Hsveteb II be e e eh e egjebetekne ebegtehe

A scuba diver is sitting on a boat while waiting to go on a dive and sees light reflected from the water's surface. At what angl

LUCKY_DIMON [66]

Answer:

θ_p = 53.0º

Explanation:

For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º

Let's write the transmission equation

n1 sin θ₁ = ne sin θ₂

The angle to normal (vertcal) is

180 = θ2 + 90 + θ_p

θ₂ = 90 - θ_p

Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray

We replace

n1 sin θ_p = n2 sin (90 - θ_p)

Let's use the trigonometry relationship

Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p

In the law of reflection incident angle equals reflected angle,

ni sin θ_p = ns cos θ_p

n₂ / n₁ = sin θ_p / cos θ_p

n₂ / n₁ = tan θ_p

θ_p = tan⁻¹ (n₂ / n₁)

Now we can calculate it

The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water

n₂ = 1.33

θ_p = tan⁻¹ (1.33 / 1)

θ_p = 53.0º

n₂ = 1.40

θ_p = tan⁻¹ (1.40 / 1)

Tep = 54.5º

4 0

3 years ago

Consider the system consisting of two blocks connected by a rope and a pulley. The coefficient of static friction between the ra

RSB [31]

Answer:

1.2 kg

Explanation:

Let UP ramp be the positive direction

F = ma

T - Wt || - Ff = m(0)

mg - Μgsinθ - μΜgcosθ = 0

m(9.8) - 13sin35 - 0.36(13)cos35 = 0

m = 13(sin35 + 0.36cos35) / 9.8

m = 1.15205... ≈ 1.2 kg

5 0

3 years ago

(6.9 * 10-6)(770 * 102)

meriva

Answer:

4948020

Explanation:

(6.9*10-6)(770*102)

Multiply 6.9 by 10 .

( 69 − 6 ) ( 770 ⋅ 102 )

Subtract 6 from 69 .

63 ( 770 ⋅ 102 )

Multiply 770 by 102 .

63 ⋅ 78540

Multiply 63 by 78540 .

4948020

5 0

3 years ago

Light of wavelength 648.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 57.5 cm from the slit.

boyakko [2]

Answer:

$71.0 \mu m$

Explanation:

The formula for the single-slit diffraction is

$y=\frac{n\lambda D}{d}$

where

y is the distance of the n-minimum from the centre of the diffraction pattern

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light

In this problem,

$\lambda=648.0 nm=6.48\cdot 10^{-7}m$

$D=57.5 cm=0.575 m$

$y=1.05 cm=0.0105 m$ , with n=2 (this is the distance of the 2nd-order minimum from the central maximum)

Solving the formula for d, we find:

$d=\frac{n\lambda D}{y}=\frac{2(6.48\cdot 10^{-7} m)(0.575 m)}{0.0105 m}=7.10\cdot 10^{-5} m= 71.0 \mu m$

6 0

3 years ago

What is the final temperature in degrees centigrade of 100g of water at 30 c if it is mixed with 50 g of water at 0?

natka813 [3]

Answer:

The final temperature = 293 K or 20 °C

Explanation:

Heat gained by water at 0°C = heat lost by water at 30°C

c₁m₁(T₃ - T₁) = c₂m₂(T₂-T₃).......................... Equation 1

Making T₃ the subject of the equation,

T₃ = (c₂m₂T₂ + c₁m₁T₁)/(c₁m₁+c₂m₂)............. Equation 2

Where m₁ =mass of water at 0°C, c₁ = specific heat capacity of water at 0°C , c₂ = specific heat capacity of water at 30°C, m₂ = mass of water at 30°C, T₁ = initial temperature of water at 0°C, T₂ = initial temperature of water at 30°C, T₃ = final temperature.

Given: m₁ = 50 g = (50/1000) kg = 0.05 kg, m₂ = 100 g = (100/1000) kg = 0.1 kg., T₁ = 0°C = 273 K, T₂ = 30°C = (30+273 )= 203 K

Constants: c₁ = c₂ = 4200 J/kg.K

Substituting these values into equation 2,

T₃ = (4200×0.1×303 + 4200×0.05×273)/(4200×0.1 + 4200×0.05)

T₃ = (127260 + 57330)/(420+210)

T₃ = 184590/630

T₃ = 293 K.

Therefore the final temperature = 293 K or 20 °C

3 0

4 years ago