In a free-fall experiment in the lab, trainee found out that the average time of fall of the teel ball i 0. 550. What i fall height?

The average force acting on the car is 1000 N.

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

Given that: A 600-kg car is moving at 5 m/s then comes to a stop in 3 seconds.

Now according to Newton's 2nd law of motion:

Magnitude of force applied on the body = change in momentum/time interval

= I final momentum - initial momentum I/time interval

= I 0 - 600×5 I /3 N

= 10000 N

Hence, average force acting on the car is 1000 N.

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The period of a wave that has a frequency of 57Hz is 0.018s.

A wave in physics is a moving disturbance in the energy level of a field.

Period refers to length of time and is denoted by T. The period of a wave can be calculated by finding the inverse of the frequency as follows;

T = 1/f

Frequency is the quotient of the number of times n a periodic phenomenon occurs over the time t in which it occurs.

T = 1/57

T = 0.018s

Therefore, 0.018s is the period of the wave.

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The calculated value is 453 g. density is the mass of a material substance per unit volume. d = M/V, where d is density, M is mass, and V is volume, is the formula for density.

Grams per cubic centimeter are a typical unit of measurement for density.

8.5 g/mL Brass volume, total mass is brass plus wood:

M = 8.5b+100 (grams) (grams).

Volume of wood: 100/0.2 = 500 mL

Brass and wood volumes add up to the total:

V = b+500

Total density equal to water for simple floatation is 1

Total mass equals 1 * V + 500.

Calculate M using [1] above:

b+500 = 8.5b+100

400 = 7.5b

53.33 mL is the brass volume.

Brass should never be measured by volume. 8.5*53.33 = 453 g is the best; make careful to fasten it to the underside of the wood.

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At an angle of 33 degrees, the ray leaves the second mirror.

As per the below-attached figure:

The light beam is reflected at A, incident on the first plane at XA, and then incident on the second plane at XB, where it is again reflected as BY.

The law of reflection is the idea that when light rays strike a smooth surface, the angle of incidence and the angle of reflection are both identical. Additionally, the incident ray, the reflected ray, and the surface normal all reside in the same plane.

From the law of reflection of angle:

angle : BAO = θ = 34 degree

It is also given that,

Angle ABO = φ

Angle AOB = 113 degree

As in a triangle,

The sum of all three angles must be 180 degrees.

So, we can write it as:

∴θ+φ+113=180

φ=180-113-34= 33 degree.

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The required diagram is attached below:

The fraction of the neutron's kinetic energy transferred to the atomic nucleus is 2.05

If the initial kinetic energy of the neutron is

4.2 × 10−13 J, its final kinetic energy  will be 2.50 joules

kinetic energy of an object is described as the energy that it possesses due to its motion which is also the work needed to accelerate a body of a given mass from rest to its stated velocity.

In elastic head-on collision, the energy of the system and total momentum is conserved.

For the neutron,

m - mass

Vi - initial velocity

Vf - final velocity

For the atom ,

M- mass

Mi - initial velocity

Mf - final velocity

Conserved momentum on the head-on collision

mvi = mvf + MVF

vi - vf = MmVf

Vi- vf = 11.1 vf...... (i)

The kinetic energy is

Vi²- vf² = 11.1 v²f.......(ii)

Combining both equations

Vi + Vf = Vf

Solve equation (I) and (III) for Vi

vi = 5.4vf

Now, the fraction of the neutron's kinetic energy transferred to the atomic nucleus,

1/2 mv²f/ 1/2 mv²i = 11.1/ 5.4

If the initial kinetic energy of the neutron is

4.2 × 10−13 J, its final kinetic energy  will be 2.50 joules

Hence, we can conclude that the fraction of the neutron's kinetic energy transferred to the atomic nucleus is 2.05

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The maximum transverse acceleration of a point on a wire is closest to 47,458.2 m/s²

The angular speed of the transverse wave is calculated as follows;

ω = 2πf

where;

ω = 2πf

ω = 2π ( 570 )

ω = 3,581.4 rad/s

The maximum transverse acceleration of a point on a wire is calculated as follows;

a = ω²A

where;

a = ω²A

a = (3581.4)²(0.0037)

a = 47,458.2 m/s²

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Answer:

The angle at which a projectile is launched is important because it determines the initial velocity of the projectile and the direction in which it will travel. The angle also affects the projectile's trajectory, or the path that it follows through the air. The steeper the launch angle, the higher the projectile will go and the farther it will travel horizontally. A shallower launch angle will result in a flatter trajectory that does not go as high or travel as far. Therefore, the angle at which a projectile is launched is a key factor in determining how far and how high it will go.

Explanation:

The angle is important when launching projectile motion because it in the calculation of the maximum height, range and time of flight, angle of launching  is as important as the magnitude of initial velocity.

When a particle is hurled obliquely near the surface of the earth, it travels along a curved path while accelerating continuously in the direction of the planet's center (we assume the particle stays close to the surface of the globe). Such a particle's motion is known as projectile motion, and its route is referred to as a projectile.

Two distinct rectilinear motions occur simultaneously in a projectile motion:

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A and B are both false.

When the highest and lowest frequency of a circuit is given then bandwidth is  calculated as 9100 Hz.

Maximum amount of data transmitted over an internet connection in given amount of time is known as bandwidth. Bandwidth is actually the volume of information that can be sent over a connection in a measured amount of time.

In signal processing, bandwidth describes difference between upper and lower frequencies in transmission signals. The measurement of bandwidth of a signal is done in hertz (Hz).

Given highest frequency of circuit is 10 kHz

= 10000 Hz

and lowest frequency of circuit is 900 Hz

Bandwidth = 10000 Hz -900 Hz

So, Bandwidth = 9100 Hz.

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To help you recall the direction of magnetic forces, use the right-hand rule. However, we can also modify the magnetic flux by introducing a wire loop into a magnetic field.

A magnetic field is created when current travels through a conductor. Individual magnetic fields are created as free electrons flow in one direction.

There are three things that can change the voltage's size: how big the magnetic field is. There are more flux lines, hence there are more flux lines for the conductor to cut.

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A boy who is 1 meter tall requires a mirror that is at least 0.50 m tall in order to see his entire height.

The mirror must be at least half as long as the individual is tall in order for them to see their entire reflection. In order to make it simpler for the person to see themselves in the mirror.

Given that the boy's height (h) = 1 m,

the mirror's height should be h/2 = 1/2 = 0.50 m.

Behind the glass of a mirror, people can see an image of themselves. This image is created when light rays hit the shining surface and reflect or bounce back, creating a "mirror image." Glass can be made into a mirror that reflects light by having a bright metal coating on one side. Mirrors can reflect 95% of the light that hits them, whereas window glass can only reflect 8% of the light that hits it.

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The fraction of a piece of silicon will be submerged when it floats in mercury is 17%

Density silicon = 2.33 g/cm³,

Density mercury 13.6 g/cm³.

2.33=/13.6 = 0.17 = 17% silicon submerged and 83%  above surface of mercury.

With the symbol Si, silicon is an atomic number 14 chemical element. It is a hard, brittle crystalline solid with a blue-grey metallic sheen that is a tetravalent metalloid and semiconductor. Because silicon is abundant in the Earth's crust, natural goods with silicon as their main component have been used for thousands of years.

The most abundant substance found in the crust of the Earth is silicon dioxide (SiO₂).

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Since the second energy level has more energy than the first, the electron must gain energy to travel from n = 1 to n = 2. To increase to the second energy level, it must gain (-3.4) - (-13.6) = 10.2 eV of energy.

By absorbing light, the electron can obtain the energy it requires. The electrons in an atom revolve around the nucleus. Small, negatively charged particles known as electrons move in a circular orbit around the nucleus. They are unable to move around at will in any location. Depending on their energy levels, their revolution is limited to certain orbits. Energy levels are nothing more than the electrons' set distances from an atom's nucleus. The electron shell is another name for the energy levels. An electron can migrate across energy levels, but it cannot remain in one energy level while in another.

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A hydraulic lift's input pressure is 1200 Pa. If the output force on the other side is 800 N, the output surface area is 0.67m^2

Given the input pressure (P1) = 1200Pa

Output force (F2) = 800N

We know that P= F/A where A is area of cross section

Let the input force be F1 and area is A1

Then by pascals law we say that:

F1/A1 = F2/A2

1200 = 800/A2 (since P1= F1/A1)

A2 = 8/12 = 0.67m^2

Hence the area of other side = 0.67m^2

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1/2tan(tetha) is the minimal coefficient of static friction required for the hoop to roll without slipping, and 1/2g sin(tetha) is the acceleration of the hoop's centre.

In mechanics, acceleration is the rate at which an object's velocity with respect to time changes. The vector quantity of accelerations (in that they have magnitude and direction). The direction of the net force imposed on an object determines its acceleration in relation to that force. In other words, static friction is the force that maintains an object's resting position.

acceleration A = 1/2g sin where an Ihoop = m*r2 ma = mg sin (o) -Fr ma = mgsin(o) -ma (0)

The least possible static friction coefficient

Alpha = r(mug cos) (tetha)

Mu - a/g cos(tetha) = 1/2 (tan(tetha)

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The work done by the force when the sled moves from x=0 to x=8 m is -2048 J

Force is a fundamental concept in physics that describes an interaction between two or more objects that results in a change in motion, shape, or energy of one or both of the objects.

Work done = W

Force = F

Distance = x

Area under the curve = A

Here, the distance moved is 8.0 m and the force is given by the graph as a function of x. Thus, we can use the area under the curve to calculate the work done from x = 0 to x = 8.

= W = F X x

= W = F ∫ x  (from x = 0 to x = 8.)

= W = - (4/3) X x³

= W = -2048 J

Therefore, the work done by the force F is -2048 J.

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Fraction of the energy density of the universe is dark matter is 24% .

It turns out that roughly 68% of the universe is dark energy. Dark matter makes up about 27%.

A mystery, unseen element termed dark matter (25%) and a force that defies gravity called dark energy seem to make up the rest of the cosmos (70 percent). Since the mean energy density of the universe is equal to the critical density (within a 0.5% margin of error), WMAP's discovery that the universe is flat implies. This is comparable to a mass density of just 5.9 protons per cubic meter, or 9.9 x 10-30 g/cm3.

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Answer:

Question 1.  50 m

Question 2.  80 kg

Question 3.  750 N

Question 4.  80 m

Explanation:

Here are some equations that we might have to use.

[tex]F_{Cnet} =m*a_c\\F_{Cnet} =\frac{m*v^2}{r}\\a_c=\frac{v^2}{r}\\a_c=\frac{F_{Cnet} }{m}\\m=\frac{F_{Cnet} }{a_c}\\m=\frac{F_{Cnet} *r}{v^2} \\r=\frac{m*v^2}{F_{Cnet} }\\r=\frac{v^2}{a_c}[/tex]

Question 1

We are given

[tex]m=700\\v=20\\F_{Cnet}=5600[/tex]

We will use [tex]r=\frac{m*v^2}{F_{Cnet} }[/tex]

[tex]r=\frac{700*20^2}{5600}[/tex]

[tex]r=50[/tex]

Question 2

We are given

[tex]r=25\\v=15\\F_{Cnet} =720[/tex]

We will use [tex]m=\frac{F_{Cnet} *r}{v^2}[/tex]

[tex]m=\frac{720*25}{15^2}[/tex]

[tex]m=80[/tex]

Question 3

We are given

[tex]m=900\\v=5\\r=30[/tex]

We will use [tex]F_{Cnet} =\frac{m*v^2}{r}[/tex]

[tex]F_{Cnet} =\frac{900*5^2}{30}[/tex]

[tex]F_{Cnet} =750[/tex]

Question 4

We are given

[tex]m=800\\v=40\\F_{Cnet}=16000[/tex]

We will use [tex]r=\frac{m*v^2}{F_{Cnet} }[/tex]

[tex]r=\frac{800*40^2}{16000}[/tex]

[tex]r=80[/tex]

Hopefully this helps! A rating/review would be much appreciated! :)

A slower-than-normal power seat is being tested by a technician. When the power seat motors are running, the dome light noticeably dims. Most likely, the seat mechanism's binding is the issue.

Instead of using manual levers, cranks, or knobs to alter the position or recline of the seat, power seats in automobiles use electric motors. They make it simpler for both the driver to select the ideal driving posture and for the passengers to settle into a comfortable position for the ride. A power seat motors doesn't have a predetermined lifespan. The switch will wear down more quickly the more frequently you change your seat. The scenario is identical to that of power door locks or windows. The switch's contacts deteriorate more quickly the more you use it. The price of an aftermarket seat motor can range from $50 to $340. According to a number of factors, including

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The momentum of the system before the object drops is 50 kg m/s.

The momentum of the system after the object drops is 50 kg m/s.

The final speed of the cart and the mass is 3.57 m/s.

We know that, momentum is the product of mass and velocity of the system.

Mathematically,

P = m* v

where,

m is the mass

v is the velocity

Given that,

Mass of the cart m₁ = 10 kg

Velocity of the cart before the object drops v₁ = 5 m/s

Mass of the object dropped m₂ = 4 kg

Velocity of the cart after the object drops v₂ = ?

According to the principle of conservation of linear momentum,

Total momentum of the system before the object drops = total momentum of the system after the object drops

So, mathematically,

m₁* v₁ = (m₁ + m₂)* v₂

10 * 5 = (10 + 4) v₂

14 v₂ = 50

v₂ = 3.57 m/s

Thus, Final speed of the cart = 3.57 m/s

Final mass of the cart = 10 + 4 = 14 kg

The momentum of the system before the object drops = m₁* v₁ = 10* 5 = 50 kg m/s

The momentum of the system after the object drops = (m₁ + m₂)* v₂ = 14* 3.57 = 49.98 ≈ 50 kg m/s

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Answer:

Approximately 2.4 x 10^-8 N.

Explanation:

The force acting on the smaller object can be calculated using the formula for gravitational force:

F = G * (m1 * m2) / d^2

Where F is the force, G is the gravitational constant (6.674 x 10^-11 N * m^2 / kg^2), m1 is the mass of the smaller object (1.5 kg), m2 is the mass of the larger object (6.0 x 10^24 kg), and d is the distance between the two objects (6.4 x 10^6 m).

Substituting these values into the formula, we get:

F = 6.674 x 10^-11 * (1.5 * 6.0 x 10^24) / (6.4 x 10^6)^2

We can simplify this expression by dividing both sides by 6.0 x 10^24 to get:

F / 6.0 x 10^24 = 6.674 x 10^-11 * (1.5 / (6.4 x 10^6)^2)

Then we can simplify the right-hand side by performing the calculations in parentheses:

F / 6.0 x 10^24 = 6.674 x 10^-11 * (1.5 / (6.4 x 10^6)^2)

               = 6.674 x 10^-11 * (1.5 / 41.6 x 10^12)

               = 6.674 x 10^-11 * 3.6 x 10^-13

Finally, we can multiply both sides by 6.0 x 10^24 to get the value of the force acting on the smaller object:

F = 6.0 x 10^24 * (6.674 x 10^-11 * 3.6 x 10^-13)

 = 2.4 x 10^-8 N

Therefore, the size of the force acting on the smaller object is approximately 2.4 x 10^-8 N.

The speed of spacecraft relative  exam after 88.0 min elapsed on her clock is 73.65 min.

What is speed ?

A direction or speed at which an object's position changes.

What is space craft ?

An object that orbits the Earth, the moon, or another celestial body is referred to as a satellite, whereas a spacecraft is a vehicle or technology designed for travel or operation outside the Earth's atmosphere.

Given speed of space craft perpendicular is v₁= 0.68 c

speed of space craft 2 is V₂= 0.24 c

They travel towards each other .

time elapsed in spacecraft 1 is = 54 mins

SO, speed of space craft ⊥ relative to space craft 2 is

v¹= V₁+V₂/1+V₁ V₂/C²

∴ V¹= (0.68+ 0.24) C/ 1+(0.68* 0.24 ) C²/C² = 0.791 C

∴ Time eclipsed in spacecraft 2 is, t¹ = t/[tex]\sqrt{1-v12/c2}[/tex] = 54/sqrt{1- (0.7911)2}[tex]\sqrt{1- 0.791}[/tex]

= 88.25 min

speed of spacecraft relative to earth is = 0.68c

∴time on earth t¹¹= t/[tex]\sqrt{r-v1/l1}[/tex]

= 54/ [tex]\sqrt{1- (0.08) 2}[/tex] = 73.65 min

Therefore, speed of spacecraft relative  exam after 88.0 min elapsed on her clock is 73.65 min.

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The instantaneous Bof change of the mass of the colony is t=2.059277.

The immediate charge of exchange is the exchange inside the rate at a specific immediate, and it's miles identical because of the trade inside the spinoff cost at a specific point. For a graph, the immediate fee of trade at a specific point is similar to the tangent line slope. that is, it's miles a curve slope.

Calculation:-

P(t)=2+5tan−1(t2)

=6P(t)=2+5tan−1(t2)=6.

The solution is t=2.059277.

Because the spinoff P′P′ may be interpreted because of the immediate fee of alternate.PP with respect to time, the calculator is used to numerically calculate the by-product of PP at this fee of tt.

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complete question is -

Where tt is measured in days. what is the instantaneous rate of change of the mass of the colony, in grams per day, at the moment the colony reaches a mass of 6 grams?

The Milky Way is near the top of the luminosity range for blue cloud galaxies. Spiral galaxies usually have a luminosity of a few hundred billion Suns and Milky Way's luminosity is about 1.5 x 10¹⁰, making it near the top of the blue cloud galaxies.

Generally, a galaxy's absolute magnitude (a measure of luminosity) is related to its mass. The relationship can be visualized in a galaxy color-magnitude diagram with three populations:

The Milky Way is a spiral galaxy. Spiral galaxies usually produce more stars because the spiral arms generate dense compression areas that provide a good environment for the formation of stars.

Your question seems incomplete, but most likely the completed version is as follows:

The luminosity of our Milky Way galaxy is about 1.5 x 10¹⁰ Lsun. Describe how the Milky Way's light output compares with those of the most luminous galaxies in the blue cloud, based on the information in Figure 1.

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The unstretched length so that it does not allow the block to leave the surface until θ= 60° is L = 4.574 ft.

Given:

The weight of the block W = 2 lb

The initial velocity of the block v_i = 0

The stiffness of the spring k = 2 lb/ft

The radius of the cylindrical surface r = 2 ft

Compute the velocity of the block at θ= 60°. Use Newton's second equation of motion in direction normal to the surface.

= F_n = m X a_n

Where, a_n is the centripetal acceleration or normal component of acceleration as follows:

= a_n = v²_2 / r

Substitute:

= F_n = m*v²2 / r

Where, F_n normal force acting on block by the surface is:

= F_n = W X cos(θ)

Substitute:

W*cos(θ) = m*v^2_2 / r

v_2 = sqrt ( r*g*cos(θ) )

Plug in the values:

v²_2 = 2 X 32.2 X cos(60)

v²_2 = 32.2 (ft/s)²

Apply the conservation of energy between points A and B where θ= 60° :

T_A + V_A = T_B + V_B

Where,

T_A : Kinetic energy of the block at initial position = 0

V_A: potential energy of the block initial position

V_A = 0.5 X k X x_A²

x_A = 2 X π- L           ( L is the original length )

V_A = 0.5X 2 X (2 X πi - L)² =

= (2 X pi - L)²

T_B = 0.5 X W/g X v_2² =

= 0.5 X 2 / 32.2 X 32.2 = 1

V_B = 0.5 X k X x_B² + W X 2 X cos(60)

and, x_B = 2 X 0.75 X π - L             ( L is the original length )

V_B = 0.5 X 2 X (1.5 X π - L)² + 2 X 1

= 2 + ( 1.5 X π- L )²

Input the respective energies back in to the conservation expression:

= 0 + (2π - L)² = 1 + 2 + ( 1.5π - L )²

= 4π² - 4πL + L² = 3 + 2.25π² - 3πL + L²

= πL = 1.75π² - 3

= L = 4.574 ft

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The complete question;

the 2-lb brick slides down a smooth roof, such that when it is at a it has a velocity of 5 ft/s. The 2-lb block is released from rest at A and slides down along the smooth cylindrical surface. Of the attached spring has a stiffness k = 2 lb/ft. Determine its unstretched length so that it does not allow the block to leave the surface until θ= 60°.

The electric force between two or more charged objects depends on the charge and the distance between the charged objects

Coulomb's law states that the force of attraction or repulsion between two charged bodies is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Coulomb's law shows the relationship between the force , charge and the distance between the bodies

therefore F= kq1q2/r²

where q1 is the charge of body 1 and q2 is the charge of body 2. r is the distance between them and k is called electrostatics constant which have a value of 9,×10⁹ Nm²/C²

Therefore what can affect the force between two charges are the product of the charge of the two bodies and the distance between the two charge.

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The total work done on the box is found to be 24 Joules.

A force of 20 newton is applied on the box where the friction force of 8 newton is also working the box is slide to a distance of 2m.

The total work done on the box can be calculated by using the relation,

W = (F-Fr)d

Where,

W is the work done,

F is the applied force,

Fr is the frictional force and,

d is the distance to which the box slides.

Here, we are subtracting the applied force and the frictional force because they are in the opposite directions.

Putting values,

W = (20-8)2

W = (12)2

W = 24 J.

So, the work done on the box is 24 Joules.

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The coherence qualities of various light sources are explored, as well as how they affect the image quality of a holographic display.

Temporal coherence is connected to the intrinsic spectrum bandwidth of the light source, but spatial coherence is determined by light source size and propagation distance. These two coherence properties, as well as the quality of the holographic reconstructed images, are measured for various light sources of diode-pumped solid-state (DPSS) laser, laser diode (LD), light emitting diode (LED), super luminescent light emitting diode (sLED), and micro light emitting diode (mLED) in various settings. Although image sharpness and speckle are both related to coherence parameters, our findings and subsequent research reveal that spatial coherence is directly connected to image sharpness.

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Due to the fundamental differences between these two galaxy types, galaxies cannot evolve from elliptical to spiral (or spiral to elliptical).

For example, in contrast to spiral galaxies, which include gas and dust as well as hot, bright O and B type stars, elliptical galaxies are devoid of visible gas and dust and hot, brilliant stars.

A galaxy with an elliptical shape and a uniform brightness distribution is called an elliptical galaxy (no spiral arms). Elliptical galaxies are stretched galaxies with an elliptical shape, according to astronomical physics. The light is distributed evenly and smoothly throughout this type of galaxy, which is typically composed of old stars.

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A cartridge is removed from the chamber by the extractor mechanism, and is then ejected by the ejector.

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