a farsighted person has a near-point distance greater than the normal distance of 25 cm. neglecting the eye-lens distance, which approach will yield an appropriate corrective lens for this condition?

Answer: The book responds to applied or tensioned force. Your hand or the work you are doing against gravity or weight force provides this force.

Explanation:

The ionization energy of an atom is the amount of energy required to completely remove an electron from its ground state. In the case of a hydrogen atom in a state with quantum number n, the ionization energy is given by the following equation: Ionization energy = -13.6 eV / n^2

For a hydrogen atom in a state with quantum number n=3, the ionization energy can be calculated as follows:
Ionization energy = -13.6 eV / 3^2 = -13.6 eV / 9 = -1.51 eV
Note that the negative sign indicates that energy is required to remove the electron. Therefore, the largest possible ionization energy of the atom in this state is 1.51 eV.
Based on the given answer choices, the correct answer is (b) 1.51 eV.
In the quantum mechanical description, the ionization energy of a hydrogen atom is given by the formula:
Ionization Energy (IE) = -13.6 eV * (1/n²)

where n is the principal quantum number. In this case, n = 3.
IE = -13.6 eV * (1/3²) = -13.6 eV * (1/9) = -1.51 eV
Since the ionization energy is negative, the largest possible ionization energy is the least negative value. Therefore, the answer is (b) 1.51 eV.

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ans =6.081X10 14 Hz

No electrons injected = 0K.E

1eV=1.6×10^-19J

h=6.63 x 10^-34js

Wo=2.52 eV / 4.032x10 -19 J

fo = ?

Wo= h fo

(6.63 x 10 - 34) fo = 4.032 x 10-19

fo = 4.032 X 10-19/6.63x10-34

fo = 6.081X10 14 Hz

The length of a vibrating string or column of air, there are a few other ways to double the fundamental frequency of a sound wave. These include:

Sound waves are a type of mechanical wave that propagates through a medium, such as air or water. They are longitudinal waves, meaning that they oscillate in the same direction as the direction of energy transfer. Sound waves are produced by the vibrations of an object, such as a musical instrument or a speaker. These vibrations cause the molecules in the surrounding medium to oscillate, creating a wave that propagates through the medium.

The frequency of the wave determines the pitch of the sound, while the amplitude determines the loudness. Sound waves can be characterized by several properties, including wavelength, frequency, amplitude, and velocity. The speed of sound varies depending on the medium through which it travels, with faster speeds in denser materials like solids.

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Throwing your keys northward would give you momentum in the north direction. When you throw something in a certain direction, the opposite reaction occurs and you are propelled in the opposite direction. So if you throw something northward, you will be propelled southward. In this scenario, you want momentum in the north direction, so throwing your keys northward would be the best option.

Bending down and throwing your keys straight up would not give you any momentum in the north direction, as the keys would simply come back down and you would stay in the same place.

To gain momentum on roller skates in the north direction, you should choose option A: Throwing your keys southward. By doing so, you apply a force in the south direction, and according to Newton's Third Law of Motion, an equal and opposite force (northward) will be applied to you. This northward force will give you momentum in the north direction. Throwing your keys northward (option B) or straight up (option C) would not provide the necessary force in the opposite direction to propel you northward.

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To determine if an AC circuit is at resonance based on the graph of current and generator EMF as functions of time, you need to look for specific characteristics in the graph. Here's how you can analyze the graph to identify resonance:

1. Determine the frequency of the AC signal: Resonance occurs when the frequency of the AC signal matches the natural frequency of the circuit. If you know the frequency of the generator EMF, you can compare it to the frequency of the current in the circuit.

2. Look for a phase shift: At resonance, the current and generator EMF should be in phase, meaning they reach their peak values at the same time. On the graph, this is represented by the peaks of both the current and generator EMF occurring at the same points in time.

3. Check for maximum amplitude: At resonance, the current in the circuit will have the maximum amplitude. This means that the current waveform should have the highest peaks compared to other frequencies. On the graph, the peaks of the current waveform should be higher than at other frequencies.

4. Analyze the current response: At resonance, the current response in the circuit should be maximized. This means that the current should be sustained at a high level without significant decay or distortion. On the graph, the current waveform should show a sustained and stable amplitude without excessive damping or distortion.

By examining these characteristics in the graph of current and generator EMF, you can identify whether the circuit is at resonance. If all these conditions are met, the circuit is likely at resonance. However, if any of these conditions are not satisfied, the circuit is not at resonance.

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The speed of the projectile at t=2.0s is 14.4 m/s.

The projectile will be moving parallel to the y-axis when the y-component of its velocity is zero. Using the kinematic equation vf=vi+at, we can find the y-component of the velocity at any time t. Differentiating this with respect to time gives us the acceleration in the y-direction, which is simply the y-component of the force. Setting this to zero and solving for t, we get t=1.0 s. At t=1.0 s, the projectile is moving parallel to the y-axis.

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The wavelength of an AM station radio wave of frequency 600 kHz is 500 meters. The relationship between frequency and wavelength can be described by the equation: wavelength (in meters) = speed of light (in meters per second) / frequency (in Hertz).

The speed of light is a constant value of approximately 3 x 10^8 meters per second. To find the wavelength of a radio wave with a frequency of 600 kHz, we can substitute these values into the equation: wavelength = 3 x 10^8 / 600,000 = 500 meters. Therefore, the wavelength of an AM station radio wave of frequency 600 kHz is 500 meters.

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The speed of the sphere when it reaches the bottom of the inclined plane is (B) 4.4 m/s.

The speed of the sphere when it reaches the bottom of the inclined plane can be found using conservation of energy:

mgh = (1/2)mv² + (1/2)Iω²

where m is the mass of the sphere, g is the acceleration due to gravity, h is the height of the inclined plane, v is the speed of the sphere at the bottom of the inclined plane, I is the moment of inertia of the sphere, and ω is the angular velocity of the sphere.

Since the sphere is rolling without slipping, we can also relate v and ω using:

v = ωR

where R is the radius of the sphere.

The moment of inertia of a solid sphere is (2/5)mr², so we can substitute this into the first equation and solve for v:

mgh = (1/2)mv² + (1/2)(2/5)mr²(ω/R)²

Simplifying and substituting ω = v/R, we get:

v = sqrt(10gh/7)

Substituting in the given values, we get:

v = sqrt(10(9.81 m/s²)(1.0 m)/7) = 4.4 m/s

Therefore, the answer is (B) 4.4 m/s.

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A 9.0-volt battery is connected to a bulb with a resistance of 2.0 ohms. To find the number of electrons leaving the battery per minute, we can use Ohm's Law and the formula for electric current. Ohm's Law states that voltage (V) equals current (I) times resistance (R). Therefore, I = V/R.

In this case, I = 9.0V / 2.0Ω = 4.5A (amperes). The electric current tells us the rate of flow of electric charge, which is carried by electrons. One electron has a charge of approximately 1.6 x 10^-19 coulombs.

Now, we can calculate the number of electrons per second by dividing the current by the charge of one electron: 4.5A / (1.6 x 10^-19 C) ≈ 2.81 x 10^19 electrons/second.

To find the number of electrons per minute, we multiply by 60 seconds: (2.81 x 10^19 electrons/s) x 60s = 1.69 x 10^21 electrons/minute. So, approximately 1.69 x 10^21 electrons leave the battery per minute when connected to the 2.0 ohm bulb.

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The ground state energy of an alpha particle confined to a one-dimensional box the size of the uranium nucleus is approximately 10.1 MeV.

[tex]E_n[/tex] = (n² * h²) / (8 * m * L²)

[tex]E_1[/tex]= (1² * h²) / (8 * m * L²)

[tex]E_1[/tex]= (1² * (6.626 x [tex]10^{-34[/tex] J s)²) / (8 * (6.64 x [tex]10^{-27[/tex] kg) * (1.5 x [tex]10^{-14[/tex] m)²)

[tex]E_1[/tex] = 1.62 x [tex]10^{-12[/tex] J

To convert this energy to electron volts (eV), we can use the conversion factor:

1 eV = 1.602 x [tex]10^{-19[/tex] J

[tex]E_1[/tex]= (1.62 x [tex]10^{-12[/tex] J) / (1.602 x [tex]10^{-19[/tex] J/eV)

[tex]E_1[/tex] = 1.01 x [tex]10^7[/tex]eV or approximately 10.1 MeV

The ground state is the lowest possible energy state that an atom or molecule can have. This state is the state of the system when all of its electrons are in their lowest possible energy level or orbit. The ground state is important because it determines the chemical and physical properties of the atom or molecule.

When an electron is excited, it absorbs energy and moves to a higher energy level or orbit. The electron then releases energy and returns to its ground state. The energy released can be in the form of light or heat. At room temperature, most atoms and molecules are in their ground state, but they can be excited by adding energy to the system.

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The tension in the string is greatest at the bottom most point of the vertical circle.

When a ball is whirled in a vertical circle with constant speed, the tension in the string varies throughout the motion. At the topmost point of the circle, the tension is less than the weight of the ball, as it provides the necessary centripetal force.

As the ball moves downward, the tension gradually increases, reaching its maximum at the bottommost point of the circle. Here, the tension in the string is the sum of the weight of the ball and the additional centripetal force required to maintain the circular motion.

Beyond the bottommost point, as the ball moves upward, the tension decreases again until it reaches the topmost point.

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The entropy of a Carnot engine during the adiabatic expansion stage does not change.

In a Carnot engine, the adiabatic expansion stage is when the working substance expands without any heat being added or removed. During this stage, the internal energy of the substance decreases, but since no heat is added or removed, the entropy remains constant.

This is because entropy is a measure of the amount of thermal energy that is unavailable to do work, and in an adiabatic process, no thermal energy is added or removed.
In a Carnot engine, the efficiency is determined by the temperature difference between the hot and cold reservoirs. The Carnot cycle consists of four stages: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.

During the isothermal expansion stage, the working substance absorbs heat from the hot reservoir and expands while doing work. During the adiabatic expansion stage, the working substance expands without any heat being added or removed. During the isothermal compression stage, the working substance releases heat to the cold reservoir and compresses while doing work.

Finally, during the adiabatic compression stage, the working substance is compressed without any heat being added or removed.
During the adiabatic expansion stage, the internal energy of the working substance decreases due to the work being done on it. However, since no heat is added or removed, the entropy remains constant. This is because entropy is a measure of the amount of thermal energy that is unavailable to do work, and in an adiabatic process, no thermal energy is added or removed. Therefore, the entropy of a Carnot engine during the adiabatic expansion stage does not change.
The entropy of a Carnot engine during the adiabatic expansion stage does not change. This is because the adiabatic expansion stage is an adiabatic process, which means that no heat is added or removed and therefore the entropy remains constant.

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The correct answer is D. Transparent to radiation. The photosphere appears sharp because it is relatively cooler than the layers below it, leading to a distinct boundary between the photosphere and the hotter layers of the Sun's interior.

The photosphere is the outermost visible layer of the Sun, and it is where the majority of the Sun's light and heat are emitted. It is not actually a solid surface but rather a layer of gas. The reason why the surface of the Sun appears sharp is that the photosphere is transparent to radiation. This means that the light and heat generated in the Sun's interior can easily pass through the photosphere and reach our eyes without significant scattering or absorption.

Unlike the layers below it, such as the convective zone and the radiative zone, the photosphere is not as dense or as opaque. This allows the radiation to pass through it relatively easily, without significant scattering or absorption. As a result, the photosphere acts as a "surface" where the radiation is emitted, giving the appearance of a sharp boundary when observed from a distance.

The photosphere of the Sun appears sharp because it is transparent to radiation, allowing the light and heat generated in the Sun's interior to pass through without significant distortion.

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The potential difference between the ends of the wire is 0.003 V.

Potential difference, also known as voltage, is a measure of the electrical energy required to move an electric charge between two points in an electric circuit. It is the difference in electric potential energy per unit of charge between two points in an electric circuit, and is measured in volts (V).

When a potential difference is applied across a conductor, an electric current flows in response to the electric field generated by the potential difference. Potential difference is a fundamental concept in the study of electricity and plays a crucial role in the functioning of electronic devices.

The electric field inside a copper wire is related to the potential difference between its ends by the formula:

E = V/L

where E is the electric field, V is the potential difference, and L is the length of the wire.

Rearranging this equation to solve for V, we get:

V = EL

Substituting the given values, we get:

V = (0.010 V/m) (0.30 m) = 0.003 V

Therefore, the potential difference between the ends of the wire is 0.003 V.

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The speed of the block after falling through a distance of 6.0 m is approximately 10.85 m/s.

The block falls under the influence of gravity, which generates a force that accelerates it downwards. The acceleration of the block is given by the formula a = g, where g is the acceleration due to gravity, which is approximately 9.81 m/s².

Since the block falls through a distance of 6.0 m, we can use the formula for the work done by gravity, which is W = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height through which the block falls. Thus, we have:

W = mgh

W = (3.0 kg)(9.81 m/s²)(6.0 m)

W = 176.58 J

The work done by gravity is equal to the kinetic energy gained by the block, which is given by the formula KE = 1/2mv², where v is the velocity of the block. Thus, we have:

KE = 1/2mv²

176.58 J = 1/2(3.0 kg)v²

v² = 117.72 m²/s²

Taking the square root of both sides, we get:

v = 10.85 m/s

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The gravitational force between two objects can be calculated using Newton's law of universal gravitation. According to this law, the gravitational force (F) between two objects is given by the equation:

F = (G * m1 * m2) / r^2,

where G is the gravitational constant (approximately 6.674 × 10^-11 N*m^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.

In this case, you and your friend have masses of 84.0 kg each and are located 5.00 m away from each other. Plugging these values into the equation, we get:

F = (6.674 × 10^-11 N*m^2/kg^2 * 84.0 kg * 84.0 kg) / (5.00 m)^2.

Calculating this expression gives us the gravitational force exerted between you and your friend. However, it's important to note that the force is mutual and acts on both of you equally due to Newton's third law of motion, which states that every action has an equal and opposite reaction. Therefore, the gravitational force you are exerting on your friend is the same as the gravitational force your friend is exerting on you.

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The distance along the central axis of the disk where the electric field is equal to two-fifth (2/5) of the field at the center is approximately 0.694 meters.

The electric field at a point on the central axis of a uniformly charged disk can be calculated using the following formula:

E = (σ / 2ε₀) × (1 - (z / √(R² + z²)))

where σ is the surface charge density of the disk, ε₀ is the permittivity of free space, R is the radius of the disk, and z is the distance of the point from the center of the disk along the central axis.

To find the distance along the central axis where the electric field is two-fifths (2/5) of the field at the center, we can set up the following equation:

(2/5) × E₀ = (σ / 2ε₀) × (1 - (z / √(R² + z²)))

where E₀ is the electric field at the center of the disk.

We know that E₀ = σ / (2ε₀), so we can simplify the equation to:

(2/5) × (σ / (2ε₀)) = (σ / 2ε₀) × (1 - (z / √(R² + z²)))

Simplifying further:

2/5 = 1 - (z / √(R² + z²))

2/5 = √(R² + z²) - z / √(R² + z²)

2√(R² + z²) / 5 = √(R² + z²) - z

Multiplying both sides by 5:

2√(R² + z²) = 5√(R² + z²) - 5z

3√(R² + z²) = 5z

Squaring both sides:

9R² + 9z² = 25z²

9R² = 16z²

z = (3/4)R

here R = 0.90 m

Therefore, the distance along the central axis of the disk where the electric field is equal to two-fifth (2/5) of the field at the center is approximately 0.694 meters.

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To calculate the magnitude of the force required to accelerate upward at 1.6 m/s2 with an inertia of 59 kg, we can use the formula F=ma, where F is the force, m is the mass (in this case, the inertia), and a is the acceleration.

Rearranging the formula, we get F=ma, which becomes F=59 kg x 1.6 m/s2, giving us a force of 94.4 N.

Therefore, to accelerate upward at 1.6 m/s2 with an inertia of 59 kg, you must exert a force of 94.4 N on the rope.

It is important to note that this calculation assumes no external forces acting on the system and that the rope is massless.

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Polaris, also known as the North Star, is located near the north celestial pole and appears almost directly above the Earth's geographic North Pole. Therefore, its position in the sky varies depending on the observer's location on the Earth's surface.

For an observer in the Northern Hemisphere, Polaris will be higher in the sky the closer they are to the North Pole. Therefore, cities located at higher latitudes, such as Anchorage in Alaska or even more extreme, the North Pole itself, will have Polaris at the highest point in the sky.

In contrast, cities closer to the equator, such as Singapore, will have Polaris near the horizon and not very high in the sky. Therefore, the answer to this question is not all the same, depending on the time of night, but rather depends on the observer's location.

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The second harmonic of a plucked guitar string is likely to be stronger than the first harmonic or fundamental when the string is plucked in the usual position because of the way the string vibrates. When the string is plucked, it vibrates not only as a whole but also in segments.

The second harmonic occurs when the string vibrates in two equal segments, which produces a higher frequency and a higher pitch.

In contrast, the fundamental frequency occurs when the string vibrates as a whole, which produces a lower frequency and a lower pitch. Because the string vibrates in segments more strongly than as a whole, the second harmonic is usually stronger than the fundamental frequency.

The second harmonic of a plucked guitar string is stronger than the first harmonic or fundamental when the string is plucked in the usual position because the string vibrates more strongly in segments than as a whole. When the string is plucked, it vibrates in segments and as a whole. The second harmonic occurs when the string vibrates in two equal segments, which produces a higher frequency and a higher pitch. In contrast, the fundamental frequency occurs when the string vibrates as a whole, which produces a lower frequency and a lower pitch. As the string vibrates in segments more strongly, the second harmonic is usually stronger than the fundamental frequency.

In conclusion, the second harmonic of a plucked guitar string is usually stronger than the fundamental frequency when the string is plucked in the usual position because the string vibrates more strongly in segments than as a whole. This occurs because the string vibrates in segments and as a whole when it is plucked, and the second harmonic occurs when the string vibrates in two equal segments, producing a higher frequency and a higher pitch.

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Terrestrial planets have a solid, rocky surface, low average density, small size, and are located within the inner solar system, while Jovian planets have extensive ring systems, numerous orbiting moons, are primarily composed of hydrogen, helium, and hydrogen compounds, and are located farther from the sun.

Terrestrial planets are typically smaller, denser, and located closer to the sun, whereas Jovian planets are larger, less dense, and located farther from the sun.Solid, rocky surface, located within the inner solar system, small size, and low average density are all characteristics of terrestrial planets.

These planets, including Earth, Mercury, Venus, and Mars, have a solid, rocky surface and a relatively small size compared to Jovian planets. They are located within the inner solar system, which means they are closer to the sun and experience higher temperatures.

On the other hand, extensive ring systems, primarily composed of hydrogen, helium, and hydrogen compounds, and numerous orbiting moons are characteristics of Jovian planets.

These planets, including Jupiter, Saturn, Uranus, and Neptune, are primarily composed of hydrogen, helium, and hydrogen compounds. They have extensive ring systems and numerous orbiting moons, which are a result of their strong gravitational fields.

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Joule heating, also known as resistive heating, is the process of heat production when an electric current passes through a conductor that has resistance.

Two electrical applications where joule heating is desirable are heating elements in electric stoves and toasters. In these applications, heating is the primary purpose, and joule heating provides an efficient way to produce the desired heat.

On the other hand, joule heating is undesirable in applications where it can cause overheating and damage to electrical components. One such application is electrical wiring, where excessive heating can lead to fire hazards. Another example is electric motors, where joule heating can reduce the efficiency and lifespan of the motor.

In summary, joule heating is desirable in applications where heating is the primary purpose, but it can be detrimental in applications where it can cause damage or reduce the efficiency of electrical components.

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The voltage across the second lamp in the circuit is 8 V.

In a series circuit, the current remains the same throughout. Therefore, if the current through one lamp is 1 A, the current through the other lamp is also 1 A.

Given that the voltage across the circuit is 9 V and the voltage across the first lamp is 1 V, we can use the concept of voltage division. Since the lamps are in series, the voltage across the second lamp can be calculated as follows:

Total voltage = Voltage across the first lamp + Voltage across the second lamp

9 V = 1 V + Voltage across the second lamp

Voltage across the second lamp = 9 V - 1 V

Voltage across the second lamp = 8 V

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The complete question is:

In a circuit of two lamps in series, if the current through one lamp is 1 A, The current through the other lamp is also 1 A. If a voltage of 9 V is impressed across the circuit and the voltage across the first lamp is 1 V, what is the voltage across the second lamp?

The photon flux on the small screen at a distance 2.3 m from the 150 W lamp emitting light of wavelength 590 nm is 2.66 x 10¹⁴ photons / (m² s).

To find the photon flux on the small screen, we can use the formula:
Photon flux = (power / energy per photon) × (number of photons / unit time) × (unit area / distance²)
First, let's calculate the energy per photon using the formula:
Energy per photon = (Planck's constant × speed of light) / wavelength
Energy per photon = (6.626 x 10⁻³⁴ J s ₓ 3 x 10⁸ m/s) / (590 x 10⁻⁹ m)
Energy per photon = 3.364 x 10⁻¹⁹ J
Next, let's calculate the number of photons emitted per unit time using the formula:
[tex]\frac{Number of photons}{unit time}   =\frac{power }{energy per photon}[/tex]
Number of photons / unit time = 150 W / 3.364 x 10⁻¹⁹ J
Number of photons / unit time = 4.46 x 10²⁰ photons/s
Now, let's calculate the photon flux on the small screen:
Photon flux = (power / energy per photon) ₓ (number of photons / unit time) ₓ (unit area / distance²)
Photon flux = (150 W / 3.364 x 10⁻¹⁹ J) ₓ (4.46 x 10²⁰ photons/s) ₓ (1 m² / (2.3 m)²)
Photon flux = 2.66 x 10¹⁴ photons / (m² s)

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A. The free-body diagram for the book would show the force of gravity, or weight, pulling the book downwards and the force of the student's hand pushing the book upwards. As the book rises at steady speed, the force exerted by the student on the book is equal to the force of gravity, or mg, because the book is not accelerating.

B. 1. The book lifted at constant speed gains potential energy because its height above the ground is increasing. It does not gain kinetic energy because its speed does not change.
2. The potential energy gained by the book is equal to the work done on it, which is 25 joules.
C. 1. The work done by the student is W = Fd = mgd, where d is the distance the book is lifted. Since the book is lifted through a height h, we have d = h, so the work done is W = mgh.
2. The potential energy gained by the book is also equal to mgh, since the work done on the book is converted into potential energy.
D. Parts B and C provide a conceptual and mathematical understanding of how work, potential energy, and the force of gravity are related in the process of lifting an object. They demonstrate that the work done on an object is converted into potential energy, which is directly proportional to the object's mass, the acceleration due to gravity, and the height it is lifted. The equation U = mgh is a simplified version of the relationship between work and potential energy, but understanding the derivation of this equation through the formal definition of work provides a deeper understanding of the underlying physics.

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If the terminal potential difference of the generator is 2000 v, the generator delivers a current of 448.8 A to the electrical grid.

Assuming that the generator is operating under ideal conditions, the electrical power output can be calculated using the formula:

Power = Voltage x Current

We can rearrange this equation to solve for the current:

Current = Power / Voltage

First, we need to calculate the electrical power output of the generator, which is equal to the mechanical power input multiplied by the generator efficiency:

Electrical Power Output = Mechanical Power Input x Generator Efficiency

= 1500 hp x 0.80

= 1200 hp

To convert horsepower (hp) to watts (W), we need to multiply by a conversion factor of 746 W/hp:

Electrical Power Output = 1200 hp x 746 W/hp

= 895,200 W

Finally, we can calculate the current delivered by the generator using the formula above:

Current = Electrical Power Output / Terminal Voltage

= 895,200 W / 2000 V

= 448.8 A

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The pendulum's period on the moon would increase compared to its period on Earth.

The period of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity. On the moon, the acceleration due to gravity is about 1/6th of that on Earth. Therefore, the pendulum's period on the moon will increase. The reason for this is that the force of gravity pulling on the pendulum is much weaker, causing it to swing more slowly and take longer to complete one cycle.

To calculate the new period, we can use the equation T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. If we assume that the length of the pendulum remains the same, but the value of g decreases by a factor of 1/6, the new period would be approximately 8.0 s.

Therefore, the pendulum's period on the moon would increase compared to its period on Earth.

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A 75.0 kg person stands on a analog scale in an elevator that accelerates upwards from rest to 30.0 m/s in 2.00 seconds. The scale exerts an upward force on the person equal to the net force, the scale reading in Newtons is 1125 N.

To calculate the scale reading in Newtons, we need to first determine the net force acting on the person while they are standing on the scale. We can use Newton's second law, which states that the net force is equal to the mass of an object times its acceleration

Fnet = ma

In this case, the person's mass is 75.0 kg, and the elevator's acceleration is

a = (vf - vi) / t = (30.0 m/s - 0 m/s) / 2.00 s = 15.0 m/[tex]s^{2}[/tex]

Where vf is the final velocity of the elevator, vi is its initial velocity (which is zero), and t is the time it takes to reach the final velocity.

Substituting the values into the equation, we get

Fnet = (75.0 kg)(15.0  m/[tex]s^{2}[/tex]) = 1125 N

Since the scale exerts an upward force on the person equal to the net force, the scale reading in Newtons is 1125 N.

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Answer:

Explanation:The force that an astronaut applies to space trash is equal in magnitude and directed in the opposite direction to the force that space junk applies to the astronaut, according to Newton's third law of motion.

Given that the astronaut applies a force of 10 N to the 5 kg of space debris, we can use the same magnitude to determine the force that the space debris will apply to the astronaut:

The force of the space debris on the astronaut is 10 N.

As a result, the astronaut was subjected to a force of 10 N from the space debris.