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statuscvo [17]
1 year ago
7

Light with wavelength in air ( lambdaair ) is incident on a oil slick ( noil = 1. 25) floating on the ocean ( nwater = 1. 33). w

hat is the thinnest thickness of oil that will brightly reflect the light?
Physics
1 answer:
crimeas [40]1 year ago
4 0

<u>26mm</u> is the thinnest thickness of oil that will brightly reflect the light.

What is wavelength ?

The distance over which a periodic wave's shape repeats is known as the wavelength in physics.  It is a property of both traveling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings.  The spatial frequency is the reciprocal of wavelength. The Greek letter lambda () is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

To learn more about wavelength visit:

brainly.com/question/16051869

#SPJ4

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La superficie de unas botas suman 400 cm2 y la persona que las usa tiene 45 kg de masa, calcula la presión que ejerce sobre el p
liubo4ka [24]

Answer:

11025 N / m²

Explanation:

Los siguientes datos se obtuvieron de la pregunta:

Área (A) = 400 cm²

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Presión (P) =?

A continuación, determinaremos la fuerza aplicada. Esto se puede obtener de la siguiente manera:

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Fuerza (F) =.?

F = m × g

F = 45 × 9,8

F = 441 N

A continuación, convertiremos 400 cm² a m². Esto se puede obtener de la siguiente manera:

1 cm² = 0,0001 m²

Por lo tanto,

400 cm² = 400 cm² × 0,0001 m² / 1 cm²

400 cm² = 0,04 m²

Por tanto, 400 cm² equivalen a 0,04 m².

Finalmente, determinaremos la presión ejercida de la siguiente manera:

Área (A) = 0.04 m².

Fuerza (F) = 441 N

Presión (P) =?

P = F / A

P = 441 / 0,04

P = 11025 N / m²

Por tanto, la presión ejercida es 11025 M / m²

4 0
3 years ago
Considering the factors that affect gravitational pull, in which location would the gravitational pull from the earth be SMALLES
OleMash [197]
C. The higher the altitude the less gravity affects you
4 0
3 years ago
Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0
2 years ago
Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k
Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

\frac{GM_{ns}}{R^2}= \omega^2 R

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0
3 years ago
Se aplica una fuerza neta de 25 N a una masa de 10 kg. ¿Cuál es la aceleración dada a la masa?
lys-0071 [83]

Answer:

Se queda

Explanation:

Porque 10 kg es más que 25N

4 0
3 years ago
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