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3 years ago

There is a reflection at the boundary between two materials if there is a change of __________ at the boundary.

Physics

1 answer:

melisa1 [442]3 years ago

8 0

<span>The speed of the light in the materials. Hope this helps!!</span>

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What is the method of heat transfer sunlight melts a wax crayon left outside.

Mars2501 [29]

Answer:

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Explanation:

This is ur answer....

<h3><em>Transfer of energy by electromagnetic waves. Radiation doesn't require matter. Sunlight melting a wax crayon left outside is radiation.</em></h3><h3><em /></h3>

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2 years ago

In an electromagnetic wave in free space, the electric and magnetic fields are.

Vlad [161]

Answer:

PERPENDICULAR

Explanation:

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2 years ago

what is the gravitational potential energy of a 150-kg object that is suspended 5-m above the earth's surface

77julia77 [94]

150x5x9.7(gravity on earth)=7275

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3 years ago

What is the gravitational force between Mars and the sun? 7.43 × 1030 N 1.79 × 1026 N 1.65 × 1021 N 3.76 × 1032 N

VMariaS [17]

The gravitational force between Mars and the Sun is $1.65\cdot 10^{21} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we have:

$m_1 = 1.99\cdot 10^{30} kg$ is the mass of the Sun

$m_2 = 6.39\cdot 10^{23} kg$ is the mass of Mars

$r=229\cdot 10^6 km = 229\cdot 10^9 m$ is the average distance Mars-Sun

Substituting into the equation, we find the gravitational force:

$F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(6.39\cdot 10^{23})}{(229\cdot 10^9)^2}=1.62\cdot 10^{21} N$

So, the closest answer is

$1.65\cdot 10^{21} N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm

lord [1]

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: $n_0$ = 1.52

Wave length of light = λ = 570 nm = $570\times10^{-9}\ m$

$\text{ Thickness}=\dfrac{\lambda}{4n}$

$=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}$

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

5 0

3 years ago