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poizon [28]
3 years ago
10

Problem 18.9 an underwater diver sees the sun 60 ∘ above horizontal. part a how high is the sun above the horizon to a fisherman

in a boat above the diver?
Physics
1 answer:
julsineya [31]3 years ago
3 0

The angles in the equation are not the angles relative to the horizon but are relative to the "normal" which means that the line that is perpendicular to the surface.

The angle under the water is 90 – 60 = 30. 

n1 for water is 1.33, n2 for air is 1 Which you seem to understand. 

(1.33)(sin (30)) = (1.00)(sin (x2)) 

Rearranging the equation above, will give us: x^2 = sin^-1((1.33 sin 30)/1) = 41.68

But remember that that is the angle relative to the normal so you have to deduct it from 90 to get the angle relative to the horizon and you get (90 – 41.68) = 48.32 degrees.

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Energy is force * distance.

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What state of matter takes the shape and volume of its container, like the steam coming from a pot A) gas B) liquid Eliminate C)
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6 0
3 years ago
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Like the moon's orbit, Earth's orbit is not a perfect circle, but an oval How do you think this might affect Earth's tides?
ExtremeBDS [4]
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3 years ago
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a t
lyudmila [28]

Answer:

  x = -3 cm

Explanation:

The electrical potential is the sum of the potentials of each charge

       V = k ∑ q_{i} / r_{i}

let's apply this to our case where the potential is V = 0 for x = 0

         0 = k (q₁ / (x₁-0) + q₂ / (x₂-0) + q₃ / (x₃-0))

in our case

q₁ = + 2.0 10⁻⁶ C

q₂ = - 6.0 10⁻⁶ C

q₃ = + 3.0 10⁻⁶ C

x₁ = -1.0 cm = 1.0 10⁻² m

x₂ = +2.0 cm = 2.0 10⁻² m

we substitute in the equation

          0 = k (2 10⁻⁶ / 1 10⁻² - 6 10⁻⁶ / 2 10⁻² + ​​3 10⁻⁶ / x)

          3 10⁻⁶ / x = 2 10⁻⁴ - 3 10⁻⁴

          3 10⁻⁶ / x = -1 10⁻⁴

           x = - 3 10⁻² m

           x = -3 cm

7 0
3 years ago
A runner traveled 7.7 miles an hour and a half what was their average speed
Alenkasestr [34]

Answer:5.13333333...

Explanation: 7.7 is the distance and an hour and a half is the average speed. You would have to divide the total distance by the total time. So it would be 7.7 divided by 1.5 which would equal 5.13333333333...

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