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vladimir2022 [97]
3 years ago
14

An opaque cylindrical tank with an open top has a diameter of 3.25 m and is completely filled with water. When the afternoon Sun

reaches an angle of 22.8° above the horizon, sunlight ceases to illuminate the bottom of the tank. How deep is the tank
Mathematics
1 answer:
kicyunya [14]3 years ago
6 0

Answer:

10.69

Step-by-step explanation:

We must first know that the angle of refraction is given by the following equation:

n_water * sin (A_water) = n_air * sin (A_air)

where n is the refractive index, for water it is 1.33 and for air it is 1.

The angle (A) in the air is 22.8 °, and that of the water is unknown.

Replacing these values we have to:

1.33 * sin (A_water) = 1 * 0.387

sin (A_water) = 1 * 0.387 / 1.33

A_water = arc sin (0.2907) = 16.9 °

now for the tank depth:

h = D / tan (A_water)

D = 3.25

Replacing

h = 3.25 / tan 16.9 °

h = 10.69

Therefore the depth is 10.69 meters.

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y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
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