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An opaque cylindrical tank with an open top has a diameter of 3.00 m and is completely filled with water. When the afternoon Sun reaches an angle of $28.0^{\circ}$ above the horizon, sunlight ceases to illuminate the bottom of the tank. How deep is the tank?

$7.97$ m

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Cornell University

Rutgers, The State University of New Jersey

Numerade Educator

Hope College

we can ride from a Snell's law. Perfect index off water times that off water is he called to end off fact in its off air air, Uh, times effective of signed single air. Oh, when they're some, is it 28 degrees horizon? Oh, Tony, Anti grease. Then this angle of that air wouldn't be 90 minus 28. This will be that air will be, 0 60 two degrees. All substitute 62 degrees here, and that's all for theta water so that the water will be substituting values. One is the effect index of air charms. Oh, 7 62 are so sign in. Worse off. We just dropped this one signing war. So for sixth tour, reflecting dicks off water 1.23 Now this gives us is a sign of water to be 41.6 degrees. We'll use this and then find the height. How deep is the tank? So let's say edges the depth off the tank. This would be D divided by Ting. 41.6 were D's and 33 d by butane, 41.6 So uses height or for 3.37 rejects