An object of uniform density applies its gravitaitonal force at __

MACHOs, or Massive Compact Halo Objects, are a type of dark matter candidate consisting of large, non-luminous celestial bodies.

MACHOs are thought to be made up of baryonic matter (protons, neutrons, and electrons), but they do not emit or reflect enough light to be easily detected.

They are theorized to reside in the halo region surrounding galaxies like the Milky Way.

Examples of MACHOs include black holes, neutron stars, and brown dwarfs. Due to their massive size and gravitational influence, they are considered as potential contributors to the unaccounted mass in the universe, known as dark matter.

Hence, MACHOs are massive, non-luminous celestial bodies that serve as a dark matter candidate, possibly contributing to the unexplained mass in the universe. They are comprised of baryonic matter and can be found in the halo region of galaxies.

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To find how much the fully extended cable stretches when 1500 kg of ore is loaded into the elevator, we need to use the following formula for elongation:

ΔL = (F × L) / (A × Y)

Where:
ΔL = elongation (stretch) of the cable
F = force applied (weight of the ore)
L = initial length of the cable (800 m)
A = cross-sectional area of the cable
Y = Young's modulus of the cable (10 × 10^10 N/m^2)

First, we need to calculate the force (F) applied by the ore. Since F = mass × acceleration due to gravity (g), we have:

F = 1500 kg × 9.81 m/s^2 = 14,715 N

Next, we need to find the cross-sectional area (A) of the cable. Since it's a circular cable, we use the formula A = π × r^2, where r is the radius of the cable. We have:

Diameter = 2.5 cm = 0.025 m
Radius (r) = Diameter/2 = 0.025 m / 2 = 0.0125 m
A = π × (0.0125 m)^2 ≈ 4.91 × 10^-4 m^2

Now, we can plug all the values into the elongation formula:
ΔL = (14,715 N × 800 m) / (4.91 × 10^-4 m^2 × 10 × 10^10 N/m^2) ≈ 0.2396 m

So, the fully extended cable stretches by approximately 0.2396 meters when 1500 kg of ore is loaded into the elevator.

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Therefore, the displacement of the gazelle during the first 12.0 s is 24 m to the right (in the positive x-direction).

Since the velocity of the gazelle is given as a function of time in the graph, we can find the distance moved and displacement during the first 12 seconds using the area under the velocity-time curve.

To find the total distance moved, we need to calculate the area under the velocity-time curve between t = 0 s and t = 12.0 s. We can divide the area into two regions: a triangle and a rectangle.

The triangle has a base of 6 s and a height of 12 m/s, so its area is:

(1/2) x 6 s x 12 m/s = 36 m

The rectangle has a width of 6 s and a height of 8 m/s, so its area is:

6 s x 8 m/s = 48 m

Therefore, the total distance moved is:

36 m + 48 m = 84 m

To find the displacement during the first 12.0 s, we need to calculate the area between the velocity-time curve and the t-axis. The triangle below the t-axis has a negative area, while the rectangle above the t-axis has a positive area. So the displacement is:

(-1/2) x 6 s x 8 m/s + 6 s x 8 m/s = 24 m

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1.

a) Hydrogen chloride (HCl) - 1 hydrogen atom, 1 chlorine atom

b) Sulfur dioxide (SO₂) - 1 sulfur atom, 2 oxygen atoms

c) Ammonia (NH₃) - 1 nitrogen atom, 3 hydrogen atoms

d) Carbon monoxide (CO) - 1 carbon atom, 1 oxygen atom

2.

a) Hydrogen chloride (HCl) :

H

|

Cl--C--

|

H

b) Sulfur dioxide (SO₂) :

O

//

O=S

\\

O

c) Ammonia (NH₃) :

H

|

H--N--H

|

H

d) Carbon monoxide (CO) :

O

//

C=O

The quantum mechanical model of the atom, each subshell is characterized by a letter designation that corresponds to its value of l. The subshell with l=1 is the p subshell, which can hold a maximum of 6 electrons.

Therefore, the element to which this atom belongs must have its highest-energy valence electrons in the 5p subshell. There are several elements that have their valence electrons in the 5p subshell, including antimony (Sb), tellurium (Te), and iodine (I).

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the third one is the brightest, the second one is the second brightest, the first one is the second dimmest, and last but not least, the last one is the dimmest.

Explanation:

hope this helps

The phase change that a reflected light wave experiences upon reflection from a denser medium is equivalent to 1/2 (or 0.5) of a wavelength.

When a light wave reflects off a denser medium, it undergoes a phase change of 180 degrees (or pi radians) due to a change in the direction of the wave's electric field vector. The phase change can also be described as a shift of one-half wavelength. This means that if the incident wave has a wavelength of λ, the reflected wave will have a phase difference of π (or 180 degrees) with respect to the incident wave, which is equivalent to a shift of one-half wavelength or λ/2. Therefore, the phase change that a reflected light wave experiences is equivalent to one-half of a wavelength.

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The logic leading to the conclusion that the Milky Way is a spiral galaxy is based on observations of dust and gas disks, hot and bright stars, and a central bright area surrounded by a flat disk. If we lived in an elliptical galaxy, the image would show stars distributed spherically around us without dust and massive star formation, and the view would not be confined to a particular plane.

1. Spiral Galaxy:
- Disks of dust and gas, with hot, bright (massive) stars
- Roughly confined to a single plane of view
- Central bright area surrounded by a flat disk

2. Elliptical Galaxy:
- Not confined to a particular plane
- No dust or formation of massive stars
- Stars distributed spherically around us

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The electron volt is a unit of:
A) energy.

An electron volt (eV) is defined as the amount of kinetic energy gained or lost by an electron when it is accelerated through an electric potential difference of one volt.

It is a convenient unit to express the energy of subatomic particles, such as electrons and photons.

The electron volt (eV) is a unit of energy that is defined as the amount of energy gained or lost by an electron when it moves through a potential difference of one volt.

The formula for calculating the energy in electron volts is:

E(eV) = q × V

where E(eV) is the energy in electron volts, q is the electric charge of the particle in coulombs, and V is the potential difference in volts.

For example, let's say we have an electron with a charge of [tex]-1.6 *  10^-19[/tex] coulombs that moves through a potential difference of 5 volts.

The energy gained by the electron can be calculated as:

[tex]E(eV) = (-1.6 * 10^-19 C) * (5 V) = -8 * 10^-19 joules[/tex]

This energy can also be expressed as -5 eV, since one electron volt is equivalent to[tex]1.6 * 10^-19[/tex] joules.

Note that the negative sign in the result indicates that the electron lost energy, rather than gaining it.

In atomic and subatomic physics, the electron volt is a useful unit of energy for describing the energies of particles like electrons, protons, and photons, which typically have very small energies.

For example, the binding energies of electrons in an atom are typically measured in electron volts.

The ionization energy of an atom, which is the minimum energy required to remove an electron from the atom, is also measured in electron volts.

A) energy is correct.

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The two observations that allow for a determination of the Milky Way's mass are the object's orbital speed and its distance from the center of the galaxy. By measuring the orbital speed of an object and its distance from the center of the galaxy, astronomers can use the laws of gravity to calculate the mass of the galaxy.

The two observations that allow for a determination of the Milky Way's mass are the object's orbital speed and its distance from the center of the galaxy. By measuring the orbital speed of an object and its distance from the center of the galaxy, astronomers can use the laws of gravity to calculate the mass of the galaxy. This is known as the Galactic Mass Problem, and it is a challenging problem because much of the mass of the galaxy is dark matter, which cannot be directly observed. Nonetheless, careful observations of the motions of stars, gas, and other objects in the Milky Way have allowed astronomers to make increasingly precise measurements of the galaxy's mass over time.
Hi! To determine the Milky Way's mass, the two observations of an object that can be used are its position (distance from the galactic center) and its orbital velocity. By applying Newton's laws of gravitation and motion, one can calculate the mass of the Milky Way within the object's orbit. Here's a step-by-step explanation:

1. Measure the object's position, specifically its distance from the galactic center.
2. Measure the object's orbital velocity, which is its speed as it orbits around the galactic center.
3. Use Newton's law of gravitation, which states that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.
4. Apply Newton's law of motion to relate the gravitational force to the object's orbital velocity and distance from the galactic center.
5. Solve for the mass of the Milky Way within the object's orbit, taking into account the position and orbital velocity measurements.

By using these observations and steps, one can determine an estimate of the Milky Way's mass.

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Without the charge, velocity, and angle information, it's not possible to calculate the magnitude of the magnetic force on the charged particles in a 4-tesla magnetic field pointing in the positive-x direction.

The magnetic force (F) on a charged particle can be calculated using the formula F = q(v × B), where q is the charge of the particle, v is its velocity, and B is the magnetic field.

The cross product (v × B) takes into account the angle between the velocity and magnetic field vectors.

Hence,  Without the charge, velocity, and angle information, it's not possible to calculate the magnitude of the magnetic force on the charged particles in a 4-tesla magnetic field pointing in the positive-x direction.

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Answer:

C

Explanation:

The option (C) "The period of oscillation for an object" does not involve an assumption about the equivalence of inertial and gravitational mass.

The option that does not involve an assumption about the equivalence of inertial and gravitational mass is At a point on the earth's surface, the freefall acceleration of all objects is the same.(B)


A) The centripetal acceleration of a satellite given by G implies that the gravitational force (which depends on gravitational mass) provides the necessary centripetal force (which depends on inertial mass) for the satellite's orbit. Thus, it involves the equivalence of inertial and gravitational mass.

C) The period of oscillation for an object, such as a pendulum or a spring-mass system, also depends on both gravitational and inertial mass. The relationship between these masses is necessary for predicting the period of oscillation.

Option B, on the other hand, does not involve this equivalence.

The freefall acceleration (g) at a point on Earth's surface being the same for all objects simply states that all objects fall towards the Earth with the same acceleration due to gravity, regardless of their mass. It doesn't require any assumption about the equivalence of inertial and gravitational mass.

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The time taken by the electric motor to lift the object is 2.205 s.

Power of the electric motor, P = 24 W

Mass of the object to be lifted, m = 1.5 kg

Distance to which it is to be lifted, d = 3.6 m

Power of the electric motor is the work done by it per unit time.

The expression for power of the electric motor can be written as,

Power, P = mgd/t

Therefore, the time taken by the electric motor to lift the object,

t = mgd/P

t = 1.5 x 9.8 x 3.6/24

t = 52.92/24

t = 2.205 s

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The speed of the satellite will increase by a factor of 8 when it is moved from a circular orbit of radius "r" to a circular orbit of radius "4r".

The speed of the satellite will change when it moves from a circular orbit of radius "r" to a circular orbit of radius "4r".

According to Kepler's laws of planetary motion, the speed of an object in circular motion is proportional to the radius of the circular path. Specifically, the formula for the speed of an object in circular motion is:

v = (2πr) / T

where "v" is the speed of the object, "r" is the radius of the circular path, and "T" is the period of the motion.

Since the satellite is moving in a circular orbit around the Earth, its speed is determined by the radius of its circular path.

When the satellite is moved into a circular orbit of radius 4r, its speed will change. To calculate the new speed, we can use the same formula for circular motion, but with the new radius "4r":

v' = (2π(4r)) / T

Simplifying the expression, we get:

v' = 8v

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The shape that your arm describes is a cone and the shape of your finger sweeping out on the wall is a circle

When you point to a wall with your arm extended at a 42-degree angle to the normal of the wall and rotate your arm in a full circle while maintaining the same angle, the shape that your arm describes is a cone. The cone is formed by the movement of your arm around an axis perpendicular to the wall, with the vertex of the cone at your shoulder and the base at your fingertip.

As your arm rotates, your fingertip sweeps out a circle on the wall. This circle is parallel to the base of the cone and is formed by the intersection of the cone with the wall. The radius of the circle is equal to the distance from your shoulder to your fingertip, and the center of the circle is located at the point where your arm intersects the wall.

The cone that your arm describes is a three-dimensional shape that is formed by rotating a line segment around an axis. In this case, the line segment is your arm, and the axis is perpendicular to the wall. The cone is a familiar shape that appears in many contexts, including the geometry of circles, the construction of paper cups and traffic cones, and the design of loudspeakers. In conclusion, the shape that your arm describes is a cone and the shape of your finger sweeping out on the wall is a circle.

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Elements heavier than hydrogen (and helium) are made by stars and therefore should be located within galaxies.

So if we see absorption lines from these elements in quasar spectra, they should have the same redshifts as hydrogen lines from intervening galaxies. Absorption lines may not be present at redshifts of protogalactic clouds that are composed of hydrogen and helium only. This means that the redshifts of these elements should be similar to the redshifts of the galaxies they are associated with. However, it is possible that absorption lines may be present at redshifts of protogalactic clouds that are composed of hydrogen and helium only. In these cases, the redshifts of heavier elements would be different than the redshifts of the protogalactic clouds and therefore not be present in the spectrum.

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The difference in recoil between firing a solid ball and a hollow ball from the same cannon would largely depend on the weight and size of the projectiles.

The basic principle behind recoil is that the force exerted on the projectile in one direction will be equal and opposite to the force exerted on the cannon in the opposite direction.

Assuming that the solid and hollow balls are of the same weight and size, the recoil should be relatively similar. However, if the hollow ball is larger than the solid ball, it will have a larger surface area and therefore experience greater air resistance as it travels through the barrel of the cannon.

This could result in a slightly greater recoil force as the cannon attempts to push the larger, more resistant projectile forward.

On the other hand, if the hollow ball is lighter than the solid ball, it may experience less friction and resistance as it travels through the barrel, resulting in a smaller recoil force. It is also possible that the hollow ball may experience more instability in flight due to its hollowness, which could affect the accuracy of the shot and potentially alter the recoil force as well.

Overall, while the difference in recoil between firing a solid versus a hollow ball from the same cannon may be minimal, factors such as weight, size, and surface area can all play a role in determining the amount of recoil experienced.

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In Experiment 2, we used a spectrophotometer to measure the absorbance of the copper(ii) sulfate solutions.

The wavelength we used for the measurements was 650 nm. This specific wavelength was chosen as it is the maximum absorbance wavelength for the copper(ii) sulfate solution.

This allowed us to accurately measure the concentration of the solution using the Beer-Lambert law, which relates the absorbance of a solution to its concentration.

By using a specific wavelength, we were able to ensure that our measurements were consistent and reliable. Overall, selecting the correct wavelength is crucial in obtaining accurate and meaningful data in spectrophotometry experiments.

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In the photoelectric effect, the maximum speed of the electrons emitted by a metal surface depends on: II and III only (Frequency of the light and Nature of the photoelectric surface).

The maximum speed of emitted electrons, or the kinetic energy, is determined by the frequency of the incident light and the work function (which is a property of the photoelectric surface).

According to the equation K.E. = hν - φ, where K.E. is the kinetic energy, h is Planck's constant, ν is the frequency of the light, and φ is the work function, the kinetic energy of the emitted electrons is directly proportional to the frequency of the light and inversely proportional to the work function.

The intensity of the light only affects the number of emitted electrons, not their maximum speed.

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we cannot directly adapt the proof that there are infinitely many primes to show that there are infinitely many primes in the arithmetic progression 4k 1, k. We would need to come up with a new approach or proof to establish this result.

The proof that there are infinitely many primes (theorem 3 in section 4.3) relies on the assumption that there exists at least one prime number. This assumption is used to construct a new prime number that is larger than any previously known prime number. However, when we try to adapt the proof to show that there are infinitely many primes in the arithmetic progression 4k 1, k, we run into a problem.

In order to adapt the proof, we would need to assume that there exists at least one prime number of the form 4k 1. However, this assumption cannot be made, as it is possible that there are only finitely many primes of this form. In fact, there are infinitely many primes of the form 4k 3, but this does not necessarily imply the existence of infinitely many primes of the form 4k 1.

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The circuit below shows four identical bulbs connected to an ideal battery, which has negligible internal resistance. When the switch is closed, rank the bulbs in order from brightest to dimmest. 1. A > B = C >D 2. A > B > C >D 3. D > C > B> A 4. D > B=C > A 5. A= B=C=D 6. A=B=> 7. A=C >D>B 8. A= B > D=C 9. A= D > B = C 10. B=C > A=D

The correct answer is 5. A=B=C=D.

Assuming batteries have the same voltage and current rating, the more power available, the more power the bulb can draw from the battery since the power in a battery-powered circuit is proportional to the number of batteries used. So, the circuit with three batteries would produce the brightest light bulb.
Since all four bulbs are identical and connected in parallel to the battery, they each receive the same voltage and therefore will emit the same amount of light. Thus, they will all be equally bright.

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Answer:

8 ft

Explanation:

Types TPT and TST shall be permitted in lengths not exceeding 2.5 m (8 ft) where attached directly, or by means of a special type of plug, to a portable appliance rated at 50 watts or less and of such nature that extreme flexibility of the cord is essential.

The total work done by a conservative force, such as those described by Hooke's law or gravitation, is independent of the path taken and depends only on the initial and final positions. This means that the work done by conservative forces is path-independent and has the property of being recoverable as potential energy.

The total work done by a conservative force like Hooke's law or gravitation is zero if the initial and final positions of the object are the same. This is because conservative forces are path-independent, meaning the work done only depends on the endpoints and not the path taken between them. Therefore, any work done in one direction will be exactly cancelled out by work done in the opposite direction.

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Paper trading can only be done on symbols for which you have access to real-time data. This means that in order to paper trade on a particular market or symbol, you need to have real-time data for that market or symbol. If you don't have access to real-time data for a particular market or symbol, you won't be able to paper trade on it.

If you're interested in accessing additional real-time markets for paper trading, you can follow the link provided to subscribe to those markets. This will give you the real-time data you need to start paper trading on those markets or symbols. Just make sure to check the subscription fees and terms before signing up.
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The special VFR clearance allows the student pilot to operate in less than VFR conditions, but only within the designated airspace and with the permission of air traffic control.

Federal Aviation Regulations (FARs), a student pilot can request a special VFR clearance in less than VFR conditions. However, this is only permitted if the student pilot is operating under the supervision of a certified flight instructor and if the flight is conducted within the airspace designated for special VFR operations.

The special VFR clearance allows the student pilot to operate in less than VFR conditions, but only within the designated airspace and with the permission of air traffic control. It's important to note that special VFR clearance should only be requested if absolutely necessary, and pilots should always prioritize safety and avoid flying in poor weather conditions whenever possible.

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To determine how close together two point sources could be at a 2-million-light-year distance, like the Andromeda Galaxy, you'll need to consider the following factors:

1. The distance of the point sources: In this case, it's 2 million light-years away, which is the approximate distance of the Andromeda Galaxy from Earth.
2. The angular resolution of the observing instrument: This is the minimum angular separation between two objects that an instrument can resolve. This value depends on the specific telescope or device you are using to observe the point sources.

To calculate the minimum separation between the point sources, you can use the formula:
Minimum separation (in light-years) = Distance (in light-years) * Angular separation (in radians)

You'll need to know the angular resolution of the observing instrument to determine the minimum separation. Once you have the angular resolution, you can convert it from arcseconds to radians by dividing it by 206,265 (since 1 radian equals 206,265 arcseconds). Then, you can plug that value into the formula above to find the minimum separation in light-years.

In summary, to find how close together the point sources could be at the 2-million-light-year distance of the Andromeda Galaxy, you need to know the angular resolution of the observing instrument, convert it to radians, and then use the formula above to calculate the minimum separation in light-years.

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Based on the diagram in fig. 26955 (similar to fig. 26912, example 2698), the potential difference between points a and d can be found by summing the potential differences across each component in the circuit that lies between these two points.

Starting from point a, we first encounter a resistor with a resistance of R1. The potential difference across this resistor can be found using Ohm's law: V1 = I * R1, where I is the current flowing through the resistor.

Next, we come across a battery with an emf of E1. Since we are moving from the negative terminal to the positive terminal of the battery, the potential difference across the battery is E1.

Moving further along the circuit, we come across another resistor with a resistance of R2. Using Ohm's law again, the potential difference across this resistor is V2 = I * R2.

Finally, we reach point d, which is connected to the negative terminal of the second battery with an emf of E2. Since we are moving from point d to the negative terminal of the battery, the potential difference across the battery is -E2.

Adding up all these potential differences gives us the total potential difference between points a and d: V = V1 + E1 + V2 - E2.

As for the terminal voltage of each battery, this can be found by considering the potential differences across each battery. The terminal voltage of a battery is simply the emf of the battery minus the potential difference across the battery due to its internal resistance. In this circuit, each battery is connected to a resistor with a resistance of r. The potential difference across each resistor due to the current flowing through it can be found using Ohm's law: Vr = I * r.

Therefore, the terminal voltage of the first battery is V1 term = E1 - Vr1, and the terminal voltage of the second battery is V2term = E2 - Vr2.

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Since we know that there are also stars outside of the Sun's orbit, this gives us valuable insights into the vastness and diversity of the universe.

The Sun is just one of billions of stars in the Milky Way galaxy, and it follows an elliptical path around the galactic center. This fact highlights that our solar system is merely a small component of a much larger cosmic structure.

Observing stars outside the Sun's orbit allows us to study their unique properties and formation processes. By analyzing their spectral characteristics, we can determine their age, chemical composition, and distance from Earth. This information helps astronomers classify stars into various categories and enhances our understanding of stellar evolution.

Moreover, investigating stars beyond the Sun's orbit has led to the discovery of exoplanets, or planets that orbit stars other than the Sun. This has opened up the possibility of finding other worlds that may host life and has fueled research into the habitability of these distant planets.

Additionally, studying distant stars contributes to our knowledge of dark matter and dark energy, two mysterious forces that govern the expansion and overall structure of the universe. By observing the motion of stars and galaxies, scientists can infer the presence of these unseen forces and develop models to explain their influence on cosmic evolution.

In conclusion, the existence of stars outside the Sun's orbit highlights the incredible scope of the universe and provides invaluable information for understanding the intricacies of cosmic phenomena. By studying these distant stars, we can expand our knowledge of celestial bodies, exoplanets, and the fundamental forces shaping the cosmos.

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The output signal of the circuit below measured on resistor RL would be a sine wave with a peak amplitude of approximately 0.9 V.

This waveform is observed because the circuit is a simple half-wave rectifier with a smoothing capacitor, which filters out the negative half-cycles of the sine wave and passes only the positive half-cycles.

The resistor RL acts as a load on the circuit, and the resulting waveform across it is a smoothed version of the positive half-cycles of the input sine wave. The transformer T steps down the voltage from 115 VAC to 12 VAC, but does not modify the waveform.

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If A 0.450-kg ice puck, moving east with a speed of 5.68 m/s , has a head-on collision with a 0.900-kg puck initially at rest. Assume that the collision is perfectly elastic. the speed of the 0.450-kg puck after the collision is v₂ = (-b ± √(b² )

Since the collision is perfectly elastic, we can use the conservation of momentum and the conservation of kinetic energy to solve for the final velocities of the two pucks.

The initial momentum of the system is:

p_initial = m₁v₁ + m₂v₂

where m₁ and v₁ are the mass and velocity of the 0.450-kg puck, and m₂ and v₂ are the mass and velocity of the 0.900-kg puck. Since the 0.900-kg puck is initially at rest, we have:

p_initial = m₁*v₁ + 0

p_initial = (0.450 kg)(5.68 m/s) = 2.556 kg m/s

The initial kinetic energy of the system is:

KE_initial = (1/2)m₁v₁² + (1/2)m₂v₂²

Again, since the 0.900-kg puck is initially at rest, we have:

KE_initial = (1/2)(0.450 kg)(5.68 m/s)²+ (1/2)(0.900 kg)(0 m/s)²

KE_initial = 7.6614 J

After the collision, the momentum of the system is still conserved, so we have:

p_final = m₁v₁' + m₂v₂'

where v₁' and v₂' are the final velocities of the 0.450-kg and 0.900-kg pucks, respectively. Since the collision is head-on, we also have:

v₁' - v₂' = - (v₁ - 0)

or

v₁' = 2v₁ - v₂

Using the conservation of kinetic energy, we can also write:

KE_final = (1/2)m₁v₁'²+ (1/2)m₂v₂'²

Substituting the expression for v₁' in terms of v₂ and simplifying, we get:

KE_final = (1/2)m₁(2v₁ - v₂)² + (1/2)m₂v₂²

KE_final = (1/2)m₁(4v₁² - 4v₁*v₂ + v₂²) + (1/2)m₂v₂²

KE_final = (1/2)(4m₁v₁² - 4m₁v₁v₂ + m₁*v₂²) + (1/2)m₂v₂²

KE_final = 2m₁v₁²- 2m₁v₁*v₂ + (1/2)m₁v₂² + (1/2)m₂v₂²

Since the collision is perfectly elastic, the kinetic energy is conserved, so:

KE_final = KE_initial

Substituting the values we know and simplifying, we get:

2m₁v₁² - 2m₁v₁*v₂ + (1/2)m₁v₂² + (1/2)m₂v₂² = 7.6614 J

Plugging in the masses and velocities, we get:

2(0.450 kg)(5.68 m/s)² - 2(0.450 kg)5.68 m/sv₂ + (1/2)(0.450 kg)v₂² + (1/2)(0.900 kg)*v₂² = 7.6614 J

Solving for v₂ using the quadratic formula, we get:

v₂ = (-b ± √(b² -)

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