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I am Lyosha [343]
3 years ago
7

Which statement is true for light passing into a medium that is less optically dense than the first medium through which it pass

ed?
The index of refraction of the second medium is higher.


The index of reflection of the second medium is lower.


The index of reflection of the second medium is higher.


The index of refraction of the second medium is lower.
Physics
2 answers:
BigorU [14]3 years ago
5 0

Answer:

The correct option is;

The index of refraction of the second medium is lower

Explanation:

The index of refraction of a material indicates the magnitude of the optical density of a material. The index of refraction or the refractive index, n, are indices (ratio) of the speed of light through an optically dense medium relative to the speed of light through a vacuum.

The definition of the refractive index is the number of times light travelling through a medium would be slower than light travelling through vacuum

Therefore, the index of refraction of a second medium that is less optically dense than a first medium from which light originates and travels through it would be lower than the index of refraction of the first medium

AleksAgata [21]3 years ago
5 0

Answer:

D

Explanation:

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The area around a charged object that can exert a force on other charged objects is an electric ___
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Electric field.
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A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and pressure are 44oF and 24 psia. A wa
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V=203 ft/s

Explanation:

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4 years ago
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In a certain experiment, cylindrical samples of diameter 4 cm and length 7 cm are used. The two thermocouples in each sample are
sergeinik [125]

Answer:

K = .3941 × 10³ W/m.K

Explanation:

Qcond = K A ΔT÷ L

∴K = Qcond ×L ÷ A ΔT

J ÷ S = P

P = I × V =Qcond

∴Qcond = I × V

               = 0.6 A × 110 V

               =66 W

L = 0.12 m

ΔT = 8 °C

Qcond =33 V

Area = (πD²) ÷ 4

       = [π (4 × 10⁻² )²] ÷  4

        = 1.256 × 10⁻³ m²

∴A = 1.256 × 10 ⁻³³ m²

So K = ( Qcond × L ) ÷ A ΔT

         = (33) (0.12 ) ÷ (1.256 ×10⁻³ ) × 8

         = 0.3941 × 10³ W/m .K

7 0
3 years ago
Which type of modulation is applied to radio-controlled toys?
Alenkinab [10]

Pulse type of modulation is applied to radio-controlled toys, therefore the correct answer is option D.

<h3>What is the frequency?</h3>

It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.

Toys controlled by remote control operate by emitting infrared radiation. These infrared rays have a frequency of 34–48 kilo Hertz.

Different types of modulations, such as frequency and amplitude modulation for transmitting and receiving video and music, are employed for other reasons.

Thus, the Pulse-type of modulation is applied to radio-controlled toys, therefore the correct answer is option D.

Learn more about frequency from here

brainly.com/question/14316711

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6 0
2 years ago
Air having a pressure of 40 psig and a volume of 8 cu ft expands isothermally to a pressure of 10 psig. Find the external work p
DerKrebs [107]

Answer:

357.6 lb-ft

Explanation:

V = Volume = 8 ft³

dP = Change in pressure = (40-10) = 30 psig

Work done is given by

W=VdP\\\Rightarrow W=8\times (40-10)\\\Rightarrow W=240\ psig.ft^3

30\ psig=44.7\ psi\\\Rightarrow 1\ psi=\dfrac{30}{44.7}

So, converting to ft-lb

\dfrac{240}{\dfrac{30}{44.7}}=357.6\ lb-ft

The external work performed during the expansion is 357.6 lb-ft

7 0
4 years ago
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