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2 years ago

10

Can you explain why number 3 is correct?

1 answer:

lukranit [14]2 years ago

7 0

No waves because Q19 waves would going at the surface at regions

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A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from

Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of incidence beam striking the water is $\theta = 49.63^o$

c the angle of refraction beam striking the water is $r = 59.7^o$

d The angle the refracted beam make with respect to the horizontal is $= 30.3^o$

e The height of the target above sea level is

$h= 125.05m$

Explanation:

From the diagram we see that the angle of the beam striking the water is

$tan \theta = \frac{100}{85}$

$\theta = tan^{-1}(\frac{100}{85} )$

$= 49.63^o$

According to Snell's law

$\mu_{water} *sin(i) = \mu_{air} *sin(r)$

Where $\mu_{water }$ is the refractive index of water = 1.333

$i$ is the angle of incidence

$\mu_{air}$ is the refractive index of air = 1

r is the angle of refraction

Substituting values accordingly

$1.33 * sin (40.37) = 1 * sin(r)$

Making r the subject of the formula

$r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})$

$= 59.7^o$

looking at the diagram we can see that to obtain the angle the refraction beam makes with the horizontal by subtracting the angle refraction from 90°

i.e $90 -59.7 = 30.3$ °

From the diagram we see that the height target above sea level can be obtained by this relation

$tan \theta = \frac{h}{214}\\$

Where h is the is the height

$tan(30.3) = \frac{h}{214}$

$h = 214 * tan (30.3)$

$=125.05m$

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Here is a set of flash card that can help with this subject.

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Answer:

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