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Valentin [98]
3 years ago
7

If soap bubble shows strong colors due to interference of the light that reflects from the inner and outer boundaries of the thi

n soap and water film. Light striking the outer surface has a phase change on reflection of pi. that is, its phase is flipped when it goes from air to the water surface and back out. Light reflecting from the inner surface, where it is in water then comes to the boundary with air, does not change its phase. Suppose you see white light reflecting from a bubble as it increases its size. The film making the bubble gets thinner and thinner and the colors reflecting from it change. If the bubble glows red, and then gets even thinner, what will you see just before it bursts?
a. Blue
b. Red
c. Darkness.
Physics
1 answer:
Veronika [31]3 years ago
8 0

Answer:

the bubble appears DARK

Explanation:

This problem shows the interference by reflection of light on the surface, let's take it into account;

* There is a phase change of 180º on the surface when it stops from a lower medium to a higher index

* Within the film the wavelength changes due to the fraction index

           \lambda_n = \frac{\lambda_o}{n}

if we plug this into the expression for constructive interference we have

       2tn = (m + ½) λ₀

for destructive interference it is

       2tn = m λ₀

apply these expressions to our case

indicate that the incident wavelength is red λ₀ = 700 nm

Let's find the minimum thickness that the film must have to have a constructive interference

the refractive index of soap is n = 1.4

         

from the constructive interference equation

            t = (0+ ½) 700 10⁻⁹ / 2 1.4

            t = 1.25 10⁻⁷ m

just before the soap bubble explodes the thickness (t) of the film less than the calculated value there is no possibility of constructive interference, therefore the bubble appears DARK

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andrezito [222]

Answer:

The middle and end of the string respectively

Explanation:

Rarefaction means region of maximum displacement and compression means region of minimum displacement, hence.

We expect a maximum displacement at the middle of the string because it is said to vibrate greatest at this point and conversely vibrate least at the end point which is the region of compression.

3 0
3 years ago
A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim
Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

F_D = -\frac{1}{2}\rho V^2 C_d A

Where

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C_d= Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

F=ma=m\frac{dV}{dt}

Equating both equations we have:

m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A

m(dV)=-\frac{1}{2}\rho C_d A (dt)

\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)

Integrating

\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)

-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t

Here,

V_f = 60mph = 26.82m/s

V_i = 120.7m/s

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C_d = 0.3

\Delta t=7s

Replacing:

\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)

-0.029 = -5.4997r^2

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4 0
4 years ago
A long, straight solenoid with a cross-sectional area of 8.00cm2 is wound with 90turns of wire per centimeter, and the windings
77julia77 [94]

Answer:

the average induced emf in the second winding is 7.9168*10^{-5}V

Explanation:

The magnetic field inside the first solenoid is given by,

B= \mu_0NI

Where \mu  is the permeability of the free space

N is the number of turns of solenoid per unit length

I is the current in the solenoid

A is the cross-sectional area of the wire

Replacing we have,

B= (4*\pi*10^{-7})(90/cm (\frac{100cm}{1m}))(0.350A)

B = 3.9584*10^{-3}T

Thus average emf induced in the second windigs is,

\epsilon{avg}=\N\frac{dB}{dt}

\epsilon_{avg}=\frac{d}{dt}(AB)

\epsilon_{avg}=A\frac{dB}{dt}

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Therefore the average induced emf in the second winding is 7.9168*10^{-5}V

7 0
4 years ago
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Zanzabum

Answer:

a=5.62[m/s^2]

Explanation:

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F=m*a\\F=fuerza total aplicada [N]\\m=masa del cuerpo[kg]\\a=aceleracion del cuerpo[m/s^2]\\\\

De esta manera despejando el valor de aceleracion de la anterior ecuacion:

a=\frac{F}{m} \\a=2294/408\\a=5.62[m/s^2]

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3 years ago
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3 0
3 years ago
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