Virtual images exist where no light rays actually can be found.
O A. True
B. False

The wavelength of the light in nanometers is 321 nm.

The angular position of the nth bright fringe above the central fringe is,

θ = (nλ)/d

where,

θ = angle to the nth bright fringe

λ = wavelength of the monochromatic light

d = distance between the two slits

n = order of the bright fringe

The given values,

θ = 3.21°

n = 3

d = 0.0340mm = 3.4 × 10⁻⁵m

Substituting these values in the above equation,

3.21° = (3λ)/(3.4 × 10⁻⁵m)

λ = (3.21 × 3.4 × 10⁻⁵m)/3

λ = 3.21 × 10⁻⁵m

  = 3.21 × 10⁻⁷m

  = 321 nm

Therefore, the wavelength of the light in nanometers is 321 nm.

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Answer:

Answer is B

Explanation:

Answer:

B. Frequency

Explanation:

Answer:

handle, trigger, spindle

Transform plate tectonic boundaries slide/move past each other, rubbing along the edges causing earthquakes, tsunami, landslides

An earthquake is the shaking of the surface of the Earth resulting from a sudden release of energy in the Earth's lithosphere that creates seismic waves.

The answer is c is because when tectonic plates are rubbing against each other that’s how an earthquake is formed.

another reason the answer is C is because when the tectonic plates are moving together it can cause the land to slide.

the last reason the answer is C is because when the tectonic plates in the ocean move it can cause a tsunami.

Hence transform plate tectonic boundaries slide/move past each other, rubbing along the edges causing earthquakes, tsunami, landslides

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Answer:

a. 13.7 s b. 6913.5 m

Explanation:

a. How much time before being directly overhead should the box be dropped?

Since the box falls under gravity we use the equation

y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.

So,

y = ut - 1/2gt²

y = 0 × t - 1/2gt²

y = 0 - 1/2gt²

y = - 1/2gt²

t² = -2y/g

t = √(-2y/g)

So, t = √(-2 × 919 m/-9.8 m/s²)

t = √(-1838 m/-9.8 m/s²)

t = √(187.551 m²/s²)

t = 13.69 s

t ≅ 13.7 s

So, the box should be dropped 13.69 s before being directly overhead.

b. What is the horizontal distance between the plane and the victims when the box is dropped?

The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m

The length of the rope when the ball is moving at 5.0 m/s is approximately 0.925 meters. The calculation was based on the conservation of angular momentum .

To determine the length of the rope when the ball is moving at 5.0 m/s, we can use the conservation of angular momentum. The angular momentum of the system remains constant as the rope wraps around the pole.

The formula for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

In this case, the moment of inertia remains constant as the mass of the ball and the pole do not change. Therefore, we have:

L1 = L2

The initial angular momentum, L1, is given by:

L1 = I1ω1

The final angular momentum, L2, is given by:

L2 = I2ω2

The moment of inertia is proportional to the square of the radius of the circular motion. Since the rope shortens, the radius decreases as well.

The initial radius, r1, is given by:

r1 = 2.0 m

The final radius, r2, is unknown.

The initial angular velocity, ω1, is given by:

ω1 = 0.20 rev/s * 2π rad/rev

= 1.26 rad/s

The final angular velocity, ω2, is given by:

ω2 = v/r2

where v is the linear velocity of the ball. In this case, v = 5.0 m/s.

Thus, we can rearrange the equation to solve for r2:

r2 = v/ω2

Substituting the known values, we get:

r2 = 5.0 m/s / ω2

Now, we can substitute the equation for the angular momentum to solve for r2:

I1ω1 = I2ω2

r1^2 * ω1 = r2^2 * ω2

Solving for r2, we have:

r2^2 = (r1^2 * ω1) / ω2

Plugging in the known values:

r2^2 = (2.0 m)^2 * 1.26 rad/s / (5.0 m/s / ω2)

r2^2 = 5.04 m^2 rad/s / (5.0 m/s / ω2)

r2^2 = 5.04 m^2 rad/s * (ω2 / 5.0 m/s)

r2^2 = 5.04 m^2 rad/s * (ω2 / 5.0 m/s)

r2^2 = 5.04 m^2 rad/s * ω2 / 5.0 m/s

Simplifying the units:

r2^2 = 5.04 m^3 rad / s^2

r2 = √(5.04) m

r2 ≈ 2.244 m

Therefore, the length of the rope when the ball is moving at 5.0 m/s is approximately 0.925 meters (2.244 m - 2.0 m).

The length of the rope when the ball is moving at 5.0 m/s is approximately 0.925 meters. The calculation was based on the conservation of angular momentum, where the initial angular momentum of the system equals the final angular momentum. By considering the initial radius, initial angular velocity, final radius, and linear velocity of the ball, we derived the equation and solved for the final radius. The result was approximately 2.244 meters, and by subtracting the initial rope length (2.0 meters), we found that the shortened length of the rope is approximately 0.925 meters.

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Answer:

i dont know

Explanation:

sorry

In the Bohr model, the number of electron wavelengths that fit around a particular orbit is directly related to the principal quantum number of that orbit.

In the Bohr model of the atom, electrons are described as moving in discrete orbits around the nucleus. These orbits are quantized, meaning that only certain specific values of the electron's angular momentum are allowed.

The Bohr model also introduced the concept of electron wavelengths, which can be thought of as the "wave-like" behavior of electrons.

To determine how many electron wavelengths fit around a particular orbit in the Bohr model, we can consider the de Broglie wavelength of the electron.

According to the de Broglie hypothesis, the wavelength associated with a particle is given by the equation lambda = h/p, where h is the Planck constant and p is the momentum of the particle.

In the Bohr model, the allowed electron orbits have discrete angular momentum values, given by the equation L = n*[tex]\bar{h}[/tex], where n is the principal quantum number and [tex]\bar{h}[/tex] is the reduced Planck constant.

The momentum of an electron in an orbit is related to its angular momentum and the radius of the orbit by the equation p = L/r.

Substituting the expression for angular momentum into the de Broglie wavelength equation, we get

[tex]\lambda[/tex] = h/(n*[tex]\bar{h}[/tex]/r) = hr/(n*[tex]\bar{h}[/tex]).

The number of electron wavelengths that fit around the orbit is then given by the circumference of the orbit divided by the wavelength, i.e.,

(2*π*r)/[tex]\lambda[/tex] = (2*π*r)/(hr/(n*[tex]\bar{h}[/tex])) = (2*π*n*[tex]\bar{h}[/tex])/h.

Simplifying further, we find that the number of electron wavelengths that fit around the orbit is equal to 2*π*n, where n is the principal quantum number.

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Answer:

0.02N

Explanation:

According to Newton's 3rd Law of motion, the bee exerts the same force on the bowling ball as the bowling ball exerts the force on the bee.

We have that for the Question "A 60kg bowling ball collides with a 0.005kg bee with a force of 0.02N. The force the bee exerts is" it can be said that The force the bee exerts is

F=0.02N

From the question we are told

A 60kg bowling ball collides with a 0.005kg bee with a force of 0.02N. The force the bee exerts is

Generally the Newtons  equation for Motion  is mathematically given as

This law perceives two bodies who collide as having forces which are equal as the magnitudes in opposite directions

Therefore

V^2= 2as

Hence

The force the bee exerts is

F=0.02N

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Sz, the z-component of the Poynting vector at (x = 0, y = 0, z = 0) at t = 0 is 0.

A monochromatic electromagnetic wave with wavelength λ = 2.8 cm, is propagating through a vacuum and its magnetic field is described by B = 8.4 sin(2π(1.0 × 10^9)t - 2π(1.0 × 10^4)x) µT.

The Poynting vector can be used to explain this in more detail. The Poynting vector can be calculated using the formula [rmS = rmE times B over rm2mu], where  is the permeability of the medium through which the wave is propagating and E is the electric field.

The equation for the electric field is E = cB, where c is the vacuum speed of light.

The above formula can be used to determine S = 0 for the Poynting vector for the given wave.

The Poynting vector has a z-component of 0 at (x = 0, y = 0, z = 0) and t = 0.

The energy motion thickness of an electromagnetic field can be shown utilizing the Poynting vector, named after John Henry Poynting.

According to the definition, the Poynting vector is the result of the vector product of the electric and magnetic fields: 2.22) S → = E → × H →

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The greatest distance the person can stand from the mirror and still see their image clearly is 2.0 meters.

The far point of a person's eyes refers to the maximum distance at which they can see objects clearly without the aid of corrective lenses. If the far point is given as 2.0 m, it means that the person can see objects clearly up to a distance of 2.0 m.

To determine the greatest distance the person can stand from the mirror and still see their image clearly, we need to consider the concept of the virtual image formed by a mirror. When we look into a mirror, our eyes perceive a virtual image as if it were formed behind the mirror.

In this case, the person's eyes can focus clearly up to a distance of 2.0 m. To see their image clearly in the mirror, the person needs the virtual image formed by the mirror to be within this range.

Since the virtual image formed by a mirror is the same distance behind the mirror as the object is in front of it, the person needs to stand at a distance from the mirror equal to the maximum distance they can focus clearly.

Therefore, the greatest distance the person can stand from the mirror and still see their image clearly is 2.0 meters.

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The direction of the magnetic force on the positive charge is to the right. Therefore, the correct answer is (d) to the right.

To determine the direction of the magnetic force on a positive charge entering a uniform magnetic field, we can use the right-hand rule for magnetic force.

If we point the thumb of our right hand in the direction of the velocity of the positive charge, and align our fingers with the magnetic field lines (in the direction of the field), then the palm of our hand will indicate the direction of the magnetic force.

In the given scenario, the positive charge is moving in the downward direction, and the magnetic field is directed into the page (represented by the x's in the figure).

Using the right-hand rule, if we point our thumb downward to represent the velocity and align our fingers into the page to represent the magnetic field, the palm of our hand will face to the right. Therefore, the direction of the magnetic force on the positive charge is to the right.

Therefore, the correct answer is d) to the right.

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Answer:

C

Explanation:

Copper is a conductor because its valence electrons can move freely from atom to atom.

The attenuation from the dielectric (air) is 0 dB/m. The attenuation from both dielectric and conductor losses for the glass dielectric is approximately 0.063 dB/m.

To calculate the attenuation in decibels per meter (dB/m) for a TM1​ wave between copper planes 1.5 cm apart with an air dielectric, we can use the following formula:

Attenuation (dB/m) = α * (1/λ)

where α is the attenuation constant and λ is the wavelength.

The attenuation constant can be calculated as:

α = (2π * frequency * ε′′) / (c * ε′)

where frequency is the given frequency (12 GHz), ε′′ is the imaginary part of the relative permittivity (dielectric loss factor), ε′ is the real part of the relative permittivity, and c is the speed of light.

For air dielectric, εr (relative permittivity) is approximately 1, so ε′ = εr = 1.

Substituting the given values into the formula:

α = (2π * 12 × 10⁹ Hz * 0) / (3 × 10⁸ m/s * 1) = 0.

Now, let's calculate the attenuation for the glass dielectric with εr = 4 and ε′′/ε′ = 2 × 10⁻³.

Since the glass dielectric is introduced, the relative permittivity becomes εr = 4 and ε′ = εr - ε′′ = 4 - (2 × 10⁻³ * 4) = 3.992.

Using the same formula as before:

α = (2π * 12 × 10⁹ Hz * 2 × 10⁻³) / (3 × 10⁸ m/s * 3.992) = 0.063 dB/m

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To calculate the mass of the planet, we can use the formula for the period of a satellite in circular orbit:

T = 2π√(r³/GM)

T is the period of the satellite (in seconds)

r is the radius of the orbit (in meters)

G is the gravitational constant (approximately 6.67430 × 10^(-11) m³/kg/s²)

M is the mass of the planet (in kilograms)

First, let's convert the period of the satellite into seconds. 7 hours and 38 minutes is equal to 7 × 60 × 60 seconds + 38 × 60 seconds, which gives us a total of 27,480 seconds.

Plugging the given values into the formula, we have:

27,480 = 2π√((2.1 × 10^6)³/(6.67430 × 10^(-11) × M))

To solve for M, we can rearrange the equation:

M = ((2.1 × 10^6)³ × 6.67430 × 10^(-11))/(2π)² × (27,480)²

M ≈ 6.042 × 10^24 kg

Therefore, the mass of the planet is approximately 6.042 × 10^24 kg.

From the given information about the satellite's radius (2.1 × 10^6 m) and period (7 hours 38 minutes), we calculated the mass of the planet to be approximately 6.042 × 10^24 kg using the formula for the period of a satellite in circular orbit.

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part a: N= 4.39 x 10^10 photons/second. part b: N = 8.25 photons/second are the answers.

The energy of a photon of wavelength 550 nm can be calculated as follows:

h*c/λ = 6.626 x 10^-34 x 3 x 10^8 / (550 x 10^-9)

wavelength = 3.6 x 10^-19 J.

In one second, a 120-W lightbulb emits energy as visible light in the following amount:

120 x 0.035 = 4.2 W.

As a result, the number of photons emitted per second can be computed as follows:

N = Power/Energy of a photon

N = 4.2/3.6 x 10^-19

N = 1.17 x 10^16 photons/second

PART A: The area of a 4.0-mm diameter pupil is given by the following formula:

A = πr^2

A = π(2.0 mm)^2

A= 3.14 x 4.0 x 10^-6 m^2

The distance from the bulb to the pupil is 2.2 m.

The bulb radiates in all directions, therefore the power per unit area is given by:

P = Power/4πr^2

P = 4.2/(4π(2.2)^2)

P = 0.04 W/m^2

The number of photons striking the pupil per second is determined by the following formula:

N = P*A/E

N = (0.04)(3.14 x 4.0 x 10^-6)/3.6 x 10^-19

N  = 4.39 x 10^10 photons/second

PART B: The bulb is now 1.5 km distant from the pupil.

The power per unit area at this distance is:

P = Power/(4πr^2)

P = 4.2/(4π(1500)^2)

P  = 2.98 x 10^-10 W/m^2.

Using the equation for the number of photons striking the pupil per second,

N = P*A/E

N = (2.98 x 10^-10)(3.14 x 4.0 x 10^-6)/3.6 x 10^-19

N = 8.25 photons/second

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Answer:

Hello There!!

Explanation:

The answer is=>Conduction.

hope this helps,have a great day!!

~Pinky~

[tex]\huge{\textbf{\textsf{{\color{pink}{An}}{\red{sw}}{\orange{er}} {\color{yellow}{:}}}}}[/tex]

Conduction.

An Otto cycle is used in gasoline engines to convert the thermal energy of burning gasoline into mechanical work. It consists of four stages: compression, heat addition, expansion, and exhaust. The compression ratio is defined as the ratio of the volume of the combustion chamber at the beginning of the compression stroke to the volume of the combustion chamber at the end of the expansion stroke.

Compression ratio, r = V1/V2 Here, V1 = Initial volume of air before compression stroke = 500 cm³V2 = Volume of air after the compression stroke. As the compression ratio is given as 10, the volume of air after the compression stroke, V2 = V1/10 = 500/10 = 50 cm³.

Temperature and pressure can be calculated using the following equations.

PV¹⁴ = Constant (P₁V₁¹⁴) T₁/V₁ = T₂/V₂ (isentropic process)

a) The highest temperature can be found at the end of the combustion stroke.

Here, V3 = V2T3/T2 = V4T4/T3T1V1¹⁴ = V2V3¹⁴ Pmax = T3V3/V2 = T4V4/V3 = P4

b) Amount of heat transferred, Qn = Cn (T4 - T1) Where Cn is the specific heat at constant volume.

Qn = Cn (T4 - T1) = cv (T4 - T1)

c) Thermal efficiency,

η = (Wnet)/Qn = (Qn - Qout)/Qn = 1 - (Qout/Qn) = 1 - (cv (T3 - T2))/(cv (T4 - T1))

d) Mean effective pressure, MEP = (Pmax - Pmin)/2 = (P4V3/V4 - P2V1/V2)/2

The formula for the constant specific heat approach - k-1.4, cp= 1.005 kJ/kg.K cv=0.718 kJ/kg.K, R = 0.287 kJ/kg.K is given below.

Cn = cp/(k-1) = 1.005/(1.4 - 1) = 1.005/0.4 = 2.5125 kJ/kg.Kcv = R/(k-1) = 0.287/(1.4 - 1) = 0.287/0.4 = 0.7175 kJ/kg.K

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Answer:

B. Radiation

Explanation:

Transfer of heat energy through space by electromagnetic radiation

The initial pressure of steam (P1) = 4 MPa, the Initial temperature of steam (T1) = 300°C = 573.15 K, Final pressure of steam (P2) = 9 MPaProcess: The given process is Isentropic Process Formula: For isentropic process, P1V1^γ = P2V2^γWhere γ = Cp / Cv = 1.3 (For Steam)And, V1 / T1^γ-1 = V2 / T2^γ-1,

Where V = Specific volume of steam solution: Specific volume of steam (V1) at initial state is given by, V1 = V2 = v (From the principle of the isentropic process).

Now, from the steam table, At 4 MPa (P1) and 573.15 K (T1), V1 = 0.1006 m³/kgAt 9 MPa (P2), V2 = V1 × (P1 / P2)^(1/γ)= 0.1006 × (4 / 9)^(1/1.3)= 0.080 m³/kg.

Let's use the formula for calculating the final temperature of the steamV1 / T1^γ-1 = V2 / T2^γ-1⇒ T2 = (V2 / V1)^(1/γ-1) × T1= (0.08 / 0.1006)^(1/1.3-1) × 573.15≈ 764.5 K≈ 491.5°C.

Therefore, the final temperature of the steam is 491.5°C (rounded off to one decimal place).

Hence, option D is correct.

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In general, the sum of the squares of the voltages across individual components may not necessarily be equal to the square of the voltage across both elements combined. It would depend on the nature of the circuit and the relationship between the voltage across the components.

It would depend on the nature of the circuit and the relationship between the voltage across the components. To determine if the sum of the square of the maximum voltages across resistor (R) and inductor (L), Vr^2 and Vl^2 respectively, is equal to the square of the voltage across both elements, Vrl^2, in accordance with equation (2), we need to have more context and information about the specific equation and circuit being referred to. The relationship between Vr^2, Vl^2, and Vrl^2 would depend on the specific circuit configuration and the equations governing its behavior.

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Answer:

a.16.125 N b. The force is an attractive force. c. 2.68 × 10¹³ electrons d. 3.75 × 10¹³ electrons

Explanation:

a. What is the magnitude of the electric force between the corks?

The electrostatic force of attraction between the two corks is given by

F = kq₁q₂/r² where k = 9 × 10⁹ Nm²/C², q₁ = +6.0 μC = +6.0 × 10⁻⁶ C, q₂ = -4.3 μC = -4.3 × 10⁻⁶ C and r = distance between the corks = 0.12 m

Substituting the values of the variables into the equation, we have

F = kq₁q₂/r²

F = 9 × 10⁹ Nm²/C² × +6.0 × 10⁻⁶ C × -4.3 × 10⁻⁶ C/(0.12 m)²

= -232.2 × 10⁻³ Nm²/(0.0144 m)²

= -16125 × 10⁻³ N

= -16.125 N

So, the magnitude of the force is 16.125 N

b. Is this force attractive or repulsive?

Since the direction of the force is negative, it is directed towards the positively charged cork, so the force is an attractive force.

c. How many excess electrons are on the negative cork?

Since Q = ne where Q = charge on negative cork = -4.3 μC = -4.3 × 10⁻⁶ C and n = number of excess electrons and e = electron charge = -1.602 × 10⁻¹⁹ C

So n = Q/e = -4.3 × 10⁻⁶ C/-1.602 × 10⁻¹⁹ C = 2.68 × 10¹³ electrons

d. How many electrons has the positive cork lost?

We need to first find the number of excess positive charge n'

Q' = n'q where Q = charge on positive cork = + 6.0 μC = + 6.0 × 10⁻⁶ C and n = number of excess protons and q = proton charge = +1.602 × 10⁻¹⁹ C

So n' = Q'/q = +6.0 × 10⁻⁶ C/+1.602 × 10⁻¹⁹ C = 3.75 × 10¹³ protons

To maintain a positive charge, the number of excess protons equals the number of electrons lost = 3.75 × 10¹³ electrons

Answer:

hi. :)))))))))))))))))))))

The force on the positive charge due to the magnetic field will be directed in the y-direction.

To determine the direction of the force on a moving charge in a magnetic field, we can use the right-hand rule.

According to the right-hand rule, if we extend the thumb, index finger, and middle finger of our right hand mutually perpendicular to each other, the thumb represents the velocity vector (v), the index finger represents the magnetic field vector (B), and the middle finger represents the direction of the force (F) acting on the charge.

In this case, we have the magnetic field pointing in the z-direction, which means the magnetic field vector (B) will point directly out of the page or screen toward you. The velocity vector (v) lies in the x-z plane, so it will be perpendicular to the magnetic field vector.

Using the right-hand rule, we can place our right hand with the thumb pointing in the positive x-direction (toward the right), the index finger pointing in the positive z-direction (out of the page), and the middle finger will naturally point in the positive y-direction (upward).

Therefore, the force acting on the positive charge due to the magnetic field will be directed in the y-direction, which is upward.

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The requried force on the wire due to Earth's magnetic field will be perpendicular to both the current direction and the magnetic field direction.

According to the right-hand rule, if a current-carrying wire is held in the right hand with the thumb pointing in the direction of the current, the direction in which the fingers curl represents the direction of the magnetic field around the wire.

In this case, the wire is carrying the current vertically downward. If we apply the right-hand rule, we can curl our fingers around the wire in the direction of the magnetic field. The direction of the magnetic field is horizontally perpendicular to the wire, pointing toward the north.

Therefore, the force on the wire due to Earth's magnetic field will be perpendicular to both the current direction and the magnetic field direction. Using the right-hand rule again, if we extend our right-hand thumb in the direction of the current (downward) and curl our fingers in the direction of the magnetic field (northward), our palm will face east. This indicates that the force on the wire will be directed eastward.

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Answer:

the work done by the lawnmower is 236.14 J.

Explanation:

Given;

power exerted by the lawnmower engine, P = 19 hp

time in which the power was exerted, t = 1 minute = 60 s.

1 hp = 745.7 watts

The work done by the lawnmower is calculated as follows;

[tex]Work = Energy = \frac{Power}{time} \\\\Work = \frac{(19 \times 745.7)}{60} = 236.14 \ J[/tex]

Therefore, the work done by the lawnmower is 236.14 J.

The weight of an object would be the least at the center of the Earth. 2000 miles above earth's surface at the equator at the south pole at the north pole at the center of earth.

The weight of an object depends on the gravitational force acting on it. The gravitational force is inversely proportional to the square of the distance between the object and the center of the Earth. As the object moves farther away from the center of the Earth, the gravitational force decreases.

At the center of the Earth, the distance between the object and the center is at its minimum. Therefore, the gravitational force and consequently the weight of the object would be the least at the center of the Earth.

In contrast, as the object moves further away from the center of the Earth, such as 2000 miles above the Earth's surface, the distance between the object and the center increases, resulting in a stronger gravitational force and a greater weight.

Thus, of the options provided, the weight of an object would be the least at the center of the Earth.

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Answer:

A roller coaster:

The first law states that an object either remains at rest or continues to move at a constant velocity, unless it is acted upon by an external force.

Hope this helps!

Answer:

This

Explanation:

First Law: Newtons first law of motion explains what happens in a car crash because it basically states that the passenger will continue to travel at the same velocity until an unbalanced force acts on he or she. The force that will act upon he or she would be the window, so you should always wear a seat belt!

Second Law: In other words, it states that the force that is applied in the crash is proportional to mass of impacting cars. This means that the bigger the force of impacting cars, the bigger the force applied, which implies a greater destruction.

Third law: Newton's Laws Applied to Collisions. Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction.