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AleksAgata [21]

3 years ago

6

We wish to coat flat glass (n = 1.50) with a transparent material (n = 1.20) so that reflection of light at wavelength 529 nm is

eliminated by interference. what minimum thickness can the coating have to do this?

1 answer:

hjlf3 years ago

8 0

For full destruction of the wavelength,
2t = y/2

t = thickness of the coating material
y = wavelength in the coating material

But;
y = y_a/n; where y_a is wavelength in air

Substituting;
y = 529/1.2 = 440.83 nm

Then,
t = y/4 = 440.83.4 = 110.21 nm

Therefore, thickness of the coating material should be 110.21 nm.

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Alona [7]

Answer:

1) The girl's acceleration at time 't' is, $a=0.47$ m/s²

2) The girl's acceleration at time 't₁' is, $a_{1}=0$ m/s²

3) The girl's acceleration at time 't₂' is, $a_{2}=-0.14$ m/s²

Explanation:

Given data,

The initial walking speed of the girl, u = 0

The speed of the girl at the time 't' 3 s is, v₁ = 1.4 m/s

The time period the girl walked at the speed 1.4 m/s is, t₁ = 6 s

The girl slows down and comes to a stop during a period, t₂ = 10 s

1) The girl's acceleration at time 't'

$a=\frac{v-u}{t}$

$a=\frac{1.4-0}{3}$

$a=0.47$ m/s²

2) The girl's acceleration at time 't₁'

$a_{1} =\frac{v_{1} -v}{t_{1}}$

$a_{1}=\frac{1.4-1.4}{6}$

$a_{1}=0$ m/s²

3) The girl's acceleration at time 't₂'

$a_{2} =\frac{v_{2} -v_{1}}{t_{2}}$

$a_{2}=\frac{0-1.4}{10}$

$a_{2}=-0.14$ m/s²

4 0

3 years ago

how long does it take an acorn to hit the ground after dropping from the branch of a tree 50.0m high? remember that g=9.8m/s^2

NeX [460]

D = V1( t ) + 1/2g( t )^2
50m = 0m/s( t ) + 1/2(9.8m/s^2)*( t )^2
V1*t cancels out
50m = (4.9m/s^2)*(t)^2
50m/(4.9m/s^2) = t^2
Metres unit cancels out so we are left with s^2
10.204s^2 = t^2
Square root both sides to cancel out square
t = 3.19 s

5 0

3 years ago

Read 2 more answers

if you have a block of steel with a density of 7.8g/cm3 and a block of gold with a density of 19.3g/cm3 each with the same volum

natita [175]

Gold because 19.3 is greater than 7.8

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3 years ago

The nucleus of an atom can be modeled as several protons and neutrons closely packed together. Each particle has a mass of 1.67

Lady bird [3.3K]

Explanation:

The nucleus of an atom can be modeled as several protons and neutrons closely packed together.

Mass of the particle, $m=1.67\times 10^{-27}\ kg$

Radius of the particle, $R=10^{-15}\ m$

(a) The density of the nucleus of an atom is given by mass per unit area of the particle. Mathematically, it is given by :

$d=\dfrac{m}{V}$ , V is the volume of the particle

$d=\dfrac{m}{(4/3)\pi r^3}$

$d=\dfrac{1.67\times 10^{-27}}{(4/3)\pi (10^{-15})^3}$

$d=3.98\times 10^{17}\ kg/m^3$

So, the density of the nucleus of an atom is $3.98\times 10^{17}\ kg/m^3$ .

(b) Density of iron, $d'=7874\ kg/m^3$

Taking ratio of the density of nucleus of an atom and the density of iron as :

$\dfrac{d}{d'}=\dfrac{3.98\times 10^{17}}{7874}$

$\dfrac{d}{d'}=5.05\times 10^{13}$

$d=5.05\times 10^{13}\ d'$

So, the density of the nucleus of an atom is $5.05\times 10^{13}$ times greater than the density of iron. Hence, this is the required solution.

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4 years ago

The surface of the paper is phosphorescent. When light shines on it, some of the energy is absorbed and re-emitted slowly over t

Ad libitum [116K]

Answer:

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$V < I < B < G < Y < O < R;$ and

$E \ \alpha \ \dfrac{1}{\lambda}$

As a result, blue light has a higher energy level than green and red light.

As a result, the surface glows due to the blue LED. The green LED, on the other hand, would not allow the surface to glow as much as the red LED, which has a lower energy level when compared to the green light. As a result, the red LED would not allow the surface to glow as well.

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3 years ago