A box slides on a level floor. It is slowing with a constant
acceleration of magnitude 2.0 m/s2
. What is the coefficient of
kinetic friction between the box and the floor?

A block of 0.5 kg is placed on top of another wooden block which weighs 1.0 kg. The coefficient of static friction between the two blocks is 0.35, whereas the coefficient of kinetic friction between the lower block and the level table is 0.20.

To calculate the maximum horizontal force that can be applied to the lower block, we need to determine the limiting frictional force between the two blocks.

Since the upper block is not moving, the force of static friction is acting on it. We can calculate this force as follows:

`F_static = friction coefficient * normal force`

where, normal force = weight of upper block = 0.5 kg * 9.81 m/s^2 = 4.905 N

`F_static = 0.35 * 4.905 = 1.718 N`

Therefore, the static frictional force acting on the upper block is 1.718 N.

Now, we need to find the maximum force that can be applied to the lower block before it starts moving. This force is equal to the force of static friction acting on the lower block.

Since the upper block is not moving, the force of static friction acting on the lower block is equal to the force of static friction acting on the upper block.

`F_static(lower block) = F_static(upper block) = 1.718 N`

This means that the maximum horizontal force that can be applied to the lower block is 1.718 N.

However, if the applied force exceeds this value, the lower block will start moving and the force of kinetic friction will be acting on it, which is equal to:

`F_kinetic = friction coefficient * normal force`

`F_kinetic = 0.20 * 4.905 = 0.981 N`

Hence, if the applied force exceeds 1.718 N, the lower block will start moving and the force of kinetic friction will act on it, which is 0.981 N.

Therefore, the maximum horizontal force that can be applied to the lower block is 1.718 N.

Answer:

Explanation:

To determine the maximum horizontal force that can be applied to the lower block without causing the blocks to move, we need to calculate the maximum static friction force between the two blocks. This force is given by:

F_friction = coefficient of static friction * normal force

where the normal force is the force perpendicular to the surface of contact between the blocks. Since the blocks are resting on a level table, the normal force acting on the lower block is equal to the weight of both blocks, which is:

N = (m1 + m2) * g

where m1 is the mass of the lower block, m2 is the mass of the upper block, and g is the acceleration due to gravity (9.81 m/s^2).

Plugging in the given values, we have:

N = (1.0 kg + 0.5 kg) * 9.81 m/s^2 = 14.715 N

The maximum static friction force is then:

F_friction = 0.35 * 14.715 N = 5.15025 N

Therefore, the maximum horizontal force that can be applied to the lower block without causing the blocks to move is 5.15025 N. If a greater force is applied, the blocks will start to move and the kinetic friction force will take effect, which is given by:

F_kinetic = coefficient of kinetic friction * normal force

where the coefficient of kinetic friction is 0.20 in this case.

The tension in the swing's chain at the instant shown in the image is 441.6 N.

option D

At the bottom of the swing, the girl is traveling at a constant speed, so her acceleration is zero. Therefore, the net force acting on her is also zero.

Thus, we have:

0 = T - mg - mv²/r

where;

At the bottom of the swing, the radius is equal to the length of the chain, so we have:

r = L = 4.0 m

Substituting the values we have:

T = (32 kg)(9.8 m/s²) + (32 kg)(4 m/s)²/4.0 m

Solving for T, we get:

T = (32 kg)(9.8 m/s²) + (32 kg)(4 m/s)²/4.0 m

T = 441.6 N.

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Answer:

To solve the problem, we use the equations of equilibrium to find the tension in each cable holding up the loudspeaker. Since the loudspeaker is in equilibrium, the sum of the forces acting on it is zero. The weight of the loudspeaker is calculated first, and then we use trigonometry to find the horizontal and vertical components of the tension in one of the cables. We then apply the equation of equilibrium in the y direction to find the tension in each cable. The final answer is that the tension in each cable is approximately 81.1 N, which balances the weight of the loudspeaker.

___

Given:

Mass of loudspeaker (m) = 15.0 kg

Distance from ceiling (h) = 1.00 m

Length of cable (l) = 2.70 m

Acceleration due to gravity (g) = 9.80 m/s^2

Weight of loudspeaker:

Fg = mg

Fg = (15.0 kg)(9.80 m/s^2)

Fg = 147 N

Horizontal and vertical components of tension in one cable:

sinθ = h/l

sinθ = 1.00 m / 2.70 m

θ = sin^(-1)(1.00/2.70)

θ ≈ 21.6°

T_x = T sinθ

T_y = T cosθ

where T is the tension in the cable.

Equation of equilibrium in y direction:

ΣF_y = 2T cosθ - Fg = 0

Solving for T:

2T cosθ = Fg

T = Fg / (2 cosθ)

Plugging in the values:

T = (147 N) / (2 cos(21.6°))

T ≈ 81.1 N

Explanation:

I'm sorry, but the statement you provided is incorrect. There is no such thing as Newton's 4th law. Newton's laws of motion consist of three laws, which are:

An object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

The acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass.

For every action, there is an equal and opposite reaction.

None of these laws state that action and reaction never start from the same point. However, it is true that the action and reaction forces act on different objects, not necessarily at the same point. This is because Newton's third law states that every action has an equal and opposite reaction, which means that when one object exerts a force on another object, the second object exerts an equal and opposite force back on the first object.

Answer: When a disrupted part of the ecosystem is left alone so that nature can help restore itself what people are counting on happening is secondary succession

Explanation:

Answer:

Explanation:

We can use the following kinematic equation to solve for the time required:

v_f = v_i + at

where:

v_f = final velocity = 25.0 m/s

v_i = initial velocity = 20.0 m/s

a = acceleration = 3.0 m/s^2

t = time

Rearranging the equation to solve for t, we get:

t = (v_f - v_i) / a

Substituting the given values, we get:

t = (25.0 m/s - 20.0 m/s) / 3.0 m/s^2

t = 1.67 s

Therefore, it will take 1.67 seconds for the car to go from a speed of 20.0 m/s to a speed of 25.0 m/s, assuming a constant acceleration of 3.0 m/s^2.

Answer:  a stretch of 2

Explanation: because it 2 (x) - 3

The approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.

We can use the conservation of energy, where the potential energy stored in the compressed spring is converted into the kinetic energy of the stuntman as he is launched out of the cannon.

The potential energy stored in the spring is given by:

PE = (1/2)kx²

where k is the spring constant and x is the distance the spring is compressed.

PE = (1/2)(266 N/m)(5 m)² = 3325 J

This potential energy is then converted into kinetic energy:

KE = (1/2)mv²

where m is the mass of the stuntman and v is his velocity.

3325 J = (1/2)(55 kg)v²

v² = (2*3325 J) / 55 kg

v²  = 121 m²/s²

v = √(121 m²/s²) = 11 m/s

Therefore, the approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.

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Answer: Each macromolecule is broken down by a specific enzyme. For instance, carbohydrates are broken down by amylase, sucrase, lactase, or maltase. Proteins are broken down by the enzymes pepsin and peptidase, and by hydrochloric acid. Lipids are broken down by lipases.

Explanation:

Hope this will help

(a) The frequency ratio between the two frequencies fi = 256 Hz and f2 = 320 Hz is:

[tex]\frac{f_2}{f_i} = \frac{320}{256} = \frac{5}{4} = 1.25[/tex]

So the frequency ratio is 1.25.

(b) Adding the interval of a fifth to f2 = 320 Hz gives:

fs = f2 * (3/2) = 320 * (3/2) = 480 Hz

The frequency ratio fs/fi is:

[tex]\frac{f_s}{f_i} = \frac{480}{256} = \frac{15}{8} = 1.875[/tex]

So the frequency ratio is 1.875.

(c) To find the frequency of f3, we need to add the interval of a fourth to f2:

f3 = f2 * (4/3) = 320 * (4/3) = 426.67 Hz

Therefore, the frequency of f3 is 426.67 Hz.

Answer: increases

Explanation:

The initial mass of the rocket was 106 kg.

We can use the principle of conservation of momentum to solve this problem.

The momentum of the rocket before takeoff is zero, since it is at rest, and the momentum after takeoff is the product of the mass of the rocket and its velocity.

However, during the takeoff, the rocket ejects a mass of fuel at a certain velocity, which creates a backward force (thrust) that propels the rocket forward.

This thrust can be calculated using the equation:

Thrust = (mass flow rate) x (exhaust velocity)

mass flow rate = (mass of fuel burned) / (burn time)

The mass of the rocket at any given time can be calculated using the equation:

mass = (initial mass) - (mass of fuel burned)

Using these equations, we can solve for the initial mass of the rocket:

Calculate the thrust:

Thrust = (118 kg / 10.0 s) x 1600 m/s = 1,888 N

Calculate the mass of the rocket at the end of the burn:

mass(end) = (initial mass) - (mass of fuel burned) = (initial mass) - 118 kg

Use the principle of conservation of momentum to find the initial mass:

momentum before = momentum after

0 = (mass(end) + 118 kg) x 110 m/s

mass(end) = -118 kg / 110 m/s = -1.07 kg/s

mass(end) = (initial mass) - 118 kg

(initial mass) = mass(end) + 118 kg

(initial mass) = (-1.07 kg/s x 10.0 s) + 118 kg

(initial mass) = 106 kg

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To solve this problem, we can use the conservation of mechanical energy principle. When the blocks are released from rest, the potential energy of the system is converted to kinetic energy. Since the surface is frictionless, the mechanical energy of the system is conserved.

Using the principle of mechanical energy conservation, we can write:

m1*g*h = (m1+m2)*v^2/2

where m1 is the mass of the first block, m2 is the mass of the second block, g is the acceleration due to gravity, h is the height that the second block falls, and v is the velocity of the system after the blocks have moved a distance x.

The displacement of the first block can be found by using the time it takes the system to reach this velocity. The time t can be found using the formula:

x = (1/2) * a * t^2

where a is the acceleration of the first block.

The acceleration of the first block is equal to the acceleration of the system, which can be found by using the equation:

m1*a = m2*g - m1*g

Substituting the value of a in the previous formula, we get:

x = (1/2) * (m2*g - m1*g) * t^2 / m1

Substituting the values we get:

x = (1/2) * (2.0 kg * 9.81 m/s^2 - 3.0 kg * 9.81 m/s^2) * (1.2 s)^2 / 3.0 kg

x ≈ 7.6 m

Therefore, the correct answer is D) 7.6 m.

Answer : 0.00885 farads or 8.85 microfarads

Explanation: To calculate the capacitance required, we can use the following formula:

C = 1 / [2 * pi * f * ((V^2 - Vlamp^2)/P)]

where:

C = capacitance in farads (F)

pi = 3.14159...

f = frequency in hertz (Hz)

V = voltage in volts (V)

P = power in watts (W)

Vlamp = voltage of the lamp in volts (V)

Using the given values, we have:

C = 1 / [2 * pi * 60 * ((230^2 - 100^2)/750)]

C = 1 / [2 * 3.14159 * 60 * ((230^2 - 100^2)/750)]

C = 1 / [113.09724]

C = 0.00885 farads (F)

Therefore, the capacitance required is approximately 0.00885 farads or 8.85 microfarads.

Answer:

The period of a ticker-timer is the time interval between two consecutive dots made by the ticker.

If the frequency of the ticker-timer is 25 Hz, then it makes 25 dots in one second. Therefore, the period of the ticker-timer can be calculated as:

Period = 1/frequency = 1/25 Hz = 0.04 seconds

If the frequency of the ticker-timer is 50 Hz, then it makes 50 dots in one second. Therefore, the period of the ticker-timer can be calculated as:

Period = 1/frequency = 1/50 Hz = 0.02 seconds

So, the period of the ticker-timer is 0.04 seconds for a frequency of 25 Hz and 0.02 seconds for a frequency of 50 Hz

Explanation:

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Answer:

If people did not have the rights established in Miranda v. Arizona, they could be subjected to police coercion and forced confessions without a lawyer present. This could lead to wrongful convictions and denial of due process, which are essential components of the American justice system

The answer is A. PE becomes part of American school curriculum.

Title IX, which forbids gender discrimination in sports, was enacted in 1972.

The YMCA opened its first gym in America in Boston in 1851.

The adjustable plate-loaded barbell was invented in the late 1940s by a man named George Snyder.

Physical Education (PE) has been a part of American school curriculums for well over a century, with the first school gymnasiums being built in the 1800s.

What is an american school curriculum?

The American school curriculum is the set of educational standards and guidelines used by schools in the United States to structure their academic programs. The curriculum typically includes courses in core academic subjects such as English, math, science, and social studies, as well as elective courses in areas such as the arts, foreign languages, and physical education.

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Downslope at 0.64 m/s², m = 12.4 kg

Upslope at 0.76 m/s², m = 6.35 kg

In Physics, mass is the most basic property of matter, and it is one of the fundamental quantities. Mass is defined as the amount of matter present in a body. The SI unit of mass is the kilogram (kg). The formula of mass can be written as:

Part A)

The sum of forces on the left-hand mass

T - mgsin(angle) = ma

T - (2.6) (9.8) (sin 70) = 2.6(.64)

T = 25.6 N

m = left mass.........M = right mass

T - mg×sin70 = ma

Mg×sin16 - T = Ma

Mg×sin16 - mg×sin70 = a×(M+m)

M×g×sin16 - mg×sin70 = Ma + ma

M× (g×sin16 -a) = m× (a + gsin70)

M = m× (a + gsin70) / (g×sin16 -a)

a) a = 0.64

M = 10.98 Kg

b) M = 11.82 kg

For the right-hand mass, the sum of forces...

mgsin(angle) - T = ma

m (9.8) (sin 16) - 25.6 = m (.64)

2.7m - 25.6 = .64m

m = 12.4 kg

Part B)

For the left-hand mass

mgsin(70) - T = ma

(2.6) (9.8) (sin 70) - T = (2.6) (.76)

T = 21.97 N

Then for the right-hand mass

T - mgsin(16) = ma

21.97 - m (9.8) (sin 16) = m (.76)

21.97 - 2.7m = .76m

m = 6.35 kg

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The right-hand mass should be 3.3 kg to accelerate up the slope at 0.76 m/s². To solve this problem, we need to use the principles of Newton's laws of motion and trigonometry.

We know that the force of gravity acting on the mass is equal to its weight, which can be calculated using the formula Fg = mg, where Fg is the force of gravity, m is the mass, and g is the acceleration due to gravity (which is approximately 9.8 m/s²).

To find the force acting down the slope, we need to calculate the component of the weight that acts down the slope, which is given by Fg sin θ, where θ is the angle of the slope. Using the given angle of 52°, we can calculate the force acting down the slope for the left-hand mass as:

Fdown = Fg sin θ

Fdown = (2.5 kg)(9.8 m/s²) sin 52°

Fdown = 18.9 N

To find the required mass for the right-hand mass to accelerate at 0.64 m/s^2 down the slope, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass times its acceleration (F = ma). Therefore, we can calculate the required force for the right-hand mass as:

F = ma

F = (m)(0.64 m/s²)

Since the force acting down the slope is 18.9 N, we can set these two equations equal to each other and solve for the mass:

F = Fdown

(m)(0.64 m/s²) = 18.9 N

m = 29.5 kg

Therefore, the right-hand mass should be 29.5 kg to accelerate down the slope at 0.64 m/s².

To find the required mass for the right-hand mass to accelerate at 0.76 m/s² up the slope, we can use the same approach, but this time we need to use the component of the weight that acts up the slope, which is given by Fg cos θ, where θ is the angle of the slope. Using the given angle of 25°, we can calculate the force acting up the slope for the right-hand mass as:

Fup = Fg cos θ

Fup = (m)(9.8 m/s²) cos 25°

Setting this equal to the force required for the right-hand mass to accelerate up the slope, we get:

Fup = ma

(m)(0.76 m/s²) = (m)(9.8 m/s²) cos 25°

m = 3.3 kg

Therefore, the right-hand mass should be 3.3 kg to accelerate up the slope at 0.76 m/s².

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Answer:

Explanation:

Assuming that air resistance is negligible, we can use the following kinematic equations to solve for the peak height:

v_f^2 = v_i^2 + 2ad

where v_f = 0 m/s (at the peak height) and a = -9.8 m/s^2 (acceleration due to gravity)

and

d = v_i t + (1/2)at^2

where d is the displacement or the peak height we want to find, v_i is the initial velocity, t is the time it takes to reach the peak height.

First, we need to resolve the initial velocity into its vertical and horizontal components:

v_i_x = v_i cos(30°) = 121.1 m/s

v_i_y = v_i sin(30°) = 70.0 m/s

Next, we can use the vertical component of the initial velocity to find the time it takes to reach the peak height:

v_f = v_i_y + at

0 m/s = 70.0 m/s + (-9.8 m/s^2)t

t = 7.14 s

Finally, we can use the time we found and the kinematic equation for displacement to find the peak height:

d = v_i_y t + (1/2)at^2

d = (70.0 m/s)(7.14 s) + (1/2)(-9.8 m/s^2)(7.14 s)^2

d = 247.5 m

Therefore, the peak height of the football is 247.5 meters.

Answer:

Explanation:

Assuming negligible air resistance, the horizontal velocity of the shingle will remain constant and the vertical motion will be influenced by gravity.

We can use the kinematic equations of motion to determine the horizontal distance traveled by the shingle. The relevant equation is:

d = v * t

where d is the distance, v is the initial horizontal velocity, and t is the time of flight.

To find the time of flight, we can use the equation for the vertical displacement of an object under constant acceleration:

y = v0t + (1/2)at^2

where y is the vertical displacement, v0 is the initial vertical velocity (which is zero), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight. Solving for t, we get:

t = sqrt((2y)/a)

where sqrt means square root.

Substituting the given values, we have:

y = 9.4 m

a = -9.8 m/s^2

t = sqrt((2*9.4 m) / -9.8 m/s^2) = 1.45 s (using the positive root since time cannot be negative)

Now, we can use the horizontal velocity to find the distance traveled in this time:

d = v * t = 7.2 m/s * 1.45 s = 10.44 m

Therefore, the shingle moves a horizontal distance of 10.44 meters in this time.

Answer:

Explanation:

alot

Answer:

0.625 meters

Explanation:

We can use the work-energy that the work done on an object is equal to the change in its kinetic energy:

Work = ΔK = Kf - Ki

Where:

Work is the work done on the object

ΔK is the change in kinetic energy of the object

Kf is the final kinetic energy of the object

Ki is the initial kinetic energy of the object (which is zero since the object is at rest)

The work done on the object is equal to the force applied to the object multiplied by the distance over which the force is applied:

Work = F × d

Where:

F is the force applied to the object (50 N)

d is the distance over which the force is applied (unknown)

So we can write:

F × d = Kf - Ki

Substituting the given values:

50 N × d = 1/2 × 10 kg × (2.5 m/s)^2 - 0

Simplifying:

50 N × d = 31.25 J

Solving for d:

d = 31.25 J / 50 N = 0.625 m

Therefore, the object was moved a distance of 0.625 meters.

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Answer:

0 Newtons, to the nearest whole number

Explanation:

Using the formula attached to this answer,

F = (6.67 x 10-¹¹ • 95 • 82) / 0.6²

F = 5.19593 x 10-⁷ / 0.36

F = 1.4433 x 10-⁶ N

F = 0.000001443 N

To the nearest whole number

F = 0 N

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Answer:

It depends on how they are made.

Explanation:

Rubber balloons are not always spherical in shape. When filled with air, the inflation forms a balance between the balloon material, including its shape and thickness and its elasticity and the pressure of the air. That’s what determines its shape. Although a sphere is often the shape, it could be tubular, offset, and most other shapes.

Answer:

Wavelength will decrease but frequency will stay the same.

Explanation:

The higher the frequency, the shorter the wavelength. The relationship between wavelength and frequency is called an inverse relationship, because as the frequency increases, the wavelength decreases.

However, the frequency usually remains the same because it is like a driven oscillation and maintains the frequency of the original source. If v changes and f remains the same, then the wavelength λ λ must change. Since v = f λ v = f λ , the higher the speed of a sound, the greater its wavelength for a given frequency.

Answer:

Abundance on Earth is not necessary for calculating the age of something using radioactive dating methods.

In contrast, the amount of radioactive isotope within a sample and its half-life are crucial for determining the age of the sample using radioactive dating methods. The amount of sample, such as 100g, is also important for determining the quantity of radioactive isotopes present, which is used in the calculation of the age of the sample.

So the change in energy of the box after being pulled by Liam over 19 m is 323 J.

Energy is capable of changing its forms. For instance, electrical energy transforms into thermal and light energy when a lightbulb is turned on. A automobile transforms the gasoline's molecular bonds' stored energy into a variety of other forms. Chemical energy is converted to light in the engine through a chemical reaction.

The work done on the box by Liam's force is:

work = force x distance x cos(theta)

where theta is the angle between the force and the direction of motion. Since the force is horizontal and the motion is also horizontal, theta is 0 degrees and cos(theta) is 1. Therefore, the work done on the box is:

work = 17 N x 19 m x 1 = 323 J

The change in energy of the box is equal to the work done on it, according to the work-energy principle. Therefore, the change in energy of the box is:

change in energy = work = 323 J

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Answer:

- 0.33 m/s

Explanation:

An illustration is shown above,

In this case, since the two objects move in opposite directions before collision, then move together, the formula to be used is,

m1u1 - m2u2 = (m1 + m2)v

Where,

m1 = mass of the first object

u1 = initial velocity of the first object

v1 = final velocity of the first object

m2 = mass of the second object

u2 = initial velocity of the second object

v2 = final velocity of the second object

Therefore,

(4.0 • 3.0) - (8.0 • 2.0) = (4.0 + 8.0)v

12 - 16 = 12v

-4 = 12v

Divide both sides by 12,

-4 / 12 = 12v / 12

-1 / 3 = v

v = -0.33 m/s

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Answer:

2) 42.6

3) 1500000 m

4) no he will not be late, it will take 3s

Explanation:

 #2:

Formula for average velocity:
 ⇒  [tex]v=\frac{d}{t}[/tex]

Substitute

 ⇒ [tex]\frac{2560}{60}[/tex]

⇒ [tex]42.6[/tex]

#3:

1st convert 60km to m/s (correct si unit must always come first)

to convert kilometer to meter, multiply by 1000 (prefix 'kilo' = 1000)

60 km x 1000 = 60000 m

use the formula for speed, distance, & time

60000 = d/25

to find displacement just multiply the speed and time together

60000 x 25

1500000m is the displacement

#4:

s = d/t

given are displacement and speed/velocity

d = 25m

v = 8.5 m/s

to find the time just divide displacement by speed

8.5 = 25/t

25/8.5

≈ 3 seconds

so he will not be late for class.

The speed of the wind is 20 miles per hour.

To solve this problem, we can use the formula:

Speed = Distance/Time

Let's assume that the speed of the wind is x miles per hour.

With the wind, the plane travels at a speed of 460 miles per hour. This means that its speed relative to the ground is the sum of its airspeed and the speed of the wind:

460 = Airspeed + x

Against the wind, the plane travels at a speed of 420 miles per hour. This means that its speed relative to the ground is the difference between its airspeed and the speed of the wind:

420 = Airspeed - x

We can solve this system of equations to find the airspeed of the plane:

460 = Airspeed + x

420 = Airspeed - x

Adding the two equations gives:

880 = 2Airspeed

Dividing both sides by 2 gives:

Airspeed = 440 miles per hour

Now that we know the airspeed of the plane, we can find the speed of the wind by substituting this value into one of the equations we obtained earlier:

460 = Airspeed + x

460 = 440 + x

x = 20

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