A rotating flywheel can be used as a method to store energy. If it has 1.0 ´ 106 J of kinetic energy when rotating at 400 rad/s, and if a frictional torque of 4.0 N×m acts on the system, in what interval of time would the flywheel come to rest?

Both of the following statements are true the average speed of gas molecules decreases with decreasing temperature. the average kinetic energy of gas molecules increases with increasing temperature.

Both statements are true because the average speed of gas molecules decreases with decreasing temperature because as the temperature decreases, the molecules/ gas have less energy, resulting in slower movement. molecules have less energy, resulting in slower movement. move slower due to reduced energy.

On the other hand, the average kinetic energy of gas molecules increases with increasing temperature because as the temperature increases, the molecules move faster and have more energy. When temperature increases, the gas molecules gain more energy and move faster, leading to an increase in their average kinetic energy. The motor energy of an article is the type of energy that it has because of its movement. It is defined as the effort required to propel a mass-bearing object from rest to its stated velocity.

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Answer:

a. Spectral type and Color

Answer:

The fundamental frequency of a person's speech, also known as the pitch or F0, can vary depending on various factors such as age, gender, emotion, and cultural background. Studies have shown that the fundamental frequency tends to increase when a person is excited or joyful. This increase in pitch during intense emotions is thought to be due to changes in the tension of the vocal cords and increased respiratory activity.

Explanation:

When the experiment is carried out in air and then immersed in water, the wavelength of light remains constant but the speed of light changes. This means that the distance between bright fringes will change. Specifically, the fringes will move closer together in water than they were in air. Therefore, the correct answer is (b) They move closer together.

When the apparatus is immersed in water, the wavelength of light decreases because the speed of light is slower in water than in air. This means that the distance between successive bright fringes decreases because the fringe spacing is proportional to the wavelength of light. Therefore, the bright fringes move closer together in the water medium as compared to their spacing in air.

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The Millenium Falcon is chased by the Imperial Forces. The ship is moving at a speed of 0. 587 c. Han Solo is shooting at the imperial fighters with his newly installed proton cannon purchased at the MSU Surplus Store for $20. 00 plus 6. 00% tax. The cannon emits protons at a speed of 0. 831 c with respect to the ship.

With the use of Relativistic velocity addition formula  we will find the velocity of proton in the resting frame of the movie audience in terms of the speed of the light when the cannon is shot in the forward direction
Formula is given as
v = (u+v')/(1+u*v'/[tex]c^{2}[/tex])
Where
v = velocity of the protons in the resting frame of the movie audience

u = velocity of the Millennium Falcon with respect to the audience

v' = velocity of the protons with respect to the Millennium Falcon

c = speed of light

By putting all the values we get
v = (0.587c + 0.831c) / (1 + 0.587c*0.831c/[tex]c^{2}[/tex])

v = (1.418c) / (1 + 0.486)

v = 0.942c

Hence, the velocity of the protons in the resting frame of the movie audience is 0.942 times the speed of light.

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The weight of one book B must be less than the weight of one book (w), and the weight of book C must be greater than the weight of one book (w).

Let's assume that the weight of one book is w, then the weight of the three books stacked vertically is 3w.

According to Newton's Third Law, the force exerted by book1 on book2 is equal in magnitude and opposite in direction to the force exerted by book2 on book1.

So, we have:

Normal force of book1 on book2 = w + 2w = 3w

Since the books are identical. Therefore:

Normal force of book1 on book2 = w + 2w = 3w

This normal force is equal in magnitude to the weight of bookC.

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When a horse pulls a cart, it applies a force on the wagon in the forward direction. According to Newton's third law, the wagon also applies an equal and opposite force on the horse in the backward direction. These two forces cancel each other out, resulting in a net force of zero.

In other words, the horse is pulling on the cart, and the cart is pulling back on the horse with an equal force. These forces cancel each other out, resulting in a net force of zero. When there is no net force acting on an object, the object remains at rest or moves at a constant speed in a straight line, according to Newton's first law of motion. Therefore, if the horse pulls on the cart but the cart is stationary or immovable, the net force on the cart is zero, and the cart will not move. However, if there is an additional force acting on it, such as friction from the ground the cart will start to move.

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The capacitance of the second capacitor is most nearly Co. The correct option is C.

The capacitance of a parallel-plate capacitor is given by:

C = εA/d

where ε is the permittivity of the material between the plates, A is the area of each plate, and d is the separation between the plates.

For the second parallel-plate capacitor, the area of each plate and the separation between the plates are both doubled. Therefore, we have:

C' = ε(2A)/(2d) = εA/d = C

So the capacitance of the second capacitor is unchanged and is equal to the capacitance of the first capacitor, which is Co.

On the other hand other options:

(A) ¼Co - This is incorrect because doubling both the area and separation of the plates would decrease the capacitance of the capacitor, not increase it. So, the capacitance cannot be 1/4 Co.

(B) ½Co - This is also incorrect for the same reason as option A. Doubling both the area and separation of the plates would decrease the capacitance of the capacitor, not increase it. So, the capacitance cannot be 1/2 Co.

(D) 2Co - This is incorrect because doubling both the area and separation of the plates would decrease the capacitance of the capacitor, not double it. So, the capacitance cannot be 2 Co.

(E) 4Co - This is also incorrect for the same reason as options D. Doubling both the area and separation of the plates would decrease the capacitance of the capacitor, not quadruple it. So, the capacitance cannot be 4 Co.

Therefore, the answer is (C) Co.

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The acceleration of a projectile at its highest point is zero. The  acceleration just before and just after reaching this point is due the vertical acceleration of the projectile  due to gravity, which is approximately 9.81 m/s² downward. This value remains constant throughout the projectile's motion.

When a projectile reaches its highest point, its acceleration is zero. This is because at the highest point, the projectile has momentarily come to a stop and is about to start falling back down due to gravity. Just before reaching this point, the acceleration of the projectile is negative, as it is slowing down due to the opposing force of air resistance. Just after reaching the highest point, the acceleration of the projectile becomes positive, as it starts accelerating due to the force of gravity pulling it back down.

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The total energy may be unchanged or decreased by the collision, depending on whether it is elastic or inelastic.

In an elastic collision, the total kinetic energy of the system is conserved, meaning that the total energy before and after the collision remains the same. In this case, the objects involved in the collision bounce off each other without any loss of energy. In contrast, in an inelastic collision, the total kinetic energy of the system is not conserved because some of the energy is lost in the form of heat, sound, or deformation. In this case, the objects involved in the collision stick together, and the final kinetic energy is lower than the initial kinetic energy.

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The velocity of a wave on a string is determined by the tension in the string, the mass of the string, and the linear density of the string.

The mass of the string per unit length is known as its linear density. Since the tension in the string is typically constant, the mass and linear density define the wave's velocity.

The velocity rises when the linear density drops, the string mass rises, or the tension in the string falls. A light, thin string, for instance, will move more quickly than a heavy, thick string.

The medium through which the wave is moving, such as air or a solid, has an impact on the wave's velocity as well.

Compared to a medium like air, a solid medium has a higher wave velocity.

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You have a tube that is half a meter long, with both ends closed, and the speed of sound under current conditions is 344 m/s. You'd like to know the lowest resonant frequency (largest wavelength) with units in Hz.

To find the lowest resonant frequency, we will use the formula for a closed-closed tube:

f = (2n - 1) * (v / 4L)

Here, f is the frequency, n is the harmonic number, v is the speed of sound, and L is the length of the tube. For the lowest resonant frequency, n = 1.

Step 1: Plug in the given values.
f = (2(1) - 1) * (344 m/s / 4(0.5 m))

Step 2: Simplify the equation.
f = (1) * (344 m/s / 2 m)

Step 3: Calculate the frequency.
f = 172 Hz

The lowest resonant frequency (largest wavelength) for your half-meter-long closed-closed tube under the given conditions is 172 Hz.

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We can distinguish between a magnetic and an electric field by observing the motion of the charged particle, analyzing the direction of motion, and considering the influence of the particle's charge.

1. Observe the motion of the charged particle: When placed in an electric field, a charged particle will experience a force that is either attracted to or repelled from the source of the field, depending on the charge. In a magnetic field, the charged particle will experience a force that is perpendicular to both its velocity and the magnetic field, causing it to move in a circular or helical path.

2. Analyze the direction of motion: In an electric field, the charged particle moves in a straight line along the field lines, either towards or away from the source, depending on its charge. In a magnetic field, the charged particle moves in a curved path, with its direction determined by the right-hand rule.

3. Consider the influence of the particle's charge: In an electric field, the force experienced by the particle is directly proportional to its charge, while in a magnetic field, the force depends on the charge, the velocity of the particle, and the magnetic field strength.

By observing these differences in motion, you can distinguish between the presence of an electric field or a magnetic field acting on a charged particle.

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To find the change in potential energy of the charge, we need to use the formula:

ΔU = qΔV

where ΔU is the change in potential energy,

q is the charge that flows, and

ΔV is the change in electric potential.

In this case, 10.5 coulombs of positive charge flow from a +12 V potential to a 0 V potential. So the change in electric potential is:

ΔV = Vf - Vi = 0 - 12 = -12 V

The negative sign indicates that the potential difference is in the opposite direction to the direction of the flow of charge.

Now we can use the formula to find the change in potential energy:

ΔU = qΔV = (10.5 C)(-12 V) = -126 J

Therefore, the change in potential energy of the charge is -126 J (joules). The negative sign indicates that the potential energy of the charge has decreased as it flowed from the higher potential to the lower potential.

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It requires 0.375j of work to stretch the spring an additional 5 cm beyond its natural length.

To stretch the spring 10 cm beyond its natural length requires 15j of work. To stretch it an additional 5 cm requires additional work.

We can use the formula for work done on a spring:

Work = 0.5kx²

Where k is the spring constant and x is the distance the spring is stretched or compressed from its natural length.

Since we know that 15j of work is required to stretch the spring 10 cm, we can solve for k:

15j = 0.5k(0.1m)²

k = 300 N/m

Now we can use this value of k to find the work required to stretch the spring an additional 5 cm:

Work = 0.5(300 N/m)(0.05m)²

Work = 0.375j

It requires 0.375j of work to stretch the spring an additional 5 cm beyond its natural length.

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The sugar molecules that results in the formation of new compounds, including caramel.

A physical change is a change that occurs without altering the chemical composition of a substance. In other words, the chemical identity of the substance remains the same before and after the change. Among the options provided, the correct answer is (b) sand is filtered from salt water. This is a physical change because the salt water and sand are physically separated, but their chemical composition remains unchanged.

Option (a) is incorrect because the process of converting glucose into energy involves chemical reactions that alter the chemical composition of glucose.

Option (c) is incorrect because the process of rusting involves a chemical reaction between iron, oxygen, and water that results in the formation of a new compound, iron oxide.

Option (d) is incorrect because the process of burning propane involves a chemical reaction between propane and oxygen that results in the formation of new compounds, carbon dioxide and water.

Option (e) is incorrect because the process of heating sugar into caramel involves a chemical reaction between the sugar molecules that results in the formation of new compounds, including caramel.

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The total electric field strength along the line connecting the midpoints of a charged glass and plastic rod decreases with distance and is directed towards the plastic rod at distances of 2.0 cm and 3.0 cm.

The negative sign in E1', E2', and E3' indicates that the electric field created by the plastic rod is in the opposite direction to that of the glass rod.

To find the total electric field at each distance, we can use the principle of superposition, which states that the total electric field at a point is the vector sum of the electric fields created by each charged object at that point.

At each distance, the direction of the electric field created by the glass rod is the same, while the direction of the electric field created by the plastic rod is opposite. Therefore, the total electric field E1 to E3 along the line connecting the midpoints of the two rods is given by:

E1 = E1' + E = 0

E2 = E2' + E = -4.86 x 10³ N/C

E3 = E3' + E = -1.61 x 10³ N/C

where E is the electric field created by the glass rod alone.

Thus, at a distance of 1.0 cm from the glass rod along the line connecting the midpoints of the two rods, the total electric field is zero. At distances of 2.0 cm and 3.0 cm, the total electric field is directed towards the plastic rod and decreases in magnitude as the distance increases.

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The question is -

A 10-cm-long thin glass rod uniformly charged to 7.00 nC and a 10-cm-long thin plastic rod uniformly charged to -7.00 nC are placed side by side, 4.20 cm apart. What are the electric field strengths of E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?

As it is a result of external forces acting upon it rather than a failure to maintain its state of motion.

The observation that a ball rolling down a bowling alley moves slightly slower with time does not violate Newton's first law of motion, also known as the law of inertia.

Newton's first law states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity (which includes moving at a constant speed in a straight line or moving with a constant speed in a curved path) unless acted upon by an external force. In other words, an object will maintain its state of motion (or lack thereof) unless an external force is applied to it.

In the case of a ball rolling down a bowling alley, the ball is subject to various external forces that act upon it and cause it to slow down. These forces include friction between the ball and the lane, air resistance, and deformation of the ball itself. These forces act in the opposite direction of the ball's motion and cause it to lose speed over time, as the ball's kinetic energy is dissipated into other forms of energy.

Therefore, the ball's decrease in speed over time is not a violation of Newton's first law, as it is a result of external forces acting upon it rather than a failure to maintain its state of motion.

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A radioactive substance has a mean-life τ of 16 years. 35.5 kg of the substance will be left after 27 years. So after 27 years, there will be approximately 30.6 kg of the radioactive substance left.


We can use the formula for radioactive decay:
N(t) = N0 * e^(-λt)
where N(t) is the amount of the substance remaining after time t, N0 is the initial amount, λ is the decay constant (related to the mean-life τ by λ = ln(2)/τ), and e is Euler's number (approximately 2.71828).
We know N0 = 166 kg, τ = 16 years, and t = 27 years. Therefore:
λ = ln(2)/τ = ln(2)/16 = 0.04355 (rounded to five decimal places)
N(27) = 166 * e^(-0.04355 * 27) ≈ 35.5 kg (rounded to one decimal place)
To find out how many kg of a radioactive substance with a mean-life τ of 16 years will be left after 27 years, given that we start with 166 kg, we can use the decay formula:
Remaining amount = Initial amount * e^(-t/τ)
where Remaining amount is the mass left after t years, Initial amount is the initial mass (166 kg), t is the time elapsed (27 years), and τ is the mean-life (16 years)

Step 1: Insert the values into the formula:
Remaining amount = 166 kg * e^(-27/16)
Step 2: Calculate the exponent:
Exponent = -27/16 ≈ -1.6875
Step 3: Calculate e^(exponent):
e^(-1.6875) ≈ 0.1847
Step 4: Multiply the initial amount by the calculated value:
Remaining amount = 166 kg * 0.1847 ≈ 30.6 kg
So after 27 years, there will be approximately 30.6 kg of the radioactive substance left.

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When object 1 and object 2 collide, the forces being applied by object 1 to object 2 and by object 2 to object 1 are equal in strength but directed in different directions.

According to Newton's third law of motion, for every action, there is an equal and opposite reaction.

We can suppose that both objects feel equal and opposite forces during the collision because their masses are the same and they are traveling toward one another at the same speed. This is because the force the item experiences are equal to the rate at which its momentum changes, and since the two objects share the same mass and are traveling at the same speed toward one another, their momentum will change in an opposite manner during the collision.

Therefore, The force that objects 1 and object 2 applied to each other when they collided was of similar magnitude, as was the force that objects 2 applied to item 1.

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The magnitude of the resultant force acting on the object during this time interval is approximately 2.8 N. The correct option is 3.

To find the magnitude of the resultant force acting on the 1.5-kg object, we first need to determine the acceleration. The initial velocity (v0) is given as 5j m/s, and the final velocity (vf) is (6i + 12j) m/s after 5 seconds (t). We can find the acceleration (a) using the formula:

a = (vf - v0) / t

a = [(6i + 12j) - (0i + 5j)] / 5
a = (6i + 7j) / 5
a = 1.2i + 1.4j m/s²

Now that we have the acceleration, we can find the net force (F) using Newton's second law of motion, F = ma:

F = (1.5 kg) × (1.2i + 1.4j) m/s²
F = 1.8i + 2.1j N

To find the magnitude of the resultant force, we use the Pythagorean theorem:

|F| = √[[tex](1.8)^2 + (2.1)^2[/tex]]
|F| = √(3.24 + 4.41)
|F| = √(7.65)
|F| ≈ 2.8 N

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The lighter cart will experience greater impulse, and is opposite to that experienced by the heavier one.

If the cart 1 and cart 2 are having unequal masses, the cart with greater mass will exert more impulse on the lighter cart.

This is because, according to laws of motion, impulse is equal to the change in momentum of an object.

So, during the collision, the lighter cart will have a greater velocity than the heavier one. As a result, it will experience a greater impulse.

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The magnitude of the other force acting on the object is approximately 13N. Therefore, option (2) is closest and correct.

To determine the magnitude of the other force acting on the object, we can use Newton's second law of motion, which states that the net force ([tex]F_{net}[/tex]) acting on an object is equal to the product of its mass (m) and acceleration (a):

[tex]F_{net}[/tex]= m * a

Given:

Mass of the object (m) = 4.0 kg

Acceleration (a) = (5i + 3j) m/s^2

Known force (F₁) = (12i + 22j) N

To find the unknown force (F₂), we can use the equation:

[tex]F_{net}[/tex] = F₁ + F₂

Substituting the given values, we have:

(12i + 22j) N + F₂ = (4.0 kg) * (5i + 3j) m/[tex]s^2[/tex]

Now, we can equate the corresponding components of the vectors:

12 + F₂ᵢ = 20

22 + F₂ⱼ = 12

Solving these equations, we find:

F₂ᵢ = 20 - 12 = 8

F₂ⱼ = 12 - 22 = -10

Thus, the magnitude of the other force (F₂) can be calculated using the Pythagorean theorem:

|F₂| = [tex]\sqrt{{(F_{2i})^2 + (F_{2j})^2}} = \sqrt{{8^2 + (-10)^2}} = \sqrt{{64 + 100}} \approx 12.81 , \text{N}[/tex]

Rounding to one decimal place, the magnitude of the other force acting on the object is approximately 12.8 N. Therefore, the closest option in the provided choices is 13 N (option 2).

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The given statement is true. Assuming the same velocity at take-off, with the exception of a take-off angle of 45 degrees, there are two take-off angles that will give identical range.

These two take off angles are complementary angles, which means their sum equals 90 degrees. If one angle is θ, the other angle will be (90 - θ). This is because the range of a projectile is determined by both its initial velocity and launch angle, and the range equation has a sine function, which exhibits symmetry for complementary angles.Hence, there are two take-off angles that will give identical range.

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The kinetic energy of the bullet as it leaves the block compared to when it entered is 25% (or 0.25 times).

To find the kinetic energy of the bullet as it leaves the block compared to when it entered, we can use the formula for kinetic energy:

KE = 0.5 * m * v², where m is the mass of the bullet and v is its velocity.

When the bullet enters the block, its velocity is v. After it penetrates the block, its velocity is reduced by half, meaning the new velocity is 0.5v.

Now, let's find the ratio of the kinetic energy after leaving the block to the kinetic energy when it entered:

KE_after/KE_before = (0.5 * m * ((0.5v)²) / (0.5 * m * v²)

Notice that the mass and the 0.5 constant factor will cancel out:

= ((0.5v)²) / (v²)

Now, we can square the term in the numerator:

= (0.25v²) / (v²)

Finally, the v² terms cancel out:

= 0.25

So, the kinetic energy of the bullet as it leaves the block is 25% (or 0.25 times) of the kinetic energy when it entered the block.

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To convert Celsius to Kelvin, you simply add 273.15. Therefore, 15°C is equivalent to 288.15 K and 35°C is equivalent to 308.15 K. The incremental change in Kelvins would be 20 K (308.15 K - 288.15 K = 20 K).

To determine the incremental change in kelvins when a substance is heated from 15°C to 35°C, follow these steps:
1. Find the temperature difference in Celsius: 35°C - 15°C = 20°C
2. Convert the temperature difference to Kelvin: Since 1°C = 1K, the incremental change in kelvins is the same as in Celsius. The incremental change when registered in kelvins would be 20K.

Kelvin 0 degrees is the temperature or kinetic energy of zero. Because these scales don't start at zero, changes in Celsius or Fahrenheit don't directly relate to kinetic energy or volume. The worth of one degree on the Kelvin scale is indistinguishable from the worth of one degree on the Celsius scale that is the temperature differential or change is indistinguishable on the two scales.

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When two identical arrows are fired at a target, the arrow that is moving twice as fast will have more kinetic energy than the slower one.  The faster arrow will penetrate the target four times as much.

This means that it will be able to penetrate the target deeper than the slower arrow.

However, the amount of penetration will not be exactly half or four times as much, as there are several factors that come into play such as the weight and design of the arrows, the type of target, and the angle of impact.

In general, when a projectile is moving faster, it will have more kinetic energy and momentum, which will allow it to overcome the resistance of the target more easily.

However, the depth of penetration will also depend on the density and toughness of the target material, as well as the angle at which the arrow hits it.

Therefore, it is difficult to predict the exact amount of penetration without more specific information about the arrows and the target.

In summary, the arrow that is moving twice as fast will generally penetrate the target deeper than the slower arrow, but the actual amount of penetration will depend on various factors. The right answer will be four times as much.

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As a spinning star collapses into a neutron star of a smaller radius, retaining essentially all of its original mass, its angular velocity increases, following the conservation of angular momentum.

This is due to the conservation of angular momentum, which states that the total angular momentum of a system remains constant unless an external torque acts upon it. Since the star's radius decreases during the collapse, its moment of inertia also decreases, causing an increase in its angular velocity to conserve angular momentum.

This effect is known as the "spin-up" phenomenon and is observed in many astrophysical phenomena, including neutron star mergers and black hole formation.

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The value of m at the end of the spring is approximately 12.33 kg.

The frequency of a spring-mass system is given by:

f = 1 / (2π) * √(k / m)

where f is the frequency, k is the spring constant, and m is the mass.

Let's call the mass at the end of the spring m1, and the additional mass added to m₁ m₂. We can write two equations for the system, one for each scenario:

f₁ = 1 / (2π) * √(k / m₁) -- Equation 1

f₂ = 1 / (2π) * √(k / (m₁ + m₂)) -- Equation 2

We know that the frequency of the system is reduced to f/3 when m2 is added, so we can write:

f₂ = f/3 -- Equation 3

Now we can substitute Equation 3 into Equation 2:

f/3 = 1 / (2π) * √(k / (m₁ + m₂))

Multiplying both sides by 3 and squaring both sides, we get:

9f² = (k / π²) * (m₁ + m₂)

Similarly, we can substitute Equation 1 into the same equation and get:

f² = (k / π²) * m₁

Dividing the two equations, we get:

9 = (m₁ + m₂) / m₁

Simplifying and rearranging, we get:

m₁ = 9m₂

Therefore, the mass at the end of the spring, m₁, is 9 times the additional mass added, m₂.

Substituting this relationship into Equation 3, we get:

f/3 = 1 / (2π) * √(k / (10m₂))

Solving for m₂:

m₂ = k / (4π²f²) * (10/9)²

We do not have a value for k, so we cannot solve for m₂ directly. However, we do know that m₁ = 9m₂. If we assume that k is constant between the two scenarios, then we can write:

f₁ / f₂ = √(m₁+ m₂) / √m₁

Substituting the relationship m₁ = 9m₂ and f₁ / f₂= 3, we get:

3 = √(10m₂) / 3√m₂

Squaring both sides and simplifying, we get:

m₂ = 27m₂ / 100

Solving for m₂, we get:

m₂ = 100 / 73 kg

Finally, we can calculate m₁:

m₁ = 9m₂ = 900 / 73 kg ≈ 12.33 kg

Therefore, the value of m at the end of the spring is approximately 12.33 kg.

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The type of wave that is purely longitudinal among the options provided is sound waves emitted from a speaker.

Longitudinal waves involve particles oscillating parallel to the direction of the wave's propagation, and this is the case for sound waves. Sound waves emitted from a speaker are an example of a purely longitudinal wave. In a longitudinal wave, the oscillations of the particles or medium are in the same direction as the direction of the wave propagation. Sound waves travel through a medium such as air, and as they move, they cause the air particles to vibrate in a parallel direction to the direction of the wave propagation. This results in the compressions and rarefactions of air particles, which we perceive as sound. Therefore, sound waves are classified as longitudinal waves. Other examples of longitudinal waves include seismic waves and ultrasonic waves.

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