Find the magnetic flux through a circular loop 4.6 cm in diameter oriented with the loop normal at 56 ∘ to a uniform 75 mt magnetic field.

Answers

Answer 1

The magnetic flux through the circular loop is 7.95 x 10⁻⁴ Wb.

To find the magnetic flux through a circular loop with a diameter of 4.6 cm, oriented at 56 degrees to a uniform 75 mT magnetic field, follow these steps:

1. Calculate the area (A) of the circular loop using the formula A = πr², where r is the radius (half the diameter): A = π(0.023 m)²≈ 1.66 x 10⁻³ m².

2. Convert the magnetic field (B) from mT to T: B = 75 mT × 10⁻³ = 0.075 T.

3. Calculate the angle (θ) in radians: θ = 56° × (π/180) ≈ 0.977 radians.

4. Use the magnetic flux (Φ) formula: Φ = B × A × cos(θ).

5. Calculate the magnetic flux: Φ = 0.075 T × 1.66 x 10⁻³ m² × cos(0.977) ≈ 7.95 x 10⁻⁴ Wb.

To know more about magnetic flux click on below link:

https://brainly.com/question/24214683#

#SPJ11

Related Questions

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.

To know more about voltage, visit https://brainly.com/question/2364325

#SPJ11

What eclipse occurs when the Moon is in between the Sun and the Earth and the Moon partially or completely blocks out the Sun?

Answers

The eclipse that occurs when the Moon is in between the Sun and the Earth and partially or completely blocks out the Sun is known as a solar eclipse.

During a solar eclipse, the Moon casts a shadow on the Earth's surface, creating a path of totality where the Sun is completely blocked out, and a partial eclipse where only part of the Sun is covered. Solar eclipses occur only during a new moon when the Moon passes between the Sun and the Earth.

During a solar eclipse, the Moon's shadow falls on the Earth's surface, causing the Sun to appear as if it is being covered or "eclipsed" by the Moon. Depending on the alignment of the Sun, Moon, and Earth, a solar eclipse can be either partial or total.

Therefore, A total solar eclipse occurs when the Moon completely covers the Sun, and only the Sun's corona (outer atmosphere) is visible as a glowing ring around the Moon.

To know more about solar eclipse refer here :

https://brainly.com/question/1077760

#SPJ11

You will need to design a cascode amplifier, which should be built and tested to meet the following requirements:1. Magnitude of the voltage gain = 12*SQRT(Z+35):± 10%, where Z is the sum of the last 3 digits of yourstudent number.2. The load resistance RL = 6*(Z+40)2 Ω, rounded up to the nearest standard value stocked in the lab, i.e.,decade multiples of 1.0, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, and 8.2 kΩ. As an example, if yourRL is 67.4 kΩ, you will round this to 68 kΩ.3. The high frequency cutoff fH is to be maximized. It must exceed 1 MHz.4. The output voltage should be able to get to 2 V peak-peak without appreciable distortion[1]. To ensurethis, the AC base-emitter voltage must be kept under 10 mV peak-peak for such an output.5. No DC current may flow in RL and no DC current may flow into or out of the signal generator.6. The low frequency fL must be less than 200 Hz.7. The input and output impedances are left to the discretion of the designer, but their magnitudes at 1kHz are to be determined by calculation and then measured.8. Total circuit power is not to exceed 50 mW.9. The transistors are all to be 2N3904.10. Collector currents in the transistors are to be 1.0 mA ±10%.11. Power-supply voltages are to be limited to +5 volts and/or +15 volts and/or -15 volts.12. No adjustable components, e.g. a trimmer potentiometer, will be allowed.13. Available capacitors are limited to: 1 x 100 µF, 1 x 33 µF, 1 x 10 µF, 2 x 1 µF, 1 x 0.1 µF [2].14. The choice of all other components is left up to the designer.

Answers

To design a cascode amplifier meeting the specified requirements, follow these steps:

1. Calculate the voltage gain (Av) using the formula Av = 12 * SQRT(Z + 35), where Z is the sum of the last 3 digits of your student number.
2. Determine the load resistance (RL) using the formula RL = 6 * (Z + 40)² Ω, and round up to the nearest standard value stocked in the lab.
3. Use a high-pass filter at the input to achieve the low frequency cutoff (fL) of less than 200 Hz.
4. Design the cascode amplifier using 2N3904 transistors with collector currents set to 1.0 mA ±10%. The power supply voltages should be limited to +5V, +15V, or -15V.
5. Maximize the high frequency cutoff (fH) to exceed 1 MHz by carefully selecting component values and minimizing parasitic capacitances.
6. Ensure the output voltage can reach 2 V peak-peak without distortion by keeping the AC base-emitter voltage below 10 mV peak-peak.
7. Prevent DC current from flowing in RL and the signal generator by using coupling capacitors.
8. Calculate and measure the input and output impedances at 1 kHz.
9. Limit the total circuit power to 50 mW.
10. Use the available capacitors (1 x 100 µF, 1 x 33 µF, 1 x 10 µF, 2 x 1 µF, 1 x 0.1 µF) in the design.
11. Choose all other components as needed to achieve the desired performance while adhering to the constraints.

By following these steps, you will design a cascode amplifier that meets the given requirements. Remember, no adjustable components are allowed, and all transistors must be 2N3904.

For more such questions on cascode amplifier, visit:

brainly.com/question/14869225

#SPJ11

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.

To know more about inertia, here

brainly.com/question/3268780

#SPJ1

The question is often asked: Can an airfoil fly upside-down? To answer this, make the following calculation. Consider a positively cambered airfoil with a zero-lift angle of -2°. The lift slope is 0.1 per degree.a/ Calculate the lift coefficient at an angle of attack of 6º. b/ Now imagine the same airfoil turned upside-down, but at the same 6° angle of attack as part (a). Calculate its lift coefficient. c/ At what angle of attack must the upside-down airfoil be set to generate the same lift as that when it is right- side-up at a 6° angle of attack?

Answers

Yes, an airfoil can fly upside-down. To calculate the lift coefficient of a positively cambered airfoil with a zero-lift angle of -2° at an angle of attack of 6º, we use the lift slope of 0.1 per degree.

a/ The lift coefficient at 6º angle of attack would be 0.1 x (6-(-2)) = 0.8

b/ When the same airfoil is turned upside-down, the lift coefficient at the same 6° angle of attack would still be 0.8 because the lift coefficient only depends on the angle of attack and the shape of the airfoil, not its orientation.

c/ To generate the same lift as when the airfoil is right-side-up at a 6° angle of attack, the upside-down airfoil must be set to an angle of attack of -6º because the lift coefficient is proportional to the angle of attack.

Learn more about airfoil here:

https://brainly.com/question/30042281

#SPJ11

Name Period_Date Rotation: Worksheet 9 Angular Momentum The following masses are swung in horizontal circles at the end of a thin string at constant speed. a. A 2.0 kg mass moving at 2.0 on the end of a 2.0 m long thin string. b. A 3.0 kg mass moving at 1.om, on the end of a 2.0 m long thin string. c. A 1.0 kg mass moving at 3.0 on the end of a 2.0 m long thin string. d. A 2.0 kg mass moving at 1.0 on the end of a 4.0 m long thin string. e. A 2.0 kg mass moving at 2.0 on the end of a 4.0 m long thin string

Answers

The length of string is 16.0 kg·m²/s

What is Mass?

Mass is a fundamental property of matter that quantifies the amount of substance or material present in an object. It is a scalar quantity, meaning it only has magnitude and no direction. Mass is commonly measured in units such as kilograms (kg), grams (g), or other appropriate units depending on the context.

a. Mass: 2.0 kg

Velocity: 2.0 m/s

Length of string: 2.0 m

Angular momentum: 8.0 kg·m²/s

b. Mass: 3.0 kg

Velocity: 1.0 m/s

Length of string: 2.0 m

Angular momentum: 6.0 kg·m²/s

c. Mass: 1.0 kg

Velocity: 3.0 m/s

Length of string: 2.0 m

Angular momentum: 6.0 kg·m²/s

d. Mass: 2.0 kg

Velocity: 1.0 m/s

Length of string: 4.0 m

Angular momentum: 2.0 kg·m²/s

e. Mass: 2.0 kg

Velocity: 2.0 m/s

Length of string: 4.0 m

Angular momentum: 8.0 kg·m²/s

Learn more about Mass from the given link

https://brainly.com/question/86444

#SPJ1

For ease of installation, a cabin that is used occasionally is supplied with baseboard electric heating. These are 30 amp circuits powered with 220 volts. To supply 70,000Btu/h (about 20kW)1. How many circuits are needed, and2. What is the resistance of each baseboard "strip" in a single circuit?

Answers

To supply 70,000 BTU/h (about 20 kW) of baseboard electric heating to the cabin, you would need 4 circuits, and the resistance of each baseboard strip in a single circuit would be approximately 7.3 ohms.

To determine how many circuits are needed and the resistance of each baseboard strip in a single circuit for a cabin with 70,000 BTU/h (about 20 kW) of baseboard electric heating, we'll need to follow these steps:

1. Convert the desired heating capacity to watts:
70,000 BTU/h * (1 kW / 3412.14 BTU/h) ≈ 20,500 W

2. Calculate the power per circuit:
Power per circuit = Voltage x Current = 220 V x 30 A = 6,600 W

3. Determine the number of circuits needed:
Number of circuits = Total Power / Power per circuit = 20,500 W / 6,600 W ≈ 3.1
Since you can't have a fraction of a circuit, you'll need 4 circuits to supply the required power.

4. Calculate the total resistance for each circuit:
Resistance (R) = Voltage² / Power = (220 V)² / 6,600 W ≈ 7.3 ohms.

Learn more about resistance:

https://brainly.com/question/17563681

#SPJ11

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

To learn more about conductors click here

brainly.com/question/8426444

#SPJ11

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.

Learn more about transition matrix at:

https://brainly.com/question/28026040

#SPJ11

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.

Learn more about maximum force: https://brainly.com/question/14434333

#SPJ11

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.

Learn more about wavelength https://brainly.com/question/31503451

#SPJ11

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.

Learn more about work here: brainly.com/question/31050706

#SPJ11

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.

Learn more about electric hair dryer here;

https://brainly.com/question/29086609

#SPJ11

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].