Near the surface of earth an electric field points radially downward and has a magnitude of 100 N/C. What charge would have to be placed on a pollen that has a mass of 4 x 10^9 kg so the pollen can be in equilibrium ( zero acceleration)

The electric potential energy of the system is

The acceleration of sphere A isa = [tex]F/m= (5.63\times10^{-2})/7.11\times10^{-3}= 7.92[/tex]m/s²

The acceleration of sphere B will be the same as the acceleration of sphere A, but in the opposite direction.

The speed of sphere A isv = u + at= 0 + 7.92×0.166= 1.31 m/s

The speed of sphere B will be the same as the speed of sphere A, but in the opposite direction.

(a) Electric potential energy of the system: We can calculate the electric potential energy of the system using the formula, U = q²/(4πε₀d)Where, q is the charge of each sphere, d is the separation distance between two spheres, and ε₀ is the permittivity of free space.

So, the electric potential energy of the system isU = [tex]6.67^{2} \times(10^{-6})^{2} /(4\times3.14\times8.85\times10^{-12}\times0.876)= 2.41\times10^{-3} J[/tex]

(b) Acceleration of sphere A:When the string is cut, the spheres will experience a force of attraction towards each other, so they will move towards each other.

We can calculate the acceleration of sphere A using the formula,F = maWhere, F is the force acting on sphere A due to the attraction with sphere B.

m is the mass of sphere A.a is the acceleration of sphere A.

So, the force acting on sphere A isF = k(q²/d²)= [tex](9\times10^9)\times(6.67\times10^-6)^{2} /(0.876)^{2} = 5.63\times10^{-2} N[/tex]

Thus, the acceleration of sphere A isa = F/m=  [tex]F/m= (5.63\times10^{-2})/7.11\times10^{-3}= 7.92[/tex] m/s²

(c) Acceleration of sphere B:The acceleration of sphere B will be the same as the acceleration of sphere A, but in the opposite direction.

(d) Speed of sphere A:We can calculate the speed of sphere A using the formula, v = u + at Where, u is the initial velocity of sphere A, which is zero.

v is the final velocity of sphere A.

t is the time taken by sphere A to reach the final velocity.a is the acceleration of sphere A, which we have already calculated.

So, the time taken by sphere A to reach the final velocity ist = √(2d/a)= √(2×0.876/7.92)= 0.166 s

Thus, the speed of sphere A isv = u + at= 0 + 7.92×0.166= 1.31 m/s(e) Speed of sphere B:

The speed of sphere B will be the same as the speed of sphere A, but in the opposite direction.

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As per the given conditions of the problem, A child sprays her sister with water from a garden hose.

The water is supplied to the hose at a rate of 0.117 L/s, and the diameter of the nozzle is 5.95 mm.

We need to determine the speed at which the water exits the nozzle.

To determine the speed v at which the water exits the nozzle, we need to apply Bernoulli's principle.

According to Bernoulli's principle, the pressure of a fluid decreases as its velocity increases, keeping the total energy constant.

The Bernoulli equation is written as:

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²

where:

P₁ = Pressure at Point 1v₁ = Velocity at Point 1h₁ = Height at Point 1ρ = Density

g = Acceleration due to gravity

P₂ = Pressure at Point 2v₂ = Velocity at Point 2h₂ = Height at Point 2

Assuming that the hose is horizontal, the height at both points 1 and 2 are equal, so the terms ρgh₁ and ρgh₂ can be ignored.

Also, assuming that the fluid is incompressible, the term P₁ can be considered negligible in comparison to P₂.

Hence the Bernoulli equation can be written as:

1/2ρv₁² = 1/2ρv₂²

The mass flow rate is given by:

ṁ = ρAvwhere,

ṁ = mass flow rate

ρ = density of the fluid

A = cross-sectional area of the nozzle

v = velocity of the fluid

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Newton's First Law of Motion states that an object will remain at rest or move at a constant speed in a straight line unless acted upon by an unbalanced force. This means that if there are no forces acting on an object,

it will stay in its current state of motion. In order for an object to satisfy this law, there must be a balance of forces acting upon it.

One example of an object satisfying Newton's First Law of Motion is a ball rolling on a flat surface with no external forces acting upon it. In this scenario, the force of gravity is acting upon the ball, pulling it towards the Earth. However, the normal force of the surface on the ball is equal and opposite to the force of gravity, resulting in a net force of zero. As a result, the ball remains at a constant speed in a straight line. The absence of any unbalanced forces acting on the ball satisfies Newton's First Law of Motion.

Another example of an object satisfying Newton's First Law of Motion is a satellite in orbit around the Earth. In this scenario, the gravitational force of the Earth is pulling the satellite towards it. However, the satellite is also moving at a constant speed in a straight line, which results in a balance of forces. The gravitational force acting on the satellite is exactly balanced by the satellite's centripetal force, which keeps it in orbit. Therefore, the satellite satisfies Newton's First Law of Motion, as there are no unbalanced forces acting upon it.

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The two waves with identical amplitude A and wavelength, and a phase shift of half a period between them travel in the same direction. The amplitude of the resultant wave is 2A.

Waves are defined as a disturbance that transfers energy through matter or space. These can be classified into two categories- mechanical waves and electromagnetic waves. Mechanical waves are those waves that need a medium for propagation, such as sound waves or water waves. Electromagnetic waves, on the other hand, do not require a medium for propagation.Wave parameters:Waves have various parameters, which are used to define them and to differentiate one type of wave from another. Some of the important wave parameters are:Amplitude

Wavelength

Frequency

Period

Velocity

Phase

These parameters are used to calculate other wave properties. Wave interference is a phenomenon in which two waves superimpose to form a resultant wave of greater or lower amplitude. Interference between two waves:When two waves interfere, their amplitudes can add up or cancel each other out, depending on the phase difference between them. There are two types of interference- constructive interference and destructive interference.In constructive interference, the waves add up to produce a wave with a higher amplitude. In destructive interference, the waves cancel each other out, resulting in a wave with zero amplitude.

The amplitude of the resultant wave:

When two waves interfere, the amplitude of the resultant wave can be calculated using the following formula:

A = 2Acosθ

where A is the amplitude of each wave and θ is the phase difference between the waves. In this case, the phase difference between the two waves is half a period. Therefore, θ = π. Substituting this value in the above equation, we get:

A = 2Acosπ= 2A(-1)= -2AThe amplitude of the resultant wave is -2A. However, since amplitude is a scalar quantity and cannot be negative, we take the magnitude of the amplitude, which is equal to 2A. Therefore, the amplitude of the resultant wave is 2A.Option C, which states "2A" is the correct answer.

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The best explanation of how baseball outfielders catch a fly ball is: Players watch the ball and position themselves so that it appears the ball is neither speeding up nor slowing down.

(Option B).

Fly balls are difficult to catch because they often come off the bat at unpredictable angles. Players must run quickly and judge the ball's distance and trajectory to get in the best position to catch it. The outfielders must consider several factors, such as the ball's speed, the wind, and the sun's position, to make the catch. Players must be able to read the ball's trajectory and determine where it will land. They must understand that the ball will move in a parabolic path, which means that it will start high and gradually lose height as it approaches the ground.

 The player must track the ball's movement and position themselves accordingly. Players watch the ball and position themselves so that it appears the ball is neither speeding up nor slowing down. They will often look over their shoulder, running toward the spot where the ball will land. This allows them to stay underneath the ball and make the catch at the right time.The player must also be aware of their surroundings to avoid colliding with other fielders. They should also communicate with their teammates to avoid confusion and maximize the chances of catching the ball.

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The electric potential difference between two points a distance d apart in a uniform electric field E is -Ed.

Therefore, the correct option is: −Ed.

What is electric potential difference?

Electric potential difference is the difference in electric potential between two points in an electric field.

The electric potential difference is also referred to as voltage and is calculated in volts.

The formula to calculate electric potential difference is:

V = W/Q

where

V = electric potential difference (volts)

W = work done (joules)

Q = charge (coulombs)

The electric potential difference can be calculated by dividing the work done by the charge of the particle moving in the electric field.

The electric potential difference is negative for the electric field directed from the higher to lower potential and is positive for the electric field directed from the lower to higher potential.

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The horsepower transmitted by the belt is 0.187 HP.

To calculate the horsepower transmitted by the belt when a pulley on a shaft of an electric motor has a pulley with a diameter of 8" attached and is rotating at 1800rpm and when the belt drive is connected, the tension on the slack side is 20lb and that in the tight side is 110lb, we can use the following formula:

Horsepower (HP) = (T₁ - T₂) x V / 33000

Where T₁ is tension in tight side (lb),T₂ is tension in slack side (lb),V is the belt velocity (ft/s).

Given data:

Diameter of the pulley = 8 "Radius of the pulley = diameter / 2 = 4 " = 1/3 ft

Speed of rotation = 1800 rpm

Belt tension on slack side = 20 lb

Belt tension on tight side = 110 lb

From the speed of rotation, we can calculate the velocity of the pulley as follows:

Velocity (V) = 2 π N R / 60

where N is the speed of rotation and R is the radius of the pulley.

Substituting the given values, we get:V = 2 π x 1800 x 1/3 / 60 = 56.55 ft/s

Now, substituting the given values in the formula for horsepower, we get:

Horsepower (HP) = (110 - 20) x 56.55 / 33000 = 0.187 HP

Therefore, the horsepower transmitted by the belt is 0.187 HP.

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The density of iron is 7.86 g cm −3 and the density of sea water is 1.10 g cm −3. Can iron float in sea water? The answer is No because the density of iron is more than the density of sea water.

Since the density of iron is greater than the density of seawater, it will sink in seawater; therefore, it cannot float in seawater.

Therefore, iron can not float in seawater, because of the buoyancy force acting on iron is lower than its weight. So, the buoyancy force is not strong enough to keep the iron object on the surface of the seawater.

The buoyancy of an object is determined by the mass and the volume of the object. If the mass of the object is greater than the mass of the volume of water that it displaces, then the object will sink.

However, it depends on the shape and surface area of the iron object. If the shape of the iron object is designed in such a way that the weight is distributed over a large surface area, it may float on the surface of the seawater. For example, think of anchor and cruise ships.

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Minimum power of the motor is  16.67 W. A reducer must be attached to decrease the motor speed from 1400 rpm to approximately 3.183 rpm.

To calculate the minimum power of the motor for the crane, we need to consider the lifting speed and the load.

The power of the motor can be calculated using the formula: Power = Force x Velocity. In this case, the force is the weight being lifted, which is 1 ton or 1000 kg, and the velocity is given as 1 m/min. However, we need to convert the velocity to m/s for consistent units. Since 1 min = 60 s, the velocity is 1 m/60 s = 1/60 m/s.

Therefore, the minimum power of the motor can be calculated as:

Power = 1000 kg x (1/60 m/s) = 16.67 W

To determine the reducer required for a 1400 rpm electric motor, we need to consider the relationship between the motor speed and the cable reel diameter.

The cable reel's diameter is given as 100 mm, which is equivalent to 0.1 m. We know that the circumference of the reel is 2πr, where r is the radius of the reel. Since the diameter is 0.1 m, the radius is 0.05 m.

The linear speed of the cable reel can be calculated using the formula: Speed = 2πr x Motor Speed.

Rearranging the formula to solve for the required motor speed, we have:

Motor Speed = Speed / (2πr) = 1 m/min / (2π x 0.05 m) = 3.183 rpm

Therefore, a reducer must be attached to decrease the motor speed from 1400 rpm to approximately 3.183 rpm.

It's worth mentioning that the additional parameters provided, such as the coefficient of elongation, densities, and heat capacities, seem unrelated to the given information about the crane and are not necessary for calculating the minimum motor power or determining the required reducer

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Schematic diagram of a heat engine:A heat engine is a device that uses thermal energy to produce mechanical work. Heat engines operate on the principle of the Carnot cycle, which involves four thermodynamic processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. The working principle of the heat engine is described as follows:

A gas that is initially at the temperature T1 and volume V1 is isothermally expanded to volume V2 by supplying heat from a high-temperature heat source, resulting in a decrease in pressure from P1 to P2. The gas is then adiabatically expanded from pressure P2 to pressure P3 while also losing heat to the low-temperature sink, resulting in a decrease in temperature from T1 to T2 and a decrease in volume from V2 to V3. The gas is then isothermally compressed to volume V4, releasing heat to the low-temperature sink, causing the pressure to rise from P3 to P4. Finally, the gas is adiabatically compressed back to volume V1,

raising the temperature from T2 to T1 and restoring the pressure from P4 to P1.The schematic diagram of a heat engine is given below:Schematic diagram of a refrigerator:A refrigerator is a device that uses work to transfer heat from a low-temperature environment to a high-temperature environment. A refrigeration cycle consists of four thermodynamic processes: isentropic compression, isobaric heat rejection, isentropic expansion, and isobaric heat absorption. The working principle of a refrigerator is as follows:

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Velocity of a car moves along an east-west road is shown in the given graph. We need to determine the correct time intervals during which the car is slowing down, moving with a constant speed, magnitude of the car's acceleration during time interval E, and during which the car is moving east.

Given graph represents the variation of velocity with time of a car moving on an east-west road. We need to find out the following things:(a) During which one or more time intervals is the car slowing down?From the graph, we can observe that the car is slowing down during the following intervals:- From A to B- From C to D- From E to FTherefore, the correct answer is ABCEF(b) During which one or more time intervals is the car moving with a constant speed?From the graph, we can observe that the car is moving with a constant speed during the following intervals:- From B to C- From D to ETherefore, the correct answer is ADF(c) Wh

Change in velocity of the car, Δv = 10 - 20 = -10 m/sTime interval, Δt = 5 sTherefore, acceleration, a = (Δv) / (Δt) = (-10) / 5 = -2 m/s²Magnitude of the acceleration, |a| = 2 m/s²Therefore, the magnitude of the car's acceleration during time interval E is 2 m/s²(d)  From the graph, we can observe that the car is moving east during the following intervals:- From A to DTherefore, the correct answer is AD

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The dielectric constant of the material is 2.

The dielectric constant (k) of a material can be determined by comparing the electric field (E) before and after inserting the dielectric between the plates of a capacitor.

The dielectric constant is given by the formula k = E₀ / E, where E₀ is the electric field without the dielectric and E is the electric field with the dielectric. In this case, the initial electric field (E₀) is 3000 V/m, and the electric field with the dielectric (E) is 1500 V/m.

Substituting these values into the formula, we find k = 3000 V/m / 1500 V/m = 2. Therefore, the dielectric constant of the material is 2.

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The period of the satellite is equal to the square root of [[tex](4\pi ^2 * (5.47 * 6.38 * 10^6)^3) / (6.67430 * 10^-11 * 5.98 * 10^24)[/tex]] seconds.

calculate the period of the satellite, we can use Kepler's third law, which states that the square of the period (T) of an object in circular orbit is proportional to the cube of the radius (r) of the orbit.

this can be expressed as:

[tex]T^2 = (4\pi ^2 / GM) * r^3[/tex]

where T is the period, G is the gravitational constant ([tex]6.67430 * 10^{-11} m^3 kg^{-1} s^{-2[/tex]), M is the mass of the Earth, and r is the radius of the orbit.

Radius of the satellite's orbit (r) = 5.47 * (radius of the Earth) = 5.47 * 6.38 × 1[tex]0^6[/tex] m

Mass of the Earth (M) = 5.98 ×[tex]10^{24[/tex] kg

Substitute these values into the formula and solve for the period (T):

[tex]T^2 = (4\pi ^2 / GM) * r^3[/tex]

[tex]T^2 = (4 * \pi ^2 / (6.67430 * 10^{-11} m^3 kg^{-1} s^{-2} * 5.98 * 10^{24} kg)) * (5.47 * 6.38 * 10^6 m)^3[/tex]

[tex]T^2 = 4\pi ^2 * (5.47 * 6.38 * 10^6 m)^3 / (6.67430 * 10^{-11} m^3 kg^{-1} s^{-2} * 5.98 * 10^{24} kg)[/tex]

Calculate the value of T:

T = [tex]\sqrt {[4\pi ^2 * (5.47 * 6.38 * 10^6 m)^3 / (6.67430 × 10^{-11} m^3 kg^{-1} s^{-2} * 5.98 * 10^{24} kg)][/tex]

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The vector sum of the two forces is approximately 0.00 N in the x-direction.

To find the vector sum of the two forces, we can break each force into its x-component and y-component and then add them together. Since both forces have the same magnitude of 38.0 N and are directed at angles of 60° above and below the x-axis, their y-components cancel each other out.

The y-component of the first force (F1) is given by F1 * sin(60°) = 38.0 N * sin(60°) = 32.9 N in the positive y-direction. The y-component of the second force (F2) is F2 * sin(60°) = 38.0 N * sin(60°) = -32.9 N in the negative y-direction.

Since the y-components cancel each other out, the net y-component is 32.9 N - 32.9 N = 0 N.

The x-component of the first force (F1) is given by F1 * cos(60°) = 38.0 N * cos(60°) = 19.0 N in the positive x-direction. The x-component of the second force (F2) is F2 * cos(60°) = 38.0 N * cos(60°) = 19.0 N in the negative x-direction.

Adding the x-components together, we have 19.0 N - 19.0 N = 0 N.

Therefore, the vector sum of the two forces is approximately 0.00 N in the x-direction.

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To calculate the base current (IB) required to switch a resistive load of 10mA, we can use the formula:

IB = (IL) / (Beta)

where IL is the load current and Beta is the DC current gain of the transistor.

Given that IL = 10mA and Beta = 100, we can substitute these values into the formula:

IB = (10mA) / (100)

IB = 0.1mA

So, the base current required to switch the load is 0.1mA.

To find the value of the base resistor (RB) required to switch the load fully "ON" when the input terminal voltage exceeds 5V, we can use Ohm's Law:

RB = (Vin - VBE) / IB

where Vin is the input terminal voltage and VBE is the base-emitter voltage of the transistor.

Given that Vin > 5V and assuming VBE = 0.7V for a silicon BJT, we can substitute these values into the formula:

RB = (Vin - 0.7V) / 0.1mA

Let's say Vin = 10V, then

RB = (10V - 0.7V) / 0.1mA

RB = 9.3V / 0.1mA

RB = 93kΩ

So, the value of the base resistor required to switch the load fully "ON" when the input terminal voltage exceeds 5V is 93kΩ.

(ii) To find the value of the base current into the transistor with a DC base bias voltage (VB) of 20V and an input base resistor (RB) of 50kΩ, we can use Ohm's Law:

IB = (VB - VBE) / RB

Given that VB = 20V and RB = 50kΩ, and assuming VBE = 0.7V for a silicon BJT, we can substitute these values into the formula:

IB = (20V - 0.7V) / 50kΩ

IB = 19.3V / 50kΩ

IB = 0.386mA

So, the value of the base current into the transistor is 0.386mA.

To find the minimum base current required to turn the transistor fully‐ON (saturated) for a load that requires 500mA of current when the input voltage is increased to 25.0V, we can use the formula:

IB = (IL) / (Beta)

Given that IL = 500mA and Beta = 100, we can substitute these values into the formula:

IB = (500mA) / (100)

IB = 5mA

So, the minimum base current required to turn the transistor fully‐ON (saturated) is 5mA.

Finally, to calculate the new value of RB, we can use Ohm's Law:

RB = (Vin - VBE) / IB

Given that Vin = 25V and assuming VBE = 0.7V for a silicon BJT, we can substitute these values into the formula:

RB = (25V - 0.7V) / 5mA

RB = 24.3V / 5mA

RB = 4.86kΩ

So, the new value of RB is 4.86kΩ.

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The force that Nancy exerts on Tonya is -910 Newtons. Here's the long answer:When Nancy and Tonya collide in the air, we have to use the Law of Conservation of Momentum. Here's the formula for this law:M₁V₁ + M₂V₂ = M₁V₁' + M₂V₂'

Where:M₁ is the mass of the first objectV₁ is the velocity of the first object before the collisionM₂ is the mass of the second objectV₂ is the velocity of the second object before the collisionV₁' is the velocity of the first object after the collisionV₂' is the velocity of the second object after the collision We are given that:Tonya has a mass of 51 kg and a velocity of 6 m/s Nancy has a mass of 50 kg and a velocity of 6 m/sTonya exerts a force of 910 N on NancyFirst, we have to calculate the momentum of each skater before the collision. The momentum formula is:p = mvwhere:p is momentum m is mass v is velocity For Tonya:p₁ = m₁v₁p₁ = 51 kg x 6 m/sp₁ = 306 kg⋅m/sFor Nancy:p₂ = m₂v₂p₂ = 50 kg x 6 m/sp₂ = 300 kg⋅m/sNow we can use the Law of Conservation of Momentum to find their velocities after the collision. We know that the total momentum before the collision must be equal to the total momentum after the collision:p₁ + p₂ = p₁' + p₂'306 kg⋅m/s + 300 kg⋅m/s = 51 kg x V₁' + 50 kg x V₂'606 kg⋅m/s = 51 kg x V₁' + 50 kg x V₂'Now we can solve for V₁':V₁' = (606 kg⋅m/s - 50 kg x V₂') / 51 kg

Now we can use Newton's Second Law of Motion to find the force that Nancy exerts on Tonya. This law states that the force acting on an object is equal to its mass times its acceleration:F = maWhere:F is force (in Newtons)m is mass (in kilograms)a is acceleration (in meters per second squared)We know that Tonya exerts a force of 910 N on Nancy, so the force that Nancy exerts on Tonya must be equal in magnitude but opposite in direction. Therefore:F₂ = -910 N Now we need to find the acceleration of each skater. We can use the following formula:a = (V₁' - V₁) / t where: a is acceleration (in m/s²)V₁' is the final velocity of TonyaV₁ is the initial velocity of Tonyat is the time it takes for the collision to occurWe know that V₁ = 6 m/s and we can assume that t is very small, so we can say that V₁' = 0 m/s. Therefore:a = (0 - 6 m/s) / t = -6/tWe can use the same formula to find the acceleration of Nancy. We know that V₂ = 6 m/s and V₂' = 0 m/s, so:a = (0 - 6 m/s) / t = -6/tNow we can use Newton's Second Law of Motion to solve for F₁:F₁ = m₁a₁F₁ = 51 kg x (-6/t)F₁ = -306/tWe know that F₁ + F₂ = 0, so:-306/t + (-910) = 0-306/t = 910t = 306/910t = 0.33626 seconds Now we can solve for F₁:F₁ = -306/tF₁ = -306 / 0.33626F₁ = -910 N Therefore, the force that Nancy exerts on Tonya is -910 Newtons.

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According to the question the magnitude of q is approximately 1.24 × 10^-5 C.

The equation that describes the electric force between two charges is F=kq1q2/r2
where F is the force between the charges, k is Coulomb's constant (k = 8.9875 × 109 N · m2/C2), q1 and q2 are the magnitudes of the charges, and r is the distance between the centers of the charges.
In the given figure, two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L.
Assume that θ is so small that tanθ can be replaced by its approximate equal, sinθ. If L=120 cm, m=12 g, and x=6.7 cm, we need to find the magnitude of q.
Let's consider the forces acting on the balls. The gravitational force acting on each ball is mg, where g is the acceleration due to gravity.
There are also electric forces acting on each ball due to the charges. Since the balls are identical, the magnitudes of these forces are the same. Let this force be F.
So, the net force acting on each ball is given by
Fnet = mg - F
For a ball to hang in equilibrium, the net force acting on it must be zero.
Therefore,Fnet = 0 => mg = F
Substituting the given values,
m*g = k*q^2/x^2
Solving for q, we get
q = √(m*g*x^2/k)
Substituting the given values, we get
q = √(0.012 kg × 9.8 m/s^2 × (0.067 m)^2/(8.9875 × 10^9 N · m^2/C^2))≈ 1.24 × 10^-5 C
Therefore, the magnitude of q is approximately 1.24 × 10^-5 C.

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At r = 0, the electric field is 0 N/C (inside the sphere).

At r = 9.2 cm, the electric field is approximately 51.2 N/C.

At r = 18.4 cm, the electric field is approximately 12.8 N/C.

At r = 27.6 cm, the electric field is approximately 7.10 N/C.

To find the magnitude of the electric field at different distances from a uniformly charged spherical distribution, we can use the formula for the electric field at a point outside the sphere:

E = (k * Q) / r^2,

where E is the electric field,

k is the electrostatic constant (approximately 8.99 x 10^9 N m^2/C^2),

Q is the total charge of the sphere, and

r is the distance from the center of the sphere.

Given that the total charge of the spherical distribution is 42.3 mC (milliCoulombs) and the radius is 18.4 cm, we need to convert the charge and radius to standard SI units (Coulombs and meters).

Total charge, Q = 42.3 mC = 42.3 x 10^(-3) C.

Radius, r = 18.4 cm = 18.4 x 10^(-2) m.

Now, we can calculate the electric field at different distances:

At r = 0:

E = (k * Q) / r^2 = (8.99 x 10^9 N m^2/C^2 * 42.3 x 10^(-3) C) / (0)^2.

The denominator is 0, which makes the electric field undefined. However, it is important to note that the electric field inside a uniformly charged sphere is zero. Therefore, at the center of the sphere (r = 0), the electric field is indeed zero.

At r = 9.2 cm = 9.2 x 10^(-2) m:

E = (8.99 x 10^9 N m^2/C^2 * 42.3 x 10^(-3) C) / (9.2 x 10^(-2) m)^2.

Calculating the expression:

E ≈ 51.2 N/C.

Therefore, at a distance of 9.2 cm from the center of the sphere, the magnitude of the electric field is approximately 51.2 N/C.

At r = 18.4 cm = 18.4 x 10^(-2) m:

E = (8.99 x 10^9 N m^2/C^2 * 42.3 x 10^(-3) C) / (18.4 x 10^(-2) m)^2.

Calculating the expression:

E ≈ 12.8 N/C.

Therefore, at a distance of 18.4 cm from the center of the sphere, the magnitude of the electric field is approximately 12.8 N/C.

At r = 27.6 cm = 27.6 x 10^(-2) m:

E = (8.99 x 10^9 N m^2/C^2 * 42.3 x 10^(-3) C) / (27.6 x 10^(-2) m)^2.

Calculating the expression:

E ≈ 7.10 N/C.

Therefore, at a distance of 27.6 cm from the center of the sphere, the magnitude of the electric field is approximately 7.10 N/C.

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Heat is 87.89°F

Airflow rate calculation:

Given: Cross-sectional area of the duct (A) = 2 sq ft, Velocity of the air (V) = 300 ft/min

Airflow rate (Q) = A * V = 2 * 300 = 600 cfm (cubic ft/min)

Heat added to the air calculation:

Given: Mass flow rate (m) = 0.083 lb/ft³, Temperature differential (t2 - t1) = 5°F, Specific heat (q) = 0.24 Btu/lb.°F

Mass flow rate (m) = (0.083 * 2 * 300) = 49.8 lb/hr

Heat added to the air (Q) = m * A * (t2 - t1) = 49.8 * 5 * 0.24 = 59.76 BTU/hr

Wattage input to the heater:

Given: Voltage (V) = 110 volts

Wattage input to the heater (W) = V * A = 2 * 110 = 220 watts

Conversion of heat added to watts:

Conversion: 1 kW = 3412 BTU/hr, 1 hour = 3600 seconds

Heat added to the air in watts = 59.76 * (1 kW/3412 BTU/hr) * 3600 sec/hr = 0.052 kW

Heat added per unit mass of air calculation:

Heat added per unit mass of air (q) = Q/m = 59.76/49.8 = 1.2 BTU/lb.°F

Calculation of temperature difference at different velocities:

For V = 0 ft/min:

Heat added (Q) = 0 cfm, Power input (W) = 0 watts, Heat per unit mass (q) = 0.69 Btu/lb.°F

Temperature difference (t2 - t1) = W/[(m)(q)] = 0/[(49.8)(0.69)] = 0°F

For V = 150 ft/min:

Heat added (Q) = 300 cfm, Power input (W) = 330 watts, Heat per unit mass (q) = 0.69 Btu/lb.°F

Temperature difference (t2 - t1) = W/[(m)(q)] = 330/[(49.8)(0.69)] = 8.94°F

t2 = t1 + (t2 - t1) = 70 + 8.94 = 78.94°F

For V = 300 ft/min:

Heat added (Q) = 600 cfm, Power input (W) = 660 watts, Heat per unit mass (q) = 0.69 Btu/lb.°F

Temperature difference (t2 - t1) = W/[(m)(q)] = 660/[(49.8)(0.69)] = 17.89°F

t2 = t1 + (t2 - t1) = 70 + 17.89 = 87.89°F

The graph of Temperature difference vs. Velocity at constant amperage of 3 is shown below:The increase in temperature difference is directly proportional to the increase in velocity.

As the airflow rate increases, so does the temperature difference across the heater. Since we're using a constant amperage of 3, the power input to the heater remains constant.

As a result, the rate of heat transfer remains constant. The temperature difference between the air and the heater will increase as the rate of airflow increases because more air is being heated over the same period. The accuracy of the system is affected by several factors such as the thermocouple's accuracy, heat losses, etc.

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As given, the power radiated by the cell phone is 1.12 W. The distance between the phone and point where the intensity is to be calculated is 3 cm.

Now, the average intensity of the waves can be calculated as:

Average Intensity = Power radiated per unit area= Power radiated / (4πr²)

Where, r is the distance between the phone and point where the intensity is to be calculated.

Thus, the average intensity of the radio waves is 0.83 W/m² (approximately).The peak intensity of the wave is four times the average intensity. Thus, the peak intensity of the wave is 3.32 W/m².The peak electric field strength can be calculated as:

E₀ = √(2I/μ₀c)

Where, I is the average intensity, c is the speed of light and μ₀ is the permeability of free space.

Thus, substituting the values of I, c and μ₀, we get:

E₀ = 9.4×10⁻⁴ N/C (approximately).The peak magnetic field strength can be calculated as:

B₀ = √(2I/ε₀c)Where, ε₀ is the permittivity of free space.

Thus, substituting the values of I, c and ε₀, we get:

B₀ = 2.4×10⁻⁵ T (approximately).

Therefore, the average intensity of these waves is 0.83 W/m², the peak intensity of the wave at this distance is 3.32 W/m², the peak electric field strength E₀ in these waves is 9.4×10⁻⁴ N/C and the peak magnetic field strength B₀ in these waves is 2.4×10⁻⁵ T.

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The speed of the river current is approximately 0.805 m/s, the swimmer's speed relative to the shore is approximately 1.62 m/s, and the swimmer should aim at an angle of approximately 30.18 degrees relative to a direct line across the river to arrive directly opposite her starting point.

To solve this problem, we can use the concept of vector addition and trigonometry.

Swimmer's maximum speed relative to the water (v_swimmer) = 1.40 m/s

Displacement downstream (d_downstream) = 46.0 m

Width of the river (d_river) = 80.0 m

To find the speed of the river current (v_current), we can use the equation:

v_current = d_downstream / t

where t is the time taken by the swimmer to cross the river.

To find the time taken by the swimmer, we can use the equation:

t = d_river / v_swimmer

Substituting the given values, we can calculate the time:

t = 80.0 m / 1.40 m/s ≈ 57.14 s

Now, we can calculate the speed of the river current:

v_current = 46.0 m / 57.14 s ≈ 0.805 m/s

The speed of the river current is approximately 0.805 m/s.

To find the swimmer's speed relative to the shore, we can use the Pythagorean theorem. The swimmer's speed relative to the shore (v_shore) can be calculated as:

v_shore = √(v_swimmer² + v_current²)

Substituting the given values, we can calculate the speed relative to the shore:

v_shore = √(1.40² + 0.805²) ≈ 1.62 m/s

The swimmer's speed relative to the shore is approximately 1.62 m/s.

To determine the direction the swimmer should aim to arrive directly opposite her starting point, we can use trigonometry. The angle (θ) can be calculated as:

θ = tan^(-1)(v_current / v_swimmer)

Substituting the given values, we can calculate the angle:

θ = tan^(-1)(0.805 / 1.40) ≈ 30.18 degrees

Therefore, the swimmer should aim at an angle of approximately 30.18 degrees relative to a direct line across the river to arrive at the point directly opposite her starting point.

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To solve this problem, we'll use the kinematic equations of motion.

We'll assume that the acceleration due to gravity is approximately 9.8 m/s².

(a) To calculate the speed of the rock at launch, we can use the equation:

v = u + at,

where:

v = final velocity (speed)

u = initial velocity (speed)

a = acceleration

t = time

Since the rock is launched vertically, the final velocity at the peak of its trajectory is 0 m/s. Therefore, the equation becomes:

0 = u - 9.8 * 2.2.

Solving for u:

u = 9.8 * 2.2

u ≈ 21.56 m/s.

Therefore, the speed of the rock at launch is approximately 21.56 m/s.

(b) To calculate the speed of the rock 5.45 s after launch, we can again use the equation:

v = u + at.

Given that t = 5.45 s and a = -9.8 m/s² (negative because it opposes the direction of motion), we substitute these values into the equation:

v = 21.56 - 9.8 * 5.45.

Calculating:

v ≈ 21.56 - 53.51

v ≈ -31.95 m/s.

Therefore, the speed of the rock 5.45 seconds after launch is approximately 31.95 m/s in the downward direction.

Note: The negative sign indicates that the rock is moving downward after reaching its peak and starting to descend.

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The net force on the point charge at x=8 cm is -15.88N to the left.

The force acting on a point charge by another point charge is given by Coulomb's law, which states that the force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

If two charges have the same sign, the force will be repulsive, and if they have opposite charges, the force will be attractive.

So, in the given charge configuration to the right, the net force (magnitude and direction) on the point charge at x=11 cm is given as follows:

At point x = 11 cm, there is a 6 μC charge 5 cm to the left and a 4 μC charge 3 cm to the right. So, the distance between the 2 charges will be 8 cm. Therefore, the net force (magnitude and direction) on the point charge at x=11 cm is given by:

[tex]����=��1�2�2=��(�1(�1−�)2+�2(�−�2)2)F net​[/tex]

[tex]= r 2 kq 1​ q 2​ ​ =kq( (x 1​ −x) 2 q 1​ ​ + (x−x 2​ ) 2 q 2​ ​ )����[/tex]

[tex]=9×109⋅2��⋅6��(0.05+0.11)2−9×109⋅2��⋅4��(0.11−0.08)2[/tex]

[tex]=15.88�,to the rightF net​ = (0.05+0.11) 2 9×10 9 ⋅2μC⋅6μC​ − (0.11−0.08) 2 9×10 9 ⋅2μC⋅4μC​ =15.88N,to the right[/tex]

Where k = 9 x 10^9 Nm^2/C^2 is Coulomb's constant; q1 = 6 μC; q2 = 4 μC; x1 = 0.05 m and x2 = 0.08 m.

Next, let's find the net force on the point charge at x=8 cm:

At point x = 8 cm, there is a 4 μC charge 3 cm to the left and a 6 μC charge 5 cm to the right. So, the distance between the 2 charges will be 8 cm. Therefore, the net force (magnitude and direction) on the point charge at x=8 cm is given by:

[tex]����=��1�2�2=��(�1(�1−�)2+�2(�−�2)2)F net​[/tex]

[tex]= r 2 kq 1​ q 2​ ​ =kq( (x 1​ −x) 2 q 1​ ​ + (x−x 2​ ) 2 q 2​ ​ )����[/tex]

[tex]=9×109⋅2��⋅4��(0.08−0.05)2−9×109⋅2��⋅6��(0.05+0.08)2[/tex]

[tex]=−15.88�,to the leftF net​ = (0.08−0.05) 2 9×10 9 ⋅2μC⋅4μC​ − (0.05+0.08) 2 9×10 9 ⋅2μC⋅6μC​ =−15.88N,to the left[/tex]

Where k = 9 x 10^9 Nm^2/C^2 is Coulomb's constant; q1 = 4 μC; q2 = 6 μC; x1 = 0.05 m and x2 = 0.08 m.

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a) The mass of the star is approximately [tex]1.57 * 10^{30}[/tex] kg.

b) The orbital period of the faster planet is approximately 10.74 years.

To solve this problem, we can use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

The formula for Kepler's Third Law can be expressed as:

[tex]T^2 = (\frac{4\pi^2}{GM}) * a^3[/tex]

Where:

T is the orbital period of the planet,

G is the gravitational constant,

M is the mass of the star,

a is the semi-major axis of the planet's orbit.

(a) To find the mass of the star, we can use the slower planet's orbital period and speed.

Given:

T₁ = 8.21 years (orbital period of slower planet)

v₁ = 46.8 km/s (orbital speed of slower planet)

Using the formula for the orbital speed of a planet:

v = 2πa / T

We can rearrange the equation to solve for the semi-major axis a:

a = (vT) / (2π)

Substituting the values for v₁ and T₁:

a = (46.8 km/s * 8.21 years) / (2π) = 150.029 km

Now, we can use the slower planet's semi-major axis to find the mass of the star (M).

Using Kepler's Third Law, we have:

[tex]T_1^2 = (\frac{4\pi^2}{GM}) * a_1^3[/tex]

Simplifying the equation and solving for M:

[tex]M = (\frac{4\pi^2}{G}) * (\frac{a_1^3}{T_1^2})[/tex]

Substituting the known values:

[tex]M = (\frac{4\pi^2}{G}) * \frac{(150.029)^3}{(8.21)^2}[/tex]

Now we need to convert the units to SI units:

1 km = 1000 m, 1 year = 365.25 days, and G = 6.67430 × 10^¹¹ m³ kg⁻¹ s⁻²

[tex]M = (\frac{4\pi^2}{G}) * \frac{(150029000)^3}{(8.21 * 365.25 * 24 * 3600)^2} \approx 1.57 * 10^{30}[/tex]

Therefore, the mass of the star is approximately [tex]1.57 * 10^{30}[/tex] kg.

(b) To find the orbital period of the faster planet, we can use the mass of the star we just calculated and the given orbital speed of the faster planet.

Given:

v₂ = 61.6 km/s (orbital speed of faster planet)

Using the same method as before, we can calculate the semi-major axis a₂ of the faster planet's orbit using its orbital speed:

a₂ = (v₂T₁) / (2π)

Substituting the values:

a₂ = (61.6 km/s * 8.21 years) / (2π) ≈ 198.494 km

Now, we can use Kepler's Third Law to find the orbital period of the faster planet (T₂):

[tex]T_2^2 = (\frac{4\pi^2}{GM}) * a_2^3[/tex]

Solving for T₂:

[tex]T_2 = \sqrt{(\frac{4\pi^2}{GM}) * a_2^3}[/tex]

Substituting the known values:

[tex]T_2 = \sqrt{(\frac{4\pi^2}{G}) * \frac{(198.494)^3}{M}}[/tex]

Converting the units to SI units:

[tex]T_2 = \sqrt{\frac{4\pi^2}{G} * \frac{(198494000)^3}{(1.57 * 10^{30})}}[/tex]

Calculating the value:

T₂ ≈ 10.74 years

Therefore, the orbital period of the faster planet is approximately 10.74 years.

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A person who weighs 120 pounds on earth would weigh approximately 20 pounds on the moon. This is because the force of gravity is weaker on the moon due to its lower mass and smaller size. The radius of the moon is about 1,737 km (1,080 miles) while its mass is approximately 7.342 × 10²² kg (81 billion tons).

Using this data, we can calculate the acceleration of gravity on the moon's surface which is about 1.62 m/s².A person who weighs 120 pounds on earth would weigh approximately 20 pounds on the moon.

Here's how to calculate the acceleration of gravity on the moon's surface:G = GM / R² where G is the acceleration of gravity, M is the mass of the moon, and R is the radius of the moon.

We know that M = 7.342 × 10²² kg and R = 1,737 km = 1,737,000 meters so we can plug in these values to find G.G = (6.67 × 10⁻¹¹ N(m/kg)²) (7.342 × 10²² kg) / (1,737,000 m)²G = 1.62 m/s²

Therefore, the acceleration of gravity on the moon's surface is 1.62 m/s².

A person who weighs 120 pounds on earth would weigh approximately 20 pounds on the moon. This is because the force of gravity is weaker on the moon due to its lower mass and smaller size.

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a) The required expression for the magnitude of the force between the plates is ∣F∣ = 1/2 q²/ε₀A².

b) The magnitude of the force between the plates is 4.83 × 10⁻⁵ N.

c)  The required expression for the stress on either plate is ∣F∣/A = q²/ε₀A³.

d) The stress is 3.05 × 10⁸ N/m² or 305 MPa.

Given relationship is, ∣F∣=dU/dx

Where, F is the force between the plates,

U is the potential energy of the system,

x is the separation of the plates of a parallel plate capacitor.

(a) The potential energy of a parallel-plate capacitor can be written as

U=1/2 q²/ε₀A

Where, q is the charge of the capacitor,

ε₀ is the permittivity of the free space

A is the area of the plates.

Substituting the value of U in the given relationship, we get:

∣F∣ = dU/dx∣F∣ = d/dx [1/2 q²/ε₀A]

∣F∣ = 1/2 q²/ε₀A² ...[1]

This is the required expression for the magnitude of the force between the plates.

(b) Substituting the given values in equation [1],

q = 6.00μ

C = 6.00 × 10⁻⁶ A s

A = 2.50 cm² = 2.50 × 10⁻⁴ m²ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²

Thus, ∣F∣ = 1/2 × (6.00 × 10⁻⁶)²/8.85 × 10⁻¹² × (2.50 × 10⁻⁴)²

∣F∣ = 4.83 × 10⁻⁵ N

Therefore, the magnitude of the force between the plates is 4.83 × 10⁻⁵ N.

(c) The electric field between the plates can be expressed as,

E = σ/ε₀

Where,σ is the surface charge density of the plates.

Substituting σ = q/A, we get,

E = q/ε₀A

We know, force per unit area is electrostatic stress and is expressed as,∣F∣/A = ε₀E²

Thus, ∣F∣/A = ε₀(q/ε₀A)²

∣F∣/A = q²/ε₀A³

This is the required expression for the stress on either plate.

(d) Substituting the given values in the above equation, we get,

q = 110 Vx = 2.00 mm = 2.00 × 10⁻³ m

Thus, ∣F∣/A = (110)²/8.85 × 10⁻¹² × (2.50 × 10⁻⁴)³

∣F∣/A = 3.05 × 10⁸ N m⁻²

Therefore, the stress is 3.05 × 10⁸ N/m² or 305 MPa.

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The air resistance force will balance the force of gravity, resulting in zero net force and zero acceleration. At this point, the projectile reaches terminal , where its velocity remains constant.

Set up a projectile launcher with a fixed initial velocity.

Measure and record the range (horizontal distance traveled) of the projectile for different launch angles.

Keep the other factors, such as initial velocity and projectile mass, constant throughout the experiment.

Vary the launch angle systematically, covering a range of angles from 0° to 90°.

Repeat each launch angle multiple times to ensure consistency and accuracy of measurements.

Plot the range of the projectile as a function of the launch angle.

Prediction and Reasoning:

Changing the initial launch angle will affect the path of the projectile.

For a fixed initial velocity, increasing the launch angle (up to 45°) will result in an increased range because the vertical component of the initial velocity contributes more to the time of flight, allowing the projectile to stay in the air longer and cover a greater horizontal distance.

However, if the launch angle exceeds 45° and approaches 90°, the range will decrease because the vertical component of the initial velocity becomes dominant, causing the projectile to spend more time in the air but covering a shorter horizontal distance.

Factors affecting range with air resistance:

Air resistance has a significant effect on the range of a projectile.

Factors such as projectile shape, cross-sectional area, and mass will impact the range when air resistance is present.

Increased air resistance will cause the projectile to experience more drag, slowing it down and reducing its range.

Behavior of velocity and acceleration vectors with air resistance:

With air resistance, the velocity vector of the projectile will gradually decrease in magnitude over time.

The acceleration vector will include both the force due to gravity and the opposing force of air resistance.

As the projectile moves, the acceleration vector will be directed downward due to gravity and opposing the direction of motion due to air resistance.

Spacing of black dots on the projectile's path:

The closer spacing of black dots near the top of the projectile's path indicates a slower horizontal velocity at that point.

As the projectile descends, it accelerates due to gravity and its horizontal velocity increases, causing the dots to be further apart.

Terminal velocity situation:

To create a situation where the projectile reaches terminal velocity, consider a projectile with a large surface area, such as a flat sheet.

Initially, the projectile will experience acceleration due to gravity, and its velocity will increase.

As the projectile gains speed, the air resistance force acting on it will also increase.

Eventually, the air resistance force will balance the force of gravity, resulting in zero net force and zero acceleration. At this point, the projectile reaches terminal velocity, where its velocity remains constant.

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In the given question, a particle having a charge of \(+2.1\ \mu C\) moves from point A to point B, which is a distance of 0.21 m. The particle experiences a constant electric force and its motion is... The electric force that is exerted on a particle is proportional to the charge of the particle and the magnitude of the electric field acting on it.

Mathematically, the electric force F on a particle with charge q is given by[tex]F = qE[/tex] where E is the electric field and q is the charge on the particle. In the given question, the force on the particle is constant. Therefore, the electric field is constant. The distance that the particle moves is also given. Therefore, we can use the formula for the work done by a constant force, which is given by [tex]W = Fdcosθ[/tex]

where d is the distance moved by the particle, θ is the angle between the direction of the force and the direction of motion of the particle, and F is the force acting on the particle.

In this case, the force is in the direction of the motion of the particle, so the angle θ between the force and the direction of motion is 0.

Therefore, cos θ = 1 and W = Fd

Substituting the values given in the question, we get

[tex]W = (2.1 × 10-6 C)(0.21 m)[/tex]

W = 4.41 × 10-7 J

Therefore, the work done on the particle is 4.41 × 10-7 J.

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(a) The acceleration of the car is 6.67 mi/hr/s.

To find the acceleration, we can use the equation:

acceleration = (change in velocity) / time

The change in velocity is the final velocity minus the initial velocity, which is 70 mi/hr - 30 mi/hr = 40 mi/hr. The time is given as 6 seconds. Converting the velocities to the consistent unit of miles per hour per second, we have:

acceleration = (40 mi/hr) / (6 s) = 6.67 mi/hr/s.

Therefore, the acceleration of the car is 6.67 mi/hr/s.

(b) It takes approximately 500 seconds for light to reach the Earth from the Sun.

To calculate the time it takes for light to reach the Earth, we need to consider the distance between the two and the speed of light. The distance from the Earth to the Sun is given as 1.5 * 10^8 km, which can be converted to meters as 1.5 * 10^11 m. The speed of light is 3.0 * 10^8 m/s.

Using the equation:

time = distance / speed

Plugging in the values, we have:

time = (1.5 * 10^11 m) / (3.0 * 10^8 m/s) = 5.0 * 10^2 s.

Therefore, it takes approximately 500 seconds for light to reach the Earth from the Sun.

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