of course.)

The supply plane should drop the package 90 meters short of the target.

The package will take 1.1 seconds to fall to the ground. During this time, the plane will travel another 110 meters. Therefore, the package will land 90 meters short of the target.

Here is the calculation:

Time = (distance / speed) = (110 m / 100 m/s) = 1.1 s

Distance traveled by plane = (speed * time) = (100 m/s * 1.1 s) = 110 m

Distance between package and target = (distance traveled by plane - distance to target) = (110 m - 110 m) = 90 m

Therefore, the supply plane should drop the package 90 meters short of the target.

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1. You need to equip your telescope with: b. spectrometer

A spectrometer is a device that separates light into its component wavelengths. This allows scientists to study the composition of objects in space, such as planets and stars.

The spectrometer would be used to analyze the light from the new planet, which would reveal the presence of different gases in the planet's atmosphere.

To determine the atmosphere of a planet, you need to use a spectrometer. A spectrometer separates light into its component wavelengths, which allows scientists to study the composition of objects in space.

The spectrometer would be used to analyze the light from the new planet, which would reveal the presence of different gases in the planet's atmosphere.

The different gases would absorb different wavelengths of light, so by studying the spectrum of the light from the planet, scientists could determine which gases are present in the atmosphere.

2. Which of the following choices best describes the suitable kind of telescope one needs to use while observing such stars?

b. a telescope with biggest aperture

The aperture of a telescope is the diameter of the main lens or mirror. The bigger the aperture, the more light the telescope can collect. This is important for observing faint objects, such as stars.

A telescope with the biggest aperture will be able to collect the most light, which is important for observing faint objects.

The amount of light that a telescope can collect is proportional to the square of the aperture. This means that a telescope with twice the aperture will collect four times as much light. For faint objects, such as stars, it is important to collect as much light as possible.

This is why a telescope with the biggest aperture is the best choice for observing such objects.

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Answer: I agree and it’s really gross

Explanation:

Given that the mass of the body attached to the spring is 2.00 kg and the time period of oscillation is 1.00 s.

We can substitute these values in the above equation to obtain:

[tex]1.00 s = 2π√(2.00 kg/k)[/tex]

Squaring both sides, we get: [tex]1.00 s^2 = 4π^2(2.00 kg/k)[/tex]

Simplifying, we get: [tex]k = (4π^2)(2.00 kg)/(1.00 s^2)k = 4π^2 × 2.00 kN/m[/tex]

Therefore, the force constant of the spring is [tex]4π^2 × 2.00 kN/m.[/tex]

This value can be further simplified to 125.66 kN/m.

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The answer is feedback.

Answer:

the rocket rises approximately 104.11 meters above the ground.

Explanation:

v = u + at

0 = 45.0 m/s + (-9.8 m/s²) * t

t = -45.0 m/s / -9.8 m/s²

t ≈ 4.59 seconds

To find the height the rocket rises above the ground, we can use the kinematic equation:

s = ut + (1/2) * a * t²

h=ut+(1/2) * g * [tex]t^{2}[/tex]

h = 45.0 m/s * 4.59 s + (1/2) * (-9.8 m/s²) * (4.59 s)²

h ≈ 104.11 meters

The wave speed when the suspended mass is m = 2.00 kg is  31.7 m/s. To solve this problem, we'll use the wave equation:

v = √(T/μ)

where:

v is the wave speed,

T is the tension in the string, and

μ is the mass per unit length of the string.

(a) To find the mass per unit length of the string, we'll rearrange the equation:

μ = T / [tex]v^2[/tex]

Given:

v = 24.0 m/s

m = 3.00 kg

We need to find the tension T. The tension in the string is equal to the weight of the suspended mass, which is given by:

T = m * g

where g is the acceleration due to gravity (approximately 9.8[tex]m/s^2[/tex]).

Substituting the values, we have:

T = (3.00 kg) * (9.8[tex]m/s^2[/tex])

T ≈ 29.4 N

Now, we can calculate the mass per unit length:

μ = (29.4 N) / (24.0[tex]m/s^2[/tex])

μ ≈ 0.06125 kg/m

Therefore, the mass per unit length of the string is approximately 0.06125 kg/m.

(b) To find the wave speed when the suspended mass is m = 2.00 kg, we can use the same equation:

v = √(T/μ)

We already know the tension T from part (a), which is 29.4 N. We can substitute this value and the new mass into the equation:

m = 2.00 kg

v = √((2.00 kg * 9.8 [tex]m/s^2[/tex]) / (0.06125 kg/m))

v ≈ 31.7 m/s

Therefore, the wave speed when the suspended mass is m = 2.00 kg is

31.7 m/s.

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The arrow's speed after 3 seconds is approximately 43.091 m/s.

The horizontal and vertical components of the velocity can be determined using trigonometry. The initial vertical velocity is given by v₀y = v₀ sin θ, and the horizontal velocity is given by v₀x = v₀ cos θ.


Given, θ = 30°, v₀ = 20 m/s, t = 3 s. The acceleration of gravity is g = 9.81 m/s².Let's now use the kinematic equation vf = v₀ + at. Since the arrow is launched at an angle above the horizontal, we need to calculate the horizontal and vertical components of the velocity.

The initial vertical velocity of the arrow is:

v₀y = v₀ sin θ= (20 m/s) sin 30°= 10 m/s

The horizontal velocity of the arrow is:

v₀x = v₀ cos θ= (20 m/s) cos 30°= 17.3205 m/s

Using the kinematic equation vf = v₀ + at, we can calculate the vertical velocity of the arrow after 3 seconds:

vfy = v₀y + gt

= (10 m/s) + (9.81 m/s²)(3 s)

= 39.43 m/s

The horizontal velocity doesn't change during the flight of the arrow, so vf = v₀x = 17.3205 m/s.

The final velocity of the arrow is given by the vector sum of the horizontal and vertical velocities:

v = √(vf² + v₀x²)

= √(39.43² + 17.3205²)

= 43.091 m/s

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The two horizontal forces that are present as a car is pushed down the driveway are push and friction.

What is force?

When an object is in motion, there are two main factors that affect its kinetic energy: the push or force applied to it and the friction it experiences.

What is friction?

What is velocity?

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Answer:

C. POSITIVE

(Sorry if wrong)

Answer:

C positive

Explanation:

It keeps moving in a positive way over and over

The main answer to the question is as follows: Given,Weight of skydiver, W = 967 NMass of skydiver, m = 98.7 kgDrag force, Fd = 1041 NNow, the skydiver is under two forces,Weight of the skydiver, W = 967 NUpward force, FUp = - 1041 N (opposite in direction to the weight of the skydiver)

Net force acting on the skydiver, F = FUp + W= - 1041 N + 967 N= - 74 NAs acceleration, a = F/mSubstituting the given values, we get,a = -74 N/98.7 kg= -0.749 m/s² :When a parachute opens, the air exerts a large drag force on it. The upward force initially greater than the weight of the sky diver and, thus, slows him down. The skydiver is under two forces: the weight of the skydiver and the upward force. The weight of the skydiver is always directed towards the ground, whereas the upward force on the skydiver is always directed away from the ground.

The net force acting on the skydiver is the sum of the weight of the skydiver and the upward force.The acceleration of the skydiver can be determined as a = F/m, where F is the net force acting on the skydiver and m is the mass of the skydiver. The acceleration of the skydiver is negative because the net force is in the opposite direction to the upward force.

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The flow rate of water is 0.0056 L/sec (rounded to 3 significant figures). To determine the flow rate in L/sec, we can utilize the concept of continuity equation.

The continuity equation is described as: A₁V₁ = A₂V₂

where A₁ and V₁ are the cross-sectional area and velocity of fluid respectively at the first point, while A₂ and V₂ are the cross-sectional area and velocity of fluid respectively at the second point.

Since we have only one point, we can solve for flow rate using the formula for volume flow rate as:

Q = Av, where Q is the volume flow rate, A is the cross-sectional area of the pipe, and v is the velocity of fluid.

Substituting the known values in the formula:

We are given that velocity of water, v = 1.30 m/s

and internal diameter of hose, d = 4.30 cm.

To obtain the cross-sectional area, A in m², we need to convert the diameter from centimeter (cm) to meter (m) using the formula:

1 cm = 0.01 m

Substituting the values of d and v:

Thus, the flow rate of water is 0.0056 L/sec (rounded to 3 significant figures).

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The strength of the electric field Ep1 at a distance of 1.3mm from a proton is 1.05 × 10^14 N/C.

The electric field is defined as the force that one proton exerts on another proton per unit charge. Coulombs or N/C are the units of electric field.

A proton is an atomic particle with a positive charge.

The strength of the electric field is determined by the distance between the charges and the charge itself.

The distance between the electric field and the proton is 1.3 mm.

Proton charges are positive, and the electric field direction is from the positive charges to the negative charges. The electric field intensity is determined by the Coulomb's law.

The formula is as follows;

E= kq/r²Where k = 9 × 10^9 N⋅m²/C² is the Coulomb constant.

q is the electric charge, and r is the distance between the two charges.

Ep1 = k * q/r1^2 = (9 × 10^9 N⋅m²/C²) * (1.6 × 10^-19 C)/(1.3 × 10^-3 m)²= 1.05 × 10^14 N/C

So, the strength of the electric field Ep1 at a distance of 1.3mm from a proton is 1.05 × 10^14 N/C.

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