In a double-slit experiment, the slits are illuminated by a monochromatic, coherent light source having a wavelength of 527 nm. An interference pattern is observed on the screen. The distance between the screen and the double-slit is 1.54 m and the distance between the two slits is 0.102 mm. A light wave propogates from each slit to the screen. What is the path length difference between the distance traveled by the waves for the fifth-order maximum (bright fringe) on the screen

Answer:

[tex]\lambda_s =6.43*10^-4m[/tex]

Explanation:

From the question we are told that:

Diffraction grating [tex]N=1250slits/cm[/tex]

Distance b/w Screen and grating length [tex]d_{sg}=17.5 cm[/tex]

Distance b/w neighboring maxima and Screen [tex]d_{ms}=8.44[/tex]

Generally the equation for grating space is mathematically given by

[tex]d(g)=\frac{1}{N}[/tex]

[tex]d(g)=\frac{100}{1250}[/tex]

[tex]d(g)=0.08[/tex]

Generally the equation for small angle approximation is mathematically given by

[tex]\triangle y=\frac{\lambda d}{L}[/tex]

Therefore for longest wavelength

[tex]\lambda _l=\frac{8.44*10^{-3}*(0.08)}{0.175m}[/tex]

[tex]\lambda _l=3.858*10^{-3}[/tex]

Therefore the third order maximum equation for the shorter wavelength as

[tex]\lambda_s =\frac{1}{6} \lambda_l[/tex]

[tex]\lambda_s =\frac{1}{6} (3.858*10^-^3)[/tex]

[tex]\lambda_s =6.43*10^-4m[/tex]

The wavelengths in the light is given as

[tex]\lambda_s =6.43*10^-4m[/tex]

When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.

Answer:

please give me brainlist and follow

Explanation:

Using locally available resources for art help in the preservation of environment. A significant and practical aspects of art is material significance. The items used by artists while making an art piece affects both the form and the material. Every material delivers something special in the creative process.

Answer:

239 rpm

Explanation: So the distance covered in one minute is 75,000 centimeters. The diameter of the wheel is 100 cm, so the radius is 50 cm, and the circumference is 100π cm. How many of these circumferences (or wheel revolutions) fit inside the 75,000 cm? In other words, if I were to peel this wheel's tread from the cart and lay it out flat, it would measure a distance of 100π cm. How many of these lengths fit into the entire distance covered in one minute? To find out how many of (this) fit into so many of (that), I must divide (that) by (this), so:

100πcm/rev

75,000cm/min

​750 min rev≈238.7324146RPM

Answer:

The force will be "9.8 N".

Explanation:

The given values are:

mass,

m = 0.7 kg

M = 2

g = 9.8

Now,

⇒  [tex]\tau = T \alpha[/tex]

then,

⇒  [tex]\frac{1}{2}mR^2(\frac{1}{R}\frac{dv}{dt}) =M(g-a_t)R[/tex]

⇒  [tex]\frac{1}{2}m \ a_t=m(g-a_t)[/tex]

⇒  [tex]a_t=\frac{2g}{(\frac{m}{M} +2)}[/tex]

On substituting the values, we get

⇒      [tex]=\frac{2\times 9.8}{\frac{0.7}{2} +2}[/tex]

⇒      [tex]=8.34 \ m/s[/tex]

hence,

⇒  [tex]T=mg+M(g-a_t)[/tex]

On substituting the values, we get

⇒      [tex]=0.7\times 9.8+2(9.8-8.34)[/tex]

⇒      [tex]=6.86+2(1.46)[/tex]

⇒      [tex]=6.86+2.92[/tex]

⇒      [tex]=9.8 \ N[/tex]

Answer:

the wave represents the second harmonic.

Explanation:

Given;

length of the cord, L = 64 cm

The first harmonic of a cord fixed at both ends is given as;

[tex]f_o = \frac{V}{2L}[/tex]

The wavelength of a standing wave with two antinodes is calculated as follows;

L = N---> A -----> N    +   N ----> A -----> N

Where;

N is node

A is antinode

L = N---> A -----> N    +   N ----> A -----> N =  λ/2  + λ/2

L = λ

The harmonic is calculated as;

[tex]f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic[/tex]

Therefore, the wave represents the second harmonic.

L = λ

Answer:

0.54 sec

Explanation:

Answer:

Explanation:

Speed = 14.2 m/s

Distance = 514 m

Time = Distance / Speed

Time = 514 / 14.2

Time = 36.19 seconds

Answer:

C. Amount of oxygen

Explanation:

Options A and D are invalid as they aren't affecting factors.

Option B is false as the altitude increases as you go up a mountain.

Option C is true as the air pressure (atmospheric pressure) is inversely proportional to the height/altitude of the mountain.

Answer:

the net work done on the object is 80 J.

Explanation:

Given;

force applied on the object, F = 100 N

mass of the object, m = 20 kg

speed of the object, v = 0.2 m/s

distance moved by the object, d = 0.8 m

The net work done on the object is calculated as follows;

W = F x d

W = 100 N x 0.8 m

W = 80 J

Therefore, the net work done on the object is 80 J.

Answer:

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Explanation:

isos

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A  = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]

Therefore, the length of the wire is 134.62 m.

1: Ming

2: Chloe

3: 40m

Answer:

[tex]65.6\ \text{J/m}^3[/tex]

[tex]0.11\ \text{J}[/tex]

Explanation:

B = Magnetic field = [tex]\mu_0 \dfrac{N}{l}I[/tex]

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]

N = Number of turns = 1270

[tex]l[/tex] = Length of solenoid = 93.9 cm = 0.939 m

[tex]I[/tex] = Current = 7.8 A

A = Area of solenoid = [tex]17.3\ \text{cm}^2[/tex]

Energy density of a solenoid is given by

[tex]u_m=\dfrac{B^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{(\mu_0 \dfrac{N}{l}I)^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{\mu_0N^2I^2}{2l^2}\\\Rightarrow u_m=\dfrac{4\pi\times 10^{-7}\times 1230^2\times 7.8^2}{2\times 0.939^2}\\\Rightarrow u_m=65.6\ \text{J/m}^3[/tex]

The energy density of the magnetic field inside the solenoid is [tex]65.6\ \text{J/m}^3[/tex]

Energy is given by

[tex]U_m=u_mAl\\\Rightarrow U_m=65.6\times 17.3\times 10^{-4}\times 0.939\\\Rightarrow U_m=0.11\ \text{J}[/tex]

The total energy in joules stored in the magnetic field is [tex]0.11\ \text{J}[/tex].

Answer:

c. Double Replacement

Explanation:

As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.

Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2

and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.

Answer:

isopropyl alcohol would form smaller droplets, because it has lower surface tension than water has

Explanation:

Ap3x

The droplets made of isopropyl alcohol instead of water be smaller due to surface tension.

The single drop of a liquid in the form of sphere is called droplet.

Water can form large dewdrops in nature. Isopropyl alcohol would form smaller droplets, because it has lower surface tension than water.

Surface tension is the property of the liquid to acquire minimum surface area.

Thus, droplets made of isopropyl alcohol instead of water be smaller.

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Answer:

[tex]4.47\ \text{km/h}[/tex]

Explanation:

[tex]\dfrac{da}{dt}[/tex] = Rate at which the distance between A and starting point of B is changing = -20 km/h

[tex]\dfrac{db}{dt}[/tex] = Rate at which the distance of B is changing = 15 km/h

[tex]\dfrac{dc}{dt}[/tex] = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = [tex]20\times 4=80\ \text{km}[/tex]

a = Distance remaining to the start point of B = [tex]110-80=30\ \text{km}[/tex]

b = Distance covered by B in 4 hours = [tex]15\times 4=60\ \text{km}[/tex]

Distance between A and B after 4 hours

[tex]c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}[/tex]

[tex]c^2=a^2+b^2[/tex]

Differentiating with respect to time we get

[tex]c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}[/tex]

The rate at which the distance between the ships is changing at 4 PM is [tex]4.47\ \text{km/h}[/tex].

Answer:

 K_f = 1881.6 J

Explanation:

To solve this exercise, let's start by finding the velocities of the bodies.

We define a system formed by the initial object and its parts, with this the forces during the explosion are internal and the moment is conserved

initial instant. Before the explosion

        p₀ = M v₀

final instant. After the explosion

        p_f = m₁ v + m₂ 0

the moeoto is preserved

         p₀ = p_f

         M v₀ = m₁ v

         v = [tex]\frac{m_1}{M}[/tex]  v₀

in the exercise they indicate that the most massive part has twice the other part

         M = m₁ + m₂

         M = 2m₂ + m₂ = 3 m₂

         m₂ = M / 3

so the most massive part is worth

        m₁ = 2 M / 3

we substitute

        v = ⅔ v₀

with the speed of each element we can look for the kinetic energy

initial

         K₀ = ½ M v₀²

Final

         K_f = ½ m₁ v² + 0

         K_f = ½ (⅔ M) (⅔ v₀)²

         K_f = [tex]\frac{8}{27}[/tex] (½ M v₀²)

         K_f = [tex]\frac{8}{27}[/tex]  K₀

the energy added to the system is

         ΔK = Kf -K₀

         ΔK = (8/27 - 1) K₀

         ΔK = -0.7 K₀

         K_f = K₀ + ΔK

         K_f = K₀ (1 -0.7)

         K_f = 0.3 K₀

let's calculate

         K_f = 0.3 (½ 64 14²)

         K_f = 1881.6 J

Answer:

 P and S waves slow down when they reach this layer. The asthenosphere, also known as the magma chamber, is the uppermost component of the mantle. This layer is partially molten and is a ductile zone in a tectonically poor state.

It's almost hard and seismic waves move through the asthenosphere at a slow rate. The fragile lithosphere and the uppermost portion of the asthenosphere are assumed to be rigid.

seismic waves travel more quickly through denser materials and therefore generally travel more quickly with the depth it moves more slowly through a liquid than a solid. Molten areas within the Earth slow down P waves and stop S waves because their shearing motion cannot be transmitted through a liquid. Partially molten areas may slow down the P waves and attenuate or weaken S waves.

hope this helps...

S and P wave slow down and stop in  the uppermost part of the mantle. - For this, scientists have concluded that the uppermost part of the mantle is partially-molten.

A planetary body's mantle is a layer that is surrounded by the crust on top and the core underneath. The largest and most substantial layer of a planetary body, mantles are often comprised of rock or ice. Planetary bodies that have undergone density differentiation typically have mantles. Mantles are found on all terrestrial planets (including Earth), many asteroids, and a few planetary moons.

Between the crust and the outer core, there is a silicate rock layer known as the Earth's mantle. Despite being mostly solid, it behaves like a viscous fluid over geological time. Oceanic crust is created by the partial melting of the mantle at mid-ocean ridges, and continental crust is created by the partial melting of the mantle at subduction zones.

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Answer:

The border is 8km above sea level.

Explanation:

We know that:

Density = 1.25 kg/m^3

Pressure = 10^5 N/m^2

g = 10m/s^2

Now, suppose that we have a virtual rectangle, such that its bases have an area of 1m^2 and the rectangle has a height equal to H.

This virtual figure has a volume V = 1m^2*H, and it is filled with air (which we know that has a density 1.25 kg/m^3)

Then the total mass inside that volume is:

M = (1.25 kg/m^3)*V = (1.25 kg/m^3)*(1m^2*H)

The weight of this mass is:

W = g*M = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)

And if we divide the weight in a given surface, let's say 1 m^2, we get the pressure per square meter, which we know is equal to  10^5 N/m^2

then:

P = 10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)

Whit this equation we can find the value of H.

10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)

10^5 N =  (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)

(10^5 N)/(10 m/s^2) = (1.25 kg/m^3)*(1m^2*H)

(10^4 kg) = (1.25 kg/m^3)*(1m^2*H)

(10^4 kg)/( 1.25 kg/m^3) = 1m^2*H

8,000 m^3 = 1m^2*H

(8,000 m^3)/(1m^2) =H

8,000 m = H

And we want this answer in km, knowing that 1,000m = 1km

8,000m = 8km = H

The border is 8km above sea level.

Height of boundaries is 8.2 km

Normal density = 1.25 kg/m³

1 atm = 101325 N/m²

Height of boundaries

Pressure = Height × Density × Gravitational acceleration

101325 = Height × 1.25 × 9.8

101325 = Height × 12.25

Height of boundaries = 101325 / 12.25

Height of boundaries = 8271.42 m

Height of boundaries = 8.2 km

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Answer:

The correct answer is -  91.4 nm

Explanation:

According to Bohr's model, the minimum wavelength to ionize Hydrogen atom from n= 1 state is expressed as:

(h×c)/λ=13.6eV

here,

h - Planck constant

c - the speed of light

λ - wavelength

Placing the value in the formula for the wavelength

(6.626×10^−34J.s × 3×10^8 m/s)/λ  =  13.6 ×1.6 × 10^−19 J

λ≈91.4nm

Thus, the correct answer would be = 91.4 nm

Answer:

C) 413 Hz

Explanation:

For destructive interference, the path difference ΔL = (n + 1/2)λ where ΔL = L₂ - L₁ where L₁ = person's distance from one speaker (the closer one) = 5.0m and L₂ = person's distance from other speaker (the farther one) = 6.2 m and λ = wavelength = v/f where v = speed of sound = 330 m/s and f = frequency

So, ΔL = (n + 1/2)λ

L₂ - L₁  = (n + 1/2)v/f

f = (n + 1/2)v/(L₂ - L₁)

At the second lowest frequency that results in destructive interference at the point where the person is standing, n = 1.

So,

f = (1 + 1/2)v/(L₂ - L₁)

f = 3v/2(L₂ - L₁)

Substituting the values of the variables into the equation, we have

f = 3v/2(L₂ - L₁)

f = 3(330 m/s)/2(6.2 m - 5.0 m)

f = 3(330 m/s)/2(1.2 m)

f = 990 m/s ÷ 2.4 m)

f = 412.5 Hz

f ≅ 413 Hz

Answer:

t = 0.437 s

Explanation:

The speed of sound is a constant that is worth v = 343 m / s

           v = d / t

            t = d / v

the time it takes for the sound to reach Clark at d = 150 m is

           t = 150/343

           t = 0.437 s

This same sound takes much longer to reach you

          t₂ = 127 10³/343

          t₂ = 370 s

Answer:

The equation says that due to variation in temperature is

delt T = .59 m/s / C = 16 C * .59 m/s = 9.44 m/s

So v = 332 m/s + 9.44 m/s = 341 m/s (to three significant figures)

The force 1 is tension force.

To find the correct statement among all the options, we need to know more about friction, tension, normal force and weight.

Thus, we can conclude that the correct option is (B).

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Answer:

Explanation:

mass per unit length ρ = .100 / 1.65 = .0606 . kg /m

length of wire L = 1.65 m

For fundamental frequency , the expression is as follows

n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

L = 1.65 , T = 16 n and m = .0606

n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]

= 4.9 /s .

This is fundamental frequency .

other mode of vibration ( first three ) will be as follows

4.9 x 2 = 9.8 /s ,

4.9 x 3 = 14.7 /s .