You have explored constructive interference from multi-layer thin films. It is also possible for interference to be destructive, a phenomenon exploited in making antireflection coatings for optical elements such as eyeglasses. In order to allow the lenses to be thinner (and thus lighter weight), eyeglass lenses can be made of a plastic that has a high index of refraction (np = 1.70). The high index causes the plastic to reflect light more effectively than does glass, so it is desirable to reduce the reflection to avoid glare and to allow more light to reach the eye. This can be done by applying a thin coating to the plastic to produce destructive interference.

a. Consider a plastic eyeglass lens with a coating of thickness d with index nc . Light with wavelength is incident perpendicular to the lens. If nc < n p , then determine an equation for d in terms of the given variables (and an integer m) in order for there to be destructive interference between the light reflected from the top of the coating and the light reflected from the coating/lens interface.
b. Repeat part a assuming that nc > n p .
c. Choose a suitable value for nc and calculate a value for d that will result in destructive interference for 500 nm light. Note that materials to use for coatings that have nc < 1.3 or nc > 2.5 are difficult to find.
d. Does the index of refraction n p of the eyeglass lens itself matter? Explain.

Answer:

Explanation:

Potential energy stored in the spring = 25 J

This energy is converted into gravitational potential energy . If h be the height attained

gravitational potential energy = mgh

mgh = 25

.15 x 9.8 x h = 25

h = 17 m

Answer:

q = C V    charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 *  10E-4  C per capacitor

N = Q / q = 1 / 1.1 * 10E-4  = 9091 capacitors

9.09 × 10³ capacitors must be connected in parallel.

Given C = 1.00μF = 1 × 10⁻⁶ F

          q = 1.00 C

          V = 110 V

The equivalent capacitance is given by

Ceq = q/V

where q = total charge on all the capacitors

             V = potential difference

For N number of identical capacitors in parallel,

Ceq = NC

Therefore,

NC = q/V

N = q/VC

Putting on the values in the above formula,

N = 1/ (110)(1 × 10⁻⁶)

   = 1 / 110 × 10⁻⁶

   = 9.09 × 10³

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Answer:

    I₂ = 0.25 I₀

Explanation:

To know the light transmitted by a filter we must use the law of Malus

          I = I₀ cos² θ

In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.

        I₁ = I₀ cos² 45

        I₁ = I₀ 0.5

this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂

       I₂ = I₁ cos² 45

       I₂ = (I₀ 0.5) 0.5

       I₂ = 0.25 I₀

this is the intensity of the light transmitted by the set of polarizers

Answer:

polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.

the cup helps it become more insulated

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

Answer:

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

Answer:

Pencil is to pen

Step by step explanation:

They are similar items, as they are both tools, but are different as to how they function.

Answer:

The  velocity is [tex]v = 37 .46 \ m/s[/tex]

Explanation:

From the question we are told that

    The start distance above the ground is  [tex]h = 71.6 \ m[/tex]

Generally according to the  law of energy conservation we have that

     [tex]PE_{top} = KE_{bottom }[/tex]

Where [tex]PE_{top}[/tex] is potential energy at the top which is mathematically represented as

      [tex]PE_{top} = m * g * h[/tex]

And  [tex]KE_{bottom }[/tex] is the kinetic energy at the bottom which is mathematically represented as

     [tex]KE_{bottom } = \frac{1}{2} * m * v^2[/tex]

Therefore  

       [tex]m * g * h = \frac{1}{2} * m * v^ 2[/tex]

=>    [tex]v = \sqrt{2 * g * h }[/tex]

substituting value

     [tex]v = \sqrt{2 * 9.8 * 71.6 }[/tex]

     [tex]v = 37 .46 \ m/s[/tex]

Answer:

C. The magnitude of the electron's acceleration inside the field

D. The radius of the circular path the electron travels

Explanation:

The radius of the electron's motion in a uniform magnetic field is given by

[tex]R = \frac{MV}{qB}[/tex]

where;

m is the mass of the electron

q is the charge of the electron

B is the magnitude of the magnetic field

V is speed of the electron

R is the radius of the electron's

Thus, the radius of the of the electron's motion will change since it depends on speed of the electron.

The magnitude of the electron's acceleration inside the field  is given by;

[tex]a_c = \frac{V^2}{R}[/tex]

where;

[tex]a_c[/tex] is centripetal acceleration of electron

Thus, the magnitude of the electron's acceleration inside the field will change since it depends on the electron speed.

The time the electron is in the magnetic field is given by;

[tex]T = \frac{2\pi M}{qB}[/tex]

The time of electron motion will not change

The magnitude of the net force acting on the electron inside the field will not change;

[tex]qVB = \frac{MV^2}{R} \\\\qVB - \frac{MV^2}{R} = 0[/tex]

Therefore, the correct options are "C" and "D"

Answer:

50000 turns

Explanation:

Vp / Vs = Np / Ns

240 / 12000 = 1000 / Ns

Ns = 50000 turns

Answer:

Explanation:

In a single slit experiment, the distance on a screen from the centre point is expressed as y = [tex]\frac{\delta m \lambda d}{a}[/tex] where;

[tex]\delta m[/tex] is the first two diffraction minima = 1

[tex]\lambda[/tex] is light wavelength

d is the distance of diffraction pattern from the screen

a is the width of the slit

Given [tex]\lambda[/tex] = 460-nm = 460*10⁻⁹m

d = 5.0mm = 5*10⁻³m

a = 1.4mm = 1.4*10⁻³m

Substituting this values into the formula above to get width of the central maximum y;

y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³

y = 2300*10⁻¹²/1.4*10⁻³

y = 1642.86*10⁻⁹

y = 1.643*10⁻⁶m

Converting the final value to cm,

since 100cm = 1m

x = 1.643*10⁻⁶m

x = 1.643*10⁻⁶ * 100

x = 1.643*10⁻⁴cm

Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is  1.643*10⁻⁴cm

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

λ = 5.2 x 10⁻⁷ m = 520 nm

Answer:

2.78 cm³/s

Explanation:

From the question,

Q = v/A'.................... Equation 1

Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.

Given: v = 100 km/h, A' = 10 km/L

Substitute this value into equation 1

Q = 100/10

Q = 10 L/h.

Now, we convert L/h to cm³/s.

Since,

1 L = 1000 cm³, and

1 h = 3600 s

Therefore,

Q = 10(1000/3600) cm³/s

Q = 2.78 cm³/s

Answer:

fully describes the concave mirror is + f

Explanation:

A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f

This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di

Answer:

+f

Explanation:

because you have to be really dumb to get an -f

Answer:

2

Explanation:

since the second object was in it stationary, it velocity is 0 m/s

Answer:

Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)

Where the charge of a single proton is:

C = 1.6x10^-19 C

Now, you need to remember that when we are working with charges, we are working with discrete math:

What means that?

If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)

So the consecutive charge will be:

C = 2*1.6x10^-19 C = 3.2x10^-19 C.

So, because we are working in discrete math, we can not have any object that has charge between  1.6x10^-19 C and 3.2x10^-19 C.

Particularly, 2.0x10^-19 C is in that range, so we can conclude that:

No, an object can not carry a charge of 2.0x10^-19 C.

Answer:

Option (A)

Explanation:

A 20 kg boy chases the butterfly with a speed of 2 meter per second.

Angle at which he runs is 70° North of West.

Therefore, Horizontal component (Vx) directing towards West will be,

Vx = v(Cos70°)

Vy = v(Sin70°)

Since momentum of a body is defined by,

Momentum = Mass × Velocity

Therefore, Westerly component of the momentum will be,

Momentum = 20 × (v)(Cos70°)

                   = 20 × 2Cos70°

                   = 13.68

                   ≈ 13.7 kg-meter per second

Therefore, Option (A) will be the answer.

Answer:

Anti-Particle

Answer:

Explanation:

We shall apply the theory of

heat lost = heat gained .

heat lost by water = mass x specific heat x temperature diff

= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J  

heat gained by ice to attain temperature of zero

= m x 2100 x 22.8 = 47880 m

heat gained by ice in melting = latent heat x mass

= 334000m

heat gained by water at zero to become warm at 32 degree

= m x 4190 x 32 = 134080 m

Total heat gained = 515960 m

So

515960 m = 51587.28

m = .1 kg

= 100 gm

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

c. The incident light must have at least as much energy as the electron work function

Answer:

[tex]1.357\times 10^{-12}[/tex]

Explanation:

Relevant Data provided

Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2

Distance which indicates d = 1.50 x 10^-3 m

Voltage which indicates V = 12 V

According to the requirement, the computation of value of its capacitance is shown below:-

[tex]Capacitance, C = \frac{\epsilon oA}{D}[/tex]

[tex]= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}[/tex]

= [tex]1.357\times 10^{-12}[/tex]

Therefore for computing the capacitance we simply applied the above formula.

Answer:

Electric field strengh is a measure of the strength of an electric field at a given point in space, equal to the field would induce on a unit electric charge at that point.

Electric field strength is also known as Electric Field Intensity .

Explanation:

Electric Field is also defined as force per charge. The unit will be force unit divided by charge unit. In this case, it will be Newton/Coulomb or N/C.

Please mark me as the brainliest!!!

Thanks!!!

Answer:

Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.

a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Explanation:

Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of  [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)

Construct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.

The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:

[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].

The question described another equation for the net force on this mass. This equation is the sum of two parts:

Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:

[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].

Combine the two equations for the net force on this mass to obtain:

[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].

From the question:

Hence, the equation will become:

[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].

Rearrange to obtain:

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:

[tex]m^2 + 11\, m + 18 = 0[/tex].

The two roots are:

Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:

[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].

For this particular ODE, that general solution would be:

[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].

Note, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time

For section [tex]\rm a.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm a.[/tex] will be:

[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

Similarly, for section [tex]\rm b.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm b.[/tex] will be:

[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Answer:

Explanation:

According to first law of thermodynamics:

∆U= q + w

= 10kj+(-70kJ)

-60kJ

, w = + 70 kJ

(work done on the system is positive)

q = -10kJ ( heat is given out, so negative)

∆U = -10 + (+70) = +60 kJ

Thus, the internal energy of the system decreases by 60 kJ.

Answer:

(A)

Explanation:

We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.

Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.

Answer:

120000    kgxm/s

Explanation:

momentum is mass times velocity so just multiply 1600 kg times 75 m/s and you get 120000  kgxm/s

Explanation:

The statement "our sun is an 'average' sun" is  true when it is used to describe or characterize some unique physical properties of stars generally in the universe. 'Average' in this sense is used to define a typical sun such as, "stars should glow like our sun an average star."

The statement is used wrongly when used to in quantifying other stars in the universe, based on calculated values from our sun. In this case, we cannot truly say if our sun is a true representative average of other stars in the universe.

Yes! it is useful to characterize the milky way by simply citing the average sun. Properties like their ability to glow and radiate heat can be defined by citing an average star like our sun, so long as we don't translate it into citing quantitative properties of the sun as an average of our Milky Way Galaxy like the mass, temperature, etc.

Answer:

The extension is  [tex]\Delta L = 0.033 \ m[/tex]

Explanation:

From the question we are told that  

     The length of the wire on a winter day is    [tex]L_w = 36.0 \ m[/tex]

      The temperature on the winter day is  [tex]T_w = -20.0 ^o C[/tex]

      The temperature on a summer day is  [tex]T_s = 34.0 ^0 C[/tex]

The the extension of the wire on a summer day is mathematically represented as

          [tex]\Delta L = \alpha L_w [T_s - T_w][/tex]

Where  

      [tex]\alpha[/tex] is the  coefficient of linear expansion of copper with a values [tex]\alpha = 17 *10^{-6}[/tex]

substituting value  

      [tex]\Delta L = 17 *10^{-6} * 36.0 [34 - [-20]][/tex]

      [tex]\Delta L = 0.033 \ m[/tex]

The two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

Electromagnetic waves are components of the electromagnetic spectrum, which is made up of the following:

Each electromagnetic wave have a specific frequency and wavelength.

However, the two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

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Answer:

gamma rays and X-rays

Explanation:

d on edge I got 100%

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

Answer:

i was going to but its to late

Explanation: