A light train made up of two cars is traveling at 90 km/h when the brakes are applied to both cars. Knowing that car A has a mass of 25 mg and car B a mass of 20 mg, and the braking force is 30 kn on each car, determine (a) the distance traveledby the train before it comes to a stop (b) the coupling force between the cars as the is slowing down.

The image of the mascara brush is located 12 cm from the mirror, on the same side as the object. This is a positive value since the image is on the same side as the object. Therefore, the distance of the image from the mirror is 12 cm.

The location of the image of the mascara brush held upright 18 cm from the concave cosmetic mirror with a focal length of 36 cm using ray tracing, we can use the following steps:

1. Draw a ray from the top of the mascara brush parallel to the principal axis of the mirror.
2. Draw a ray from the top of the mascara brush through the focal point of the mirror.
3. Where the two rays intersect is the location of the image of the mascara brush.

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The gravitational potential energy of the person standing on the roof of the 10-story building varies depending on the height above a reference point.

The formula to calculate gravitational potential energy is:

GPE = mgh

where m is the mass of the object, g is the acceleration due to gravity (9.81 m/s^2), and h is the height above a reference point.

(a) The gravitational potential energy of the 61.2 kg person standing on the roof of the 10-story building relative to the tenth floor would be:

GPE = (61.2 kg) x (9.81 m/s^2) x (7 x 2.50 m)

GPE = 10736.94 J

(b) The gravitational potential energy of the person relative to the sixth floor would be:

GPE = (61.2 kg) x (9.81 m/s^2) x (4 x 2.50 m)

GPE = 5328.36 J

(c) The gravitational potential energy of the person relative to the first floor would be:

GPE = (61.2 kg) x (9.81 m/s^2) x (9 x 2.50 m)

GPE = 14726.86 J

Therefore, the gravitational potential energy of the person standing on the roof of the 10-story building varies depending on the height above a reference point.

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When anything or a system is at rest and neither tilts nor rotates, it is said to be in static equilibrium.

Thus, The phrase "static" denotes that the body is not moving, whereas the term "equilibrium" denotes the balance of all opposing forces.

So, if an object is at rest and all external forces and other influences are in balance, the system is said to be in static equilibrium. A rigid body is an idealized object in physics whose deformation is negligible or, at the absolute least, extremely close to zero.

The rigid body definition states that even if an object's neighbouring points change, the thing remains its original shape. When all of the points in a body move consistently in a single direction or line, this is known as translational motion. It describes an object's linear motion in one to three dimensions.

Thus, When anything or a system is at rest and neither tilts nor rotates, it is said to be in static equilibrium.

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Answer:

20W

Explanation:

As P = W/t

P = FS/

P = 20x5/5

= 100/5

=20J/s or W

therefore, P = 20W

The correct statement describing the observed characteristics of gamma-ray bursts is that they are brief, very intense, nonrepeating, sudden releases of energy that appear to come from very large distances outside our galaxy.

GRBs are among the most energetic and luminous events that occur in the universe, and they are associated with the deaths of massive stars. GRBs are usually accompanied by sudden releases of X-ray energy from the buildup of hydrogen around neutron stars near the central regions of our galaxy and the centers of some globular star clusters. GRBs are thought to be caused by the core collapse of a massive star at least eight times heavier than our Sun when it exhausts the processes that release energy by nuclear fusion in its core.

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The speed of the combined mass after the collision is 4.73 m/s when a 90-kg player is running with a velocity of 4 m/s in the x direction.

To solve this problem, we need to use the law of conservation of momentum, which states that the total momentum of a system remains constant in the absence of external forces.
Before the collision, the momentum of player 1 is given by:
p1 = m1*v1 = 90 kg * 4 m/s = 360 kg*m/s in the x direction
Before the collision, the momentum of player 2 is given by:
p2 = m2*v2 = 80 kg * 5 m/s = 400 kg*m/s in the y direction
Since the collision is inelastic, the players tangle together and move as a single mass after the collision. Let's call the combined mass "M" and the velocity of the combined mass "v". The law of conservation of momentum tells us that:
p1 + p2 = M*v
Substituting in the values we calculated earlier, we get:
360 kg*m/s + 400 kg*m/s = M*v
Simplifying: 760 kg*m/s = M*v
To find the speed of the combined mass, we need to divide the momentum by the total mass of the players:
v = 760 kg*m/s / (90 kg + 80 kg) = 4.73 m/s

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it would take light approximately 25,000 years to travel from the center of the Milky Way galaxy to us.

It would take approximately 26,000 years for light to travel from the center of the Milky Way galaxy to us here on Earth. This is because the distance between us and the center of the galaxy is about 26,000 light years.
To determine how long it would take light to travel from the center of the Milky Way galaxy to us, you will need to consider the distance and the speed of light.

Step 1: Determine the distance
The distance from the center of the Milky Way galaxy to our solar system is approximately 25,000 light-years.

Step 2: Determine the speed of light
The speed of light is approximately 299,792 kilometers per second (km/s) in a vacuum.

Step 3: Calculate the time
Since the distance is given in light-years and the speed of light is already factored into this unit, the time it takes for light to travel from the center of the Milky Way to us is the same as the distance.

So, it would take light approximately 25,000 years to travel from the center of the Milky Way galaxy to us.

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Nucleation and growth processes can be controlled to tailor the microstructure of a material to meet specific performance requirements.

Nucleation and growth are important processes in the control of microstructure in materials science.

Nucleation refers to the initial formation of a new phase, while growth refers to the increase in size and volume of that phase. These processes can be controlled to tailor the microstructure of a material to meet specific performance requirements.

In general, nucleation and growth can be controlled by manipulating factors such as temperature, pressure, and chemical composition. For example, in the production of metals, the rate of cooling can be controlled to control nucleation and growth.

Rapid cooling will result in small nucleation sites and slow growth, leading to a fine-grained microstructure. Slower cooling rates result in larger nucleation sites and faster growth, leading to a coarser-grained microstructure.

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The potential energy (PE) of the 79 kg cinder block sitting on a 20 m high platform is 15,444 J (joules).

To calculate the potential energy of the cinder block, you can use the formula PE = mgh, where "m" is the mass, "g" is the acceleration due to gravity (9.8 m/s²), and "h" is the height.

First, find the product of the mass (79 kg) and the acceleration due to gravity (9.8 m/s²): 79 kg * 9.8 m/s² = 774.2 kg*m/s² (also known as newtons, N).

Next, multiply this value by the height (20 m): 774.2 N * 20 m = 15,444 J. Therefore, the potential energy of the cinder block is 15,444 J.

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In matter-antimatter annihilation, 100% of the mass is converted to energy, making it much more efficient than fusion. In fact, matter-antimatter annihilation is considered the most efficient energy source known to exist. However, the practical application of harnessing this energy source is still limited by the difficulty and expense of producing and containing antimatter.

Fusion reactions occur when two nuclei come together to form one atom. The reaction that happens in the sun fuses two Hydrogen atoms together to produce Helium. It looks like this in a very simplified way: H + H → He + ENERGY. This energy can be calculated by the famous Einstein equation, E = mc2.

Matter-antimatter reactors would be orders of magnitude more efficient at producing energy than fusion, the next best reaction mechanism. However, it still isn't possible to fully capture the released energy from a matter-antimatter event.

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The correct answer is A. The frequency of a standing wave formed on a string with fixed boundaries at either end, of harmonic order n, is related to the distance between the boundaries L and the wave speed v by the formula fn = n v/4L where n = 1,3,5...

This formula is derived from the fundamental frequency equation, which states that the frequency of a standing wave is proportional to the wave speed divided by the wavelength. In the case of a standing wave on a string with fixed boundaries at either end, the wavelength is twice the distance between the boundaries. Therefore, the frequency equation becomes fn = n v/2L where n is an odd integer. However, since we are only considering harmonic orders of n = 1,3,5..., we divide by 2 to get the equation fn = n v/4L. This formula is important in understanding the behavior of standing waves on strings and helps to determine the natural frequencies at which the string will vibrate.

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1.3 × 10¹⁹ electrons are flowing past any point in the wire per second when a current of 2.10A flows in a wire.

Current (I) is the flow of charge (Q) per unit time (t). Mathematically, it can be represented as I = Q/t.

Current, I = 2.10 A. The elementary charge of an electron, e = 1.6 × 10⁻¹⁹ C.

To find the number of electrons per second, we will first calculate the charge flowing per second.

Since I = Q/t, we have Q = I × t.

For a 1-second time interval, t = 1 s. Therefore, Q = 2.10 A × 1 s = 2.10 C.

To find the number of electrons, we divide the charge flowing per second by the charge of an electron:

Number of electrons = Q/e = 2.10 C / (1.6 × 10⁻¹⁹ C/electron) ≈ 1.31 × 10¹⁹ electrons.
When expressing the answer with two significant figures, we get 1.3 × 10¹⁹ electrons.

So, when a current of 2.10 A flows in a wire, approximately 1.3 × 10¹⁹ electrons are flowing past any point in the wire per second.

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The given statement " If you have a solid object that has a nonzero charge density throughout its volume, you know it must be an insulator" is False.

The charge density of a solid object, whether it is an insulator or a conductor, does not determine the type of material.

Charge density refers to the amount of charge per unit volume of a material. It can be either positive or negative, depending on the type of charge present (e.g., positive or negative ions) and their distribution within the material.

A solid object with a nonzero charge density throughout its volume can be either an insulator or a conductor, An insulator is a material that does not allow the free flow of electric charges, such as electrons. Insulators have tightly bound electrons and do not readily conduct electricity. Examples of insulators include rubber, plastic, and wood.

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The radius of the planet TrES-1 is approximately 92,450 km.

To calculate the radius of the planet TrES-1, we can use the information given about the blocked light during the transit and the radius of the star. Since the planet blocks 2% of the star's light, we can use the following formula:

Blocked light (%) = (Area of planet's disk) / (Area of star's disk) = (π × (planet's radius)²) / (π × (star's radius)²)

Given that the star TrES-1 has a radius of about 85% of our sun's radius, we can substitute this information into the formula:

0.02 = (π × (planet's radius)²) / (π × (0.85 × sun's radius)²)

Now, we can solve for the planet's radius. First, we can simplify the equation by canceling out the π:

0.02 = (planet's radius)² / (0.85 × sun's radius)²

Next, we can multiply both sides by (0.85 × sun's radius)² to isolate the planet's radius squared:

0.02 × (0.85 × sun's radius)² = (planet's radius)²

To find the planet's radius, we simply take the square root of both sides:

planet's radius = √(0.02 × (0.85 × sun's radius)²)

Now, we can plug in the sun's radius (approx. 696,340 km) and calculate the radius of the planet TrES-1:

planet's radius ≈ √(0.02 × (0.85 × 696,340 km)²) ≈ 92,450 km

Therefore, the radius of the planet TrES-1 is approximately 92,450 km.

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When an electron initially reaches the field zone, the magnetic force acting on it will be directed either away from the page or in your direction.

One of the four fundamental forces of nature, the electromagnetic force, results in the magnetic force, which is brought about by the motion of charges.

If an electron is moving perpendicularly into the plane of the page, and there is a uniform magnetic field B directed downward parallel to the plane of the page, then the direction of the magnetic force on the electron will be perpendicular to both the direction of motion of the electron and the direction of the magnetic field.

According to the right-hand rule, if the index finger of your right hand points in the direction of the magnetic field (downward in this case) and the middle finger points in the direction of the electron's motion (perpendicular into the page in this case), then the thumb will point in the direction of the magnetic force. In this case, the direction of the magnetic force will be out of the page or towards you.

Therefore, the direction of the magnetic force on the electron when it first enters the field region will be out of the page or towards you.

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The complete question is:

A uniform magnetic field B directed downward parallel to the plane of the page exists in the shaded region shown below: An electron can enter the field from different directions Suppose that the electron is moving perpendicularly into the plane of the page when it enters the region of the magnetic field What is the direction of the magnetic force , if any on the electron when it first enters the field region?

The process involved in the formation of hail pellets includes the formation of ice crystals directly from water. These ice crystals are tossed up and down in clouds, where they grow by accumulating layers of ice. Typically, hail pellets can grow to more than 5 mm in diameter.

One part of the process in the formation of hail pellets is that they grow to less than 5 mm in diameter as they are tossed up and down in clouds.

As they continue to grow, they lose layers of ice and may become smaller. However, it's important to note that hail pellets do not form ice crystals directly from water, as they require supercooled water droplets to form.

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When the plates of a parallel-plate capacitor are moved apart while connected to a battery, "the amount of charge on the plates decreases".

When the plates of a parallel-plate capacitor are connected to a battery, the potential difference between the plates is fixed, which means the voltage across the capacitor remains constant.

According to the capacitance equation,

C = Q/V,

where C is the capacitance, Q is the charge on the plates, and V is the voltage across the capacitor, the charge on the plates is proportional to the capacitance.

When the plates of the capacitor are moved apart, the distance between the plates increases, which means the capacitance of the capacitor decreases. This is because the capacitance of a parallel-plate capacitor is given by:

C = εA/d

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

As the distance between the plates increases, the capacitance decreases, and since the voltage remains constant, the charge on the plates must also decrease.

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The magnitude of the emf induced in coil 2 is -0.0471 V.

To find the magnitude of the emf induced in coil 2, we can use Faraday's law of electromagnetic induction.

According to this law, the emf induced in a coil is equal to the rate of change of magnetic flux through the coil.

In this case, the changing current in coil 1 produces a magnetic field that varies with time. This changing magnetic field induces a changing magnetic flux through coil 2, which is located nearby.

The magnitude of the emf induced in coil 2 can be calculated as:
emf = -N2 dΦ/dt
where N2 is the number of turns in coil 2, and dΦ/dt is the rate of change of magnetic flux through coil 2.

The magnetic flux through coil 2 depends on the magnetic field produced by coil 1 and the area and orientation of coil 2 with respect to the magnetic field.

Assuming that the two coils are closely coupled and aligned such that the magnetic field produced by coil 1 passes through coil 2, we can write:
Φ = B A
where B is the magnetic field at the location of coil 2, and A is the area of coil 2 perpendicular to the magnetic field.

Since the magnetic field produced by coil 1 is proportional to the current in coil 1, we can write:
B = μ0 N1 I1 / (2π r)
where μ0 is the permeability of free space, N1 is the number of turns in coil 1, I1 is current in coil 1, and r is the distance between the two coils.

Substituting this expression for B into the equation for Φ, we get:
Φ = μ0 N1 N2 I1 A / (2π r)

Taking the derivative with respect to time, we obtain:
dΦ/dt = μ0 N1 N2 A (-dI1/dt) / (2π r)

Substituting this expression for dΦ/dt into the equation for emf, we finally get:
emf = μ0 N1 N2 A (-dI1/dt) / (2π r)

Plugging in the given values, we get:
emf = (4π × 10⁻⁷ T·m/A) × (100 turns) × (50 turns) × (π × 0.05 m²) × (-3.00 A/s) / (2π × 0.10 m)
emf = -0.0471 V

Therefore, the magnitude of the emf induced in coil 2 is -0.0471 V. Note that the negative sign indicates that the emf induces a current in coil 2 that opposes the decrease in current in coil 1, in accordance with Lenz's law.

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In finding the net electric potential due to several charged particles, it is true that we add the contributions as scalars, complete with the algebraic signs. This means we consider the magnitude and sign of each particle's electric potential contribution and sum them up to find the net electric potential at a specific point.

When finding the net electric potential due to several charged particles, we add the contributions as scalars, complete with the algebraic signs. This means that we add the electric potentials due to each charged particle using simple addition, taking into account the direction of each electric potential (positive or negative). The resulting net electric potential at any point in space is the algebraic sum of the individual electric potentials.

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Answer:

The answer for his displacement is 583m,S59°W

Explanation:

hyp²=opp²+adj²

x²=300²+500²

x²=340000

take square root of both sides

x≈583

using SOH CAH TOA

tany=opp/adj

tany=500/300

y=tan‐¹(5/3)

y≈59°

A star with a mass like our Sun cannot fuse elements beyond carbon and oxygen because it does not have sufficient temperature and pressure in its core to overcome the strong electrostatic repulsion between the atomic nuclei of heavier elements.

Stars generate energy through nuclear fusion, where lighter elements combine to form heavier elements in their cores. In the case of a star like our Sun, it primarily fuses hydrogen into helium. As the hydrogen in the core depletes, the core contracts and heats up, allowing for helium fusion to occur. This process creates elements like carbon and oxygen.
However, fusing elements beyond carbon and oxygen requires much higher temperatures and pressures. This is due to the strong electrostatic repulsion between the positively charged atomic nuclei of these heavier elements, also known as the Coulomb barrier. To overcome this barrier, the core needs to have enough energy for the nuclei to come close enough together for the strong nuclear force to bind them.
In summary, a star with a mass like our Sun cannot produce elements beyond carbon and oxygen because its core does not reach the necessary temperature and pressure required to overcome the electrostatic repulsion between the nuclei of heavier elements.

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If a soccer ball is falling with constant velocity, the net force acting on it is zero.

If a soccer ball is falling with constant velocity, it means that the ball is not accelerating. Therefore, the net force acting on the soccer ball is zero. This concept is described by Newton's first law of motion, which states that an object in motion will remain in motion with a constant velocity unless acted upon by an external force.

In this scenario, the force of gravity is acting on the soccer ball, pulling it towards the ground. However, there must be an opposing force that is equal in magnitude and opposite in direction to balance out the force of gravity and maintain a constant velocity.

This opposing force could be air resistance, which is caused by the friction between the soccer ball and the air molecules surrounding it. The amount of air resistance depends on factors such as the size, shape, and speed of the soccer ball.

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Safely de-energizing the debarker requires that six electrical sources and one pneumatic energy source need to be disconnected and locked out. This statement is true.

Lock out, tag out (LOTO) is a safety practise used to make sure that dangerous equipment is turned off completely and cannot be started up again until maintenance or repair work has been finished. Before any work is done on the equipment in question, dangerous energy sources must be "isolated and rendered inoperative".

The isolated power sources are then locked, and a tag identifying the worker and the explanation for the LOTO is attached to the lock. The employee then keeps the lock's key, making sure that only they can open the lock and turn on the machine. This stops equipment from accidentally starting up while it's in a dangerous state or when a worker is in close proximity to it.

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In the z-direction, the form's centre of mass is 1.54 inches distant from the origin.

This task requires us to locate the centroid of a three-dimensional object made by rotating a coloured triangle around the z-axis using the z-coordinate. Either the disc method or the centroid formula for a three-dimensional object can be used to calculate the shape's volume.

The cross-sectional area of the shape at a distance x from the origin is calculated using geometry. By using the integration restrictions and the understanding that in cylindrical coordinates, z = x, it is able to conduct an integral and determine the z-coordinate of the centroid.

The answer to the problem is roughly 1.54 inches. Therefore, in the z-direction, the form's centre of mass is 1.54 inches distant from the origin.

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The maximum electric potential at the surface of the raindrop can be found using the formula V = E * d, where V is the electric potential, E is the electric field, and d is the distance from the surface to a point infinitely far away where the electric potential is zero. Since the electric potential is zero infinitely far away, we can assume that d is equal to infinity.

Therefore, the maximum electric potential at the surface of the raindrop is:

V = E * infinity

Since infinity is not a real number, we cannot directly calculate this value. However, we can say that the electric potential at the surface of the raindrop is proportional to the electric field just outside the raindrop. So, we can write:

V_surface = k * E

where k is a constant of proportionality.

Therefore, the maximum electric potential at the surface of the raindrop is:

V_surface = k * E_max

where E_max is the maximum electric field just outside the raindrop.

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At room temperature, the alloy containing 60wt%pb-40wt%sn is expected to have two phases: α-phase (tin-rich solid solution) and β-phase (lead-rich solid solution).

The amount of each phase present in the material at room temperature can be determined using the lever rule. Assuming complete solidification, the α-phase will have a weight fraction of 40/100 or 0.4, while the β-phase will have a weight fraction of 60/100 or 0.6.

The expected microstructure of the alloy at room temperature will consist of two distinct phases, with the α-phase appearing lighter in color than the β-phase. The two phases will be intermixed throughout the material, with varying grain sizes and orientations depending on the cooling rate and other processing conditions. A typical microstructure may show a network of α-phase surrounded by a continuous β-phase matrix.

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Star B is further away, and it is approximately 0.398 times farther away than Star A.

To determine how many times further away it is than star A, we will follow these steps:

Step 1: Recall the distance modulus formula which relates absolute magnitude (M), apparent magnitude (m), and distance (d) in parsecs:
m - M = 5 * log10(d/10)

Step 2: Since both stars have the same apparent magnitude, let's call it 'm' for both.

We can set up two equations for the two stars:
m - 5 = 5 * log10(d_A/10)
m - 7 = 5 * log10(d_B/10)

Step 3: We want to find the ratio of the distances (d_B/d_A).

Subtract the first equation from the second to eliminate 'm':
-2 = 5 * log10(d_B/d_A) - 5 * log10(1)

Step 4: Simplify and solve for the ratio of distances:
-2/5 = log10(d_B/d_A)
[tex]10^{(-2/5)}[/tex] = d_B/d_A

Step 5: Calculate the value of the ratio:
d_B/d_A ≈ 0.398

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The best assessment for Mr. Vasquez to use is: having students explain why shadow lengths change during the day.

This assessment option allows students to demonstrate their understanding of the relationship between shadow length and time by explaining the underlying factors that cause the change.

Students will need to consider the position of the sun throughout the day, how it affects the angle of the light, and how this results in varying shadow lengths. This assessment goes beyond merely observing or describing shadows, as it requires students to apply their knowledge of the concept and engage in critical thinking.

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The best sketch of several rays from the sun incident on the magnifying glass on a clear, sunny day would be Figure (A).

In Figure (A), we can assume that parallel rays of sunlight are incident on the magnifying glass. This is because the sun is far away, and its rays can be considered parallel when they reach the Earth's surface.

The magnifying glass, being a convex lens, focuses these parallel rays of light onto a single point called the focal point.

This is the basic principle of a magnifying glass, which allows it to concentrate sunlight and create a focused image. The other figures might not represent the correct behavior of sunlight and the magnifying glass.

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"Opacity" The opacity of the gas in the star can affect the passage of light through it, causing the light emitted by the star to vary periodically.

Essentially, when the gas in the star becomes more opaque, it blocks more light from passing through, resulting in a dimming effect.

Conversely, when the gas becomes less opaque, more light can pass through, resulting in a brightening effect.

Hence, changes in a star's opacity can cause variations in the light emitted by the star.

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