the planet that has the lowest density is _____ while the planet that is tilted 90° to the orbital plane is _____.

The 2018 Ford Explorer gets 22 miles per gallon (mpg), slightly better gas mileage than previous year’s models. In the case of the 2018 Ford Explorer, engineers and designers likely implemented design improvements and optimizations to enhance its fuel economy.

The 2018 Ford Explorer achieves a fuel efficiency of 22 miles per gallon (mpg), which represents a slight improvement compared to the gas mileage of previous year's models. This measurement indicates the distance in miles that the vehicle can travel on one gallon of fuel. With 22 mpg, the 2018 Explorer demonstrates enhanced fuel economy, which can be attributed to various factors such as advancements in engine technology, aerodynamics, and efficiency optimizations. These may include the use of lightweight materials to reduce vehicle weight, aerodynamic enhancements to reduce drag, and engine advancements such as improved combustion efficiency and transmission optimization. Collectively, these efforts contribute to the slight increase in gas mileage compared to previous models.

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The q3q3 is located at approximately x = -0.119 m on the x-axis. when the net force on q1q1 is 7.00 N.

Given:

Charge q1 = +3.00 μC at the origin (x = 0 m).

Charge q2 = -5.00 μC at x = 0.200 mm = 0.0002 m.

Charge q3 = -8.00 μC (location unknown).

We need to determine the location of q3 such that the net force on q1 is 7.00 N in the -x direction.

The force between two charges can be calculated using Coulomb's law:

F = k * |q1 * q2| / r^2

Where:

F is the force between the charges.

k is Coulomb's constant, approximately 8.99 × 10^9 N m^2/C^2.

|q1| and |q2| are the magnitudes of the charges.

r is the distance between the charges.

Let's first calculate the force between q1 and q2. Since q1 and q2 have opposite charges, the force will be attractive:

F12 = k * |q1 * q2| / r12^2

Substituting the given values:

F12 = (8.99 × 10^9 N m^2/C^2) * |3.00 × 10^-6 C| * |-5.00 × 10^-6 C| / (0.0002 m)^2

F12 = -0.67425 N

The negative sign indicates that the force is in the -x direction.

Now, let's consider the force between q1 and q3. The net force on q1 is given as 7.00 N in the -x direction. Therefore, the force between q1 and q3 should be:

F13 = -7.00 N - F12

Substituting the values:

-7.00 N = -7.00 N - (-0.67425 N)

-7.00 N = -7.00 N + 0.67425 N

-7.00 N = -6.32575 N

The force between q1 and q3 is approximately -6.32575 N.

We can calculate the distance between q1 and q3 using the formula for force:

F13 = k * |q1 * q3| / r13^2

Substituting the known values:

-6.32575 N = (8.99 × 10^9 N m^2/C^2) * |3.00 × 10^-6 C| * |-8.00 × 10^-6 C| / r13^2

Simplifying the equation:

r13^2 = (8.99 × 10^9 N m^2/C^2) * |3.00 × 10^-6 C| * |-8.00 × 10^-6 C| / -6.32575 N

r13^2 = 0.4048 m^2

Taking the square root of both sides:

r13 = √0.4048 m^2

r13 ≈ 0.6367 m

The distance between q1 and q3 is approximately 0.6367 m.

Since q3 has a negative charge and the net force on q1 is in the -x direction, q3 must be located to the left of q1. Therefore, the position of q3 is approximately x = -0.6367 m.

The q3q3 is located at approximately x = -0.119 m when the net force on q1q1 is 7.00 N.

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Answer:

When you go bald

Explanation:

(a) The magnitude of the impulse experienced by the soccer ball during the bounce is 0.6923 kg·m/s, (b) the magnitude of the constant force acting on the soccer ball during the bounce is 13.846 N and (c) the soccer ball transferred approximately 12.026 joules of energy to the environment during the bounce.

What is energy and how is it measured?

Energy is a fundamental concept in physics that refers to the ability of a system to do work or cause a change. It is a scalar quantity and is associated with various forms, such as kinetic energy, potential energy, thermal energy, and others.

The SI unit of energy is the joule (J).

Part (a):

The magnitude of the impulse (J) experienced by the soccer ball during the bounce can be calculated using the equation:

J = Δp,

where Δp is the change in momentum.

The change in momentum is given by:

Δp = m * Δv,

where m is the mass of the soccer ball and Δv is the change in velocity.

The initial velocity of the soccer ball is zero as it is dropped from rest. The final velocity can be calculated using the equation for final velocity in free fall:

v² = u² + 2gh,

where v is the final velocity, u is the initial velocity (zero in this case), g is the acceleration due to gravity, and h is the height.

Calculating the final velocity:

v²  = 0² + 2 * 9.8 m/s² * 2.12 m,

v ≈ 6.02 m/s.

Substituting the values into the equation for change in momentum:

Δp = m * (v - u),

Δp = 0.115 kg * (6.02 m/s - 0 m/s).

Calculating:

Δp ≈ 0.6923 kg·m/s.

Therefore, the magnitude of the impulse experienced by the soccer ball during the bounce is approximately 0.6923 kg·m/s.

Part (b):

The magnitude of the constant force (F) acting on the soccer ball can be calculated using the equation:

F = Δp / Δt,

where Δp is the change in momentum and Δt is the time interval.

Given that the soccer ball was in contact with the ground for Δt = 0.05 s, we can substitute the values into the equation:

F = 0.6923 kg·m/s / 0.05 s.

Calculating:

F = 13.846 N.

Therefore, the magnitude of the constant force acting on the soccer ball during the bounce is 13.846 N.

Part (c):

The energy transferred to the environment during the bounce can be calculated as the work done by the force of the ball on the ground.

The work done is given by:

W = F * d,

where F is the magnitude of the force and d is the distance over which the force acts.

In this case, the force acts over the distance between the initial and final heights, which is h₁ - h₂.

Substituting the values:

W = 13.846 N * (3.05 m - 2.12 m).

Calculating:

W ≈ 12.026 J.

Therefore, the soccer ball transferred approximately 12.026 joules of energy to the environment during the bounce.

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(a) Yes, it is possible for a large force to produce only a small or zero torque when the line of action of the force does not create a moment arm or when the force is applied directly through the axis of rotation.

(b) Yes, it is possible for a small force to produce a large torque when the force is applied at a greater distance from the axis of rotation, creating a larger moment arm.

Torque is the rotational equivalent of force and is calculated by multiplying the force by the distance from the axis of rotation. If the line of action of the force passes through the axis of rotation, the moment arm becomes zero, resulting in no torque being generated. This occurs when the force is applied directly on the axis or when the force is balanced by an equal and opposite force that cancels out the rotational effect.

Similarly, even if the force is not directly on the axis of rotation, if the moment arm is very small, the torque produced will also be small. The moment arm is the perpendicular distance between the axis of rotation and the line of action of the force. If the force is applied very close to the axis, the moment arm will be small, resulting in a smaller torque.

If a small force is applied at a considerable distance from the axis of rotation, the moment arm becomes larger, resulting in a larger torque. This is similar to using a wrench or a long lever to apply a small force to loosen a tight bolt. The length of the wrench or lever increases the moment arm, allowing a small force to produce a large torque.

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The angle at which the second-order maximum is produced in the diffraction grating is 36.97°.

Angle at which the first-order maximum is produced, θ₁ = 17.5°

The equation for the diffraction grating is given by,

nλ = 2d sinθ

Given that the same light is used for both diffraction grating procedures. The equation for wavelength can be given as,

λ = 2d sinθ/n

So, we can say that,

λ₁ = λ₂

2d sinθ₁/n₁ = 2d sinθ₂/n₂

sinθ₁/n₁ = sinθ₂/n₂

So,

sinθ₂ = n₂sinθ₁/n₁

sinθ₂ = 2 x sin(17.5)/1

sinθ₂ = 2 x 0.30071

sinθ₂ = 0.60142

Therefore, the angle at which the second-order maximum is produced is given by,

θ₂ = sin⁻¹(0.60142)

θ₂ = 36.97°

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The decrease in wave speed in layer B can be explained by the change in the properties of the medium through which the wave is propagating. Generally, the speed of a wave depends on the properties of the medium it is traveling through, such as the density and elasticity.

There are several factors that can lead to a decrease in wave speed in layer B:

Change in Density: If the density of the medium increases in layer B compared to layer A, it will result in a decrease in wave speed. This is because a denser medium tends to slow down the propagation of waves.

Change in Elasticity: If the elasticity (or stiffness) of the medium decreases in layer B compared to layer A, it can cause a decrease in wave speed. A less elastic medium offers more resistance to wave propagation, resulting in slower wave speed.

Change in Temperature: In some cases, temperature variations can affect the properties of the medium. For example, in the case of sound waves, as temperature increases, the speed of sound generally increases due to an increase in the elasticity and average kinetic energy of the molecules. Conversely, a decrease in temperature can lower the wave speed.

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For an otto cycle:

a) Highest temperature and pressure are 1000K and 2.5 MPa.

b) amount of heat transferred, 150.2 kJ/kg

c) thermal efficiency, 56.5%

d) mean effective pressure is 1.31 MPa.

Given:

Initial conditions T1 = 27C = 300K, P1 = 100 kPa = 100 × 10³ Pa, V1 = 500 cm³ = 500 × 10⁻⁶ m³

Compression ratio (r) = V1/V2 = 10

End of isentropic expansion T3 = 1000 K

Specific heat ratio (k) = 1.4

Specific heat at constant pressure (cp) = 1.005 kJ/kg.K = 1005 J/kg.K

Specific heat at constant volume (cv) = 0.718 kJ/kg.K = 718 J/kg.K

Gas constant (R) = 0.287 kJ/kg.K = 287 J/kg.K

a) Highest temperature and pressure in the cycle:

At the end of the isentropic compression (point 2), we use the relation T2 = T1 × (V1/V2)^(k-1)

⇒ T2 = 300 K × 10^(1.4-1) = 300 K × 10^0.4 = 509 K

The pressure at the end of the compression stroke (point 2) is given by P2 = P1 × (V1/V2)^k

⇒ P2 = 100 × 10³ Pa × 10^1.4 = 2.5 MPa

The maximum temperature T3 is given in the problem as 1000K.

The maximum pressure in the cycle is the pressure at point 2, P2 = 2.5 MPa.

b) Amount of heat transferred:

The heat input is during the constant volume process 2-3, given by Q_in = m × cv × (T3 - T2)

But we do not have the mass (m) of the gas, we can calculate the change in internal energy per unit mass as ΔU = cv × (T3 - T2) = 718 J/kg.K × (1000K - 509K) = 352.6 kJ/kg

The heat rejected is during the constant volume process 4-1, given by Q_out = m × cv × (T4 - T1)

Using the adiabatic process, we know that T4 = T1 × (V2/V1)^(k-1) = 300 K × 10^0.4 = 509 K

ΔU = cv × (T4 - T1) = 718 J/kg.K × (509K - 300K) = 150.2 kJ/kg

c) Thermal efficiency:

The thermal efficiency of an Otto cycle is given by η = 1 - 1/(r^(k-1))

⇒ η = 1 - 1/(10^0.4) = 0.565 or 56.5%

d) Mean effective pressure (mep):

The thermal efficiency can also be expressed as η = 1 - V2/V1 = mep/(P2 - P1)

⇒ mep = η × (P2 - P1) = 0.565 × (2.5 MPa - 100 kPa) = 1.31 MPa

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Complete question:

An Otto cycle with compression ratio of 10.The air is at 100 kPa,27C,and 500 cm prior to the compression stroke. Temperature at the end of isentropic expansion is 1000 K. Determine the followings

a) Highest temperature and pressure in the cycle b) amount of heat transferred, c) thermal efficiency, and d) mean effective pressure. Use 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 10 points

Answer:

1. sand and water

2. suspension

mark me as brainliest plz

The block's speed after it has traveled 2.0 m up the inclined plane, ignoring friction, is approximately 6.19 m/s.

To determine the block's speed after it has traveled 2.0 m up the inclined plane, we can use the principles of kinematics. We'll consider the initial speed, distance traveled, and the angle of the inclined plane.

Using the kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

Given that the initial speed (u) is 8.0 m/s and the distance traveled (s) is 2.0 m, we need to find the acceleration (a).

The component of gravity acting down the inclined plane is given by:

mg sin(θ)

where m is the mass of the block and θ is the angle of the inclined plane.

Since there is no friction, the net force along the incline is equal to the component of gravity acting down the incline:

ma = mg sin(θ)

Canceling out the mass (m) on both sides:

a = g sin(θ)

Using the known values of the angle of the inclined plane (θ = 32°) and the acceleration due to gravity (g = 9.8 m/s^2):

a = (9.8 m/s^2) sin(32°)

a ≈ 5.27 m/s^2

Now we can substitute the values into the kinematic equation:

v^2 = u^2 + 2as

v^2 = (8.0 m/s)^2 + 2(5.27 m/s^2)(2.0 m)

v^2 ≈ 64.0 m^2/s^2 + 21.08 m^2/s^2

v^2 ≈ 85.08 m^2/s^2

Taking the square root of both sides:

v ≈ √(85.08 m^2/s^2)

v ≈ 9.23 m/s

Therefore, the block's speed after it has traveled 2.0 m up the inclined plane, ignoring friction, is approximately 9.23 m/s.

The block's speed after it has traveled 2.0 m up the inclined plane, ignoring friction, is approximately 9.23 m/s. This calculation is based on the initial speed, distance traveled, and the angle of the inclined plane, using principles of kinematics.

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Answer:

[tex]55.64\ \text{nm}[/tex]

Explanation:

[tex]\lambda[/tex] = Wavelength falling on film = 520 nm

n = Refractive index of film = 2.62

T = Thickness of film

m = Order

We have the relation

[tex]2T=\dfrac{m\lambda}{n}\\\Rightarrow T=\dfrac{m\lambda}{2n}\\\Rightarrow T=\dfrac{m\times 520}{2\times 2.62}\\\Rightarrow T=99.24m[/tex]

The thickness should be greater than 1036 nm. This means [tex]m=11[/tex]

[tex]T=99.24\times 11=1091.64\ \text{nm}[/tex]

Thickness of the film to be added would be

[tex]\Delta T=1091.64-1036=55.64\ \text{nm}[/tex]

Thickness of the film to be added is [tex]55.64\ \text{nm}[/tex].

Answer:

Explanation:

The ray of light is passing from high refractive index medium to low refractive index medium so condition for cancellation of reflected light is as follows .

2μt = (2n+1) λ/2

where μ is refractive index of the medium , t is thickness , λ is wavelength of light and n is a integer .

Putting n = 10

2x 2.62 x t = 21 x 520 / 2 nm

5.24 t = 5460 nm

t = 1042 nm

Thickness required to be added

= 1042 - 1036 = 6 nm .

Answer: The resistance in the circuit is 250 ohms

Explanation:

According to Ohm's law:

[tex]V=IR[/tex]

where V = voltage  = 55 V

I = current in Amperes = 0.22 A

R = Resistance = ?

Putting in the values we get:

[tex]55V=0.22A\times R[/tex]

[tex]R=250ohm[/tex]

Thus the resistance in the circuit is 250 ohms

Answer:

Period = 0.68 seconds

Explanation:

Given the following data;

Number of oscillation = 22

Time = 14.9 seconds

To find the period;

Method I.

Period = time/number of oscillation

Period = 14.9/22

Period = 0.68 seconds.

Method II.

We would find the frequency of the wave;

Frequency = time/number of oscillation

Frequency = 22/14.9

Frequency = 1.48 Hertz

Next, we find the period;

Period = 1/frequency

Period = 1/1.48

Period = 0.68 seconds

Answer:

the answer is B I promise

The force F2 will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod.

To determine which force produces a torque about an axis perpendicular to the plane of the diagram at the left end of the rod, we need to consider the concept of torque and the position of the forces relative to the axis.

Torque is the rotational equivalent of force and is given by the equation: Torque = Force × Distance × sin(θ), where Force is the magnitude of the force, Distance is the perpendicular distance from the axis of rotation to the line of action of the force, and θ is the angle between the force and the lever arm.

In the given diagram, we have two forces acting on the rod, F1 and F2. The lever arm for each force is the distance from the left end of the rod to the line of action of the force.

For force, F1, the line of action passes through the left end of the rod. Therefore, the lever arm is zero, and sin(θ) is also zero since the angle between the force and the lever arm is 0 degrees. Consequently, the torque produced by force F1 is zero.

For force, F2, the line of action is not passing through the left end of the rod. The lever arm for force F2 is the perpendicular distance from the left end of the rod to the line of action of F2. Since this distance is non-zero and the angle between the force and the lever arm is non-zero, both the distance and sin(θ) are non-zero.

Therefore, the torque produced by force F2 is non-zero and will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod.

Out of the two forces pictured, the only force F2 will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod. Force F1 will not produce any torque since its line of action passes through the left end of the rod, resulting in a lever arm of zero.

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Option (A) because a force acts on it , is the correct option .

A space probe in remote outer space continues moving because a force acts on it.

According to Newton's first law of motion, an object will continue to move in a straight line at a constant velocity unless acted upon by an external force. In the case of a space probe in remote outer space, several forces can act on it to maintain its motion.

One of the significant forces at play is gravity. While space is mostly empty, gravitational forces from celestial bodies can still influence the probe's trajectory. If the probe is near a massive object like a planet or a star, the gravitational force exerted by that object can provide the necessary force to keep the probe moving. In this scenario, the probe would move in a curved path around the massive object due to the gravitational force acting as a centripetal force.

Additionally, other forces such as propulsion systems, solar radiation pressure, or gravitational assists from planetary flybys can also act on the space probe, ensuring its continued motion and trajectory adjustments.

A space probe in remote outer space continues moving due to the presence of external forces acting on it. These forces, such as gravity, propulsion systems, solar radiation pressure, or gravitational assists, provide the necessary force to counteract any potential deceleration or deviation from its intended path.

While the probe may move in a curved path due to gravitational forces, it ultimately remains in motion because forces act upon it. Therefore, option A) is the correct choice.

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The image distance from the second lens is approximately 36.4 cm.

What is a lens?

A lens is a transparent optical device that has the ability to refract (bend) and focus light. It consists of a piece of transparent material, such as glass or plastic, that has curved surfaces.

1/f = 1/v - 1/u

Given:

The focal length of each lens (fff) is -16 cm (since it's concave, the focal length is negative).

The lenses are separated by 8.5 cm.

The object distance (u) is 35 cm.

Let's denote the image distance from the first lens as v₁ and the image distance from the second lens as v₂.

From the first lens:

1/f₁ = 1/v₁ - 1/u

Substituting the values:

1/-16 = 1/v₁ - 1/35

Simplifying:

-1/16 = (35 - v₁) / (35v₁)

Cross-multiplying and rearranging:

35v₁ - v₁^2 = -16 * 35

Simplifying further:

v₁^2 - 35v₁ - 560 = 0

We can solve this quadratic equation to find the value of v₁. Using the quadratic formula:

v₁ = (-b ± √(b^2 - 4ac)) / 2a

For the given equation:

a = 1, b = -35, c = -560

v₁ = (-(-35) ± √((-35)^2 - 4 * 1 * -560)) / (2 * 1)

v₁ = (35 ± √(1225 + 2240)) / 2

v₁ = (35 ± √3465) / 2

We take the positive value since v₁ represents a real image. Using a calculator, we find:

v₁ ≈ 44.9 cm (rounded to two significant figures)

Now, we can find the image distance (v₂) from the second lens:

v₂ = v₁ - 8.5 cm

v₂ ≈ 44.9 cm - 8.5 cm

v₂≈ 36.4 cm (rounded to two significant figures)

Therefore, the image distance from the second lens is approximately 36.4 cm.

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The value of the load resistor (R) is approximately 7.47 Ω. The internal resistance of the battery is approximately 2.64 Ω.

To determine the value of the load resistor (R) and the internal resistance of the battery, we can use Ohm's law and the power formula. Let's break down the calculations for each part:

(a) Finding the value of R:

The power (P) delivered to the load resistor can be calculated using the formula P = V²/R, where V is the voltage and R is the resistance. Given that the power delivered is 24.0 W and the voltage is 13.4 V, we can rearrange the formula to solve for R:

R = V²/P = (13.4 V)² / 24.0 W ≈ 7.47 Ω.

(b) Determining the internal resistance of the battery:

The total voltage (V_total) across the battery can be calculated by adding the voltage drop across the load resistor (V_load) to the voltage drop across the internal resistance of the battery (V_internal).

We know that V_total = V_load + V_internal = 13.4 V.

Since V_load = IR (Ohm's law), where I is the current flowing through the circuit, we can substitute I = P/V_load = 24.0 W / 13.4 V.

Substituting these values into the equation, we have 13.4 V = (24.0 W / 13.4 V)R + V_internal.

To solve for V_internal, we rearrange the equation as follows:

V_internal = 13.4 V - (24.0 W / 13.4 V)R.

Substituting the values of V, P, and R, we find:

V_internal ≈ 13.4 V - (24.0 W / 13.4 V)(7.47 Ω) ≈ 2.64 Ω.

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Answer:

since the charges are of different nature it's a attractive force

Explanation:

magnitude of force=

9*10^9*5*10^-6*3*10^-6/0.04

= 3.375N answer

Answer:

because they are underaged and prob dont care and also the gov thinks that the youth cant make a reasonable decision for them selves for sum like that and the media influnces them to by saying whats going on and who supports who

The volume of water in the 250 mL cylinder is approximately 249.975 mL.

To calculate the volume of water, we multiply the density of water by the volume of the cylinder.

Given:

Density of water = 0.9999 g/mL

Volume of the cylinder = 250 mL

The formula for calculating the volume of a substance is:

Volume = Mass / Density

Since we are given the density and we want to find the volume, we rearrange the formula as:

Volume = Mass / Density

The mass can be obtained by multiplying the density by the volume:

Mass = Density * Volume

Substituting the given values:

Mass = 0.9999 g/mL * 250 mL

Simplifying, we get:

Mass = 249.975 g

Therefore, the volume of water in the 250 mL cylinder is approximately 249.975 mL.

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The voltage of the generator in the series RCL circuit at resonance is approximately 122.64 volts.

To determine the voltage of the generator in a series RCL circuit, we need to use the power and current values. In a series RCL circuit at resonance, the power dissipated is equal to the power supplied by the generator.

In this case:

Power dissipated (P) = 65.0 W

Current (I) = 0.530 A

The formula for power in an electrical circuit is:

P = VI

Where:

P is the power in watts

V is the voltage in volts

I is the current in amperes

Rearranging the formula to solve for voltage (V), we get:

V = P / I

Substituting the given values:

V = 65.0 W / 0.530 A

V ≈ 122.64 volts

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Answer:

The resistor has a resistance of 10.667 ohms.

Explanation:

By Ohm's Law, voltage ([tex]V[/tex]), in volts, is directly proportional to the current ([tex]i[/tex]), in amperes, and by definition of power ([tex]\dot W[/tex]), in watts, we have the following formula:

[tex]\dot W = i^{2}\cdot R[/tex] (1)

Where [tex]R[/tex] is the resistance, in ohms.

If we know that [tex]\dot W = 24\,W[/tex] and [tex]i = 1.5\,A[/tex], then the resistance of the resistor is:

[tex]R = \frac{\dot W}{i^{2}}[/tex]

[tex]R = 10.667\,\Omega[/tex]

The resistor has a resistance of 10.667 ohms.

Solution :

As the charge is stationary, hence

[tex]$F_m= qvB \sin \theta$[/tex]

[tex]$F_m=0$[/tex]

Hence, no torque at all.

When the charge is moving in positive x direction and the field will be in the negative y direction outside the bar, then :

[tex]$F = q(V \hat i \times B(- \hat j))$[/tex]

    [tex]$= -qV B (\hat i \times \hat j)$[/tex]

    [tex]$=qVB(- \hat k)$[/tex]

Hence, the force have direction [tex]$(- \hat k)$[/tex].

When instead of charge, an iron nail is used, then there will be induced magnetic field in the soft iron. The nature of the pole induced will be opposite near tot he bar. That is the north pole will be induced near the south pole and vice versa. That is why whichever be the pole of magnet closest to iron will be attracted by iron.

Answer:

B

Explanation:

Answer:

b.

Explanation:

Answer:

Attractive force and r = 0.399 m

Explanation:

One charge is positive and the other charge is negative. Opposite charges attract, so there has to be a force that attracts between them.

q1 = 10.0 x 10^-6 C

q2 = -3 x 10^-6 C

F = 1.7 N

Plug those values into Coulomb's Law:

[tex]F = k\frac{q1q2}{r^{2} } \\1.7 = \frac{(9x10^{9})(10.0 x 10^{-6})(-3 x 10^{-6})}{r^{2} }[/tex]

Solve for r

r = 0.399 m

The image distance from the converging lens is 10.2 cm.

The focal length of the converging lens, f = 28 cm

The distance of the object from the converging lens, u = -16 cm

The optical center or axis of a convergent lens serves as the focal point for light, a lens that generates a real image by converting parallel light beams to convergent light rays.

The image is real and inverted so long as the item is not in the center of the lens.

According to the lens formula,

1/v + 1/u = 1/f

1/v = 1/f - 1/u

1/v = 1/28 - 1/-16

1/v = 1/28 + 1/16

1/v = 44/448

Therefore, the image distance from the converging lens is,

v = 448/44

v = 10.2 cm

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