If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed

Answer:

The acceleration is   [tex]a =2\ m/s^2[/tex]

Explanation:

From the question we are told that

       The  mass of the jumbo jet is  [tex]m_j = 100000\ kg[/tex]

        The thrust is  [tex]F_k = 50000 \ N[/tex]

Generally given that the jet has four engines the total thrust is  

        [tex]F_t = 4 * F_k[/tex]

substituting values

       [tex]F_t = 4 * 50000[/tex]

      [tex]F_t = 200000 \ N[/tex]

Generally the acceleration of the is mathematically represented as

         [tex]a = \frac{F_t}{m}[/tex]

substituting values

       [tex]a =2 \frac{N}{kg}[/tex]

Now  

        [tex]N = kg \cdot m/s^2[/tex]

Hence

         [tex]a =2 \frac{kg * \cdot m/s^2}{kg}[/tex]

        [tex]a =2\ m/s^2[/tex]

Answer:

Explanation:

The lens equation is expressed as;

1/f = 1/u+1/v where;

f is the focal length of the lens

u is the object distance

v is the image distance

Since the near sighted person wants focus the starts at nigt, the stars at night are the images located that infinity. Hence the image distance v = ∞.

The object distance u = 130cm

Substituting the given parameters in the formula to get the focal length f

[tex]\frac{1}{f} = \frac{1}{\infty} + \frac{1}{130} \\\\As \ x \ tends \ to \ \infty, \, \frac{a}{x} \ tends \ to \ 0 \ where\ 'a' \ is \ a\ constant \\\\} \\\\[/tex]

[tex]\frac{1}{f} = 0+ \frac{1}{130}\\\\[/tex]

[tex]\frac{1}{f} =\frac{1}{130}\\cross\ multiply\\\\f = 130*1\\\\f = 130cm[/tex]

Hence the focal length of contact lenses that would allow a nearsighted person with a 130 cm far point to focus on the stars at night is 130cm

Answer: no u

Explanation: no u

Answer:

a)   Δt = 24.96 s , b)  τ = 0.078 N m

Explanation:

This is a rotational kinematics exercise

        θ = w₀ t - ½ α t²

Let's reduce the magnitudes the SI system

       θ = 60 rev (2π rad / 1 rev) = 376.99 rad

       w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s

      α = (w₀ t - θ) 2 / t²

let's calculate the annular acceleration

      α = (43.98 10 - 376.99) 2/10²

      α = 1,258 rad / s²

Let's find the time it takes to reach zero angular velocity (w = 0)

        w = w₀ - alf t

         t = (w₀ - 0) / α

         t = 43.98 / 1.258

         t = 34.96 s

this is the total time, the time remaining is

         Δt = t-10

         Δt = 24.96 s

To find the braking torque, we use Newton's law for angular motion

        τ = I α

the moment of inertia of a circular ring is

       I = M r²

we substitute

         τ = M r² α

we calculate

        τ = 0.625  0.315²  1.258

        τ = 0.078 N m

The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

Given data:

The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex]   (rps means rotation per second).

The time interval is, t' = 10.0 s.

The number of rotations made by wheel is, n = 60.0.

The mass of bike wheel is, m = 0.625 kg.

The radius of wheel is, r = 0.315 m.

The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,

[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]

Here, [tex]\theta[/tex] is the angular displacement, and its value is,

[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]

And, angular speed is,

[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]

Solving as,

[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]

Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.

[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]

Then total time is,

T = t - t'

T = 35.18 - 10

T = 25.18 s

Now, use the standard formula to obtain the value of braking torque as,

[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]

Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

Learn more about the rotational motion here:

https://brainly.com/question/1388042

Answer:

Hey there!

PE=mgh, so as height decreases, so does the potential energy.

KE=mv^2, so as velocity increases, kinetic energy increases.

Thus, the correct answer would be Decreases, Increases.

Let me know if this helps :)

Answer:

d²x/dt² = - 4dx/dt - 4x is the required differential equation.

Explanation:

Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,

F = mg

kx = mg

k = mg/x

= 4 kg × 9.8 m/s²/2.45 m

= 39.2 kgm/s²/2.45 m

= 16 N/m

Now the drag force f = 16v where v is the velocity of the mass.

We now write an equation of motion for the forces on the mass. So,

F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass

-kx - 16v = 4a

-16x - 16v = 4a

16x + 16v = -4a

4x + 4v = -a where v = dx/dt and a = d²x/dt²

4x + 4dx/dt = -d²x/dt²

d²x/dt² = - 4dx/dt - 4x which is the required differential equation

Answer:

The  rotational kinetic energy is  [tex]K = 2116.3 \ J[/tex]

Explanation:

From the question we are told that

    The moment of inertia  is  [tex]I = 0.35 \ kg \cdot m^2[/tex]

    The number of revolution is N  =  70 revolution

     The  time taken is  t  =  4.0  s

Generally the angular velocity is mathematically represented as

      [tex]w = \frac{2 \pi N }{t }[/tex]

substituting values

      [tex]w = \frac{2* 3.142 * 70 }{4 }[/tex]

       [tex]w = 109.97 \ rad/s[/tex]

The rotational kinetic energy K i mathematically represented as

       [tex]K = \frac{1}{ 2} * I * w^2[/tex]

substituting values

       [tex]K = \frac{1}{ 2} * 0.35 * (109.97)^2[/tex]

       [tex]K = 2116.3 \ J[/tex]

Answer:

The period of the motion will still be equal to T.

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

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

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.

Answer:

Using a process of testing ideas and gathering evidence

Explanation:

Answer:

196 nm

Explanation:

Given that

Value of wavelength, = 490 nm

Time spent in air, t(a) = 17.5 ns

Thickness of glass, th = 0.8 m

Time spent in glass, t(g) = 21.5 ns

Speed of light in a vacuum, c = 3*10^8 m/s

To start with, we find the difference between the two time spent

Time spent on glass - Time spent in air

21.5 - 17.5 = 4 ns

0.8/(c/n) - 0.8/c = 4 ns

Note, light travels with c/n speed in media that has index of refraction

(n - 1) * 0.8/c = 4 ns

n - 1 = (4 ns * c) / 0.8

n - 1 = (4*10^-9 * 3*10^8) / 0.8

n - 1 = 1.2/0.8

n - 1 = 1.5

n = 1.5 + 1

n = 2.5

As a result, the wavelength of light in a medium with index of refraction would then be

490 / 2.5 = 196 nm

Therefore, our answer is 196 nm

Answer:

[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.

Explanation:

According to current SI unit conversions, 1 atmosphere is equal to 760 millimeters of mercury. The current pressure is determined by simple rule of three:

[tex]p = \frac{760\,mm\,Hg}{1\,atm} \times (1\times 10^{-6}\,atm)[/tex]

[tex]p = 7.6\times 10^{-4}\,mm\,Hg[/tex]

[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.

Answer:

Distance= 4 miles

Time = 36.3 seconds

Explanation:

80 mph = 178.95 m/s

90 mph = 201.32 m/s

V = u +at

201.32= 0+a(4.5)

201.32/4.5= a

44.738 m/s² = a

Acceleration of the cop car

= 44.738 m/s²

Distance traveled at 4.5seconds

For the cop car

S= ut + ½at²

S= 0(4.5) + ½*44.738*4.5

S= 100.66 meters

Distance traveled at 4.5seconds

For the speeding car

4.5*178.95=805.275

The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal

X/178.95= (704.675+x)/201.32

X-0.89x= 626.37

0.11x= 626.37

X= 5694.3 meters

The time = 5694.3/178.95

Time =31.8 seconds

So the distance they meet

= 5694.3+805.275

= 6499.575 meters

= 4.0 miles

The Time = 4.5+31.8

Time = 36.3 seconds

Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{1}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{1}=0.01021\ Wb[/tex]

We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{2}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{2}=0.00127\ Wb[/tex]

We need to calculate the magnitude of the average induced emf

Using formula of emf

[tex]\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})[/tex]

Put the value into t5he formula

[tex]\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})[/tex]

[tex]\epsilon=0.016867\ V[/tex]

[tex]\epsilon=16.87\ mV[/tex]

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

[tex]I=\dfrac{\epsilon}{R}[/tex]

Put the value into the formula

[tex]I=\dfrac{0.016867}{2.5}[/tex]

[tex]I=0.00675\ A[/tex]

[tex]I=6.75\ mA[/tex]

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Answer:

y = 20.99 V / A

there is no friction    y = 20.99 h

Explanation:

Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy

when the block of ice is submerged it is subjected to two forces its weight  hydrostatic thrust

              F_net= ∑F = B-W

the expression stop pushing is

              B = ρ_water g V_ice

where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice

We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³

             W = ρ-ice g V

              F_net = (ρ_water - ρ_ ice) g V

this is the net force directed upwards, we can find the potential energy with the expression

            F = -dU / dy

            ΔU = - ∫ F dy

            ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy

            ΔU = - (ρ_water - ρ_ ice) g A y² / 2

we evaluate between the limits y = 0,  U = 0, that is, the potential energy is zero at the surface

             U_ice = (ρ_water - ρ_ ice) g A y² / 2

now we can use the conservation of mechanical energy

starting point. Ice depth point

             Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2

final point. Highest point of the block

             [tex]Em_{f}[/tex] = U = m g y

as there is no friction, energy is conserved

            Em₀ = Em_{f}

            (ρ_water - ρ_ ice) g A y² / 2 = mg y

let's write the weight of the block as a function of its density

            ρ_ice = m / V

            m = ρ_ice V

we substitute

             (ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y

              y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A

let's substitute the values

             y = 0.913 / (1 - 0.913) 2 V / A

             y = 20.99 V / A

This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction

                V / A = h

where h is the height of the block

                 y = 20.99 h

Answer:

Explanation:

We can conclude that the forces of the two teams are equal and opposite and hence they cancel each other. Therefore none of the teams won as the rope did not move.

hope this helps

plz mark as brainliest!!!!!!!

Answer:

The question is not clear enough. So i have attached a copy of the correct question.

A) B = V/c

B) c = Lv

Explanation:

A) we know that formula for magnetic flux is;

Φ = BA

Where B is magnetic field and A is area

Now,

Let's differentiate with B being a constant;

dΦ/dt = B•dA/dt

From faradays law, the EMF induced is given as;

E = -dΦ/dt

However, we want to express it in terms of V and E.M.F is also known as potential difference or Voltage.

Thus, V = -dΦ/dt

Thus, we can now say that;

-V = B•dA/dt

Now from the question, we are told that dA/dt = - c

Thus;

-V = B•-c

So, V = Bc

Thus, B = V/c

B) according to Faraday's Law or Lorentz Force Law, an electromotive force, emf, will be induced between the two ends of the sidelength:

Thus;

E =LvB or can be written as; V = LvB

Where;

V is EMF

L is length of bar

v is velocity

From the first solution, we saw that;

V = Bc

Thus, equating both of the equations, we have;

Bc = LvB

B will cancel out to give;

c = Lv

Explanation:

Answer:

The number of  fringe pattern shift is   m  = 40

Explanation:

 From the question we are told that

      The  Michelson interferometer is moved 20 wavelengths i.e  [tex]20 \lambda[/tex]

Generally the distance which the  Michelson interferometer is moved is mathematically represented as

         [tex]d = \frac{m * \lambda}{2}[/tex]

Here [tex]m[/tex] is the number of  fringe pattern shift

     So

         [tex]20 \lambda = \frac{m * \lambda}{2}[/tex]

         [tex]40 \lambda = m * \lambda[/tex]

        m  = 40

Answer:

352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.

Explanation:

The decay of radioisotopes are represented by the following ordinary differential equation:

[tex]\frac{dm}{dt} = -\frac{t}{\tau}[/tex]

Where:

[tex]t[/tex] - Time, measured in seconds.

[tex]\tau[/tex] - Time constant, measured in seconds.

[tex]m[/tex] - Mass of the radioisotope, measured in grams.

The solution of this expression is:

[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]

Where [tex]m_{o}[/tex] is the initial mass of the radioisotope, measured in kilograms.

The ratio of current mass to initial mass is:

[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]

The time constant is now calculated in terms of half-life:

[tex]\tau = \frac{t_{1/2}}{\ln2}[/tex]

Where [tex]t_{1/2}[/tex] is the half-life of the radioisotope, measured in seconds.

Given that [tex]t_{1/2} = 88\,s[/tex], the time constant of the radioisotope is:

[tex]\tau = \frac{88\,s}{\ln 2}[/tex]

[tex]\tau \approx 126.957\,s[/tex]

Now, if [tex]\frac{m(t)}{m_{o}(t)} = \frac{1}{16}[/tex] and [tex]\tau \approx 126.957\,s[/tex], the time is:

[tex]t = -\tau \cdot \ln\frac{m(t)}{m_{o}}[/tex]

[tex]t = -(126.957\,s)\cdot \ln \frac{1}{16}[/tex]

[tex]t \approx 352\,s[/tex]

352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.

Answer:

D. evacuated tube collector due to reduced heat loss

Explanation:

Evacuated tube collectors has vacuum which reduces the loss of heat and increase the efficiency of the collector. It has a major application in solar collector, and converts solar energy to heat energy. It can also be used for heating of a definite volume of water majorly for domestic purpose.

During cold weather, the conservation and efficient use of heat is required. Therefore, evacuated tube collector is preferred so as to reduce heat loss and ensure the maximum use of heat energy.

Answer:

μ = 1

F = P√2

Explanation:

The parabola equation is: y = ½ x².

The slope of the tangent is dy/dx = x.

The angle between the tangent and the x-axis is θ = tan⁻¹(x).

At x = 1, θ = 45°.

Draw a free body diagram of the block.  There are three forces:

Weight force P pulling down,

Normal force N pushing perpendicular to the surface,

and friction force Nμ pushing up tangential to the surface.

Sum of forces in the perpendicular direction:

∑F = ma

N − P cos 45° = 0

N = P cos 45°

Sum of forces in the tangential direction:

∑F = ma

Nμ − P sin 45° = 0

Nμ = P sin 45°

μ = P sin 45° / N

μ = tan 45°

μ = 1

Draw a new free body diagram.  This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.

Sum of forces in the tangential direction:

∑F = ma

F − Nμ − P sin 45° = 0

F = Nμ + P sin 45°

F = (P cos 45°) μ + P sin 45°

F = P√2

Answer: (V1+V2)/2

Explanation: This is because basically with the question they are trying to say u(initial velocity) is V1 and v(final velocity) is V2 as the journey starts off with V1 and ends with V2 so therefore we know an equation where average velocity=(u+v)/2. So here it’s (V1+V2)/2

Answer: 156.02 metre.

Explanation:

Give that a certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied.

Let us use third equation of motion,

V^2 = U^2 + 2as

Since the car is decelerating, V = 0

And acceleration a will be negative.

U = 33 mph

S = 39 m

Substitute both into the formula

0 = 33^2 - 2 × a × 39

0 = 1089 - 78a

78a = 1089

a = 1089 / 78

a = 13.96 m/h^2

If we assume that the car decelerate at the same rate.

the distance the car will stop had it been going 66.0mph will be achieved by using the same formula

V^2 = U^2 + 2as

0 = 66^2 - 2 × 13.96 × S

4356 = 27.92S

S = 4356 / 27.92

S = 156.02 m

Therefore, the car would stop at

156.02 m

Answer:

C. planetary accretion

Explanation:

Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.

Answer:

[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]

Explanation:

Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.

Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.

Answer:

The width of Center bright fringe is 10.2mm

Explanation:

Given that if

Y/ L << 1 then

Sin theta will be approx Y/L

So sin theta approx Y/L = lamda/a

Y= a x lambda/a

By substituting

1.9x 10^ -3m x 550*10^-9/ 0.2 x 10^-3m

= 5.2mm

But

Change in y = 2y = 10.4mm

Answer:

9.82 × [tex]10^{-35}[/tex] Hz

Explanation:

De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:

λ = [tex]\frac{h}{mv}[/tex]

where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.

Given that: h = 6.63 ×[tex]10^{-34}[/tex] Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;

λ = [tex]\frac{h}{mv}[/tex]

  = [tex]\frac{6.63*10^{-34} }{2.5*2.7}[/tex]

 = [tex]\frac{6.63 * 10^{-34} }{6.75}[/tex]

 = 9.8222 × [tex]10^{-35}[/tex]

The wavelength of the object is 9.82 × [tex]10^{-35}[/tex] Hz.

Answer:

The current is  [tex]I = 1420 \ A[/tex]

Explanation:

From the question we are told that

   The  diameter of the wire is  [tex]d = 5.68 \ mm = 0.00568 \ m[/tex]

    The  magnetic field is  [tex]B = 0.100 \ T[/tex]

Generally the radius of the wire is mathematically evaluated as

       [tex]r = \frac{d}{2}[/tex]

substituting values

     [tex]r = \frac{ 0.00568}{2}[/tex]

     [tex]r = 0.00284 \ m[/tex]

Generally the magnetic field is mathematically represented as

       [tex]B = \frac{\mu_o * I}{ 2 \pi r }[/tex]

=>    [tex]I =\frac{ B * 2 \pi r }{\mu_o}[/tex]

Here [tex]\mu_o[/tex] is the permeability of free space  with value [tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]

substituting values

=>     [tex]I =\frac{ 0.100 * 2 * 3.142 * 0.00284 }{ 4 \pi * 10^{-7}}[/tex]

=>     [tex]I = 1420 \ A[/tex]

Answer:

1.1micro meter

Explanation:

Given that

Constructive interference is

ma = alpha x sin theta

Alpha = 1 x 594 x10^ -9/ sin 32.8°

= 1.1 x 10^ -6m

Explanation:

Answer:

C. 450v

Explanation:

Using

Voltage= B*distance of separation*velocity

3mm x 0.3T x 5E5m/s

= 450v

Answer:

(a). The largest value of θ is 71.9°.

(b). The second largest value of θ is 57.7°.

(c). The third largest value of θ is 47.7° .

Explanation:

Given that,

Width of diffraction grating [tex]w= 19.2\ mm[/tex]

Number of rulings[tex]N=6010[/tex]

Wavelength = 337 nm

We need to calculate the distance between adjacent rulings

Using formula of distance

[tex]d=\dfrac{w}{N}[/tex]

Put the value into the formula

[tex]d=\dfrac{19.2\times10^{-3}}{6010}[/tex]

[tex]d=3.19\times10^{-6}\ m[/tex]

We need to calculate the value of m

Using formula of constructive interference

[tex]d \sin\theta=m\lambda[/tex]

[tex]\sin\theta=\dfrac{m\lambda}{d}[/tex]

Here, m = 0,1,2,3,4......

[tex]\lambda[/tex]=wavelength

For largest value of  θ

[tex]\dfrac{m\lambda}{d}>1[/tex]

[tex]m>\dfrac{d}{\lambda}[/tex]

Put the value into the formula

[tex]m>\dfrac{3.19\times10^{-6}}{337\times10^{-9}}[/tex]

[tex]m>9.46[/tex]

[tex]m = 9[/tex]

(a). We need to calculate the largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{9\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=71.9^{\circ}[/tex]

(b). We need to calculate the second largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{8\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=57.7^{\circ}[/tex]

(c). We need to calculate the third largest value of θ

Using formula of constructive interference

[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]

Now, put the value of m in to the formula

[tex]\theta=\sin^{-1}(\dfrac{7\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]

[tex]\theta=47.7^{\circ}[/tex]

Hence, (a). The largest value of θ is 71.9°.

(b). The second largest value of θ is 57.7°.

(c). The third largest value of θ is 47.7° .