6. The electric field has a magnitude of 3. 0 N/C at a distance of 30 cm from a point charge. What is the charge?
A) 1. 4 nC B) 30 pC C) 36 mC D) 12 mC

Answers

Answer 1

The charge is 30 pC, when electric field has a magnitude of 3.0 N/C at the distance of 30 cm from point charge. Option B is correct.

We will use Coulomb's law to solve this problem;

Coulomb's law states that the magnitude of the electric field E created by a point charge Q at a distance r from the charge is given by;

E = k × Q / r²

where k is Coulomb's constant, which is approximately 9 x 10⁹ Nm²/C².

In this problem, we are given electric field having a magnitude of 3.0 N/C at a distance of 30 cm from a point charge. Converting 30 cm to meters, we have;

r = 0.3 m

Plugging the given values into Coulomb's law, we have;

3.0 N/C = (9 x 10⁹ Nm²/C²) × Q / (0.3 m)²

Solving for Q, we get;

Q = (3.0 N/C) × (0.3 m)² / (9 x 10⁹ Nm²/C²)

Q = 30 x 10⁻¹² C

Q = 30 pC

Therefore, the charge is 30 pC.

Hence, B. is the correct option.

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Related Questions

An automobile dealer calculates the proportion of new cars sold that have been returned a various numbers of times for the correction of defects during the warranty period. The results are shown in the following table. Number of returns 0 Proportions 0.28 0.36 0.23 0.09 0.04 (a) Graph the probability distribution function. (b) Calculate and graph the cumulative probability distribution. (b) Calculate and graph the cumulative probability distribution. Find the mean of the number of returns of an automobile for corrections for defects during the warranty period. (d) Find the variance of the number of returns of an automobile for correc- tions for defects during the warranty period.

Answers

(a) Bar graph with x-axis showing the number of returns and y-axis showing the proportion of new cars sold with that number of returns.

(b) Line graph with x-axis showing the number of returns and y-axis showing the cumulative proportion of new cars sold with up to that number of returns.

(d) Variance = 1.715 returns^2, calculated as [(0-0.95)^20.28]+[(1-0.95)^20.36]+[(2-0.95)^20.23]+[(3-0.95)^20.09]+[(4-0.95)^20.04].

In part (a), we represent the probability distribution function using a bar graph. The x-axis shows the number of returns, and the y-axis shows the proportion of new cars sold with that number of returns. In part (b), we plot the cumulative probability distribution using a line graph. The x-axis shows the number of returns, and the y-axis shows the cumulative proportion of new cars sold with up to that number of returns.

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What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, if the cell’s internal resistance is 2.00 Ω ?

Answers

Output Voltage of the lithium cell = 2.9994 V

Output Voltage is the maximum voltage that a cell can offer after overcoming its own internal resistance that arises from the construction of the cell.

To find the output voltage of the lithium cell, you need to consider the voltage drop across the internal resistance due to the current draw. You can use Ohm's Law for this calculation: Voltage drop = Current × Resistance.

In this case, the current draw is 0.300 mA, and the internal resistance is 2.00 Ω. First, convert the current to amperes: 0.300 mA = 0.0003 A.

Now, calculate the voltage drop: Voltage drop = 0.0003 A × 2.00 Ω = 0.0006 V.

Finally, subtract the voltage drop from the initial cell voltage: Output voltage = 3.0000 V - 0.0006 V = 2.9994 V.

The output voltage of the lithium cell is approximately 2.9994 V.

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show that the transition matrix is regular and find its steady-state vector.

Answers

To show that a transition matrix is regular by raising the matrix to different powers until you find a power with all positive elements and its steady-state vector is equation πP = π while ensuring the elements of π sum to 1.

A transition matrix is regular if some power of the matrix has only positive elements, a steady-state vector is a probability vector that remains unchanged after being multiplied by the transition matrix. To demonstrate that the transition matrix is regular, raise the matrix to different powers until you find a power with all positive elements. If such a power exists, the matrix is considered regular.

Next, to find the steady-state vector, solve the following equation: πP = π, where π is the steady-state vector, and P is the transition matrix. Additionally, ensure the elements of π sum to 1, representing the total probability, you can solve this system of linear equations using methods like Gaussian elimination, matrix inversion, or iterative techniques. In summary, to show that a transition matrix is regular, find a power of the matrix with all positive elements, then, to find the steady-state vector, solve the equation πP = π while ensuring the elements of π sum to 1.

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for a particular process at 300 k, δg δ g = -10.0 kj and δh δ h = -7.0 kj. if the process is carried out reversibly, the amount of useful work (in kj) that .

Answers

The amount of useful work that can be obtained for this particular process at 300 K is -3.0 kJ.

For a reversible process, the amount of useful work (in kJ) that can be obtained is given by the equation:

w = -δg = δh - Tδs

where δs is the change in entropy of the system.

Since the process is carried out reversibly, δs can be calculated using the equation:

δs = δqrev / T

where δqrev is the heat absorbed by the system during a reversible process.

Since δh = -7.0 kJ and δg = -10.0 kJ, we know that the process is exothermic (δh < 0) and spontaneous (δg < 0). Therefore, δqrev must be negative, indicating that heat is released from the system.

We can calculate δqrev using the equation:

δqrev = -Tδs = -T(δh / T) = -δh = 7.0 kJ

Substituting this value into the equation for work, we get:

w = -δg = δh - Tδs = -10.0 kJ - (-7.0 kJ) = -3.0 kJ

Therefore, the amount of useful work for this particular process at 300 K is -10.0 kJ.

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An astronaut wishes to visit the Andromeda galaxy, making a one-way trip that will take 30.0 years in the spaceship's frame of reference. Assume the galaxy is 2.00 million light-years away and his speed is constant. (a) How fast must he travel relative to Earth? (b) What will be the kinetic energy of his spacecraft, which has a mass of 1.00 x 10^6 kg? (c) What is teh cost of this energy if it is purchased at a typical consumer price for electric energy, 13.0 cents per kWh? The following approximation will prove useful:
For x <<1

Answers

(a)the astronaut must travel at a speed of 0.999999999998 times the speed of light relative to Earth.

(b)the kinetic energy of the spacecraft is 4.499 x 10^23 Joules.

(a) To determine the speed of the astronaut relative to Earth, we can use the formula for time dilation in special relativity: t_0 = t / sqrt(1 - v^2/c^2)

where t_0 is the proper time (i.e. the time experienced by the astronaut), t is the time measured by observers on Earth, v is the velocity of the spacecraft relative to Earth, and c is the speed of light. Solving for v, we get: v = c * sqrt(1 - (t/t_0)^2)

Plugging in the given values, we get: v = c * sqrt(1 - (30.0 years / t_0)^2)

where t_0 is the proper time experienced by the astronaut. We know that the distance to the Andromeda galaxy is 2.00 million light-years, so we can use the distance formula to find t_0: t_0 = d/v

where d is the distance to the Andromeda galaxy. Plugging in the given values, we get:

t_0 = (2.00 million light-years) / c = (2.00 million light-years) / (299,792,458 m/s) = 6.32 x 10^15 s

Substituting this value into the formula for v, we get:

v = c * sqrt(1 - (30.0 years / 6.32 x 10^15 s)^2)

v = 0.999999999998 c

Therefore, the astronaut must travel at a speed of 0.999999999998 times the speed of light relative to Earth.

(b) To find the kinetic energy of the spacecraft, we can use the formula:

K = (1/2) * m * v^2

where K is the kinetic energy, m is the mass of the spacecraft, and v is the velocity of the spacecraft relative to Earth. Plugging in the given values, we get:

K = (1/2) * (1.00 x 10^6 kg) * (0.999999999998 c)^2

K = 4.499 x 10^23 J

Therefore, the kinetic energy of the spacecraft is 4.499 x 10^23 Joules.

(c) To find the cost of this energy, we need to convert Joules to kilowatt-hours (kWh) and then multiply by the price per kWh. We can use the following conversion factor:

1 J = 2.77778 x 10^-7 kWh

Plugging in the given values, we get:

cost = (4.499 x 10^23 J) * (2.77778 x 10^-7 kWh/J) * (13.0 cents/kWh)

cost = $165,643,646,517,000,000,000,000

Therefore, the cost of this energy is $165,643,646,517,000,000,000,000 (over 165 sextillion dollars). This highlights the fact that the amount of energy required for intergalactic travel is immense, and that our current understanding of physics may not allow for such journeys to be feasible.

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A typical power for a laser used in physics labs is 0.75 mW. This laser would produce a beam that is about 1 mm in diameter I found the average intensity: to be 955 W/m2 I found the average energy density of this beam: 3.183*10^-6 J/m^3 I need help with this question. Lets say the laser beam is reflected completely off a mirror. What is the maximum force the beam can exert on the mirror? I did 955 x 2x (4pie x 10^-7)/(3*10^8) but says my answer is wrong, is this the right equation???

Answers

The maximum force of a typical power for a laser used in physics labs is 0.75 mW and would produce a beam that is about 1 mm in diameter can exert on the mirror is 5 x 10⁻⁹ N.

In physics, power is the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity.

To find the maximum force the laser beam can exert on the mirror when it is reflected completely, you should use the following formula:

Force (F) = (2 × Power (P)) / Speed of light (c)

where Power (P) = 0.75 mW (convert to Watts: 0.75 x 10⁻³ W), and Speed of light (c) = 3 x 10⁸ m/s.

F = (2 × (0.75 x 10⁻³ W)) / (3 x 10⁸ m/s)

Thus, the maximum force of the beam can exert on the mirror is approximately 5 x 10⁻⁹ N.

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A closed curve encircles several conductors. The line integral around this curve is ∮B⃗ ⋅dl⃗ = 4.25×10^-4 T⋅m .
A) What is the net current in the conductors?
B) If you were to integrate around the curve in the opposite direction, what would be the value of the line integral?

Answers

a) the net current in the conductors is 3.38 A. b) the line integral in the opposite direction will be: ∮B⃗ ⋅dl⃗ = -4.25×10^-4 T⋅m

To solve this problem, we can use Ampere's Law, which relates the line integral of the magnetic field around a closed loop to the net current passing through the loop.

A) The equation for Ampere's Law is: ∮B⃗ ⋅dl⃗ = μ0I, where μ0 is the permeability of free space and I is the net current passing through the loop. Solving for I, we get:

I = ∮B⃗ ⋅dl⃗ / μ0

Substituting the given values, we get:

I = (4.25×10^-4 T⋅m) / (4π×10^-7 T⋅m/A)

I = 3.38 A

Therefore, the net current in the conductors is 3.38 A.

B) If we integrate around the curve in the opposite direction, the value of the line integral will be negative, since the direction of the magnetic field will be opposite. Specifically, we can use the fact that reversing the direction of the line integral is equivalent to reversing the direction of the loop, which changes the sign of the enclosed current.

Therefore, the line integral in the opposite direction will be: ∮B⃗ ⋅dl⃗ = -4.25×10^-4 T⋅m

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A student wants to determine the angular speed w of a rotating object. The period T is 0.50 s +/- 5%. The angular speed ω is
ω = 2π/T
What is the percentage uncertainly of ω? A. 0.2%
B. 2.5% C. 5% D. 10%

Answers

The percentage uncertainty of the angular speed= 5%

To determine the percentage uncertainty of the angular speed ω, we can use the following formula:

Percentage Uncertainty of ω = (Percentage Uncertainty of T) * (Constant Value)

In this case, the percentage uncertainty of T is 5%, and the constant value is 2π (from the formula ω = 2π/T). Therefore:

Percentage Uncertainty of ω = (5%) * (2π)

However, since we're only interested in the percentage uncertainty, we don't need to multiply by the constant value (2π). So, the percentage uncertainty of ω is the same as the percentage uncertainty of T:

Percentage Uncertainty of ω = 5%

So, the correct answer is C i.e.5%

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I need help with my physics homework
When a thin stick of mass M and length L is pivoted about one end, its moment of inertia is I=(1/3)ML^2. When the stick is pivoted about its midpoint, its moment of inertia is

A.) (1/12)ML^2
B.) (1/6)ML^2
C.) (1/3)ML^2
D.) (7/12)ML^2
E.) ML^2

Answers

The moment of inertia of the stick when it is pivoted about its midpoint is (1/6)ML². The answer choice is (B).

When the stick is pivoted about its midpoint, we can split it into two equal pieces of mass M/2 and length L/2, each with a moment of inertia of:

I₁ = (1/3)(M/2)(L/2)² = (1/12)ML²

The parallel axis theorem tells us that the moment of inertia of the whole stick about its midpoint is equal to the sum of the moments of inertia of the two pieces plus the moment of inertia of the stick as if it were a point mass at its center of mass:

I₂ = 2I₁ + (1/12)M(L/2)²

I₂ = (2/12)ML² + (1/48)ML²

I₂ = (1/6)ML²

Option B is correct.

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the current is from left to right in the conductor showmn. the magnetic field is onto the page and point is at higher potential than point t. the charge carriers are
a. positive b. negative c. neutral d. absent e. moving near the speed of light

Answers

The charge carriers in the conductor are likely to be negative, and they are not absent, but their speed is not near the speed of light due to the effects of collisions in the conductor.

Based on the given information, we know that the conductor is experiencing a magnetic force due to the magnetic field pointing onto the page. Additionally, we know that the point labeled as "point" is at a higher potential than the point labeled as "t".

The direction of the current flowing from left to right in the conductor is indicative of the direction in which the charge carriers are moving. Given this information, we can determine that the charge carriers in the conductor must be negative because electrons are negatively charged and move opposite to the direction of conventional current flow.

Furthermore, the fact that the charge carriers are moving in a conductor does not necessarily imply that they are moving near the speed of light. The speed at which electrons move in a conductor is known as the drift velocity and is typically much slower than the speed of light due to collisions with other particles in the conductor.

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A solid has a(n) ______ volume and maintains its _____ regardless of the container in which it is placed.

Answers

Answer:

it has a fixed volume and maintains it's shape

what wavelength bands were placed into which color guns

Answers

wavelength bands were placed into which color guns,

we need to understand that color guns are components of a cathode ray tube (CRT) display, commonly used in older televisions and monitors. CRT displays have three color guns: red, green, and blue (RGB). These color guns produce different wavelengths corresponding to their respective colors.

1. Red color gun: The red color gun corresponds to the wavelength band of approximately 620-750 nm.
2. Green color gun: The green color gun corresponds to the wavelength band of approximately 495-570 nm.
3. Blue color gun: The blue color gun corresponds to the wavelength band of approximately 450-495 nm.

By combining various intensities of these three primary colors, CRT displays can produce a wide range of colors on the screen.

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When helium capture occurs with a carbon 12 nucleus, what results?
A) Nitrogen 14
B) Oxygen 16
C) Neon 20
D) Silicon 28
E) Nickel 56

Answers

When helium capture occurs with a carbon 12 nucleus, it results in Nitrogen 14.

Helium capture, also known as alpha capture, is a type of nuclear reaction in which a helium nucleus (consisting of two protons and two neutrons, denoted as an alpha particle) is captured by a target nucleus.

When a helium capture occurs with a carbon 12 nucleus (which has 6 protons and 6 neutrons), the resulting nucleus will have 8 protons and 8 neutrons. This results in the formation of a nitrogen 14 nucleus, which has 7 protons and 7 neutrons, denoted as 14N or Nitrogen-14.

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A 1.60 m tall person lifts a 1.75 kg book off the ground so it is 2.00 m above the ground.What is the potential energy of the book relative to the ground?What is the potential energy of the book relative to the top of the person's head?How is the work done by the person related to the answers in parts A and B?W=Uground−UheadW=UgroundW=Uground+UheadW=Uhead−UgroundW=Uhead

Answers

The potential energy is PE_head ≈ 6.86 J (Joules) and the work done by the person is approximately 27.43 Joules.

To find the potential energy of the book relative to the ground, we can use the formula for gravitational potential energy,
PE = m * g * h

where PE is potential energy, m is the mass of the book (1.75 kg), g is the acceleration due to gravity (9.81 m/s^2), and h is the height above the ground (2.00 m).

PE_ground = 1.75 kg * 9.81 m/s^2 * 2.00 m
PE_ground ≈ 34.29 J (Joules)

To find the potential energy of the book relative to the top of the person's head, we need to determine the height above the person's head,

height_above_head = 2.00 m - 1.60 m = 0.40 m

PE_head = 1.75 kg * 9.81 m/s^2 * 0.40 m
PE_head ≈ 6.86 J (Joules)

To relate the work done by the person to the potential energies, we can use the following equation,

W = PE_ground - PE_head

where W is the work done by the person.

W = 34.29 J - 6.86 J
W ≈ 27.43 J

The work done by the person is approximately 27.43 Joules.

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the current in an electric hair dryer is 11aa. part a how much charge flows through the hair dryer in 4.0 minmin ?

Answers

To find the charge that flows through the electric hair dryer in 4.0 minutes, we need to use the formula:

charge = current x time

Substituting the given values, we get:

charge = 11 A x 4.0 min = 44 C

Therefore, the amount of charge that flows through the hair dryer in 4.0 minutes is 44 Coulombs.
Hi! To calculate the charge that flows through the electric hair dryer in 4.0 minutes, you can use the formula Q = I * t, where Q is the charge, I is the current, and t is the time.

Given that the current (I) in the hair dryer is 11 A, and the time (t) is 4.0 minutes (or 240 seconds, since 1 minute = 60 seconds), you can plug these values into the formula:

Q = 11 A * 240 s

Q = 2640 C

So, 2640 Coulombs of charge flow through the electric hair dryer in 4.0 minutes.

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A large electromagnet coil is connected to a 130 Hz ac source. The coil has a resistance 410 omega or Ohms, and at this source frequency the coil has an inductive reactance 230 omega or Ohms.
Part A.) What is the inductance of the coil? ( answer is L= ? H)
Part B.) What must the rms voltage of the source be if the coil is to consume an average electrical power of 830 W? (answer is V(rms)= ? V)

Answers

The inductance of the coil is approximately 0.444 H. The rms voltage of the source must be approximately 291 V for the coil to consume an average electrical power of 830 W.

Part A:

We can use the equation X_L = 2πfL, where X_L is the inductive reactance, f is the frequency, and L is the inductance of the coil. Substituting the given values, we get:

230 Ω = 2π(130 Hz)L

Solving for L, we get:

L = 230 Ω / (2π × 130 Hz) ≈ 0.444 H

Therefore, the inductance of the coil is approximately 0.444 H.

Part B:

The average electrical power consumed by the coil is given by P = V(rms)I cos(φ), where V(rms) is the rms voltage, I is the rms current, and cos(φ) is the power factor. Since the coil has only inductive reactance, the power factor is zero, and cos(φ) = 0. Therefore, the equation simplifies to:

P = V(rms)I

We know that the resistance of the coil is 410 Ω, and the inductive reactance is 230 Ω. Therefore, the total impedance of the coil is:

Z = √([tex]R^{2}[/tex] + [tex]X_L^{2}[/tex]) = √([tex]410^{2}[/tex] + [tex]230^{2}[/tex]) ≈ 470 Ω

Since the current through the coil is given by I = V(rms) / Z, we can substitute this expression into the equation for power:

P = V(rms)(V(rms) / Z)

Solving for V(rms), we get:

V(rms) = √(PZ) = √(830 W × 470 Ω) ≈ 291 V

Therefore, the rms voltage of the source must be approximately 291 V for the coil to consume an average electrical power of 830 W.

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a force f is applied to a 2.0 kg, radio-controlled model car parallel to the x- axis as it moves along a straight track. the x-component of the force varies with the x-coordinate of thecar.
The work done by the force when the car moves from x=0.0m to x=7.0m is?
What is the speed of the car at x=4.0m?

Answers

To calculate the work done by the force when the car moves from x=0.0m to x=7.0m, we need to integrate the force over the distance traveled.W = ∫Fdx

Since the force varies with the x-coordinate of the car, we need to know the equation for the force as a function of x. Without that information, we can't calculate the work done.

To calculate the speed of the car at x=4.0m, we need to use the equations of motion. Assuming that the force is the only external force acting on the car, we can use:

F = ma

where F is the force, m is the mass of the car (2.0 kg), and a is the acceleration of the car.

Since the force varies with x, we need to know the equation for the force as a function of x. Without that information, we can't calculate the acceleration of the car, and therefore we can't calculate the speed at x=4.0m.

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Which of the following is not a property of light?
a. light is a form of matter less dense than air
b. light travels in straight lines
c. light has different colors
d. light has different intensities, and cam be bright or dim

Answers

Answer: Light is not a form of matter, so a is not a property of light.

Explanation:

Johnny, of mass 65 kg, and Lucy, of mass 45 kg, are facing each other on roller blades. The coefficient of kinetic friction between the roller blades and concrete surface is 0.20. When Johnny pushes Lucy from rest he applies a force for 1.0 s. Lucy then slows down to a stop in another 8.0 s. Calculate:
a. The applied force exerted by Johnny on Lucy.
b. How long it takes Johnny to come to rest.

I tried calculated the force exerted but I would need acceleration which I don't have...any tips on how to solve this one??? help is appreciated!!

Answers

Answer:

John applied a force of approximately [tex]795\; {\rm N}[/tex] (on average, rounded) on Lucy.

John slows down to a stop after approximately another [tex]5.37\; {\rm s}[/tex].

(Assuming that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)

Explanation:

Assuming that the surface is level. The normal force on Johnny will be equal to the weight of Johnny: [tex]N(\text{John}) = m(\text{John})\, g[/tex]. Similarly, the normal force on Lucy will be equal to weight [tex]N(\text{Lucy}) = m(\text{Lucy})\, g[/tex].

Multiply normal force by the coefficient of kinetic friction to find the friction on each person:

[tex]f(\text{John}) = \mu_{k}\, N(\text{John}) = \mu_{k}\, m(\text{John})\, g[/tex].

[tex]f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g[/tex].

Again, because the surface is level, the net force on each person after the first [tex]1.0\; {\rm s}[/tex] will be equal to the friction. Divide that the net force on each person by the mass of that person to find acceleration:
[tex]\displaystyle a(\text{John}) = \frac{\mu_{k}\, m(\text{John})\, g}{m(\text{John})} = \mu_{k}\, g[/tex].

[tex]\displaystyle a(\text{Lucy}) = \frac{\mu_{k}\, m(\text{Lucy})\, g}{m(\text{Lucy})} = \mu_{k}\, g[/tex].

(Note that the magnitude of acceleration is independent of mass and is the same for both John and Lucy.)

[tex]a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}[/tex].

In other words, after the first [tex]1\; {\rm s}[/tex], both John and Lucy will slow down at a rate of [tex]1.962\; {\rm m\cdot s^{-2}}[/tex].

To find the speed of Lucy immediately after the first [tex]1.0\: {\rm s}[/tex], multiply this acceleration by the time [tex]t = 8.0\; {\rm s}[/tex] it took for Lucy to slow down to [tex]0\; {\rm m\cdot s^{-1}}[/tex]:

[tex]\begin{aligned}& (8.0\; {\rm s})\, (1.962\; {\rm m\cdot s^{-2}}) \\ =\; & (8.0)\, (1.962)\; {\rm m\cdot s^{-1}} \\ =\; & 15.696\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Thus, in the first [tex]1.0\; {\rm s}[/tex], Lucy accelerated (from [tex]0\; {\rm m\cdot s^{-1}}[/tex]) to [tex]15.696\; {\rm m\cdot s^{-1}}[/tex].

The average acceleration of Lucy in the first [tex]1.0\; {\rm s}[/tex] would be [tex](15.696) / (1) = 15.696\; {\rm m\cdot s^{-2}}[/tex]. Multiply this average acceleration by the mass of Lucy to find the average net force on Lucy during that [tex]1.0\; {\rm s}[/tex]:

[tex]\begin{aligned}F_{\text{net}}(\text{Lucy}) &= m(\text{Lucy})\, a \\ &= (45)\, (15.696)\; {\rm N} \\ &= 706.320\; {\rm N}\end{aligned}[/tex].

This net force on Lucy during that [tex]1.0\; {\rm s}[/tex] is the combined result of both the push from Johnny and friction:

[tex]F_{\text{net}}(\text{Lucy}) = F(\text{push}) - f(\text{Lucy})[/tex].

Since [tex]f(\text{Lucy}) = \mu_{k}\, N(\text{Lucy}) = \mu_{k}\, m(\text{Lucy})\, g[/tex]:

[tex]\begin{aligned}F(\text{push}) &= F_{\text{net}}(\text{Lucy}) + f(\text{Lucy}) \\ &= F_{\text{net}}(\text{Lucy}) + \mu_{k}\, m(\text{Lucy})\, g \\ &= (706.320) \; {\rm N}+ (0.2)\, (45)\, (9.81)\; {\rm N} \\ &= 706.320\; {\rm N} + 88.290\; {\rm N} \\ &=794.610\; {\rm N}\end{aligned}[/tex].

In other words, Johnny would have applied a force of [tex]794.610\; {\rm N}[/tex] on Lucy.

By Newton's Laws of Motion, when Johnny exerts this force on Lucy in that [tex]1.0\; {\rm s}[/tex], Lucy would exert a reaction force on Johnny of the same magnitude: [tex]794.610\; {\rm N}[/tex].

Similar to Lucy, the net force on Johnny during that [tex]1.0\; {\rm s}[/tex] will be the combined effect of the push [tex]F(\text{push})[/tex] and friction [tex]f(\text{John}) = \mu_{k}\, m(\text{John})\, g[/tex]:

[tex]\begin{aligned}F_{\text{net}}(\text{John}) &= F(\text{push}) - f(\text{John}) \\ &= F(\text{push}) - \mu_{k}\, m(\text{John})\, g\\ &= 794.610\; {\rm N} - (0.2)\, (65)\, (9.81)\; {\rm N} \\ &= 667.080\; {\rm N}\end{aligned}[/tex].

Divide net force by mass to find acceleration:

[tex]\begin{aligned}\frac{667.080\; {\rm N}}{65\; {\rm kg}} \approx 10.2628\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].

In other words, Johnny accelerated at a rate of approximately [tex]10.5406\; {\rm m\cdot s^{-2}}[/tex] during that [tex]1.0\; {\rm s}[/tex]. Assuming that Johnny was initially not moving, the velocity of Johnny right after that [tex]1.0\; {\rm s}\![/tex] would be:

[tex](0\; {\rm m\cdot s^{-1}}) + (10.2628\; {\rm m\cdot s^{-2}})\, (1.0\; {\rm s}) = 10.2628\; {\rm m\cdot s^{-1}}[/tex].

After the first [tex]1.0\; {\rm s}[/tex], the acceleration of both John and Lucy (as a result of friction) would both be equal to [tex]a = \mu_{k}\, g= (0.2)\, (9.81\; {\rm m\cdot s^{-2}}) = 1.962\; {\rm m\cdot s^{-2}}[/tex]. Divide initial velocity of Johnny by this acceleration to find the time it took for Johnny to slow down to a stop:

[tex]\displaystyle \frac{10.2628\; {\rm {m\cdot s^{-1}}}}{1.962\; {\rm m\cdot s^{-2}}} \approx 5.23\; {\rm s}[/tex].