What is the maximum wavelength of incident light for which photoelectrons will be released from gallium

Answer:

The room with the lower temperature

Explanation:

Using

PV=nRT

Since both the rooms same volume and are connected, so they will have same pressure

PV=nRT=constant

nT=Constant/R=constant

If T is more n has to be less

Thus, lower the temperature, more the number molecules.

Answer:

A) 3.13 m/s

B) 5.34 N

C) W = 26.9 J

Explanation:

We are told that the position as a function of time is given by;

x(t) = αt² + βt³

Where;

α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³

Thus;

x(t) = 0.21t² + 0.0204t³

A) Velocity is gotten from the derivative of the displacement.

Thus;

v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)

v(t) = 0.42t + 0.0612t²

v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

a(t) = v'(t) = 0.42 + 2(0.0612t)

a(4.5) = 0.42 + 2(0.0612 × 4.5)

a(4.5) = 0.9708 m/s²

Force = ma = 5.5 × 0.9708

F = 5.3394 N ≈ 5.34 N

C) Since no friction, work done is kinetic energy.

Thus;

W = ½mv²

W = ½ × 5.5 × 3.1293²

W = 26.9 J

Answer:

Explanation:

That Universe Consists of Matter

Answer:

strong nuclear is the stronger force; it is related to mass

Explanation:

If we compare the force of gravity to strong nuclear force, we could conclude that strong nuclear is the stronger force; it is related to mass, therefore the correct answer is option C

The nuclear force is the interaction between the subatomic particles that make up a nucleus. There are two types of nuclear forces: the strong nuclear force and the weak nuclear force. Depending on the separation between the proton neutron and proton pairs, these nuclear forces can be both attracting and positive.

Both types of nuclear forces come under the four fundamental forces of nature. There are mainly four fundamental forces of nature electromagnetic force, gravitational force, strong nuclear force, and weak nuclear force.

Thus, Option C is the appropriate response since, when compared to the force of gravity, the strong nuclear force is the greater force because it is tied to mass.

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Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

Answer:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]

[tex]E_{left} = 9210.5 \ N/C[/tex]

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

[tex]E_{right} = -9210.5 \ N/C[/tex]

The electric field strength at the midpoint;

[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]

Answer:

They both have the same acceleration

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

Answer:

The electric field is [tex]E = 5.25 V/m[/tex]

Explanation:

From the question we are told that

    The peak magnitude of the magnetic field is  [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]

Generally the peak magnitude of the electric field is mathematically represented as

         [tex]E = c * B[/tex]

Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

So

       [tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]

       [tex]E = 5.25 V/m[/tex]

The peak magnitude of the electric field will be "5.25 V/m".

According to the question,

Magnetic field's peak magnitude, B = 17.5 nT or,

                                                           = 17.5 × 10⁻⁹ T

Speed of light, c = 3.0 × 10⁸ m/s

We know the relation,

→ E = c × B

By substituting the values, we get

      = 3.0 × 10⁸ × 17.5 × 10⁻⁹

      = 5.25 V/m

Thus the above approach is appropriate.

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Answer:

The average density of the rod is 1.605 kg/m.

Explanation:

The average density of the rod is given by:

[tex] \rho = \frac{m}{l} [/tex]    

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:

[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex]   (1)

Using u = x+1  →  du = dx  → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]

[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]

Therefore, the average density of the rod is 1.605 kg/m.  

I hope it helps you!    

The average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

Given data:

The length of rod is, L = 3 m.

The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].

To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3,  The expression for the average density is given as,

[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)

Using u = x+1  

du = dx

u₁= x₁+1 = 0+1 = 1

and

u₂ = x₂+1 = 3+1 = 4

By entering the values above into (1), we have:

[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]

Thus, we can conclude that the average density of the rod is  [tex]1.605 \;\rm kg/m^{3}[/tex].

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Answer:

a

   [tex]F = 67867.2 \ N[/tex]

b

  [tex]\Delta L = 2.6 \ mm[/tex]

Explanation:

From the question we are told that

      The Young modulus is  [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]

      The stress is  [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]

      The  diameter is  [tex]d = 2.40 \ cm = 0.024 \ m[/tex]

The radius is mathematically represented as

       [tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]

The cross-sectional area is  mathematically evaluated as

        [tex]A = \pi r^2[/tex]

         [tex]A = 3.142 * (0.012)^2[/tex]

        [tex]A = 0.000452\ m^2[/tex]

Generally the stress is mathematically represented as

        [tex]\sigma = \frac{F}{A}[/tex]

=>     [tex]F = \sigma * A[/tex]

=>    [tex]F = 1.50 *10^{8} * 0.000452[/tex]

=>    [tex]F = 67867.2 \ N[/tex]

Considering part b

      The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]

Generally Young modulus is mathematically represented as

           [tex]E = \frac{ \sigma}{ strain }[/tex]

Here strain is mathematically represented as

         [tex]strain = \frac{ \Delta L }{L}[/tex]

So    

       [tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]

        [tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]

=>     [tex]\Delta L = \frac{\sigma * L }{E}[/tex]

substituting values

       [tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]

       [tex]\Delta L = 0.0026[/tex]

Converting to mm

      [tex]\Delta L = 0.0026 *1000[/tex]

      [tex]\Delta L = 2.6 \ mm[/tex]

Answer:

4. Liters

5. Celsius

6. Grams

Answer:

The wavelength is  [tex]\lambda = 1805 nm[/tex]

Explanation:

From the question we are told that

    The wavelength of the light is  [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]

     The  distance of the screen is  D  =  3.0  m

     The  fringe width is  [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]

Generally the fringe width for a bright fringe  is mathematically represented as

          [tex]y = \frac{ \lambda * D }{d }[/tex]  

=>     [tex]d = \frac{ \lambda * D }{ y }[/tex]

=>     [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]

=>     [tex]d = 0.000373 \ m[/tex]

Generally the fringe width for a dark fringe  is mathematically represented as

      [tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

Here  m = 0  for  first order dark fringe

   So  

         [tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]

looking at which we see that   [tex]y_d = y[/tex]

         [tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]

=>    [tex]\lambda = 1805 *10^{-9} \ m[/tex]

=>    [tex]\lambda = 1805 nm[/tex]

Answer:

a)  q = 7.5 10⁻² C , b) 190 J , c)  I₀ = 50 A , d) 1.5 mC

Explanation:

The expression for capacitance is

            C = q / DV

            q = C DV

let's reduce the magnitudes to the SI system

            ΔV = 5 kV = 5000 V

            C = 15 μF = 15 10⁻⁶ F

              t = 90 μs = 90 10⁻⁶ s

            q = 15 10⁻⁶ 5000

            q = 7.5 10⁻² C

b) the energy in a capacitor is

             U = ½ C ΔV²

             U = ½ 15 10⁻⁶ 5000²

             U = 1,875 10² J

answer  190 J

c) At the moment the discharge begins, all the current is available and it decreases with time,

whereby

                V = I R

in the first instant I = Io

                I₀ = V / R

                I₀ = 5000/100

                I₀ = 50 A

but this is for a very short time

answer 50 A

d) The definition of current is

            i = dq / dt

in this case they give us the total current and the total time, so we can find the total charge

            i = q / t

            q = i t

            q = 17 90 10⁻⁶

            q = 1.53 10⁻³ C

answer is 1.5 mC

Answer:

Option C - she will speed up because air resistance has reduced and be less than gravity

Explanation:

We are told that Luz is skydiving with terminal velocity and her body parallel to the ground. Now, at this point she will be experiencing a gravitational force acting downwards, and also air resistance as a result of the drag force on her body

Since the downward gravitational force on Luz is constant, she will fall with a net force of;

F_net = F_g - F_d

where;

F_net is the net force on Luz acting downwards

F_g is the gravitational force on Luz

F_d is the drag force on Luz

The drag force on her body is proportional to the surface area of attack.

We are now told that Luz changes her body position to feet first toward the ground. This means that the surface area of attack is reduced because the feet will consume less space than the frontal part of her body. Thus, the drag force will be lesser then before she changed her body position due to reduced air resistance on her body.

Now, from earlier, we saw that;

F_net = F_g - F_d

So, the lesser F_d is, the higher F_net becomes.

Thus, she will speed up because air resistance has reduced and be less than gravity.

Answer:

C

Explanation:

EDGE 2020

Answer:

Binding Energy = 2.24 eV

Explanation:

First, we need to find the energy of the photon of light:

E = hc/λ

where,

E = Energy of Photon = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = 400 nm = 4 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(4 x 10⁻⁷ m)

E = (4.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 3.1 eV

Now, from Einstein's Photoelectric Equation:

E = Binding Energy + Kinetic Energy

Binding Energy = E - Kinetic Energy

Binding Energy = 3.1 eV - 0.86 eV

Binding Energy = 2.24 eV

Explanation:

Given that,

Linear speed of both disks is 5 m/s

Mass of disk 1 is 10 kg

Radius of disk 1 is 35 cm or 0.35 m

Mass of disk 2 is 3 kg

Radius of disk 2 is 7 cm or 0.07 m

(a) The angular velocity of disk 1 is :

[tex]v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s[/tex]

(b) The angular velocity of disk 2 is :

[tex]v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s[/tex]

(c) The moment of inertia for the two disk system is given by :

[tex]I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2[/tex]

Hence, this is the required solution.

Answer:

2,66

Explanation:

The refractive index= real depth/ apparent depth

real depth = refractive index * apparent depth

Let's assume index for water is 1.33

real depth = 2*1,33 = 2,66

Answer:

Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.

Explanation:

Answer:

"Endoscope" is the correct answer.

Explanation:

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.

For this "type" of motion, displacement (Δx) can be determined by:

[tex]\Delta x = v_{i}.t + \frac{a}{2}.t^{2}[/tex]

[tex]v_{i}[/tex] is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

[tex]\Delta x = 5(6) - \frac{4}{2}.6^{2}[/tex]

[tex]\Delta x = 30 - 2.36[/tex]

[tex]\Delta x[/tex] = - 42

Displacement of the boat for t=6.0s interval is [tex]\Delta x[/tex] = - 42m, i.e., 42 m to the left.

Answer:

18.5 cm

Explanation:

From;

1/u + 1/v = 1/f

Where;

u= object distance = 86cm

image height = 23 cm

Radius of curvature = 37 cm

The radius of curvature (r) is the radius of the sphere of which the mirror forms a part.

Focal length (f) = radius of curvature (r)/2 = 37cm/2 = 18.5 cm

Therefore, the focal length of the mirror is 18.5 cm

Answer:

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

Explanation:

given data

frequency f = 100 mhz = 100 × [tex]10^{6}[/tex] Hz

we consider here intensity I = 0.2 W/m²

solution

we take here plank constant is h i.e = 6.626 × [tex]10^{-34}[/tex] s

and take energy density is E

so here

E × C = I  

E = [tex]\frac{I}{C}[/tex]   ................1

here C = 3 × [tex]10^{8}[/tex] m/s

so photon density is

photon density = [tex]\frac{I}{C} \times \frac{1}{f \times h}[/tex]     ...............2

photon density = [tex]\frac{0.2}{3 \times 10^8} \times \frac{1}{100 \times 10^6 \times 6.626 \times 10^{-34} }[/tex]

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

Given that,

A charged particle enters a magnetic field with an angle theta .

If [tex]\theta=90^{\circ}[/tex]

We know that,

If the angle is 90° then the charged particle enters perpendicular to the B.

B is magnetic field.

The charged particle will be follow of the circular path.

If the angle is greater than 0 and less than 90° then the charged particle will be show the helical path.

Hence, This is required answer.

Answer:

1. Elastic collision

2. Inelastic collision    

Explanation:

Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision

the collision is described by the equation bellow

[tex]m1U1+ m2U2= m1V1+m2V2[/tex]

Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity

the collision is described by the equation bellow

[tex]m1U1+ m2U2= V(m1+m2)[/tex]

Answer:

The  displacement is  [tex]A_r = 6.071 \ cm[/tex]

Explanation:

From the question we are told that

   The initial displacement is [tex]A_o = 10 \ cm[/tex]

     The displacement at the end of first oscillation is  [tex]A_d = 9.05 \ cm[/tex]

Generally the damping constant of this damped oscillator is mathematically represented as  

           [tex]\eta = \frac{A_d}{A_o}[/tex]

substituting values

           [tex]\eta = \frac{9.05}{10}[/tex]

        [tex]\eta = 0.905[/tex]

The displacement after 4 more oscillation is mathematically represented as

       [tex]A_r = \eta^4 * A_d[/tex]

substituting values

      [tex]A_r = (0.905)^4 * (9.05)[/tex]

      [tex]A_r = 6.071 \ cm[/tex]

Answer:

Displacement reached is 6.0708 cm

Explanation:

Formula for damping Constant "C"

[tex]C^n=\frac{A_2}{A_1}[/tex]                  where n=1,2,3,........n

Where:

[tex]A_2[/tex] is the displacement after first oscillation    

[tex]A_1\\[/tex] is the initial Displacement

[tex]A_1=10\ cm\\A_2=9.05\ cm\\[/tex]

In our case, n=1.

[tex]C=\frac{9.05}{10}\\C=0.905[/tex]

After 4 more oscillation, n=4:

[tex]C^4=\frac{A_6}{A_2}[/tex]                                        

Where:

[tex]A_6[/tex] is the final Displacement after 4 more oscillations.

[tex]A_6=(0.905)^4*(9.05)\\A_6=6.0708\ cm[/tex]

Displacement reached is 6.0708 cm

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, [tex]E_p[/tex] = 95 V

the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V

the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns

the number of turns in the secondary coil, [tex]N_s[/tex] = ?

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]

[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 4145 turns.

Answer:

A. Ef/ Ei = 1/2

B. EF/ Ei = 1

C Ef / Ei = 4

Explanation:

To solve this we apply Coulomb's law which States that

E = Kq / r^2

Where

q = charge r = straight line distance from q to the point in question and

K = Coulomb's constant

Then

Ei = K Q / r^2

So

A) If Q is halved then

Ef = K Q / (2 r^2)

Ef/Ei = 1/2

B) If R is halved, the value of the E-f

at a distance r remains unchanged. So

Ef/Ei = 1

C) if r is now r/2 then

Ef = K Q / (r/2)^2 = K Q / r^2/4 = 4 K Q / r^2

Ef / Ei = 4

Answer:

Some parts of your question is missing attached below is the missing parts and the answer provided is pertaining to your question alone

answer : -6661.59 volts

Explanation:

The total electric potential can be calculated using this relation

V = k [tex](\frac{q1}{r1} + \frac{q2}{r2})[/tex]

q 1 = 1.62 uc

r1 = 4.00 m

q2 = -5.73 uc

r2 = 5.00 m  

k = 8.99 * 10^9 N.m^2/c^2

insert the given values into the above equation

V = ( 8.99 * 10^9 ) * [tex](\frac{1.62*10^{-6} }{4} + \frac{-5.73*10^{-6} }{5})[/tex]  =  -6661.59 volts

Answer:

Following are the answer to this question:

Explanation:

Formula:

[tex]D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J[/tex]

Calculating point A:

when the value is [tex]0.38[/tex]

[tex]\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\[/tex]

                   [tex]=2.632[/tex]

[tex]\to D(a.v) = \frac{1}{0.38} \times 206265\\[/tex]

               [tex]=542,802.6[/tex]

Calculating point B:

when the value is [tex]0.75[/tex]

[tex]\to D(PC)=\frac{1}{0.75}[/tex]

                [tex]=1.33[/tex]

[tex]\to D(a.v) = \frac{1}{0.75} \times 206265\\[/tex]

             [tex]=275,020[/tex]

Calculating point C:

when the value is [tex]0.28[/tex]

[tex]\to D(PC)=\frac{1}{0.28}[/tex]

                [tex]=3.571[/tex]

[tex]\to D(a.v) = \frac{1}{0.28} \times 206265\\[/tex]

               [tex]=736660.7[/tex]

Calculating point D:

when the value is [tex]0.42[/tex]

[tex]\to D(PC)=\frac{1}{0.42}[/tex]

                [tex]=2.38[/tex]

[tex]\to D(a.v) = \frac{1}{0.42} \times 206265\\[/tex]

               [tex]=490910.7[/tex]

Calculating point E:

when the value is [tex]0.31[/tex]

[tex]\to D(PC)=\frac{1}{0.31}[/tex]

                [tex]=3.226[/tex]

[tex]\to D(a.v) = \frac{1}{0.31} \times 206265\\[/tex]

               [tex]=665370.97[/tex]