which of the following are likely to play a role in determining whether a galaxy is spiral or elliptical? select all that apply. select all that apply. the density of the protogalactic cloud from which the galaxy was born the rotation rate of the protogalactic cloud from which the galaxy was born the age of the universe at the time the galaxy first formed collisions or other interactions that the galaxy has had with other galaxies in the past

a merry-go-round rotates from rest with an angular acceleration of 1.50 rad/s2. 8.67 seconds & 20.4 seconds it take to rotate through (a) the first 4.19 rev and (b) the next 4.19 rev.

To solve this problem, we need to use the equations of rotational motion. The equation we need to use is:
θ = ωi*t + 1/2*α*t^2
where θ is the angle rotated (in radians), ωi is the initial angular velocity (in radians per second), α is the angular acceleration (in radians per second squared), and t is the time (in seconds).
For part (a), we want to find the time it takes to rotate through the first 4.19 rev, which is equivalent to 4.19*2π radians. We know that the merry-go-round starts from rest (ωi = 0) and has an angular acceleration of 1.50 rad/s^2. Substituting these values into the equation above, we get:
4.19*2π = 0*t + 1/2*1.50*t^2
Simplifying, we get:
t = √(4.19*2π / 0.75) = 8.67 seconds
Therefore, it takes 8.67 seconds to rotate through the first 4.19 rev.
For part (b), we want to find the time it takes to rotate through the next 4.19 rev. At this point, the merry-go-round is already rotating with some angular velocity, which we need to find first. Using the equation:
ωf = ωi + α*t
where ωf is the final angular velocity, we get:
ωf = 0 + 1.50*8.67 = 13.00 rad/s
Now we can use the same equation as before to find the time it takes to rotate through the next 4.19 rev, but with ωi = 13.00 rad/s:
4.19*2π = 13.00*t + 1/2*1.50*t^2
Simplifying, we get a quadratic equation:
0.75t^2 + 13.00t - 26.17π = 0
Using the quadratic formula, we get:
t = (-13.00 ± √(13.00^2 + 4*0.75*26.17π)) / 1.50
t ≈ 20.4 seconds or t ≈ -34.4 seconds
We can discard the negative solution since time cannot be negative. Therefore, it takes approximately 20.4 seconds to rotate through the next 4.19 rev.
So, the answers are:
(a) 8.67 seconds
(b) 20.4 seconds

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The reasons that a promontory will be more vulnerable to wave erosion than a bay;- Waves bend around a promontory and strike it from both sides,- Powerful waves focus most of their energy at a promontory and - A promontory will receive more wave action than a bay.

A promontory is more vulnerable to wave erosion than a bay due to the following reasons:

1. Waves bend around a promontory and strike it from both sides: This phenomenon, called wave refraction, concentrates the wave energy on the promontory, making it more prone to erosion.

2. A promontory will receive more wave action than a bay: Bays are generally more sheltered and have a lower exposure to waves, whereas promontories are exposed to the full force of waves, leading to more erosion.

3. Powerful waves focus most of their energy at a promontory: Due to the shape of the coastline, waves tend to focus their energy on the headlands, like promontories, which makes them more vulnerable to erosion compared to bays.

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As light enters the diamond, it is refracted at an angle of 12.3 degrees.

According to the aforementioned assertion, the speed of light in a diamond is 1.42 times that of light in a vacuum. The speed of light in a diamond will be 2.42 times slower than it is in air due to the high refractive index of diamonds.

[tex]heta2 = 12.3 degrees[/tex]

When we simplify the equation, we obtain:

[tex](1/2.42) x Sin(30) = Sin(theta2)[/tex]

[tex]Theta2 sin(2) = 0.208[/tex]

When we take the inverse sine of both sides, we obtain:

[tex]12.3 degree theta2[/tex]

Using both sides' inverse sine,

[tex]theta2 = 12.3 degrees[/tex]

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The stability of the nucleus is maintained through the combined effects of the strong force and neutrons.

Although protons repel each other due to their positive charge, they are stable in a nucleus because of the strong force, which is a fundamental force that binds the particles together.

The strong force is the strongest force in nature and overcomes the electromagnetic force that causes the protons to repel each other. Neutrons, which have no charge, also play a significant role in stabilizing the nucleus.

The neutrons act as a buffer between the positively charged protons, separating them from each other and reducing the electrostatic repulsion. Electrons, which have a negative charge, are not involved in stabilizing the nucleus as they are located outside the nucleus in orbitals around the nucleus.

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Answer: hope it helps

Explanation:

A protostar becomes a main sequence star when its core temperature exceeds 10 million K. This is the temperature needed for hydrogen fusion to operate efficiently.

To find the distance at which the centripetal acceleration of an orbiting space station around a planet is equal to Earth's gravitational acceleration, we need to set up an equation involving the planet's mass (M), gravitational constant (G), and Earth's gravitational acceleration (g).

Given:
M = 8.00 × 10²³ kg
G = 6.67 × 10^(-11) m³·kg^(-1)·s^(-1)
g = 9.81 m/s² (Earth's gravitational acceleration)

Centripetal acceleration (a_c) is given by the formula:

a_c = (G * M) / r²

where r is the distance from the planet's center.

We want the centripetal acceleration to be equal to Earth's gravitational acceleration, so we can set them equal:

g = (G * M) / r²

Now, we need to solve for r:

r² = (G * M) / g

r² = (6.67 × 10^(-11) m³·kg^(-1)·s^(-1) * 8.00 × 10²³ kg) / 9.81 m/s²

r² ≈ 5.42 × 10¹² m²

Now, take the square root of both sides to find r:

r ≈ 2.33 × 10^6 m

So, at a distance of 2.33 x 10^6 meters from the planet's center, the centripetal acceleration of an orbiting space station will be equal to the gravitational acceleration on Earth's surface.

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The following formula can be used to determine the distance from the planet's centre at which the centripetal acceleration of an orbiting space station equals the gravitational acceleration on Earth's surface:

[tex]r = (GM/g)^(1/3)[/tex]

where the gravitational constant, G, equals 6.67 1011 m3 kg-1 s-1.

M is equal to 8.00 1023 kg (the planet's mass).

Gravitational acceleration on Earth's surface is equal to 9.81 m/s2.

When we change the values, we obtain:

[tex]r = [(6.67 × 10^-11) × (8.00 × 10^23) / 9.81]^(1/3)[/tex]

[tex]r = 2.33 × 10^6 m[/tex]

Therefore, 2.33 x 106 m is the necessary distance.

F = G (m1m2 / r2), where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them, can be used to express the gravitational force between two objects. When a planet and a satellite are involved, the centripetal force that holds the satellite in orbit around the planet is produced by the gravitational force. As a result, we may compare the centripetal force to gravity and find r. This results in the formula above, which we can use to calculate the necessary distance.

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The weight of the object out of water is 800 kg.

To solve this problem, we need to use the principle of buoyancy. When an object is placed in water, it experiences an upward force called buoyant force, which is equal to the weight of the water displaced by the object.

In this case, the object has a volume of 1.00 m³, and 20.0% of its volume is above the waterline. Therefore, the volume of the object submerged in water is:

Vsubmerged = 1.00 m3 - 0.20 x 1.00 m³ = 0.80 m³

We also know the density of water is 1000 kg/m³. Therefore, the weight of the water displaced by the object is:

Wwater = density of water x volume of water displaced
Wwater = 1000 kg/m³ x 0.80 m³
Wwater = 800 kg

This means the buoyant force acting on the object is 800 kg. In order for the object to float, the buoyant force must be equal to the weight of the object. Therefore, we can find the weight of the object as:

Weight of object = Buoyant force = 800 kg

So the object weighs 800 kg out of the water.

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The wave speed in the first question is 15 m/s, the wavelength in the second question is 2.25 meters, and the frequency 2.3 Hz.

For the first question, we can use the formula v = λf, where v is the wave speed, λ is the wavelength, and f is the frequency. Substituting the given values[tex]v = 5.0 m * 3.0 Hz = 15 m/s[/tex].

For the second question, we can rearrange formula to solve for wavelength: λ = v/f. Substituting the given values λ = 4.5 m/s ÷ 2.0 Hz = 2.25 meters. For the third question, we can again rearrange the formula to solve for frequency: f = v/λ. Substituting the given values, f = 6.9 m/s ÷ 3.0 meters = 2.3 Hz.

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The right response is resting tremor (option b). A patient may have spasticity, autonomic dysreflexia, and orthostatic hypotension following a spinal cord injury (SCI). SCI is not often linked to resting tremor.

SCI can interfere with the body's ability to communicate with the brain, leading to a variety of physical symptoms. Spasticity, which manifests as stiffness, muscle spasms, and increased muscle tone, is a frequent consequence. Patients with SCI at or above the T6 level may develop autonomic dysreflexia, a potentially fatal illness that is characterised by an abrupt rise in blood pressure. When someone stands up, their blood pressure drops, causing lightheadedness and dizziness. This condition is known as orthostatic hypotension.

While essential tremor, Parkinson's disease, and other neurological illnesses are frequently linked to resting tremor, SCI is not typically one of them.

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The right response is resting tremor (option b). A patient may have spasticity, autonomic dysreflexia, and orthostatic hypotension following a spinal cord injury (SCI). SCI is not often linked to resting tremor.

SCI can interfere with the body's ability to communicate with the brain, leading to a variety of physical symptoms. Spasticity, which manifests as stiffness, muscle spasms, and increased muscle tone, is a frequent consequence. Patients with SCI at or above the T6 level may develop autonomic dysreflexia, a potentially fatal illness that is characterised by an abrupt rise in blood pressure. When someone stands up, their blood pressure drops, causing lightheadedness and dizziness. This condition is known as orthostatic hypotension.

While essential tremor, Parkinson's disease, and other neurological illnesses are frequently linked to resting tremor, SCI is not typically one of them.

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The law of conservation of energy states that energy is neither created nor destroyed but is transformed from one form to another.

The law of conservation of energy is the law that states that energy is neither created nor destroyed but is transformed from one form to another.

Examples of activities of everyday life that shows the conservation of energy include the following:

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An example of the law of conservation of energy is a roller coaster.

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.

A roller coaster car gains kinetic energy as it moves down the track, but it also loses potential energy. At the bottom of the track, the car has the most kinetic energy and the least potential energy, while at the top of the track, it has the most potential energy and the least kinetic energy. However, the total amount of energy in the system remains constant.

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In order for a vehicle travelling at 18.0 m/s to negotiate  highway bend without sliding, curve must be banked (inclined). The radius of curve approximately 33.1 metres.

The coefficient of side friction is found to be 0.10, and the superelevation at one horizontal curve has been set at 6.0%.the formula for calculating a road curve's radiusFind the shortest curve radius necessary to ensure safe vehicle operation.

speed of the car v = 18.0 m/s

angle of banking of the curve θ = 37°

acceleration due to gravityg = 9.81 m/s²

radius of the curve = r

N = mg * cos(θ).........1

also

N = mv² / r...........2

from equation 1 and 2 we get

mg * cos(θ) = mv² / r

r = v² / (g * cos(θ))

r = (18.0 m/s)² / (9.81 m/s² * cos(37°)) ≈ 33.1 m

Therefore, radius of the curve is approximately 33.1 meters.

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Any electromagnetic radiation with a frequency lower than 9.84 x 10¹⁴ Hz (infrared, microwave, radio waves) will not cause the photoelectric effect in aluminum.

If infrared light shines on the aluminum surface, no electrons will be emitted via the photoelectric effect because the frequency of infrared light is lower than the threshold frequency of aluminum. The photoelectric effect occurs when a photon with enough energy (frequency) is absorbed by an electron in a metal, causing the electron to be emitted from the metal.

The minimum frequency or threshold frequency ([tex]f_{t}[/tex]) of a metal can be calculated using the equation:

[tex]f_{t}[/tex] = Φ ÷ h

where Φ is the work function of the metal (the minimum energy required to remove an electron from the metal) and h is Planck's constant. For aluminum, Φ = 4.08 eV.

Converting Φ to joules and using h = 6.626 x 10⁻³⁴ J s, we get:

Φ = 4.08 eV x 1.6 x 10⁻¹⁹ J/eV

Φ = 6.528 x 10⁻¹⁹ J

[tex]f_{t}[/tex] = 6.528 x 10⁻¹⁹ J ÷ 6.626 x 10⁻³⁴ J s

≈ 9.84 x 10¹⁴ Hz

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The negative sign indicates that the force is exerted in the opposite direction to the motion of the car. This force is applied over a short time interval and is relatively large, causing the car to experience a significant deceleration during the collision.

During the collision, the toy car experiences a change in momentum. Since momentum is conserved in the absence of external forces, the momentum of the car before the collision must be equal in magnitude and opposite in direction to the momentum after the collision.

The initial momentum of the car is given by:

p = mv = 0.3 kg * 0.82 m/s = 0.246 kgm/s

After the collision, the car comes to a stop, so its final momentum is zero. Therefore, the change in momentum is:

Δp = p_final - p_initial = -0.246 kg*m/s

The force exerted by the wall on the car during the collision can be calculated using the impulse-momentum theorem

J = Δp = FΔt

where J is the impulse, Δt is the time interval over which the force is applied, and F is the force

From the figure, we can see that the time interval for the collision is approximately 0.020 s. Therefore, the force exerted by the wall on the car is: F = Δp / Δt = -0.246 kg*m/s / 0.020 s = -12.3 N

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The block moves outward along the slot with a speed of [tex]$v = \sqrt{100t^2 + 25r^2}$[/tex] m/s at an angle of approximately 78.69° with the positive x-axis. The platform rotates at a constant rate of θ˙ = 5 rad/s.

The motion of the block can be analyzed by considering both its radial and tangential components of velocity.

The radial component of velocity is given by r˙, which is 10t m/s. This means that the distance between the center of rotation and the block increases with time.

The tangential component of velocity is given by θ˙r, where r is the distance between the center of rotation and the block. This means that the block moves around the center of rotation at a constant angular speed of 5 rad/s.

To find the total velocity of the block, we can use the Pythagorean theorem:

[tex]$v = \sqrt{\dot{r}^2 + (\dot{\theta}r)^2}$[/tex]

Substituting the given values, we get:

[tex]$v = \sqrt{(10t)^2 + (5r)^2} = \sqrt{100t^2 + 25r^2}$[/tex]

To find the direction of the velocity, we can use the tangent of the angle between the velocity vector and the positive x-axis:

tan(θ) = (θ˙r)/r˙ = 5

This means that the angle between the velocity vector and the positive x-axis is constant at arctan(5) ≈ 78.69°.

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The angular velocity of the coil is approximately 7.27 rad/s.

The formula for the maximum emf induced in a rotating coil is given by: emf = NABw, where N is the number of turns in the coil, A is the area of the coil, B is the strength of the magnetic field, and w is the angular velocity of the coil.

Solving for w, we get: w = emf/(NAB)

Substituting the given values, we get: w = (28.0 x 10^-3)/(16 x 16 x 16 x 0.060 x 2π) ≈ 7.27 rad/s.

Therefore, the angular velocity of the coil is approximately 7.27 rad/s.

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Answer:

The type of collision, whether elastic or inelastic, can be determined by observing the behavior of the colliding objects before and after the collision. Here are some key characteristics that can help identify whether a collision is elastic or inelastic:

Conservation of Kinetic Energy: In an elastic collision, kinetic energy is conserved, while in an inelastic collision, some of the kinetic energy may be converted into other forms of energy.

Objects' Motion After Collision: In an elastic collision, objects bounce off each other and move independently, while in an inelastic collision, objects may stick together, deform, or move as a single mass.

Restitution Coefficient: In an elastic collision, the restitution coefficient is close to 1, indicating high bounce-back, while in an inelastic collision, the restitution coefficient is less than 1, indicating less bounce-back.

Conservation of Momentum: In both elastic and inelastic collisions, momentum is conserved, but the change in velocity of the objects after the collision can indicate whether the collision is elastic or inelastic.

The nonconservative work done on the surfer is 2845.5 J.

We can use the work-energy theorem to solve this problem. The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. In this case, we can calculate the initial and final kinetic energies of the surfer and find the difference, which will give us the net work done.

The initial kinetic energy of the surfer is:

[tex]K_i = (1/2) * m * v_i^2[/tex]

[tex]K_i = (1/2) * 73.2 kg * (1.44 m/s)^2[/tex]

K_i = 75.7 J

The final kinetic energy of the surfer is:

[tex]K_f = (1/2) * m * v_f^2[/tex]

[tex]K_f = (1/2) * 73.2 kg * (8.89 m/s)^2[/tex]

K_f = 2921.2 J

The change in kinetic energy is:

ΔK = K_f - K_i

ΔK = 2921.2 J - 75.7 J

ΔK = 2845.5 J

According to the work-energy theorem, this change in kinetic energy must be equal to the net work done on the surfer. Therefore, the nonconservative work done on the surfer is:

W_nc = ΔK

W_nc = 2845.5 J

So, the nonconservative work done on the surfer is 2845.5 J.

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The net torque required to accelerate is 28496 Nm.

The net torque required to accelerate a uniform disk of radius 7.5 m and mass 29000 kg from rest to 0.63 rad/s in 27 s is needed.

The problem is asking for the net torque required to accelerate a merry-go-round from rest to a final angular velocity of 0.63 rad/s in 27 seconds. The merry-go-round is assumed to be a uniform disk, which means that its mass is evenly distributed across its entire radius. We are also given the radius of the merry-go-round (7.5 m) and its mass (29000 kg).

To solve the problem, we can use the formula:

[tex]τ = Iα[/tex]

where τ is the net torque applied to the merry-go-round, I is its moment of inertia, and α is its angular acceleration. Since the merry-go-round is initially at rest, its initial angular velocity is zero. Using the formula for angular acceleration, we can find that:

[tex]α = Δω/Δt = (0.63 rad/s - 0 rad/s) / 27 s = 0.0233 rad/s^2[/tex]

To find the moment of inertia of the merry-go-round, we can use the formula for the moment of inertia of a uniform disk:

[tex]I = (1/2)mr^2[/tex]

where m is the mass of the disk and r is its radius. Substituting the given values, we get:

[tex]I = (1/2)(29000 kg)(7.5 m)^2 = 1220625 kg m^2[/tex]

Finally, we can use the formula [tex]τ = Iα[/tex] to find the net torque required to accelerate the merry-go-round:

[tex]τ = (1220625 kg m^2)(0.0233 rad/s^2) = 28496 Nm[/tex]

Therefore, the net torque required to accelerate the merry-go-round is 28496 Nm.

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A proton, moving in a uniform magnetic field, moves in a circle perpendicular to the field lines and takes time t for each circle. if the proton's speed tripled, the time to go around each circle remains the same.

If the proton's speed is tripled, the new time period to go around each circle can be calculated using the relation between centripetal force, magnetic force, and velocity. The centripetal force is given by Fc = mv^2/r, and the magnetic force is given by Fm = qvB. Equating the two forces (Fc = Fm), we get mv^2/r = qvB, which simplifies to r = mv/qB.

When the speed is tripled (3v), the new radius (r') becomes r' = m(3v)/qB = 3r.

The circumference of the circle is proportional to the radius, so the new circumference (C') is 3 times the original circumference (C).

The time period (t') is given by t' = C'/3v, and since C' = 3C, we have t' = 3C/3v = C/v.

Comparing this to the original time period (t = C/v), we find that the new time period (t') is equal to the original time period (t). Therefore, when the proton's speed is tripled, the time to go around each circle remains the same.

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The Gibbs energy change (ΔG) is indeed a measure of the spontaneity of a process and the useful energy available from it.

Explanation:

1. Gibbs energy (G) is a thermodynamic potential that combines enthalpy (H) and entropy (S) to predict whether a process will be spontaneous or not at a constant temperature (T) and pressure (P).

2. The change in Gibbs energy (ΔG) is calculated using the formula: ΔG = ΔH - TΔS, where ΔH is the change in enthalpy and ΔS is the change in entropy.

3. If ΔG is negative, the process is spontaneous, meaning it will proceed on its own without the need for external energy input. A negative ΔG also indicates that the system releases useful energy.

4. If ΔG is positive, the process is non-spontaneous and will require external energy to proceed. The useful energy in this case is not available, as it must be supplied from an external source.

5. If ΔG is equal to zero, the process is at equilibrium, meaning the forward and reverse processes occur at the same rate, and there is no net change in the system.

In summary, the Gibbs energy change (ΔG) is an important parameter that helps determine the spontaneity of a process and the amount of useful energy that can be obtained from it.

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The standing wave's fundamental frequency is the frequency of the first harmonic, which has one node and two antinodes, whereas the number of nodes determines the standing wave's harmonic number.

A 4.35 metre long, 137 gramme wire is being pulled at 125 newtons of force. With seven nodes total, including the endpoints, a standing wave has developed.

A collection of dots and dashes can be used to represent the wave pattern. The relationship between wave speed and wavelength is used to compute the standing wave's frequency. The tension in the wire and its linear mass density are used to calculate the wave speed.

The standing wave's fundamental frequency is the frequency of the first harmonic, which has one node and two antinodes, whereas the number of nodes determines the standing wave's harmonic number.

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The intensity of a linearly polarized electromagnetic wave is directly proportional to the square of its electric field amplitude.

The electric field of a linearly polarized electromagnetic wave can be represented by a sine or cosine function, where the amplitude of the wave represents the maximum value of the electric field.

The intensity of the wave is proportional to the average power per unit area that is carried by the wave.

Mathematically, the intensity (I) of an electromagnetic wave is given by the formula:

I = (1/2)εcE0^2

where ε is the electric constant (approximately equal to 8.85 x 10^-12 F/m), c is the speed of light in a vacuum (approximately equal to 3.00 x 10^8 m/s), and E0 is the amplitude of the electric field.

From this formula, it is clear that the intensity of the wave is proportional to the square of the electric field amplitude.

Therefore, if the electric field amplitude of a linearly polarized electromagnetic wave is increased by a factor of 2, the intensity of the wave will increase by a factor of 4 (i.e., 2 squared).

Similarly, if the electric field amplitude is decreased by a factor of 2, the intensity of the wave will decrease by a factor of 4.

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Answer:

The answer is Continental Grip

If the period of the oscillator doubles, the wavelength of the waves on the string will also double and the wave speed does not change.

When an oscillator creates periodic waves on a stretched string and the period of the oscillator doubles, the following happens to the wavelength and wave speed:

1. Wavelength (λ): The relationship between the period (T), frequency (f), and wavelength (λ) is given by the equation:

T = 1/f

Since the period doubles (T becomes 2T), the frequency will halve (f becomes f/2) to maintain the relationship. The equation for the wave speed (v) is:

v = fλ

As the frequency is halved, to maintain the same wave speed, the wavelength must also double (λ becomes 2λ). So, the wavelength will increase.

2. Wave speed (v): In this situation, the wave speed remains constant. As mentioned above, when the frequency is halved, the wavelength doubles, which means the product of the frequency and wavelength (v = fλ) remains the same. Therefore, the wave speed does not change.

In summary, when the period of the oscillator doubles, the wavelength will double, and the wave speed will remain constant.

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The tension in the rope pulling the skier is 514 N.

The power produced by the motor to maintain a constant speed of 13.4 m/s is 35700 W. The power required to pull the skier at the same constant speed is 36800 W. The difference in power is due to the additional work required to overcome the frictional force between the skier and the water.

The force required to maintain a constant speed can be calculated using the formula F = P/V, where F is the force, P is the power, and V is the velocity. For the motor to maintain a constant speed of 13.4 m/s, the force required is 2672.69 N.

To find the tension in the rope pulling the skier, we need to subtract the force required to maintain the constant speed (2672.69 N) from the force required to pull the skier (which we do not know yet). The result is the tension in the rope pulling the skier.

To find the force required to pull the skier, we can use the same formula, F = P/V, with the power of 36800 W and the velocity of 13.4 m/s. This gives a force of 2746.27 N.

Subtracting the force required to maintain the constant speed (2672.69 N) from the force required to pull the skier (2746.27 N) gives a tension in the rope of 514 N.

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The period of the horizontal mass-on-a-spring system with a stiffness of 500 N/m and a mass of 25.7 kg is approximately 1.424 seconds.

We'll use the following terms in our calculation: stiffness of the spring (k), mass of the system (m), and period (T).

The formula to calculate the period of a mass-on-a-spring system is:

T = 2π √(m/k)

where:
T = period (in seconds)
m = mass of the system (25.7 kg)
k = stiffness of the spring (500 N/m)

Now, we'll plug in the values:

T = 2π √(25.7 kg / 500 N/m)

To calculate the square root:

T = 2π √(0.0514)

T = 2π × 0.2266

Finally, multiply by 2π:

T ≈ 1.424 seconds

So, the period of the horizontal mass-on-a-spring system is approximately 1.424 seconds.

More on period: https://brainly.com/question/15299167

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