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miv72 [106K]
3 years ago
13

A light wave is shone on a glass block at the boundary between air and glass which wave phenomena can been demonstrated

Physics
2 answers:
solniwko [45]3 years ago
7 0
At the boundary between air and glass, we can see mainly refraction ...
the path of the light bends as it crosses the boundary, and proceeds
from there in a different direction.

If we're very careful and make sensitive measurements, then we can also
detect transmission, some reflection, and a tiny bit of dispersion at the boundary.
Alexeev081 [22]3 years ago
4 0

Answer:

Reflection and Refraction

Explanation:

There are four wave phenomenon that can be observed while studying any wave: reflection, refraction, dispersion, and interference. This depends on the medium of interaction among other factors.

Here two of the wave phenomenon can be observed, namely:

1. Refraction: The phenomenon of bending of light when it passes from one medium to another. So when the light will travel from air to glass, it will divert from its path and bend a little towards the normal. (Normal is an imaginary line drawn perpendicular to the boundary between two mediums)

2. Reflection: Some part of the light will be reflected back from the boundary. It depends on how transparent the glass block is. If it is 100% transparent (which is very nearly impossible to make) there will be no reflection.

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If you ran 15 km/h for 20 min, how much distance would you cover?
nikitadnepr [17]
Since we have 15 kilometers per hour, and we're looking for 20 minutes, let's set up proportions.
20/60 minutes = x/15
20/60 = 1/3, so let's leave that simplified.
1/3 = x/15
Look at the denominators, 3 to 15 is a factor of 5, so multiply the numerator by 5.
1 • 5 = 5, so you will cover 5 kilometers in 20 minutes.

I hope this helps!
4 0
3 years ago
An iron railroad rail is 800 ft long when the temperature is 31°C. What is its length (in ft) when the temperature is −17°C?
natta225 [31]

Answer:

799.54 ft

Explanation:

Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given:

α = 1.2×10⁻⁵ / °C

L₀ = 800 ft

ΔT = -17°C − 31°C = -48°C

Find: ΔL

ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)

ΔL = -0.4608

Rounded to two significant figures, the change in length is -0.46 ft.

Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.

7 0
3 years ago
A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo
zvonat [6]

Answer:

13 m/s

Explanation:

I assume we are ignoring friction.

The boy's PE will all be converted to KE at the bottom of the hill.

to find PE = mgh   we need to know h

   h = 50 sin 10 = 8.68 meters

     then:    PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j

KE = 1/2 m v^2 = <u>1703 .49</u>

            v = 13 m/s

7 0
3 years ago
Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constan
vladimir2022 [97]

Answer:

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

S = k_{H} \cdot p (1)

<em>where S: is the solubility or concentration of Ar in water, k_{H}: is Henry's law constant and p: is the pressure of the Ar </em>

<u>Since the argon is 0.93%, we need to multiply the equation (1) by this percent:</u>

S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L}

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

Have a nice day!          

8 0
3 years ago
A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. Yo
aniked [119]

Answer:

a) 46.5º  b) 64.4º

Explanation:

To solve this problem we will use the laws of geometric optics

a) For this part we will use the law of reflection that states that the reflected and incident angle are equal

     θ = 43.5º

This angle measured from the surface is

     θ_r = 90 -43.5

     θ_s = 46.5º

b) In this part the law of refraction must be used

     n₁ sin θ₁ = n₂. Sin θ₂

     sin θ₂ = n₁ / n₂ sin θ₁

The index of air refraction is n₁ = 1

The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface

     θ_s = 90 - θ

     θ_s = 90- 43.5

     θ_s = 46.5º

     sin θ₂ = 1 / 1.68  sin 46.5

     sin θ₂ = 0.4318

     θ₂ = 25.6º

The angle with respect to the surface is

     θ₂_s = 90 - 25.6

     θ₂_s = 64.4º

measured in the fourth quadrant

3 0
3 years ago
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