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The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of

Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is

$\theta _d =0.19 ^o$

Explanation:

From the question we are told that

The refractive index of the transparent acrylic plastic for blue light is $n_F = 1.497$

The wavelength of the blue light is $F = 486.1 nm = 486.1 *10^{-9} \ m$

The refractive index of the transparent acrylic plastic for red light is $n_C = 1.488$

The wavelength of the red light is $C = 656.3 nm = 656.3 *10^{-9} \ m$

The incidence angle is $i = 45^o$

Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as

$r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]$

$r_F = 28.18^o$

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

$r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]$

$r_F = 28.37^o$

The angle between the blue beam and the red beam in the acrylic block

$\theta _d = r_C - r_F$

substituting values

$\theta _d = 28.37 - 28.18$

$\theta _d =0.19 ^o$

4 0

2 years ago

Heartburn is a form of indigestion felt as a burning sensation in the chest when stomach acid comes back up into the esophagus.

xenn [34]

The antacid is basic so it neutralizes acidity or lowers it. Then if it goes into the esophagus, it's not as strong and it doesn't hurt, and it also calms your stomach because the acidity in your stomach is also lower. Antacids are therefore taken by many people, especially as they grow older and things like heartburn become more common.

5 0

2 years ago

Read 2 more answers

A drag racer starts from rest and accelerates at 10 m/s squared for the

nika2105 [10]

Answer:

54 $\frac{m}{s}$

Explanation:

$v=u+at$

where $u=$ initial velocity

$v=0+(10\frac{m}{s^{2} } )$ × $(5.4s)$

$v=54\frac{m}{s}$

6 0

2 years ago

The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?

Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2$

$For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2$

The second distance, r₂, can be determined from sound intensity formula given as;

$I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m$

Therefore, the second distance of the sound from the source is 431.78 m.

7 0

2 years ago

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 7.0 m/s2. Secret agent Austin Powers jumps on ju

krok68 [10]

Answer:

a) 725.5 m

b) 630 m

Explanation:

Given data:

acceleration of Helicopter = 7.0 m/s^2

time spent upwards by helicopter = 11.0 seconds

a) Determine the maximum height above ground reached by the helicopter

h1 = at^2 /2

= 7 * 11^2 / 2

= ( 7 * 121 ) / 2 = 423.5 m

also v = a*t = 7 * 11 = 77 m/s

also we calculate h2

h2 = v^2 / 2g

= (77^2) / 2 * 9.81

= 302 m

therefore the maximum height = 302 + 423.5 = 725.5 m

b) Given that ; power deploys a jet pack strapped on his back at 7.0 s and with a downward acceleration of ; 1.0 m/s^2

Determine distance Power reaches before helicopter crashes

s = ut + 1/2 at^2

h.gt^2 - 77t - 423.5 m = 0

h.gt^2 - 77t = 423.5

t = 17. 66 secs

Yf = 423.5 + 77 *7 - 4.9 *7

Yf = 928.2

Vf = u + at

= 77 - 9.8*7 = 8.4 m/secs

t' = 17.66 - 7 = 10.66 secs

hence

Yf = 725.5 - 8.4 * 10.66 + 1/2 * -1 * 10.66

= 630 meters

6 0

2 years ago