Please help in 5 mintues
A piece of chess game called master
a. king
b. queen
c. rook
d. bishop

Explanation:

Thanx for the figure;
The force component of F   UP the ramp that  moves the crate must equal the force of the crate DOWN the ramp

75 kg = mg Newtons = 735.8 Newtons

    Downplane force is   735.8 sin 34°  = 411.4 Newtons

   Fn =The horizontal force will be found by   cos 34 = 411.4/ F                                                                       F = 411.4/cos (34) = 496 N

Normal Force =  735.8 cos 34°  = 610 N   This part is due to the mass of the crate....there is additional normal force from the force pushing the crate up the hill  (from below)

   = F sin34   =  496 sin 34 = 277.4 N

            SUM of normal forces = 610 + 277.4 = 887.4 N

Answer:

  a.  |F| ≈ 496 N

  b.  normal force ≈ 887 N

Explanation:

You want the magnitude of the horizontal force F that moves a crate up a 34° ramp at constant speed, and you want the magnitude of the normal force on the crate.

The constant speed of the crate tells you the net force up the ramp is zero. This is the sum of the component of force F in that direction and the force due to gravity in the opposite direction:

  F·cos(34°) - m·g·sin(34°) = 0

  F = mg·tan(34°) = (75 kg)(9.8 m/s²)tan(34°) ≈ 496 N

The magnitude of force F is about 496 N.

The normal force on the crate will be the sum of the component of F in that direction and the force due to gravity in the same direction:

  F·sin(34°) +m·g·cos(34°) ≈ 887 N

The magnitude of the normal force is about 887 N.

The kinematic relations allow finding that the answers for the velocity and the position of the car in a time of 2 s. are:

Kinematics studies the movement of bodies giving relationships between the position, velocity and acceleration of the body

They indicate that the initial velocity is v₀ = 1 m/s, that the car has an acceleration of a = 3 m/s² and asks about the position and velocity after a time of t = 2s

Let's used the kinematics relation

                 v = v₀ + a t

                 v = 1 + 3 2

                 v = 7 m / s

We look for the position

                  x = v₀ t + ½ a t²

                 x = 1 2 + ½ 3 2²

                 x = 8m

In conclusion, using the kinematics relations we can find the answers for the speed and the position of the car in a time of 2 S. They are

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The input force on the lever of mechanical advantage 9.5, in order for it to lift a rock of 445 N is 46.84 N.

Force can be defined as the product of mass and acceleration.

To calculate the input force on the lever, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the input force on the lever is 46.84 N.

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The x component of the force is -8.03 N.

The y component of the force is -11.47 N.

The parallel components of the force is resolved into x and y components as follows;

The x component of the force is calculated as follows;

Fx = F cosθ

where;

Fx = 14 N x cos (235)

Fx = -8.03 N

The y component of the force is calculated as follows;

Fy = F sinθ

where;

Fy = 14 N x sin (235)

Fy = -11.47 N

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Answer:

Explanation:

Answer:

\(2.50 * 10x^{10x} coulomb\)

Explanation:

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The maximum displacement = 0.0189 m

The Total energy of the system = 1.188 J

The Gravitational potential energy = 0.22 J

Let m be = Mass of the object = 1.2 kg

As given K = Spring Constant of the spring = 300 N/m

Let v denote = Maximum speed with which the body oscillates = 30 cm/s

Let 'x' be the maximum displacement of the body

x²= mv²/k=1.2* 0.3 /300= 0.0189

Total energy= kinteric energy+potential energy =kx²/2 + mv²/2

=1.188J

The Gravitational potential energy of the system is

E=Mgx=1.2*9.8*0.0189

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The 1.2 kg object's maximum displacement when it is suspended from a spring with a 300 N/m force constant oscillates at a maximum speed of 30 cm/s. This displacement is equal to 0.0189 m.

When the object is displaced to its maximum, the system's total energy is 1.188 J.

The energy of gravity is equal to 0.22 J.

Allow m to be equal to 1.2 kg, the object's mass.

K = Spring as stated. Springs have a constant of 300 N/m.

the body can oscillate at a maximum speed of 30 cm/s (let v signify this).

Suppose that "x" represents the body's largest displacement.

x²= mv²/k=1.2* 0.3 /300= 0.0189

Total energy = interstellar energy + potential energy = kx2/2 + mv2/2

=1.188J

According to the system's gravitational potential energy,

E=Mgx=1.2*9.8*0.0189

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Answer:

75%

\(\dfrac{A}{\sqrt{2}}\)

Explanation:

x = Displacement of spring = \(\dfrac{1}{2}A=\dfrac{A}{2}\)

k = Spring constant

Total energy of the spring is

\(E=\dfrac{1}{2}kA^2\)

Elastic potential energy is given by

\(U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}k(\dfrac{A}{2})^2\\\Rightarrow U=\dfrac{1}{2}k\dfrac{A^2}{4}\\\Rightarrow U=\dfrac{1}{4}\times \dfrac{1}{2}kA^2\\\Rightarrow U=\dfrac{1}{4}E\)

Total energy is given by

\(E=U+K\\\Rightarrow K=E-U\\\Rightarrow K=E-\dfrac{1}{4}E\\\Rightarrow K=\dfrac{3}{4}E\\\Rightarrow \dfrac{K}{E}=0.75\)

The percentage will be

\(\dfrac{K}{E}=0.75\times 100\%\\\Rightarrow \dfrac{K}{E}=75\%\)

The required percentage is 75%

According to the given condition

\(\dfrac{E}{2}=\dfrac{1}{2}kx^2\\\Rightarrow \dfrac{\dfrac{1}{2}kA^2}{2}=\dfrac{1}{2}kx^2\\\Rightarrow \dfrac{1}{2}kA^2=kx^2\\\Rightarrow x=\sqrt{\dfrac{1}{2}A^2}\\\Rightarrow x=\dfrac{A}{\sqrt{2}}\)

So, the energy is half when displacement is \(\dfrac{A}{\sqrt{2}}\)

Let x be the displacement of spring which is given to be  \(\frac{1}{2} A=\frac{A}{2}\)

k= spring constant

Total energy of the spring is:

\(E=\frac{1}{2} kA^2\)

Elastic potential energy is given as:

\(U=\frac{1}{2} kx^2\)

Now on substituting the value of x in above equation:

\(U=\frac{1}{2} k\frac{A^2}{4}\)

It can also be written as:

\(U=\frac{1}{4} *\frac{1}{2} kA^2 \\\\U=\frac{1}{4} E\)    (∵\(E=\frac{1}{2} kA^2\))

Total energy can be written as:

E=U+K

⇒K=E-U

⇒K=E-1/4 E

⇒K=3/4 E

⇒\(\frac{K}{E} =0.75\)

(i) Now, we need to calculate the percentage of kinetic energy:

\(\frac{K}{E} =0.75*100=75\%\)

The percentage of kinetic energy is 75%.

(ii) In second it is asked at what displacement, as a fraction of A, is the energy half kinetic and half potential?

∴\(\frac{E}{2} =\frac{1}{2} kx^2\)

On substituting the value of E in above equation we will get:

\(x=\sqrt{\frac{1}{2} A^2} \\\\x=\frac{A}{\sqrt{2} }\)

So, the energy is half when displacement is \(\frac{A}{\sqrt{2} }\) .

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When you honk your horn while driving 80 km/h, your friend standing on the sidewalk hears a frequency of 340 Hz.

This is due to the Doppler Effect, which causes a change in frequency for a sound wave as it approaches or moves away from an observer. As you drive towards your friend, the frequency of the horn increases, and as you drive away from them, the frequency decreases. This shift in frequency is proportional to the relative speed between the source of the sound and the observer. The change in frequency can be calculated using the equation: f' = f × (v ± v0) / (v ± v s), where f' is the observed frequency, f is the emitted frequency, v0 is the speed of sound in air, v is the relative speed between the source and observer, and vs is the speed of the source.

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Soft sand makes it challenging to ride a bicycle since the tires can't generate enough friction. The bicycle struggles to move ahead as a result, and its tires sink.

Riding a bicycle on soft sand poses a risk of the vehicle being stuck and challenging to move. Injuries might result if the rider loses balance and falls off the bicycle. The frame, tires, and other parts of the bicycle could also be harmed by the sand.

Furthermore, sand, dirt, wood chips, deep pea gravel, soft grass, and snow all drastically alter your speed. Sand, however, can be the most challenging of all of them because you typically enter the area quickly from grass or another hard surface.

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Answer:

\(\large \boxed{42\, \mu \text{C}}$\)

Explanation:

The formula for the force exerted between two charges is

\(F=k \dfrac{ q_1q_2}{r^2}\)

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

\(F=k\dfrac{q^{2}}{r^2}\)

\(\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}\)

Answer

To get critical pressure

We use

Pc = a/(27b²)

So

= 3.610/(27 X 0.0429²)

We have

= 72.7 atm

Critical temperaturewe

We use

Tc = 8a/27Rb

= 8 x 3.610/(27 x 0.0812 x 0.0429)

= 307 K

Critical volume

We use

Vc =3b =

3 x 0.0429

= 0.129L/mol

The resultant displacement of the free end of rod B from its present location when the temperature rises from 15°C to 25°C is -0.0012 m.

When a metal is heated, it tends to expand, and when cooled, it tends to contract.

The coefficient of linear expansion for a metal determines how much the metal changes in length per unit of temperature change.

The amount of change in length (L) per unit length (Lo) per unit of temperature change (∆T) is given by

∆L=αL ∆T, where α is the coefficient of linear expansion.

We'll use the following formula to calculate the resulting displacement of the free end of rod B from its present location when the temperature rises from 15°C to 25°C:

ΔL = LαΔT

where L is the length of the metal rod

α is the coefficient of linear expansion

ΔT is the temperature difference.

Let us first calculate the coefficient of linear expansion for aluminum and steel.

ALUMINUM:α (aluminum) = 24 × 10-6 /°C

STEEL:α (steel) = 12 × 10-6 /°C

Now, let's calculate the change in length of the rods as a result of the temperature change.

ΔLA = αA

LΔT= 24 × 10-6/°C × 1.0 m × (25°C − 15°C)

LΔT= 0.0024 m

ΔLB = αB

LΔT= 12 × 10-6/°C × 1.0 m × (25°C − 15°C)

LΔT= 0.0012 m

Now we may calculate the resultant displacement of the free end of rod B from its present location.

ΔLresultant = ΔLB - ΔLA

ΔLresultant = (0.0012 m) - (0.0024 m)

ΔLresultant = -0.0012 m

Therefore, the resultant displacement of the free end of rod B from its present location when the temperature rises from 15°C to 25°C is -0.0012 m.

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Answer:

Distance from sun to Neptune = 4.495E9 km

Time for light to travel = 4.495E9 / 3E5 sec = 14,980 sec

That is from sun to Neptune time fof light = 250 min

Time for light to travel from sun to earth is about 8 min

So the time from Neptune would be 242 to 258 min depending on position of Neptune - Note that Neptune is about 30X as far from the sun as earth and

250 min / 8 min is roughly 30

The uniform motion of kinematics allows us to find the time it takes for light to arrive from Neptune to Earth, which varies between:

          t₁ = 1.45 10⁴ s and t₂₂= 1.55 10⁴ s

depending on the relative distance of the two planets

given parameters

to find

They  velocit of an electromagnetic wave is constant, so we can use the uniform motion relationships

             v = d / t

             t = d / v

where v is the speed of light, d the distance and y time, in this case the speed of the wave is the speed of light (v = c)

We look in the tables for the distances and the rotation periods around the sun

                           distance ( m)         period (s)

Sun Neptunium     4.50 10¹²             5.2 10⁹

Sun - Earth             1.5    10¹¹              3.2 10⁷

With the data of the period it is observed that the rotation of Neptune is much greater than that of Eart rotation around the sun, for which we will assume that Neptunium is fixed in space and the Earth may be in its aphelion or perihelion, maximum approach o away distance from the sun, consequently we calculate the time for the two cases:

Maximum approach

positions relative distance from the dos Plantetas is

         Δd = \(x_{Neptuno - Sum} - x_{Earth - Sum}\)d  

         Δd = 4.50 10¹²  - 1.5 10¹¹

         Δd = 43.5 10¹¹ m

the time it takes for Neptune's light to reach Earth is

        Δt = \(\frac{ 43.5 \ 10^{11} }{3 \ 10^8}\)  

        Δt = 14.5 10³ s

        Δt = 1.45 10⁴ s

We reduce to hours

        Δt = 1.45 10⁴ s (1 h / 3600 s) = 4.03 h

Maximum away

         Δd = \(x_{Neptune - Sum} + x_{Neptune-Sum}\)  

         Δd = 4.50 10¹² + 1.5 10¹¹

         Δd = 46.5 10¹¹

The time is

         Δt = \(\frac{46.5 \ 10^{11}}{ 3 \ 10^8}\)  

         Δt = 15.5 10³

         Δt = 1.55 10⁴ s

We reduce to hours

         Δt = 1.55 10⁴ s (1 h / 3600 s) = 4.31 h

In conclusion, the time it takes for light to arrive from Neptune to Earth varies between:

          t₁ = 1.45 10⁴ s and t₂ = 1.55 10⁴ s

depending on the relative distance of the two plants

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Answer:

The value is \(y = 3.097 * 10^{-5} \ m \)

Explanation:

From the question we are told that

The diameter of the pupil is \(d_p = 4.2 \ mm = 4.2 *10^{-3} \ m\)

The distance of the page from the eye \(d = 29 \ cm = 0.29 \ m\)

The wavelength is \(\lambda = 500 \ nm = 500 *10^{-9} \ m\)

The refractive index is \(n_r = 1.36\)

Generally the minimum separation of adjacent dots that can be resolved is mathematically represented as

\(y = [ \frac{1.22 * \lambda }{d_p * n_r } ]* d\)

         \(y  = [ \frac{1.22 *  500 *10^{-9} }{4.2 *10^{-3} * 1.36} ]* 0.29\)

         \(y  = 3.097 * 10^{-5} \  m \)

Answer:

A. gravitational force always true.

B, C and D could be true under the correct conditions

The momentum of the zombie is 535.3 kg m/s.

Momentum is a fundamental concept in physics that describes the amount of motion that an object has. It is defined as the product of an object's mass and its velocity.

Here,

The momentum (p) of an object is defined as the product of its mass (m) and velocity (v)

p = m * v

In this case, the mass of the zombie is 101 kg and its velocity is 5.3 m/s. So, the momentum of the zombie can be calculated as:

p = 101 kg * 5.3 m/s

p = 535.3 kg m/s

Therefore, the momentum of the zombie is 535.3 kg m/s.

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Answer:

\(T_{1}\) = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about \(T_{1}\) we have:

(M + m) g * 0.5L - \(T_{2}\)(L - d) = 0

⇒ \(T_{2}\) = [(M + m) g * 0.5L] ÷ (L - d)

\(T_{2}\) = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

\(T_{2}\)= 59.535 ÷ 2.4

\(T_{2}\) = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

\(T_{1}\) + \(T_{2}\) = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

\(T_{1}\) + 24.81 = 26.46 + 13.23

\(T_{1}\) = 14.88 N

1 Newton force is equal to the 10⁻⁵ dynes. The unit of force will be dyne and the value of force will be 0.1 dynes.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration.

Its SI unit is Newton, MKS unit is kgm/s² and CGS unit is dyne.

The given formulae are;

F = k I l

k is a constant with unit=  N A-1 m-1

I is currently measured in mA = 10⁻³A

l is the length measured in mm= 10⁻³ m

F = k I l

F= N A-1 m-1× 10⁻³A×10⁻³ m

F= 10⁻⁶ N                 ( 1N = 10⁻⁵ dyne)

F= 0.1 dyne

Hence the unit of force will be dyne and the value of force will be 0.1 dynes.

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The pattern of lines of force around two positive charges separated from each other demonstrates the repulsive nature of like charges and the direction and strength of the electric field between them.

When two positive charges are separated from each other, the lines of force (also known as electric field lines) originate from one positive charge and terminate on the other positive charge. The lines of force follow a pattern that reflects the repulsion between the positive charges.

Here's a verbal description of the pattern:

From each positive charge, the lines of force radiate outward in all directions.

The lines of force are evenly spaced and radially symmetric around each charge, indicating that the electric field strength is the same at all points along a given line.

As the lines of force move away from each charge, they curve away from each other, reflecting the repulsion between the positive charges.

The density of lines of force is higher near the charges and decreases as they move further apart.

The lines of force never cross each other, maintaining their continuous and unbroken nature.

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There are several drawbacks to fiber-optic connectors. Due to severe restrictions on how much it can bend and the need for specialized training for employees working on the wire, its installation is expensive.

The three most widely used styles for networking and audio/video in the USA are LC, SC, and ST. LC and SC are typically the most popular styles. ST connectors are used less frequently these days. SC, ST, or MIC connectors are the most popular types of connectors for Fiber Optic networking. Networks using Token Ring use IDC/UDC. In Ethernet networking, connectors like RJ-45 and BNC are frequently used. The most widely utilised connectors in use today are ST, SC, FC, MT-RJ, & LC connectors, while less popular possibilities include Plastic FOC, Opti-Jack, LX-5, Volition, MU, and E2000.

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(a) The position-time equations for fishing boat is x = 3.4t + 40.8

(b) The time taken for the speedboat to catch up with the fishing boat is 7.1 seconds.

The position-time equations for fishing boat will be obtained by applying the principle of relative velocity as follows;

Let the time when the speedboat catches up = t

let the total time spent by fishing boat = t + 12

x = 3.4(t + 12)

x = 3.4t + 40.8

fishing boat is moving at a constant speed = 3.4 m/s

speed boat started moving 12 seconds later,

distance between the boats = 3.4 m/s  x 12 s = 40.8 m

v = u + at

v = 2.8 + 1.7t

set up the following equation to determine the time when the two boats meet.

vt - x = 40.8

t(2.8 + 1.7t) -  (3.4t + 40.8) = 40.8

2.8t + 1.7t² - 3.4t - 40.8 = 40.8

1.7t² - 0.6t  - 81.6 = 0

solve the quadratic equation using formula method as shown below;

a = 1.7, b = -0.6, c = -81.6

t = 7.1 seconds

Thus, the time taken for the speedboat to catch up with the fishing boat is 7.1 seconds.

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Answer:

A critical angle of 60.1°

Explanation:

Let's say

n1 Refractive index of rarer medium

n2 Refractive index of denser medium

So using the relation

စc= Sin^ -1(n1/n2)

So

စc = Sin^-1(1.3/1.5) = 60.1°

The statement that describes the transformation of the graph of function f onto the graph of function g is: The graph shifts 2 units right and 7 units down.

To determine the transformation of the graph of function f onto the graph of function g, we compare the two functions f(x) and g(x) and observe the changes in the equations.

The function f(x) = (1/x - 3) + 1 represents a reciprocal function that is shifted vertically 1 unit up and horizontally 3 units to the right. The reciprocal function is reflected about the line y = x.

The function g(x) = (1/(1 + 4)) + 3 simplifies to g(x) = 4 + 3 = 7, which is a constant function representing a horizontal line at y = 7.

By comparing the equations, we can see that the transformation from f(x) to g(x) involves the following changes:

The term 1/x in f(x) is replaced by the constant 1/(1 + 4) in g(x), resulting in a vertical shift of 7 units up.

The term -3 in f(x) is replaced by 3 in g(x), resulting in a vertical shift of 3 units up.

The +1 in f(x) is replaced by +3 in g(x), resulting in an additional vertical shift of 2 units up.

Therefore, the overall transformation is a shift of 2 units to the right and 7 units down.

Hence, the correct statement is: The graph shifts 2 units right and 7 units down.

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Use the kinematics formula: \(v = v_0+at\)

Your initial velocity is 10 m/s. Your acceleration is 4 m/s². The time elapsed is 4 s. And so,

\(v=10+4(4) \\v= 26 \text{ m/s}.\)

Answer:

I think you would need too time your friend and know how too divide that time by other racers to see how much faster she needs to get. I THINK.

The cost of operation for an A.C for 10 hours per day for a month will be 71,100 francs.

Power is the amount of energy transferred or converted per unit time. The unit of power is the watt, equal to one joule per second. Power is a scalar quantity.

Cost of operation for 10 hours a day;

Daily consumption = 3kW x 10 hours

Daily Consumption = 30kW

Since 1kWH = 79 francs;

Daily consumption amount = 30 x 79 francs

Daily consumption amount = 2,370 francs

Therefore, the monthly consumption (using 30days) will be;

2,370 francs x 30 = 71,100 francs

In conclusion, 71,100 francs will be spent in a month (30 days) to run the 3kW rated A.C for 10 hours a day at 1kWH.

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Jonathan needs to maintain a separation of 0.543 mm between the plates to get the desired charge, and a dielectric constant of 92.6 to achieve a separation of 5 mm with a dielectric.

(a) Using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage, we can solve for the capacitance: C = Q/V =\((8.15 x 10^-9 C) / (50 V) = 1.63 x 10^-10 F.\)

Then, using the formula for capacitance of parallel plate capacitors: C = ε0A/d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the separation distance, we can solve for the separation distance: d =\(_{3}OA/C = (8.85 x 10^-12 F/m) x (0.01 m^2) / (1.63 x 10^-10 F) = 0.543 mm.\)

(b) To find the dielectric constant, we can use the formula for capacitance of a parallel plate capacitor with a dielectric: C = εrε0A/d, where εr is the relative permittivity or dielectric constant of the material. Solving for εr, we get: εr = Cd / ε0A = \((1.63 x 10^-10 F)\) x (0.005 m) / \((8.85 x 10^-12 F/m)\) x \((0.01 m^2)\) = 92.6.

Therefore, Jonathan should use a dielectric with a relative permittivity of 92.6 to achieve a separation of 5 mm.

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