"A massive star nearing the end of its life undergoes successive cycles of energy production in its interior (2023)

physical middle School


answer 1




The correct answer is part C.

The phenomenon mentioned in the previous question is simply because the melting of iron nuclei into heavier nuclei requires energy rather than generating excess energy, so no additional gas pressure is created to stop the collapse, so the process does not go beyond the silicon melting cycle. to produce iron.

Related questions

A spring with k = 53 N/m is suspended vertically next to the ruler. The end of the spring is near the 18 cm mark on the ruler. Now, if you put a 2.4 kg weight on the end of the spring and drop it, where will the end of the spring line up with the mark on the ruler when the weight is at its lowest position?



1,07 metros


x = spring compression

k = spring constant = 53 N/m

Initial length = 18 cm

P = kinetic energy

K = kinetic energy

At the lowest mass point, energy is conserved as follows

At its lowest position, the mark on the ruler will be

The spring wire ends at 1.07 m.

When a star is on the main sequence, its equilibrium is the result of the outward pressure of hot gas and the inward pressure of ________.


Answer: When a star is on the main sequence, its equilibrium is the result of the outward pressure of hot gas and the inward pressure of _______.


Inside a star, the inward force of gravity is balanced by the outward pressure. Stars are stabilized (ie nuclear reactions are kept in check) by pressure and temperature thermostats.




When a star is on the main sequence, its equilibrium is the result of the outward pressure of hot gas and the inward pressure of gravity.

Gravity is a property of the mass of an object. The pressure of gravity is the resultant force felt by an object of a given mass.

When gravity hits the star's surface, a density gradient builds up.

Gravitational pressure is gravitational compression, which increases an object's density, compresses its mass, and reduces its size.

If the star's nuclear fusion fuel runs out, the star may go out of sequence.

The pendulum length of your grandfather clock is 0.9930 meters. Part A If the clock runs very slowly, 19 seconds per day, how would you adjust the length of the pendulum? Note: Due to the precise nature of this problem, treat the constant g as unknown (ie, do not assume it is exactly equal to 9.80 m/s2).




The period (T) of a simple pendulum depends on the length of the pendulum (l) and the acceleration due to gravity (g). The period of a simple pendulum can be calculated as follows:

t = 2

The number of hours in a day is 24 hours.

Convert 24 hours to seconds as follows,

24x60x60 = 86400 seconds

The number of cycle clocks made per day is,

86400/2 = 43200

If the clock loses 19 seconds, that's 9.5 fewer cycles per day. so,

(43200 -9.5) cycles = 43190.5 cycles

The term will be reduced. 43190.5/43200 = 0.9998

Then it's the new time period,

T ny = 0.9998 dice



NEW/LOT =0.9996

Replace LODD with 0.9930 m

LNUEVO = 0,9930 X 0,9996 = 0,9926

The length difference is

LNEW Lottery

= 0,9926- 0,9930 = -0,0004

This is the same as 0.4mm.

Therefore, the new length of the pendulum is 0.9926 m and the length of the pendulum must be subtracted. 0.4mm

The adjustment of the pendulum is given mathematically as

dL=0,4 mm

What adjustments should be made to the pendulum?

Question parameters:

The pendulum length of your grandfather clock is 0.9930 m.

If the clock is very slow, every day it goes 19 seconds slower.

In general, the equation for the period is given mathematically as

Among them is the total cycle.

Tc=(43200 -9.5) cycles

Tc = 43190.5 cycles


T ny = 0.9998 dice

Lny/Lold =0.9998^{2}


Lny = 0.9930 X 0.9996


In short, the difference is in the length.

dL = 0,9926-0,9930

dL = -0,0004

dL=0,4 mm

Read more about distance


A solid door that weighs 22 kg is 220 cm high and 91 cm wide. a) What is the moment of inertia of the door that rotates about a vertical axis inside the door, 15 cm from the edge? b) What is the moment of inertia of the door that rotates about a vertical axis inside the door, 15 cm from the edge?


The moment of inertia of a door that rotates on a vertical axis inside the door, 15 cm from an edge, is: I = 3.5648 kg.m²

it is given to us;

The mass of the door; m = 22kg

Door height; height = 220 cm = 2.2 meters

The width of the door; length = 91 cm = 0.91 m

  • Using the parallel axis theorem, the moment of rotation about a vertical axis inside the door, 15 cm from the edge, is;

I = (1/12)ML2 + M(½L - L')2

donde L' = 15 cm = 0,15 m


Yo = (1/12)(22 × 0,91²) + 22(½*0,91 - 0,15)²

yo = 1,5182 + 2,0466

I = 3,5648 kg.m²

  • In short, the moment of inertia of a door rotating about a vertical axis inside the door is 15 cm from one edge.

Read more; brainly.com/question/6956628



Mass of solid door = 22 kg

Bedroom = 220cm x 91cm

moment of inertia of the hinge

r is the distance from an edge, equal to 91 cm or 0.91 m

The moment of inertia of a rectangular door about its center is equal to

Moment of rotation about a vertical axis inside the door, 15 cm from the edge

A 4.9 kg block of wood is pushed down a slope at an angle of 27° to the horizontal. From rest, the block slid a distance of 2.7 m in 5.4 seconds. The acceleration due to gravity is 9.81 m/s 2 . Find the coefficient of kinetic friction between the block and the plane.



µk = 0.488


Newton's Second Law:

∑F = m*a forma (1)

∑F : algebraic sum of forces in Newton (N)

m : time s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the inclined plane, and the y-axis in the direction perpendicular to it.

force acting on the block

W: weight of the block: in the vertical direction

FN: normal force: normal to the slope

fk: kinetic friction: parallel to the slope

W calculation


B= 4,9 kg* 9,8 m/s² = 48,02 N

x-y weight components

Wx = Wsen θ = 48.02*sen27° = 21.8N

Wy = Wcos θ = 48.02*cos27° = 42.786 N

UN calculation

We use the formula (1)

∑Fy = m*ay ay = 0

FN-Wy = 0

FN = pregnant

FN = 42,786 women

fk calculation

fk = μk* FN= μk*42.786 Formula (1)

kinematic block

Since the block is moving at a uniform velocity, we use the following formula to calculate the block's acceleration:

d = v₀*t+(1/2)*a*t² forma (2)


d: displacement (m)

v₀: initial velocity (m/s)

t: time interval (m/s)

a: acceleration (m/s²)


d= 2.7 meters

v₀ = 0

t = 5.4 seconds

We substitute the data in equation (2) i

d = v₀*t+(1/2)*a*t²

2,7 = 0+(1/2)*a*(5,4)²

2,7 = (14,58)*a

Yo = 2,7 / (14,58)

a= 0,185 m/s²

We use formula (1) to calculate μk:

∑Fx = m*ax , ax= a : block acceleration

wx-fk= m*a i forma (1), fk=μk*42.786

21.8 - (42.786)*μk = (4.9)*(0.185)

21.8 -0.907= (42.786)*μk

20.89 = (42.786)*μk

µk = (20.89) / (42.786)

µk = 0.488

The I-beams are driven into the ground with a 2100 kg pile driver. The pile driver descends 5.00 m before hitting the top of the beam. It then drives the beam 12.0 cm toward the ground when it comes to rest. Taking energy into account, calculate the average force that the beam exerts on the pile driver when the pile driver is at rest. a) m=2100 kg b) Xi=5.00m c) Xf=12.0 cm =.12m



f = 878.080N


Pile driver mass (m) = 2100 kg

Distance from hammer to steel beam(s) = 5 m

Depth of drive steel (d) = 12 cm = 0.12 m

Acceleration due to gravity (g0 = 9.8 m/s^{2}

Calculate the average force exerted by the beam on the pile driver.

  • Work done = force x distance
  • Work done = change in potential energy of the hammer
  • Equating the two previous equations, we have

Kraft x soporte = m x g x (s - d)

f x 0,12 = 2100 x 9,8 x (5- (-0,12))

d = -0.12 because the steel beam collapsed when we disassembled it

The initial position is the origin, which is 0

f = ( 2100 x 9,8 x (5- (-0,12)) ) ÷ 0,12

f = 878.080N

234.0 g of lead plate is heated to 86.0°C and poured into a calorimeter containing 611.0 g of water with an initial temperature of 24.0°C. Ignoring the heat capacity of the container, find the final equilibrium temperature (in oC) of lead and water.






Water quality in calorimeter

initial water temperature

The initial temperature of the lead plate.

We know that the heat capacities of lead and water are and

Let's take the final temperature of the system.

energy saving

heat lost by lead = heat gained by water

The resistance R of a cylindrical wire is given by where p is the resistivity of the wire material, L is the length of the wire, and d is the diameter of the wire. What happens to the resistance of the wire as the diameter approaches 0?



its fine



Resistance = R

resistivity = ρ

Length = L

diameter = d

The resistance of the wire R is

A = area

Now set the value of A

Since d tends to infinity, d² will also tend to infinity.

Therefore, as d approaches zero, the resistance approaches infinity.

So the answer is ---

its fine




Why is Saturn almost as big as Jupiter, even though it is less massive? one. Saturn's rings make Earth appear larger. Jupiter's strong magnetic field limits its size.
b) Saturn has more hydrogen and helium than Jupiter and is therefore less dense.
c) Jupiter's greater mass compresses it further and increases its density.
d) Saturn is farther from the Sun and therefore cooler and therefore less compact.



c) Jupiter's greater mass compresses it further and increases its density.


There is more mass inside Jupiter, and this mass compresses Jupiter to a certain degree. Therefore, its density increases. Now more mass is packed into a smaller volume. Therefore, its size will not increase significantly. Saturn, on the other hand, is less massive and less dense. This gives the Saturn quite a loud volume.

Therefore, option C is correct.

Briefly explain why the intensity reflected from the back of the film (ie, the right surface where the liquid-gas interface is present) is 2.78% of the intensity of the beam transmitted through the front surface.



Let the incoming light travel in the +x direction so that it hits the left side of the film at normal incidence (called "front"). This means that the beam transmitted to the liquid is essentially as strong as the incoming beam.

Almost all the light reflected from behind passes through the front. (but only 2.78% is reflected to the right by the front surface). This means that there are two rays reflected in the -x-direction, one from the front and one from the back, and these rays are almost equal in intensity.

I hope this helps. Thank you

A baseball is thrown at an angle of 29 degrees to the ground with a velocity of 24.3 m/s. The ball is caught 51.0463 meters from the thrower. The acceleration due to gravity is 9.81 m/s2. How long was it in the air? Answer in units of p.



T = 2.4 seconds



Kastevinkel = 29°

Velocity relative to the ground = 24.3 m/s

Ball caught from distance = 51.0463

Acceleration due to gravity = 9.8 m/s²

What is the time of the ball in the air = ?


velocity of the ball in the x-direction

V_x = v cos θ

V_x = 24,3 x coseno 29°

V_x = 21,25 m/s

velocity in the y-direction

V_y = v sen θ

V_y = 24.3 x sin 29°

V_y = 11,78 m/s

The distance from the ground when the ball reaches its maximum height.

x = 51.0463/2 = 25.52 meters

At the highest point, the velocity is equal to zero.

once the ball is in the air

v = tu + a t

0 = 11.78 - 9.8 tons

t = 1,20

This time you must go half the distance

Total time = 2 x 1.20

T = 2.4 seconds

The time the ball is in the air is T = 2.4 s

A cheerleader lifts her 37.4 kg partner 0.817 m off the ground before releasing her. The acceleration due to gravity is 9.8 m/s 2. If she does this 27 times, how much work does she do? Answer in units of J



W = 8085.064J



Mass of cheerleading partner = 37.4 kg

The distance she was lifted = 0.817 m

Acceleration due to gravity = 9.8 m/s²

Number of times she was chosen = 27 times

The work you did = ?


The work done will be equal to the potential energy converted to the number of times you lift.

Work done = N m g h

Width = 27 x 37.4 x 9.8 x 0.817

W = 8085.064J

The work done by your partner is equal to W = 8085.064 J

A box weighing 150 N is pulled across a horizontal floor with a force of 110 N. The pull can be directed horizontally or at an angle θ above the horizontal. When the pull is directed horizontally, the kinetic friction acting on the box is twice that when the pull is directed at an angle θ. Find theta.





= kinetic friction

= angular traction

= Box weight = 150 N

kinetic friction

angle pull

N = tensile force

according to the question

Applying Newton's second law in the vertical direction, we get

The angle is 42.99°

Explain why the atomic radius decreases as the period moves to the right for main group elements but not for transition elements. Match the words in the left column with the corresponding spaces in the sentences on the right.





answered only the first question

In the cycle from left to right, the nuclear charge increases so that the nuclear size is compressed. Therefore, the atomic radius decreases.

In transition elements, the electrons in the ns^2 orbitals remain unchanged, that is, the outermost orbital has 2 electrons and the electrons add to the (n-1)d orbitals. Therefore, the outer orbital electrons experience almost the same nuclear gravitational force and therefore stay the same size.

2. Which of the following is an example of passive immunity? A. Anti-rabies antibodies given to a person bitten by a possibly rabid dog B. Heat-inactivated influenza antigen grown in eggs C. Live virus antigen given to sugar cubes to prevent polio D. A and B. E. B and c.



A. Giving rabies antibodies to people who have been bitten by a possibly rabid dog


Passive immunity is immunity or protection against certain infectious diseases that is obtained by injecting antigens or antibodies into the bloodstream, or certain antibodies can be passed through colostrum or breast milk.

Option A. Administer rabies antibodies to people who have been bitten by a possibly rabid dog. Immunization against rabies infection is given to humans

For option B: it only tells us the type of infection, but not the immunity given

Option c is probably completely correct as it is another method of immunization but not a passive immunization.

The electric motor does 30 kJ of work and receives 4 kJ of heat from the surrounding environment. What is the change in the internal energy of the motor?



ΔU= *-26 KJ



The work done by the motor W= 30 KJ

Motor thermal gain Q= 4 KJ

Sign the agreement:

If heat is added to the system, it is considered positive, and if heat is rejected from the system, it is considered negative.

If the system does the work, it is considered positive and if the system does the work, it is considered negative.

From the first law of thermodynamics

Q = W + ΔU

ΔU = change in internal energy

Q = heat transfer

W = work

Now send the value

4 = 30 + dig

ΔU= -26 kJ


Internal energy ΔU=-26KJ



Engine work=+30KJ

The heat gained by the engine = +4KJ

When solving thermodynamic problems, it is reasonable to use the notation convention

Heat is positive if heat is added to the system, but becomes negative if heat is rejected by the system.

If the system works, the work is positive, but if it is successfully done on the system, the work becomes negative.

Using the first law of thermodynamics


∆U= 4-30=-26KJ

Two horizontal forces act on a 1.4-kg cutting board that slides across a frictionless kitchen table that lies in the xy-plane. A fortress it is. For each of the following forces, find the acceleration of the anvil in unit vector notation.



in the

b) de

c) of


According to Newton's second law, we know

F = en

so we want

so we want

in the

b) de

c) of

A monochromatic beam is incident on a barium blank with a work function of 2.50 eV. If a potential difference of 1.00 V is required to return all the emitted electrons, what is the wavelength of the beam? a) 355 nmb) 497 nmc) 744 nmd) 1.42 pm) None of these Answer



b) 497nm



Work function, φ = 2.50 eV

Stopping potential, V₀ = 1.00 eV

Electronic charge, e = 1.6 x 10⁻¹⁹ C

Speed ​​of light, c = 3 x 10⁸ m/s

Planck's constant, h = 6.63 × 10⁻³⁴ Js = 4.14 × 10⁻¹⁵ eVs

Einstein's photoelectric equation is given by:

= E - φ ----- (1)

is the maximum kinetic energy

E is the energy of the absorbed photon: E = hf

ϕ is the work function of the surface: ϕ = hf₀

The potential difference required to support all the ejected electrons is called the stopping potential (V₀)

Braking potential (V₀)


Substitute (2) for (1)

eV0 = E - ϕ

1,6 x 10⁻19 x 1 = mi - 2,50

E = 1.6 x 10-19 + 2.50

E = 2,50 electronvoltios

But E = hf = hc/λ


λ = (4.14 × 10⁻¹⁵ × 3 × 10⁸) / 2.50

λ = 1240 x 10⁻⁹ / 2.50

λ = 496.8 x 10⁻⁹ 米

λ = 497 × 10⁻⁹ m (aprox.)

λ = 497 nm

The specific material has a refractive index of 1.25. What is the speed of light in matter as a percentage of the speed of light in a vacuum?



80% (first percentage)


The material has a refractive index (n) of 1.25

The speed of light (c) in a vacuum is 2.99792458 x 10⁸ m/s

We can find the speed of light (v) in a material using the relation

n = c/v, correspondingly

v = c/n

SAv = 2,99792458 x 10⁸ m/s ÷ (1,25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore, the percentage of the speed of light in a vacuum compared to the speed of light in material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1,25) × 100 = 0,8 × 100 = 80 %

Therefore, the speed of light (v) in material is eighty percent of the speed of light (c) in a vacuum.


The speed of light in a vacuum is 80% of the speed of light in this material.


Common values ​​for the refractive index are between 1 and 2, since nothing can travel faster than the speed of light, therefore no material has a refractive index less than 1.

According to the formula

where is the refractive index

is the speed of light in a vacuum

If the speed of light is in the material, it can be seen that n and v are inversely proportional, the higher the refractive index, the slower the speed of light.

Since we know that the speed of light in a vacuum is 300,000 km/s using the formula we obtained,

To find the percentage,


A 4.0-kg particle is forced to move along the x-axis by a single force F(x) = −cx3, where c = 8.0 N/m3. The velocity of the particle at A (xA = 1.0 m) is 6.0 m/s. What is its speed at point B, where xB = −2.0 m?


Answer: 4.58m/s



particle mass

to integrate


"A massive star nearing the end of its life undergoes successive cycles of energy production in its interior? ›

A high mass star near the end of its life undergoes successive cycles of energy generation within its core in which gravitational collapse increases the temperature to the point where a new nuclear fusion cycle generates sufficient energy to stop the collapse.

What type of stars end their lives in a supernova event? ›

Supernovae are most likely to form in stars that are at least eight times the mass of our Sun, while the most massive stars may not form at all. Stars with a mass of at least eight times that of our Sun are more likely to finish their lives as supernovae.

Which of the following will a high mass star not do at or near the end of its life? ›

A Kerr black hole. Which of the following will a high-mass star NOT do at or near the end of its life? Eject its outer layers and become a white dwarf.

What is the end product of the life of a low mass star? ›

Over its lifetime, a low mass star consumes its core hydrogen and converts it into helium. The core shrinks and heats up gradually and the star gradually becomes more luminous. Eventually nuclear fusion exhausts all the hydrogen in the star's core.

Which event takes place first in the stages before the birth of a star? ›

A protostar is the earliest stage of a star's life. A star is born when the gas and dust from a nebula become so hot that nuclear fusion starts.

What is a star that is nearing the end of its lifecycle called? ›

White Dwarf - a star that has exhausted most or all of its nuclear fuel, collapsed into a size similar to the Earth; such a star is near the final stage of its evolution.

What is the explosion at the end of a stars life called? ›

A supernova is a large explosion that takes place at the end of a star's life cycle.

What happens to a high mass star at the end of its life cycle? ›

A massive star will undergo a supernova explosion. If the remnant of the explosion is 1.4 to about 3 times as massive as our Sun, it will become a neutron star.

What is the interior of a massive star before a supernova? ›

Just before it exhausts all sources of energy, a massive star has an iron core surrounded by shells of silicon, sulfur, oxygen, neon, carbon, helium, and hydrogen. The fusion of iron requires energy (rather than releasing it).

What is a star called that is at the end of its life is very small about 10 miles across and is made up entirely of neutrons? ›

A neutron star is the result of the collapse of a star after a supernova explosion. The resulting neutron star is made up almost entirely of neutrons. Neutron stars are some of the most dense objects in the universe, having a mass up to two times our sun, but a diameter ranging only between 10 - 20 kilometers.

What are the three end stages of stars? ›

Three possible end stages of stars are:
  • White dwarf. Initial mass of the star is less than 8 times mass of the Sun. ...
  • Neutron star. Initial mass of the star is in between the 8 times mass of the Sun and 25 times mass of the Sun. ...
  • Blackhole. Initial mass of the star is greater than 25 times mass of the Sun.

What type of star will go supernova? ›

These are the elements that make up stars, planets, and everything on Earth – including us! Will our Sun go supernova? No, smaller stars like our Sun end their lives as dense hot objects called white dwarfs. Only stars that contain more than about 8 to 10 times the mass of our Sun will go supernova.

Do all stars end their life as a supernova? ›

While not every star in the main sequence will go supernova, those that do will first pass through a "giant" phase after they exhaust their hydrogen reserves and instead start fusing helium and residual heavier elements like oxygen, silicon, and carbon rather than hydrogen (limited fusing of these heavier elements ...

What type of stars end their lives? ›

A star ends its life as a neutron star when the pressure of its electrons balances its gravity.

What mass of star ends its life in a supernova? ›

Supernova being the last stages of a star, will happen only if the mass of the star exceeds the solar mass and hence a 10 solar mass star will have a high probability of becoming a supernova.


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