Question 24 of 30How does the force of gravity vary with distance?O A. As distance increases, gravity becomes zero.B. As distance increases, gravity increases.C. As distance increases, gravity decreases.D. As distance increases, gravity stays the same.SUBMIT

Answer:

Ionic bond

Explanation:

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Answer:

N = 1.42 × 10⁴ cycles

Explanation:

Given that:

frequency f = 4.9 × 10¹⁴ Hz

Time = 2.9 × 10⁻¹¹ s

Speed = 2.3 × 10⁸ m/s

Recall that:

wavelength \(\lambda = \dfrac{c}{f} \\ \\\)

Horizontal distance \(\Delta x = ct\)

Number of wavelengths \((N) = \dfrac{\Delta x}{\lambda}\)

\(N = \dfrac{ct}{c/f} \\ \\ N= ft\)

N = (4.9 × 10¹⁴ cycles/s) (2.9 × 10⁻¹¹ s)

N = 14210

N = 1.42 × 10⁴ cycles

Answer:

37.4m/s

Explanation:

since the car doesn't accelerate, we can use the formula v=ωr where v is linear speed, ω is angular speed (rads/second) and r is radius. Substitute values for equation:

v=55*0.68

v=37.40

Answer:

The amount of energy carried by a wave is related to the amplitude of the wave

Explanation:

A high energy wave is characterized by a high amplitude; a low energy wave is characterized by a low amplitude.  The energy imparted to a pulse will only affect the amplitude of that pulse.

Hope this helped!!!

Answer: Distance traveled in time t is s

Explanation: Self Explanatory

a) The speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

b) We cannot calculate the work done by the friction force.

c) The net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

(a) The work done on the 6.00 kg block by gravity can be calculated using the formula:

Work_gravity = force_gravity * displacement * cos(theta),

where force_gravity is the weight of the block, displacement is the downward displacement of the block, and theta is the angle between the force and displacement vectors (which is 0 degrees in this case).

The weight of the block is given by:

force_gravity = mass * acceleration_due_to_gravity = 6.00 kg * 9.8 m/s^2 = 58.8 N.

Plugging in the values, we get:

Work_gravity = 58.8 N * 0.800 m * cos(0) = 47.04 J.

The work done on the 6.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(180) = -29.6 J.

The negative sign indicates that the tension is in the opposite direction of the displacement.

Using the work-energy theorem, we can find the speed of the 6.00 kg block after descending 0.800 m:

Work_net = change_in_kinetic_energy.

Since the block starts from rest, its initial kinetic energy is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * mass * velocity^2.

Solving for velocity, we get:

velocity = sqrt(2 * Work_net / mass).

The net work done on the block is the sum of the work done by gravity and the tension:

Work_net = Work_gravity + Work_tension = 47.04 J - 29.6 J = 17.44 J.

Plugging in the values, we get:

velocity = sqrt(2 * 17.44 J / 6.00 kg) = 2.07 m/s.

Therefore, the speed of the 6.00 kg block after descending 0.800 m is 2.07 m/s.

(b) The total work done on the 8.00 kg block during the 0.800 m displacement can be calculated using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the 8.00 kg block is not moving vertically, its initial and final kinetic energies are zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 0.

The work done on the 8.00 kg block by the tension in the rope is given by:

Work_tension = tension * displacement * cos(theta).

Plugging in the values, we get:

Work_tension = 37.0 N * 0.800 m * cos(0) = 29.6 J.

The work done on the 8.00 kg block by the friction force can be calculated using the formula:

Work_friction = force_friction * displacement * cos(theta),

where force_friction is the frictional force on the block. However, the problem statement does not provide the value of the friction force. Therefore, we cannot calculate the work done by the friction force.

(c) The net work done on the system of two blocks during the 0.800 m displacement of the 6.00 kg block can be found using the work-energy theorem:

Work_net = change_in_kinetic_energy.

Since the system starts from rest, the initial kinetic energy of the system is zero. Therefore:

Work_net = Final_kinetic_energy - Initial_kinetic_energy = 1/2 * (6.00 kg + 8.00 kg) * velocity^2.

Simplifying, we get:

Work_net = 1/2 * 14.00 kg * velocity^2.

Using the value of velocity calculated in part (a), we get:

Work_net = 1/2 * 14.00 kg * (2.07 m/s)^2 = 29.13 J.

The work done on the system of two blocks by gravity is the sum of the work done on the individual blocks by gravity:

Work_gravity_system = Work_gravity_6kg + Work_gravity_8kg = 47.04 J + 0 J = 47.04 J.

The work done on the system of two blocks by the tension in the rope is the sum of the work done on the individual blocks by the tension:

Work_tension_system = Work_tension_6kg + Work_tension_8kg = -29.6 J + 29.6 J = 0 J.

Therefore, the net work done on the system of two blocks during the 0.800 m downward displacement of the 6.00 kg block is 29.13 J. The work done by gravity is 47.04 J, the work done by friction is unknown, and the work done by the tension in the rope is zero.

Note: The calculations for part (b) and (c) were based on the given information, but the value of the friction force was not provided in the problem statement.

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Answer:

Explanation:

The potential energy stored in the compressed spring is given by:

PE = (1/2) k x^2

where:

k = spring constant = 925 N/m

x = compression of the spring = 3.00 m

Substituting the values, we get:

PE = (1/2) (925 N/m) (3.00 m)^2 = 4162.5 J

At the bottom of the incline, the roller-coaster car has both potential energy (PE) and kinetic energy (KE). At the top of the incline, the roller-coaster car will have only potential energy, because it has come to a stop. We can therefore set the PE at the bottom equal to the PE at the top:

PE_bottom = PE_top

where:

PE_bottom = m g h, where m is the mass of the roller-coaster car, g is the acceleration due to gravity (9.81 m/s^2), and h is the height of the incline

PE_top = 4162.5 J, the potential energy stored in the compressed spring

Substituting the values, we get:

m g h = 4162.5 J

Solving for h, we get:

h = 4162.5 J / (m g) = 4162.5 J / (120.00 kg x 9.81 m/s^2) ≈ 3.54 m

Therefore, the roller-coaster car will roll up the incline to a height of approximately 3.54 meters before coming to a stop.

The roller-coaster car will roll up approximately 7.08 meters up the incline before coming to a stop.

To calculate how high up the incline the roller-coaster car will roll before coming to a stop, we can use the principle of conservation of mechanical energy. At the initial position, the roller-coaster car has potential energy stored in the compressed spring, and at the highest point on the incline, it will have only potential energy due to its height.

The total mechanical energy at the initial position is the sum of the potential energy stored in the compressed spring and the kinetic energy of the roller-coaster car at that point. At the highest point on the incline, the roller-coaster car will come to a stop, so its kinetic energy will be zero, and only potential energy due to height will remain.

The equation for conservation of mechanical energy is:

Initial Mechanical Energy = Final Mechanical Energy

The initial mechanical energy is the potential energy stored in the compressed spring:

Initial Mechanical Energy = (1/2) * k * \(x^{2}\)

where k is the spring constant (925 N/m) and x is the compression of the spring (3.00 meters).

Now, at the highest point on the incline, the final mechanical energy is the potential energy due to height:

Final Mechanical Energy = m * g * h

where m is the mass of the roller-coaster car (120.00 kg), g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the height of the incline.

Setting the initial mechanical energy equal to the final mechanical energy:

(1/2) * k * \(x^{2}\) = m * g * h

Now, let's plug in the known values and solve for h:

(1/2) * 925 N/m * \((3.00 m)^2\) = 120.00 kg * 9.81 m/s² * h

925 N/m * 9 \(m^{2}\) = 120.00 kg * 9.81 m/s² * h

8325 Nm = 1176.00 kgm²/s² * h

Now, divide both sides by 1176.00 kg*m²/s² to solve for h:

h = 8325 Nm / 1176.00 kgm²/s²

h ≈ 7.08 meters

Hence, the roller-coaster car will roll up approximately 7.08 meters up the incline before coming to a stop.

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The mass of the the object, given that the same force acted upon it to accelerate at 2 m/s² is 34 Kg

First, we shall determine the force. This can be obtained as follow:

Force (F) = mass (m) × acceleration (a)

Force = 4 × 17

Force = 68 N

Finally, we shall determine the mass of the object that will accelerate at 2 m/s² when the force of 68 N is applied to it. Details below:

Force = mass × acceleration

68 = mass × 2

Divide both sides by 2

Mass = 68 / 2

Mass = 34 Kg

Thus, from the above calculation, we can conclude that the mass is 34 Kg

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The final velocity of the red ball is 18 m/s.

The term momentum has to do with the product of the mass and the velocity of an object We know that the momentum is always conserved in accordance with the Newton third law. Also it is clear that the momentum before collision is equal to the total momentum after collision and we are going to apply this principle here.

Then;

Mass of the blue ball =  6 kg

Mass of the red ball =  1 kg

Initial velocity of the blue ball =  4 m/s

Initial velocity of the red ball = 0 m/s

Final velocity of the red ball = ??

Final velocity of the blue ball =  1 m/s

We now have;

(6 * 4) + (1 * 0) = (1 * v) + (6 * 1)

24 = v + 6

v = 24 - 6

v = 18 m/s

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270 FAST bacteria occupy the culture at time t = 10 s and 318 FAST bacteria occupy the culture at time t = 11 s.

Define FAST bacteria.

A group of bacteria with the property of acid fastness is referred to as acid-fast bacteria, sometimes known as acid-fast bacilli or simply AFB. A bacterium's capacity to withstand acid decolorization during staining methods is known as acid fastness.

Mycobacteria's high mycolic acid concentration, which causes the staining pattern of poor absorption followed by significant retention, is what makes them acid-fast.

At t = 10 s

N(t) = 10+6t+2t^2

N(t) = 10 +6*10 + 2*10*10

      = 270

At t  = 11s

N(t) =  10+6*11+2*11*11

     =  318

Difference between two will be 318 - 270 =  48

Rate of bacterial growth will be (270-10/270) *100%

i.e. 96.2%

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Ecotourism has the potential to eliminate the requirement to hunt wildlife for a living. Ecotourism generates revenue by maintaining the rainforest; deforestation is discouraged because it reduces tourist revenue.

Ecotourism firms safeguard biodiversity by preserving animals in their natural habitats and preserving natural ecosystems in biosphere reserves, wildlife sanctuaries, and national parks.

Ecotourism contributes to the preservation of a destination's ecological and biological diversity. For example, since Costa Rica's jungle is so popular with visitors, inhabitants work hard to protect it instead of attempting deforest it for a short-term profit.

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Answer:

28 m/s

Explanation:

v(final)^2=v(initial)^2+2a(delta y)

=sqrt(2*-9.8 m/s^2*-40)

28 m/s

The speed of the camera with which it hits the water is 27.93 meters per second.

The speed of an object is the rate of change of position of an object in any of the directions. Speed can be measured as the ratio of distance covered by the object to the total time taken in which the distance was covered by the object. Speed is a scalar quantity because it has only magnitude and no direction.

v = gt

where, v is the velocity,

g is the acceleration due to gravity,

and t is the time taken by the object

v = (9.8 m/s²) (2.85s)

v = 27.93 m/s

Therefore, the speed of the camera as it hits the water is 27.93 meters per second.

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Answer:

I=0.02[amp]

Explanation:

Since the resistors are connected in series, we can find the total resistance.

\(Rt = 100 + 300+50\\Rt = 450[ohm]\)

Now we can use ohm's law which tells us that the voltage is equal to the product of resistance by the current.

\(V =I*R\\I=V/R\\I=9/450\\I=0.02[amp]\)

Answer: I think this is the answer, In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

Explanation: I had a question similar to this, Hope this helps!

The total distance is 105 kilometers and the total displacement is 15 kilometers east. Option C

To calculate the total distance, we add the distances traveled in each leg of the journey: 60 kilometers (from A to B) + 45 kilometers (from B back to A) = 105 kilometers.

However, displacement refers to the change in position of an object in a straight line from its starting point to its ending point. In this case, since the boat starts and ends at the same point (A), the total displacement is zero.

Hence The total distance is 105 kilometers and the total displacement is 15 kilometers east.

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Gravitational force = G · (mass₁) · (mass₂) / (distance)

(distance²) = G · (mass₁) · (mass₂) / (Gravitational force)

G = 6.67 x 10⁻¹¹ n-m² / kg²  (the "gravitational constant")

Distance²  = (6.67 x 10⁻¹¹ n-m² / kg²) (28,500 kg) (2.2 x 10⁸ kg) / (39 N)

Distance² = (6.67 · 28,500 · 2.2 x 10⁻³ N-m²) / (39N)

Distance²  =  (418.209 N-m²) / (39N)

Distance²  =  10.72 m²

Distance = 3.275 meters

An absurd scenario, but that's by golly what the math says with the numbers provided.  I guess it's a teeny tiny planet orbiting 3.275 meters outside a teeny tiny black hole.

The primary difference is that electromagnetic waves can propagate through a vacuum or empty space, while mechanical waves require a physical medium to transmit energy.

A primary difference between an electromagnetic wave and a mechanical wave is the medium through which they propagate.

Electromagnetic waves can propagate through a vacuum or empty space without requiring a material medium. They are generated by the oscillation and interaction of electric and magnetic fields.

Examples of electromagnetic waves include radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. These waves can travel through space, air, or other materials, as they do not rely on physical particles to transmit energy.

On the other hand, mechanical waves require a physical medium to propagate. They are disturbances that travel through a material medium, transferring energy from one location to another. Mechanical waves rely on the interaction and displacement of particles within the medium to transmit energy.

Examples of mechanical waves include sound waves, water waves, seismic waves, and waves on a string. These waves cannot travel through a vacuum as they depend on the physical presence and interaction of particles within the medium.

In summary, the primary difference is that electromagnetic waves can propagate through a vacuum or empty space, while mechanical waves require a physical medium to transmit energy.

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Answer:

d. 4 km NE

Explanation:

Give me brainliest if that helped!

Answer:

Invasive plants also degrade the environment by changing the species composition and the ecology of invaded ecosystems

Explanation:

Answer:

they can harm the enviroment by changing the species composition and the ecology of invaded ecosystems !!

To calculate the energy gained when changing the temperature of 100g of solid candle wax at 20°C to liquid at 68°C, we need to consider two processes:

(1) raising the temperature of the wax from 20°C to 68°C

(2) melting the wax at its melting point of 68°C.

The first process requires an energy input of Q1 = mcΔT, where m is the mass of the wax (100 g), c is the specific heat capacity of the wax (29 J/g°C), and ΔT is the change in temperature (68°C - 20°C = 48°C). Thus, Q1 = (100 g)(29 J/g°C)(48°C) = 139,200 J.

The second process requires an energy input of Q2 = mL, where L is the specific latent heat of fusion of the wax (220 J/g). Thus, Q2 = (100 g)(220 J/g) = 22,000 J.

Therefore, the total energy gained when changing the temperature of 100g of solid candle wax at 20°C to liquid at 68°C is Q = Q1 + Q2 = 139,200 J + 22,000 J = 161,200 J.

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The East/West component of SuperPointParticleDog's force on the ball is -115.7 N. (Westward).

Isaac's force:

Magnitude: 255 N

Angle: 13 degrees North of East

x-component: 255*cos(13) = 245.1 N (Eastward)

y-component: 255*sin(13) = 58.1 N (Northward)

Newton's force:

Magnitude: 156 N

Angle: 34 degrees South of East

x-component: 156*cos(34) = 129.4 N (Westward)

y-component: 156*sin(34) = 86.5 N (Southward)

Now we can add the x- and y-components of the forces to find the net force:

Net force:

x-component: 245.1 N - 129.4 N = 115.7 N (Eastward)

y-component: 58.1 N - 86.5 N = -28.4 N (Southward)

The net force has an Eastward component of 115.7 N. Therefore, the East/West component of SuperPointParticleDog's force on the ball is -115.7 N. (Westward).

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To answer these questions, we need to consider the rate at which energy is being stored in the r-l circuit shown in the figure.

(a) At t = 0, just after the circuit is completed, the current i in the circuit will be zero. Therefore, the rate at which energy is being stored in the circuit, PL, will also be zero.

(b) At t → [infinity], a long time after the circuit is completed, the current in the circuit will have reached its final steady-state value. In an r-l circuit, the steady-state current is given by the formula:

i = e/r

The rate at which energy is being stored in the circuit is given by the product of the current and the voltage across the inductor:

PL = i*e = (e/r)*e = e^2/r

(c) At the instant when i = e/2R, one-half the current's final value, the rate at which energy is being stored in the circuit is given by:

PL = i*e = (e/2r)*e = e^2/2r

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Answer:

The right approach is "1479°C".

Explanation:

The given values are:

Mass of iron piece,

\(m_p=42 \ kg\)

Mass of iron bucket,

\(m_I=5 \ kg\)

Mass of water,

\(m_w=10 \ kg\)

Iron's specific heat,

\(C_I=470 \ J/Kg^{\circ}C\)

Water's specific heat,

\(C_w=4186 \ J/Kg^{\circ}C\)

Initial temperature,

\(t_I=23^{\circ}C\)

Final equilibrium temperature,

\(T=150^{\circ}C\)

Latent heat,

\(L_v=2260\times 10^3 \ J/Kg\)

As we know,

The heat lost by the glowing piece of iron will be equal to the heat gain by the iron bucket as well as water, then

⇒ \(m_IC_I \Delta T=m_wC_w(100-23)+m_wL_v+m_bC_I(150-23)\)

On substituting the given values, we get

⇒ \(42\times 420\times \Delta T=10\times 4186(100-23)+10(2260\times 10^3)+5\times 420(150-23)\)

⇒ \(17640 \Delta T=3.22\times 10^6+2.26\times 10^7+2.667\times 10^5\)

⇒          \(\Delta T=\frac{2.60867\times 10^7}{17640}\)

⇒          \(\Delta T=1479^{\circ}C\)

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

   The amplitude of the lateral  force is  \(F = 25 \ N\)

   The frequency is   \(f = 1 \ Hz\)

   The mass of the bridge per unit length is  \(\mu = 2000 \ kg /m\)

    The length of the central span is  \(d = 144 m\)

     The oscillation amplitude of the section  considered at the time considered is  \(A = 75 \ mm = 0.075 \ m\)

      The time taken for the undriven oscillation to decay to \(\frac{1}{e}\)  of its original value is  t = 6T

Generally the mass of the section considered is mathematically represented as

            \(m = \mu * d\)

=>        \(m = 2000 * 144\)

=>        \(m = 288000 \ kg\)

Generally the oscillation amplitude of the section after a  time period  t is mathematically represented as

                 \(A(t) = A_o e^{-\frac{bt}{2m} }\)

Here b is the damping constant and the \(A_o\) is the amplitude of the section when it was undriven

So from the question  

               \(\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }\)

=>            \(\frac{1}{e} =e^{-\frac{b6T}{2m} }\)

=>          \(e^{-1} =e^{-\frac{b6T}{2m} }\)

=>           \(-\frac{3T b}{m} = -1\)

=>         \(b = \frac{m}{3T}\)

Generally the amplitude of the section considered is mathematically represented as

           \(A = \frac{n * F }{ b * 2 \pi }\)

=>       \(A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }\)

=>       \(n = A * \frac{m}{3} * \frac{2\pi}{25}\)

=>       \(n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}\)

=>       \(n = 1810 \ people\)

Answer:

A. the average acceleration was the same

Explanation:

Acceleration is calculated by finding the difference of the initial velocity from the final velocity (on impact, usually 0) and then dividing by the amount of time that took place. If we assume that both balls were thrown at the same initial force, and ended up hitting the wall at the same time then we can say that the average acceleration was the same. If the initial velocity was not the same then we would need the initial velocity of each ball in order to calculate the acceleration of each object and determine which had a greater acceleration.

Answer:

I = 0.09[amp] or 90 [milliamps]

Explanation:

To solve this problem we must use ohm's law, which tells us that the voltage is equal to the product of the voltage by the current.

V = I*R

where:

V = voltage [V]

I = current [amp]

R = resistance [ohm]

Now, we replace the values of the first current into the equation

V = 180*10^-3 * R

V = 0.18*R (1)

Then we have that the resistance is doubled so we have this new equation:

V = I*(2R) (2)

The voltage remains constant therefore 1 and 2 are equals and we can obtain the current value.

V = V

0.18*R = I*2*R

I = 0.09[amp] or 90 [milliamps]

The height of the slope is 323.6 meters. The slope could be calculated by using the energy conservation principle.

The force of friction prevents movement between two surfaces that are in touch with one another.

It is caused by the irregularities and imperfections in the surfaces and the molecular interactions between them. Friction can act in both the direction of motion, called kinetic friction, and opposite to the direction of motion, called static friction. The magnitude of friction depends on the nature of the surfaces in contact, the force pressing them together, and the relative speed between them.

We can use conservation of energy to solve this problem, since friction is neglected. The total mechanical energy of the system (skier + Earth) is conserved:

KE_i + PE_i = KE_f + PE_f

where PE stands for potential energy and KE for kinetic energy.

At the top of the slope, the skier has only potential energy:

PE_i = mgh

where m is the skier's mass, g is the acceleration brought on by gravity, and h is the slope's height.

KE_f = (1/2)mv²

where v denotes the skier's speed at the slope's base.

Setting the initial and final energies equal to each other, we have:

mgh + 0 = (1/2)mv² + 0

Solving for h, we get:

h = (v²)/(2g) - (1/2)h

Plugging in the given values, we get:

h = (26.5²)/(2*9.81) - (1/2)h

Simplifying and solving for h, we get:

h = 323.6 m

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Answer:

Explanation:

3.  answer is d.

Displacement increases b/c slopes of both lines are positive. Spring 1 displaces more than Spring 2 per unit force applied.  Graphs are straight lines, so the increase is proportional.

4.  Fnet = 5 - 30 - 5 = -30 N       Positive direction is to the right, negative direction is to the left.

answer is d. 30 N to the LEFT

The cannon will completely stop when it collides with the barrier.

To calculate the speed at which the cannon collides with the barrier, we can follow these step-by-step calculations:

Determine the initial momentum of the system.

Since the cannon is initially stationary, the initial momentum is zero.

Apply the conservation of linear momentum.

According to the principle of conservation of linear momentum, the initial momentum of the system (zero) is equal to the final momentum of the system. The final momentum is the momentum of the cannon after firing.

Calculate the final momentum of the system.

Let's assume the mass of the cannon is represented by 'm' and the final velocity of the cannon is represented by 'v'. The final momentum of the system is given by: final momentum = m × v.

Set up the equation.

Since the initial momentum is zero, we have: 0 = m × v.

Solve for the final velocity of the cannon.

Dividing both sides of the equation by 'm', we get: v = 0.

Interpret the result.

The calculation shows that the final velocity of the cannon is zero. This means that the cannon comes to a complete stop when it collides with the barrier.

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