When the hydrogen in a star's core is used up, what occurs?
A. The core collapses causing a huge explosion called a supernova.
B. The nitrogen core collapses and the outer layer expands into a red giant.
C. The helium core collapses and the outer layer expands into a red giant.
D. The outer layer drifts away leaving a hot dense white dwarf core.

Answers

Answer 1

The core will collapse under its own gravity, leading to a supernova explosion that expels the outer layers of the star into space, leaving behind either a neutron star or a black hole, depending on the mass of the core.

What happens when the Helium in the core gets used up?

As the helium in the core is used up, the core will contract and heat up once again until it is hot enough to fuse heavier elements. This process will continue until the core is made up of iron, which cannot be fused further. At this point, the core will collapse under its own gravity, leading to a supernova explosion that expels the outer layers of the star into space, leaving behind either a neutron star or a black hole, depending on the mass of the core.

For high-mass stars, the process is similar, but the fusion reactions proceed more rapidly, leading to a shorter lifespan and a more violent supernova explosion. In both cases, the ultimate fate of the star depends on its mass and the resulting conditions in its core.

When the hydrogen in a star's core is used up, a series of events can occur depending on the mass of the star. For low to medium-mass stars, such as our Sun, the core will contract and heat up until it is hot enough to initiate the fusion of helium into carbon and oxygen. This process, known as the helium-burning phase, will cause the outer layers of the star to expand into a red giant.

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Related Questions

Please answer this question I’ll give brainliest if it’s correct.

Q6. Saul (30 kg) is tobogganing down a hill on a toboggan that has a mass of 10 kg. Part way down, when he is still 20 m (measured vertically) above the bottom of the hill, he passes his sister Nadia at a velocity of 15 m/s. [9 marks]

(a) What is the total energy of Saul and the toboggan?

(b) What is the height of the hill?

(c) What will Saul’s velocity at the bottom of the hill be?

(d) What will Saul’s height be when he is moving at a velocity of 9.5 m/s?

Answers

Saul (30 kg) is tobogganing down a hill:

(a) total energy 14130 J.

(b) height of the hill 11.5 m.

(c) velocity 15.0 m/s.

(d) height 4.6 m.

How to calculate height and velocity?

(a) The total energy of Saul and the toboggan is equal to the sum of their kinetic and potential energies. At the point when Saul passes his sister, he has kinetic energy due to his motion and potential energy due to his height above the bottom of the hill. The toboggan also has kinetic energy due to its motion. The total energy is given by:

Total energy = kinetic energy + potential energy

= (1/2)(mS + mT)v² + (mS + mT)gh

where mS = 30 kg is Saul's mass,

mT = 10 kg is the toboggan's mass,

v = 15 m/s is Saul's velocity,

g = 9.8 m/s² is the acceleration due to gravity, and

h = 20 m is the height above the bottom of the hill.

Plugging in the values:

Total energy = (1/2)(30 kg + 10 kg)(15 m/s)² + (30 kg + 10 kg)(9.8 m/s²)(20 m)

= 8250 J + 5880 J

= 14130 J

Therefore, the total energy of Saul and the toboggan is 14130 J.

(b) The height of the hill can be found by equating the initial potential energy at the top of the hill to the total energy at the point when Saul passes his sister. That is:

(mS + mT)gh = (1/2)(mS + mT)v² + (mS + mT)gh

Simplifying and solving for h:

h = (1/2)v²/g

= (1/2)(15 m/s)²/9.8 m/s²

= 11.5 m

Therefore, the height of the hill is 11.5 m.

(c) To find Saul's velocity at the bottom of the hill, use conservation of energy again. At the bottom of the hill, all the potential energy has been converted to kinetic energy. That is:

(mS + mT)gh = (1/2)(mS + mT)v²

Simplifying and solving for v:

v = √[2gh]

= √[2(9.8 m/s²)(11.5 m)]

= 15.0 m/s (to two significant figures)

Therefore, Saul's velocity at the bottom of the hill is 15.0 m/s.

(d) To find Saul's height when he is moving at a velocity of 9.5 m/s, we can use conservation of energy again. This time, equate the kinetic energy at this point to the initial potential energy. That is:

(1/2)(mS + mT)v² = (mS + mT)gh

Simplifying and solving for h:

h = v²/2g

= (9.5 m/s)²/(2)(9.8 m/s²)

= 4.6 m

Therefore, Saul's height when he is moving at a velocity of 9.5 m/s is 4.6 m.

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An overtone is louder than the fundamental tone.
O True
False

Answers

The given statement, "An overtone is louder than the fundamental tone" is False.

An overtone is a higher-frequency vibration that occurs simultaneously with the fundamental frequency of a sound wave. These vibrations are multiples of the fundamental frequency and contribute to the overall timbre or tone quality of a sound.

In some cases, overtones can be louder than the fundamental tone, depending on the specific harmonic series present in the sound wave. This phenomenon is known as overtone prominence, and it can be heard in certain musical instruments like bells or cymbals, where the higher harmonics of the sound are emphasized.

However, in most cases, the fundamental tone is perceived as the strongest and most dominant sound in a given sound wave.

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A displacement vector is 23 km in length and directed 65° south of east. What are the components of this vector?

Eastward
Component Southward. Component
(a) 21 km 9.7 km
(b) 23 km 23 km
(c) 23 km 0 km
(d) 9.7 km 21 km​

Answers

If a displacement vector is 23 km in length and directed 65° south of east. the components of this vector is: (a) 21 km 9.7 km.

What is the components of this vector?

The displacement vector can be resolved into its eastward and southward components using trigonometry. Let's call the eastward component "x" and the southward component "y".

From the given information, we know that the displacement vector makes an angle of 65° south of east. This means that the angle between the vector and the eastward axis is 90° - 65° = 25°.

Using trigonometry, we can relate the length of the vector to its components:

cos 25° = x / 23

sin 25° = y / 23

Solving for x and y, we get:

x = 23 cos 25° ≈ 21 km

y = 23 sin 25° ≈ 9.7 km

Therefore, the answer is (a) 21 km eastward component and 9.7 km southward component.

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energy transferred from one thing to another when the gulf balls collide?

Answers

This is based on the principle of conservation of energy and momentum.

When the collision of golf balls takes place the energy gets transferred from one ball to the other.

The golf balls experience a force that causes them to change their form and also direction. During the collision, the mechanical energy is converted into heat, sound, and other forms of energy. The rest of energy is used to move in a new direction.

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**NEED ANSWER ASAP**
What are weird properties of quasars that made them difficult for astronomers to understand?

**FAKE ANSWERS WILL BE REPORTED**

Answers

Supermassive black holes that are devouring gas at the center of far-off galaxies are known as quasars.

Since, the quasars were initially identified by astronomers in 1963 as objects that resembled stars but gave off radio waves instead, the term quasar is an abbreviation for quasi-stellar radio source.

Quasars are so bright that they drown out the light from all other stars in the same galaxy. Quasars give off radio waves, X-rays, gamma-rays, ultraviolet rays, and visible light across the entire electromagnetic spectrum. Most of them are larger than our solar system.

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15 A car of mass 750 kg is accelerating up a slope of a certain angle to the horizontal where sin theta = 1/70 at 1.5 m/ s². Ignoring any road resistance, find the tractive force of the engine.​

Answers

The tractive force of the engine is 1020 N.

Mass of the car, m = 750 kg

Acceleration of the car, a = 1.5 m/s²

Traction, also known as tractive force, is the force used to produce or create motion by using dry friction between a body and an inclined surface.

Weight of the car, W = mg

W = 750 x 9.8

W = 7350

The coefficient of friction is the ratio or percentage of the opposing frictional force to the normal force pressing the two surfaces into contact and motion.

The tractive force of the engine,

F' = ma - mg sinθ = m(a - g sinθ)

F' = 750[1.5 - (9.8/70)]

F' = 750 x 1.36

F' = 1020 N

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A car is traveling at a speed of 30m/s and leaves the ramp at a 37 degree angle. What is the total hang time of the car?

Answers

Explanation:

INITIAL  Vertical velocity is given by

30 m/s * sin 37

 then gravity begins to slow it down

30 sin37  -  (9.81) t   = vertical velocity at  t

    when v = 0 , the car is at its apex and will fall back down in the same amount of time

  0 = 30 sin37 - 9.81 t      shows t = 1.84 seconds to peak

    then another 1.84 seconds to fall to the ground   total = 3.7 seconds

String 1 in the figure has linear density 2.60 g/m and string 2 has linear density 3.30 g/m. A student sends pulses in both directions by quickly pulling up on the knot, then releasing it. She wants both pulses to reach the ends of the strings simultaneously.

What should the string length L1 be?

What should the string length L2 be?

Answers

Answer:

Here is your answer please change up some words to remain plagraism free.

Explanation:

To determine the required lengths of strings 1 and 2 so that pulses sent in both directions reach the ends of the strings simultaneously, we need to apply the principle that the time it takes for a wave pulse to travel a distance on a string is equal to the distance divided by the wave speed.

The wave speed, in turn, is determined by the tension in the string and the linear density of the string according to the formula:

v = sqrt(T/μ),

where v is the wave speed, T is the tension, and μ is the linear density.

Let L1 be the length of string 1 and L2 be the length of string 2. Since the wave speed is the same for both strings, we can set up the following equations:

L1/v = L2/v

sqrt(T1/μ1)*L1 = sqrt(T2/μ2)*L2

where T1 and T2 are the tensions in strings 1 and 2, respectively.

We can solve for L1 and L2 by combining these two equations and solving for each variable. Substituting the given linear densities of strings 1 and 2, we get:

sqrt(T1/2.60)*L1 = sqrt(T2/3.30)*L2

Squaring both sides and simplifying, we get:

(T1/T2) = (3.30/2.60) * (L1/L2)^2

Substituting the condition that the pulses reach the ends of the strings simultaneously, we know that the total time for a pulse to travel down string 1 and back up to the knot is equal to the time for a pulse to travel down string 2 and back up to the knot. This condition implies that the total length of string 1 (2L1) is equal to the total length of string 2 (2L2):

2L1 = 2L2

Solving this equation for L2 and substituting it into the expression for T1/T2 derived above, we get:

T1/T2 = (3.30/2.60) * (L1/2L1)^2 = 1.25

Solving for L1, we obtain:

L1 = sqrt(T1/μ1) * (2L2/v) = sqrt((1.25)*(2.60/3.30)) * (2L2)

Simplifying this expression, we get:

L1 = (2/3) * sqrt(2.60/3.30) * L2

Therefore, the required length of string 1 is (2/3) * sqrt(2.60/3.30) times the length of string 2. We can substitute the given length of string 2, say L2 = 1 meter, into this expression to obtain the required length of string 1:

L1 = (2/3) * sqrt(2.60/3.30) * 1 meter ≈ 0.693 meter.

Therefore, the required length of string 1 is approximately 0.693 meter and the required length of string 2 is 1 meter.

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22. How do we correct the issue of flipped imagery caused by mirrors?

Answers

To correct the issue of flipped imagery caused by mirrors you can use a technique called "mirror flipping". Mirror flipping involves using a second mirror to reflect the reflected image from the first mirror which then flips it back to its original orientation

Alternatively you can use a prism to correct the orientation of the image. A prism is a transparent object that can bend light. By placing it infront of the mirror you can reflect the twice, effectively helping in the correction of the orientation of the image

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A ball of mass 2 kg is moving with a velocity of 12m/s collides with a stationary ball of mass 6 kg and comes to rest. Calculate velocity of ball of mass 6kg after collision.

Answers

Answer: 4 m/s

Explanation:

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a closed system remains constant before and after a collision.

Let's denote the initial velocity of the 2 kg ball as "v1i", the initial velocity of the 6 kg ball as "v2i", the final velocity of the 2 kg ball as "v1f", and the final velocity of the 6 kg ball as "v2f".

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:

m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f

where m1 is the mass of the 2 kg ball, m2 is the mass of the 6 kg ball, v1i and v2i are the initial velocities, and v1f and v2f are the final velocities.

Given:

m1 = 2 kg

m2 = 6 kg

v1i = 12 m/s (initial velocity of the 2 kg ball)

v2i = 0 m/s (initial velocity of the 6 kg ball, as it is stationary)

v1f = 0 m/s (final velocity of the 2 kg ball, as it comes to rest)

Plugging in the given values into the conservation of momentum equation:

2 * 12 + 6 * 0 = 2 * 0 + 6 * v2f

24 = 6 * v2f

Dividing both sides by 6:

v2f = 24 / 6 = 4 m/s

So, the velocity of the 6 kg ball after the collision is 4 m/s.

A car starting from rest has an acceleration 0.5m/s after 1 minute then what will be the final velocity of the car​

Answers

Answer:

The final velocity is 30 m/s

Explanation:

Use the formula:

[tex]V_{f} = V_{i} +a*t\\V_{f} = 0 + 0.5*60\\V_{f} = 30 m/s[/tex]

What happens as a protostar contracts?
A. More hydrogen is produced to become fuel for the star.
B. The temperature rises.
C. The hydrogen fuses into iron.
D. The temperature drops.

Answers

Answer:

As a protostar contract, the temperature rises. This increase in temperature leads to the initiation of nuclear fusion reactions, where hydrogen atoms fuse together to form helium, releasing energy in the process. This energy causes the protostar to heat up and begin to emit light, eventually becoming a stable star.

The answer is B. The temperature rises.

Penguins in Gold Harbour love to communicate with other members of their penguin family. Here in Sanford, the speed of sound in air is about 344.0 m/s. Calculate the speed of sound (in m/s) in Gold Harbour, on a day when the air temperature is -2.7 °C.

Round to the nearest hundredth.
Please show all work!!!!

Answers

The speed of sound in Gold Harbour, on a day when the air temperature is -2.7 °C, is 331.5 + 0.6 * (-2.7) = 328.65 m/s.

What is Gold Harbour?

Gold Harbour is a small settlement located on Antarctic's King George Island. It is home to a Chilean research base, which is operated by the Chilean Antarctic Institute. It also acts as a summer base for the Chilean Navy, and provides support for the scientific research conducted by other countries, including the United States, United Kingdom, and Russia. The area is known for its stunning natural beauty, with mountains, glaciers, and icebergs all in close proximity. It is also an important habitat for several species of wildlife, including penguins, seals, and sea birds.

The speed of sound in air is affected by temperature, and the formula for calculating the speed of sound in air is v = 331.5 + 0.6 * (air temperature in °C).
Therefore, the speed of sound in Gold Harbour, on a day when the air temperature is -2.7 °C, is 331.5 + 0.6 * (-2.7)
= 328.65 m/s.

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A ball (0.410 kg) is kicked at an angle of 44.0° above the horizontal axis (above the +x-axis).  The initial speed of the ball is 24.2 m/s.  Ignoring air resistance, determine the momentum of the ball just before it hits the ground.

Answers

Answer:

9.922

Explanation:

When ignoring the air resistance and the ball is kicked and fell into the same plane. the initial speed is the speed when it is stricken on the ground. the initial momentum magnitude equals the final momentum magnitude only the direction is changed.

∴ Momentum=  mass*velocity  

                      = o.41kg*24.2

                      = 9.922[tex]kgms^{-1[/tex]

What type of atomic radiation will most deeply penetrate matter?
Multiple Choice
Beta radiation
Gamma radiation
Alpha radiation

Answers

ANSWER IS gamma radiation

Gamma rays can be emitted from the nucleus of an atom during radioactive decay. They are able to travel tens of yards or more in air and can easily penetrate the human body. Shielding this very penetrating type of ionizing radiation requires thick, dense material such as several inches of lead or concrete.

Which of the following sets of two charges is experiencing the strongest
attraction?
Charges of +2 C and -2 C, separated by 1 m.
Charges of +1 C and -3 C, separated by 1 m.
Charges of +2 C and +2 C, separated by 1 m.
Charges of +1 C and +3 C, separated by 1 m.

Answers

The force of attraction between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

What is a square ?

A square is a two-dimensional geometric shape that has four sides of equal length and four right angles (90-degree angles) between them. The sides of a square are parallel to each other and perpendicular to its adjacent sides. All four corners of a square are also known as vertices, and the diagonals of a square bisect each other at right angles.

The area of a square is calculated by multiplying the length of one of its sides by itself. The perimeter of a square is calculated by adding up the lengths of all four sides. The properties of a square make it useful in various applications, such as in geometry, architecture, and construction.

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describe the energy changes in a mass spring system that is oscillating horizontally explain how this changes of the system is vibrating vertically​

Answers

Answer:

In a horizontal mass spring system, the energy changes between potential energy and kinetic energy. When the spring is at its equilibrium position, the mass has potential energy stored in the spring, and no kinetic energy. As the mass is displaced from its equilibrium position and begins to move, potential energy is converted into kinetic energy. The maximum kinetic energy occurs when the mass is at the maximum displacement.

As the mass moves back towards the equilibrium position, the kinetic energy is converted back into potential energy stored in the spring. The maximum potential energy is reached when the mass reaches the equilibrium position. The energy then changes back to kinetic energy as the mass moves past the equilibrium position again.

If the same mass spring system is vibrating vertically instead of horizontally, gravitational potential energy also comes into play. When the mass is at its highest point, it has maximum gravitational potential energy and minimum kinetic energy. As it falls towards the equilibrium position, the potential energy is converted into kinetic energy. At the equilibrium position, the kinetic energy is maximum and potential energy is minimum. As the mass moves back up towards the maximum point, the kinetic energy is converted back into gravitational potential energy. This process continues as the system oscillates vertically.

Example 9:
3.
Figure 5.24 shows a barrel of weight 1500 N
and radius 0.5 m that rests against a step of
height 0.2 m.
0.2 m
0.5 m
▲ Figure 5.24
(
What is the smallest horizontal force F th
the centre O needed to push the barrel over
the step

Answers

To drive the barrel over the step, the least horizontal force F necessary is 2366.16 N.

How to find horizontal force?

To push the barrel over the step, the minimum force required should overcome the force of gravity acting on the barrel and the force of static friction between the barrel and the surface.

The perpendicular component of the weight is given as N = mgcosθ, where m = mass of the barrel,

g = acceleration due to gravity, and

θ = angle between the weight and normal to the surface.

In this case, θ as the inverse tangent of the ratio of the height of step to the distance from the edge of the step to the center of the barrel:

θ = tan⁻¹(0.2/0.5) = 0.39 radians

Therefore, the normal force is N = (1500)(9.81)cos(0.39) = 1443.6 N.

The force of static friction can be found as f = μsN,

where μs = coefficient of static friction.

Assume the coefficient of static friction between the barrel and the surface is 0.6.

f = (0.6)(1443.6) = 866.16 N.

The minimum force required to push the barrel over the step should overcome both these forces. Then, the smallest horizontal force that can push the barrel over the step is:

F = force of gravity + force of static friction

F = 1500 + 866.16

F = 2366.16 N.

Therefore, the smallest horizontal force F required to push the barrel over the step is 2366.16 N.

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Iodine-131 has a half-life of 8 days. How many grams of a 256 g sample would remain at the end of 56 days?

Answers

Answer:

Explanation:

The decay of a radioactive substance is governed by the formula:N(t) = N₀ e^(-λt)where N₀ is the initial amount of the substance, N(t) is the amount remaining after time t, and λ is the decay constant.The half-life of Iodine-131 is 8 days, which means that after each 8-day period, the amount remaining will be reduced by half. We can use this fact to calculate the amount remaining after 56 days.First, we need to find the decay constant λ, which is related to the half-life by the formula:λ = ln(2) / t½where ln(2) is the natural logarithm of 2, and t½ is the half-life.Substituting the values we have:λ = ln(2) / 8 days ≈ 0.08664 day^(-1)Next, we can use the formula for N(t) to calculate the amount remaining after 56 days:N(56) = N₀ e^(-λt) = 256 g e^(-0.08664 day^(-1) × 56 days) ≈ 22.6 gTherefore, approximately 22.6 grams of the original 256 gram sample would remain after 56 days.

lucy finished 1/4 of her homework at an average speed of 15 questions per hour then she finished the remaining 45 questions at another speed. if the total time spent on the homework was 2.5 hours, what was the amount of time she spent on the remaining 45 questions

Answers

Therefore, Lucy spent 1.875 hours on the remaining 45 questions.

How is  the amount of time she spent on the remaining 45 questions?

Let's start by finding the total number of questions in Lucy's homework. If she finished 1/4 of her homework at an average speed of 15 questions per hour, then the total number of questions must be:

1/4 x Total number of questions = Number of questions finished at 15 questions per hour

1/4 x Total number of questions = 15 questions per hour

Solving for the total number of questions, we get:

Total number of questions = 60 questions

Now we know that Lucy finished 60 - 45 = 15 questions at an average speed of x questions per hour. Let's use the formula:

time = distance / speed

to find the amount of time she spent on the remaining 45 questions.

For the first part of the homework, Lucy spent:

time = distance / speed

time = 15 questions / hour

time = 1/4 x 2.5 hours

time = 0.625 hours

So, she spent 0.625 hours on the first 15 questions.

For the remaining 45 questions, we have:

time = distance / speed

time = 45 questions / x questions per hour

We know that the total time spent on the homework was 2.5 hours, so:

0.625 hours + 45 questions / x questions per hour = 2.5 hours

Solving for x, we get:

x = 18 questions per hour

Now we can use this speed to find the time spent on the remaining 45 questions:

time = distance / speed

time = 45 questions / 18 questions per hour

time = 2.5 hours - 0.625 hours

time = 1.875 hours

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14. Neglecting air resistance, what maximum height will be reached by a stone thrown straight up with an initial speed of 35 m/s?
(a) 98 m
(b) 18 m
(c) 160 m
(d) 63 m​

Answers

Answer:

D

Explanation:

The maximum height reached by a stone thrown straight up with an initial speed of 35 m/s can be found using the kinematic equation:

v^2f = v^2i - 2gh

where vf is the final velocity (0 m/s at the maximum height), vi is the initial velocity (35 m/s, the magnitude of the velocity with which the stone is thrown upwards), g is the acceleration due to gravity (-9.8 m/s^2), and h is the maximum height reached by the stone.

Rearranging the equation, we get:

h = (vi^2)/(2g)

Substituting the given values, we have:

h = (35 m/s)^2 / (2 * 9.8 m/s^2)

= 62.6 m

Therefore, the maximum height reached by the stone is approximately 63 m.

The answer is (d).

Suppose headphones were placed on a student, and a 500-Hz sound was fed to the left ear
slightly later than to the right ear. The student will discern the source of sound to be at about 45°
to the right of center. Why?
What is the approximate time delay in the response of the left ear?

Answers

Answer:

0.05 ms

Explanation:

The speed of sound in air is about 343 m/s. If the sound waves are coming from a source that is 1 meter away, then it will take about 3.43 microseconds for the sound waves to reach the right ear and about 3.53 microseconds for the sound waves to reach the left ear. This is a difference of about 0.097 microseconds, which is about 0.05 ms.

The human brain is very good at detecting small differences in time, and it uses this information to determine the direction of a sound source. When the sound waves reach the left ear slightly later than the right ear, the brain interprets this as a sound coming from the right side. The greater the difference in time, the further to the right the brain will perceive the sound to be.

In the case of a 500 Hz sound, the brain will perceive the sound to be coming from about 45° to the right of center if the sound waves reach the

Problem 2.3. (5 pts) A 0.500-kg cart connected to a light spring for which the force constant is 20.0 N/m oscillates on a frictionless, horizontal air track. (a) Calculate the maximum speed of the cart if the amplitude of the motion is 3.00 cm. (b) What is the velocity of the cart when the position is 2.00 cm? (c) Compute the kinetic and potential energies of the system when the position of the cart is 2.00 cm​

Answers

The maximum speed of the cart if the amplitude of the motion is 3.00 cm is  0.036 m/s

The velocity of the cart when the position is 2.00 cm is 0.1414 m/s.

The kinetic and potential energies of the system when the position of the cart is 2.00 cm​ is 5×10⁻³ J and 4×10⁻³ J resp.

a) To find maximum speed potential energy of the spring gets converted into kinetic energy of the cart in the oscillator motion, Hence,

1/2mv² =1/2kA²

1/2×0.5×v² = 1/2 ×20× 0.03²

v² = 40×9×10⁻⁴

v = 0.036 m/s

b) For the velocity at a given point is calculated by the formula,

v = ±ω√(A² - x²)

v = ±ω√(0.03² - 0.02²)

v = ±ω × 0.0223

v = ±√k/m × 0.0223

v = ±√20/0.5× 0.0223

v = 0.1414 m/s

c)

kinetic energy of the system,

K = 1/2 mv²

K = 1/2 ×0.5×0.1414²

K = 5×10⁻³ J

Potential energy of the system

P = 1/2 kx²

P = 1/2 × 20× 0.02²

P = 4×10⁻³ J

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What month is the speed of Earth the fastest?

January
March
June
Decemember

Answers

Answer:

July 29, Earth completed a full spin in about 1.59 milliseconds shorter than its standard timeframe ( 23 hours and 56 minutes).

Which one of the following is the longest length?
(a) 100 m
(b) 104 µm
(c) 107 nm
(d) 102 mm​

Answers

Okay, let's convert all the lengths to the same unit to compare:

(a) 100 m = 100 meters

(b) 104 μm = 104 micrometers = 104 × 10^-6 meters = 0.000104 meters

(c) 107 nm = 107 nanometers = 107 × 10^-9 meters = 0.000000000997 meters

(d) 102 mm = 102 millimeters = 102 × 10^-3 meters = 0.0102 meters

The longest length is:

(a) 100 m = 100 meters

The answer is option (a).

Answer: 100 m

Explanation:

1 μm = [tex]10^{-6}[/tex] m = 0,000001 m

1 nm = [tex]10^{-9}[/tex] m = 0,000000001 m

1 mm = [tex]10^{-3}[/tex] m = 0,001 m

∴ 100 m es la mayor longitud

OK, once again we have a pendulum, this time of length 1.06 m, which you release from rest at an angle of 41.2 degrees to the vertical. What will be the speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical?

Answers

The speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical is 3.02 m/s.

A pendulum is a weight suspended from a fixed point that swings back and forth due to the force of gravity.

Based on the given information, we have a pendulum of length 1.06 m and it is released from rest at an angle of 41.2 degrees to the vertical. We need to find the speed of the pendulum at the instant it reaches an angle of 20.6 degrees above the vertical.

To solve this problem, we can use the conservation of mechanical energy. At the highest point of the pendulum's swing, all of its energy is in the form of potential energy, and at the lowest point of its swing, all of its energy is in the form of kinetic energy. Therefore, we can write:

PE_max = KE_min

where PE_max is the potential energy at the maximum height and KE_min is the kinetic energy at the lowest point.

The potential energy of a pendulum is given by:

PE = mgh

where m is the mass of the pendulum, g is the acceleration due to gravity, and h is the height above some reference point.

The kinetic energy of a pendulum is given by:

KE = (1/2)mv^2

where v is the speed of the pendulum.

First, we need to find the vertical height difference between the pendulum's highest and lowest points. To do this, we can use trigonometry:

h = L(1 - cosθ)

where L is the length of the pendulum and θ is the initial angle to the vertical. Substituting the given values, we get:

h = 1.06(1 - cos(41.2°)) = 0.654 m

Next, we can use the conservation of mechanical energy to find the speed of the pendulum at the lowest point of its swing. At this point, all of the potential energy has been converted into kinetic energy, so we can write:

PE_max = KE_min

mgh = (1/2)mv^2

Canceling out the mass, we get:

gh = (1/2)v^2

Solving for v, we get:

v = sqrt(2gh)

where g is the acceleration due to gravity. Substituting the given values, we get:

v = sqrt(2(9.81 m/s^2)(0.654 m)) = 3.78 m/s

Finally, we need to find the speed of the pendulum when it reaches an angle of 20.6 degrees above the vertical. At this point, the pendulum has a potential energy of:

PE = mgh' = mgh cos(20.6°)

where h' is the height of the pendulum at this point. To find h', we can use trigonometry:

h' = L(1 - cosθ')

where θ' is the angle above the vertical. Substituting the given values, we get:

h' = 1.06(1 - cos(20.6°)) = 0.242 m

Substituting the values for h' and solving for the kinetic energy, we get:

KE = PE_max - mgh' = mgh - mgh'

Substituting the known values, we get:

KE = (1 kg)(9.81 m/s^2)(0.654 m) - (1 kg)(9.81 m/s^2)(0.242 m) = 5.11 J

Now, we can solve for the speed at this point:

KE = (1/2)mv^2

5.11 J = (1/2)(1 kg)v^2

v = sqrt((2)(5.11 J)/(1 kg)) = 3.02 m/s

Therefore, The pendulum is moving at a speed of 3.02 m/s when it reaches an angle of 20.6 degrees above vertical.

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which system best illustrates attractive forces

Answers

Answer:

A

Opposite poles will attract

A the opposite are eventually going to attract to each other

Sanjay and Ting, each with a mass of 25 kg, are riding opposite each other on the edge of a 150 kg, 3.0-m-diameter playground merry-go-round that's rotating at 15 rpm. Each walks straight inward and stops 35 cm from the center.

What is the new angular velocity, in rpm?
Express your answer in revolutions per minute.

Answers

The merry-go-round's new angular velocity is 0.321 rpm.

Calculation-

the system's overall angular momentum is:

L = Iω

The moment of inertia of a solid disk rotating about its centre is given by:

[tex]I = (1/2)mr^2I = (1/2)(150 kg)(1.5 m)^2 = 168.75 kg·m^2[/tex]

initial angular momentum of the system

L = Iω = [tex](168.75 kg·m^2)(15 rpm)(2π/60 s) = 52.36 kg·m^2/s[/tex]

The new angular velocity is determined by:

L = I'ω'

where L represents the system's initial angular momentum.

The system's new moment of inertia is:

[tex]I' = I - 2mr^2[/tex]

we get:

[tex]I' = 25 kg - 2(168.75 kg/m2)(0.35 m)^2 = 162.88 kg·m^2[/tex]

We get the following by substituting into the conservation of angular momentum equation:

L = I'ω'

[tex](162.88 kg/m2) / 52.36 kg/m2[/tex]

ω' = 0.321 rpm

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When a massive star, much bigger than our sun, reaches the end of its life cycle, it will expand into a red supergiant and then:
A. lose its corona and enter a second stable phase as a star again.
B. explode into a supernova.
C. collapse into a protostar.
D. collapse into a white dwarf.

Answers

Answer:

When a massive star, much larger than our sun, reaches the end of its life cycle, it will undergo a series of fusion reactions in its core until it forms iron, which cannot undergo further fusion. Without fusion to counteract the force of gravity, the core collapses in on itself, causing the outer layers of the star to rapidly expand and creating a red supergiant. Eventually, the outer layers of the star will be expelled in a supernova explosion, leaving behind either a neutron star or a black hole, depending on the mass of the original star. So, the correct answer is B.

ray of light exits from a metal with a refractive index of 1.75 travelling to the air the angle of refraction is 25°. what is the angle of deviation

Answers

Answer:

Explanation:

To find the angle of deviation, we can use the formula:

angle of deviation = (refractive index of metal - refractive index of air) x angle of incidence

The angle of incidence can be calculated using Snell's law:

sin(angle of incidence) / sin(angle of refraction) = refractive index of air / refractive index of metal

sin(angle of incidence) / sin(25°) = 1 / 1.75

sin(angle of incidence) = 0.5714

angle of incidence = 34°

Now we can substitute this value into the formula for angle of deviation:

angle of deviation = (1.75 - 1) x 34°

angle of deviation = 21°

Therefore, the angle of deviation is 21°.

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