Physics Pp:3-Variant 2- Discussing Answers!!

[BumbleBEA]

i wrote the same thing hi5!!!
what abt the page to the right of the waves?????

What was the question? Haha!

 

[B333]

It was about how the wavefront in the shallow water changed and you had to give a reason for your answer

 

[kboss]

What was the question? Haha!

LOL...
it was the one abt what change was made so that no refraction took place

 

[Jay Jun]

i wrote the same thing hi5!!!
what abt the page to the right of the waves?????

I wrote region P and Q was the same depth for the next page question.


Isn't the refractive index 1.305? n=1/Sin(C)

 

[BumbleBEA]

Oh

I'm not wholly sure but since there was to visible change, I said that they made the depth of water in both areas equal so that no refraction would take place

That was a tough paper... hope I get an A* (call me Asian, but yeah )

 

[Chucky]

The half life actually IS 1.6 days.

The count rate started at 52, so yes you deduct background radiation which is 14.
52-14=38
The half life of the sample means that the rate count for the radiation emitted by the sample should halve
38/2=19
Thus, you must deduct 19 from the initial count rate of 52, since background radiation is always present
52-19=33
So you find the x-axis equivalent when y=33
It came out to be 1.6 days.

Note: An alternative way would be to add back the background radiation, which is what I did
19+14=33

It comes out to be the same.

yes i know, i forgot to add background radiation back again!

 

[B333]

No the refractive index was 1.56
i think it was 1/sin 40 because n=1/sin C and the critical angle was 40
Sorry

 

[BumbleBEA]

I wrote region P and Q was the same depth for the next page question.


Isn't the refractive index 1.305? n=1/Sin(C)

No... I don't think so.

Refer to my answer to some other person above/in a previous page.

Note:

n=1/sin 40=1.56 (3 significant figures)

 

[kboss]

Oh

I'm not wholly sure but since there was to visible change, I said that they made the depth of water in both areas equal so that no refraction would take place

That was a tough paper... hope I get an A* (call me Asian, but yeah )

hi5 again!!!!
regarding that asian thing....most of those on XPC are Asians....

 

[Chucky]

I wrote region P and Q was the same depth for the next page question.


Isn't the refractive index 1.305? n=1/Sin(C)

the index is something around 1.6

refractive index air/refractive index of glass = sin 90/sin 40

since it goes from an optically denser to a medium less dense the equation reverses.

and the reasoning for the region question is correct.

 

[Jay Jun]

I was using this formula for half life because I thought background radiation was excluded. A=A(o)e^-(ln(2)/half life)(time)
My answer was 3.00998 = 3.01 days

 

[B333]

What about the tidal question? What was the quality those stations had to get electricity when the tide is low or something like that?

 

[Jay Jun]

the index is something around 1.6

refractive index air/refractive index of glass = sin 90/sin 40

since it goes from an optically denser to a medium less dense the equation reverses.

and the reasoning for the region question is correct.

The refractive index question, dint they give you a new graph? the angle of incidence was 40

 

[BumbleBEA]

yes i know, i forgot to add background radiation back again!

That's a shame, but I'm sure you'll get some effort marks!

I made quite careless mistakes as well.

For the lightbulb question, I made it all the way until 40-31.7, and my idiot self calculates wrong and wrote:

40=31.7=9.3 <---FOR THE LOVE OF GOD, WHY?!

Haha... so I put 9. What a shame.

 

[kboss]

the index is something around 1.6

refractive index air/refractive index of glass = sin 90/sin 40

since it goes from an optically denser to a medium less dense the equation reverses.

and the reasoning for the region question is correct.

yo...where u frm??

 

[Jay Jun]

What about the tidal question? What was the quality those stations had to get electricity when the tide is low or something like that?

The generator can turn both ways. (clockwise,anti-clockwise)

 

[kboss]

The refractive index question, dint they give you a new graph? the angle of incidence was 40

exactly...do 1/sin 40 and ull get 1.56

 

[Jay Jun]

exactly...do 1/sin 40 and ull get 1.56

typo 50 not 40. 1/sin(50) = 1.3

 

[BumbleBEA]

typo 50 not 40. 1/sin(50) = 1.3

Nope. 40 degrees.

Measure the angle of incidence relative to the normal.

90=50=40

 

[kboss]

typo 50 not 40. 1/sin(50) = 1.3

but 50 isn't the angle of incidence...its 40!!!