physics question 6

Question

physics question 6

kittykat · Answer

I am confused which angle goes where

kittykat · Answer

Like I get 26.8 , but where is that? under it or over it ?

kittykat · Answer

what do I do after I get 26.8 as the angle ? @agent0smith

agent0smith · Answer

Just do them one at a time, starting from the top. and remember Z angles.
|dw:1470712501090:dw|

agent0smith · Answer

ie the refracted angle becomes the incident angle for the next boundary.

kittykat · Answer

so if you look at the question angleA becomes angleW?

kittykat · Answer

Can you do one angle for me ? Im kinda confused

agent0smith · Answer

No, like this:
|dw:1470716300291:dw|

kittykat · Answer

ok let me try to slove it now

kittykat · Answer

I got 23.6 but thats the wrong answer :(

agent0smith · Answer

You aren't showing any work so idk how you're getting that. Always show work, not answers. $$\Large n_1 \sin \theta_1 = n_2 \sin \theta_2 $$

kittykat · Answer

yes first I did that I got 26.8 right?

agent0smith · Answer

It's easier if you actually show the equation, with numbers plugged in, then i can tell if you did it right.

agent0smith · Answer

26.8 degrees is correct for w.

kittykat · Answer

thne I did (sin26.8)(1.33)/1.50

kittykat · Answer

but I go the wrong answer

kittykat · Answer

@agent0smith

agent0smith · Answer

You didn't say what you got... http://www.wolframalpha.com/input/?i=arcsin(1.33+sin+(26.8+degrees)%2F1.5)

agent0smith · Answer

(sin26.8)(1.33)/1.50 is not the angle

arcsin( (sin26.8)(1.33)/1.50 ) is the angle

osprey · Answer

Snell’s law of refraction
The “daisy chained version”
See enclosed for the details ...

agent0smith · Answer

@osprey your representation of Snell's law is incorrect. https://www.math.ubc.ca/~cass/courses/m309-01a/chu/Fundamentals/snell.htm

agent0smith · Answer

Apply snell's law$$\Large n_a \sin \theta_a = n_w \sin \theta_w = n_g \sin \theta_g$$so we can just skip the middle boundary entirely and use$$\Large n_a \sin \theta_a = n_g \sin \theta_g$$and plug in values$$\Large 1 \sin 36.8  = 1.5 \sin \theta_g$$solve for theta$$\Large \sin^{-1} \left(  \frac{ \sin 36.8 }{ 1.5 } \right)  =  \theta_g$$
gives the angle in glass as 23.54 degrees

agent0smith · Answer

There's no need to find the angle in water, it just takes more time that way.

kittykat · Answer

Can you explain  me the differnece in elastic and inelastic   , I m confused doing this qestion

kittykat · Answer

@agent0smith

agent0smith · Answer

Elastic collision means kinetic energy is conserved. 

Inelastic means KE is not conserved (some is turned into heat/sound etc).