Recent content by Sammiebaby966

Thank you so much! =]

Oh wow, I am having a slow night sorry! Sooo let's try this again: Sin\theta2=(1.00/1.33)Sin(76)=.730 \theta2=Sin^-1(.730) \theta2=46.9' Sooo, tan(46.9')=(5.50m)/depth Depth=(5.50m)/tan(46.9') Depth=5.15m That sounds better. Did I mess up again or did I get it right?

So, tan(10.5')=5.50m/x x=5.50m/tan(10.5') x=29.7m So the depth is 29.7m? I am confused because that seems large.

Sin\theta1n1=Sin\theta2n2 I got the angle of the index of refraction. Sin\theta2=(1.00/1.33)Sin(14')=.182 \theta2=Sin^-1(.182) \theta2=10.5' Im not sure how to find the depth tho.

We wish to determine the depth of a swimming pool filled with water without getting wet. We measure the width of the pool, which is 5.50m. We then note that the bottom edge is just visible when we stand and look at an angle of 14.0 degrees above the horizontal line. a)Calculate the depth of...