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

[kboss]

refractive index is 1.56.....
and what abt the wave questions ???
i took a random guess and put CD

 

[JOEJOE]

ITS CD because all the rest are speeding up rather than speeding down...

 

[kboss]

ITS CD because all the rest are speeding up rather than speeding down...

ok....
for the one on the right hand page...what was the change ???
some ppl in my class said tilting the tank wud work...
i just wrote to use a tank with uniform depth or something like that

 

[kareem]

refractive index is 1.56.....
and what abt the wave questions ???
i took a random guess and put CD

no it was sin50/sin90 which gives 0.6 sumthing

 

[kareem]

nope. the refractive index is correct. 1.56.

sin of angle of i / sin of angle r which is sin 50 / sin 90 which was 0.6 sumthing

 

[kboss]

sin of angle of i / sin of angle r which is sin 50 / sin 90 which was 0.6 sumthing

do this....
sin of critical angle = 1/ refractive index
i.e reflective index = 1/ sin of critical angle
and angle of refraction wasn't given too....
plus the refractive index has to be more than one if it is a denser medium than air...which i suppose glass is

 

[JOEJOE]

Critical angle was 40 right.. so 1/sin40 will give 1.56....

Guys can u answer this?

"the answer about explain the radio activity was becoz its random and slow process bla bla right?
for the dot and the charge u should draw like 5 arrows inwards right?
Plus... what was the answer for the last question about radioactivity ( why its not enuff precautions etc..)"

 

[kboss]

Critical angle was 40 right.. so 1/sin40 will give 1.56....

Guys can u answer this?

"the answer about explain the radio activity was becoz its random and slow process bla bla right?
for the dot and the charge u should draw like 5 arrows inwards right?
Plus... what was the answer for the last question about radioactivity ( why its not enuff precautions etc..)"

the answer for the radioactivity thing (q1) was correct as mentioned....cuz it was background radiation and its always random.
second and third questions no idea
though for the third one i just put that the a-particles and gamma waves would travel through the air regardless of how far you were.
which is why a suit would be better
still am not sure if that's correct

 

[sara16]

ahh i left the last whole page,the bulb question, didnt have time to do it..these questions contained atleast 18 marks :'(
the ones i did ...are most probably wrong...:'( * badly wanna go back in time and do all the questions right*

 

[BumbleBEA]

the answer is 8
let the total number of bulbs be x.
5% of original resistance is 8.4 ohms
voltage is same (source same) and the max current before the fuse melts is 0.9A
therefore current= voltage/total resistance
0.9 = 240/(x * 8.4)
the answer comes in 31.77 , round that off its 32. so the total number of bulbs is 32. difference 40-32= 8 bulbs


It's upwards, as the current induced in the coil , produces a magnetic field that OPPOSES the motion on the bar magnet.
therefore when it is entering the solenoid, it will produce north to repel
when it is leaving the solenoid, it will produce south to attract, overall all the forces exerted being upwards.



it's way more. you have to deduct background radiation which was around 14 counts per minute.
you find that from the graph (it is the point after which the counts do not decrease anymore)
52-14= 38
38/2= 19 counts per minute
find out how long it took to come to 19 counts per minute.
should be more than 3 days.


when it is low tide, meaning the sea water is at a lowe level, water moves FROM RIGHT to LEFT



I'm not sure about this, but I heard the red ray is supposed to refract more, so you draw one line refracting and a faint line for a partial reflection as well.
if you have drawn the reflection, you get one mark.



OR THEN NOT, remember to get that sequence correct in your drawing.
Y was 0, Z was 1.

hope that helped.

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.

 

[BumbleBEA]

guys, do you know the answer for the stupid bird ? does it stay stationary when rotated or it rotates back anticlockwise ? explain too please

It rotates anticlockwise, because if you measured the distances of the forces from the pivot, the anticlockwise moment of force worked out to be about 0.44Nm, whilst the clockwise force was 0.30-something Nm.

 

[BumbleBEA]

for 3 i got 2.6 days , you had to use the graph i think

I think you forgot to take into account the background radiation present. That was my first answer

 

[BumbleBEA]

how do you know that the background radiation is exactly is 14 ?? it varies.. so i dont think making any calculation with the background radiation is valid. all i did was just read from the graph and i got 2.6 days. in any textbook, when they show how to find the half life time, they just read from the graph tho. well i dont know, this is my point of view

We can hypothesize that background radiation is approximately 14 since that's the minimum y-axis measurement when the sample was no longer emitting radiation (where the graph leveled out).

 

[BumbleBEA]

isnt the answer of finding the counter weight in no.1 was 0.5 or 0.6 ? i cant recall. and the bird is like 1 kg or some shit ? i can hardly remember but yeah

The counter weight was 0.6kg according to my calculations.

 

[BumbleBEA]

Did it say 14 background radiation in the question??

That was the approximate answer to the question prior.

 

[kboss]

That was the approximate answer to the question prior.

what did u put for the waves question?????

 

[BumbleBEA]

no it was sin50/sin90 which gives 0.6 sumthing

I'm afraid not. The answer is indeed 1.56, approximately.

*refractive index of air=1*

refractive index 1 x sin theta 1= refractive index 2 x sin theta 2

The angle of incidence from the air to the glass was 90 degrees, and the angle of refraction once it entered the glass was 40 degrees.

Thus:

1 x sin 90= refractive index of glass x sin 40

refractive index of glass= sin 90 / sin 40

=1.555.... or 1.56 (3 significant figures)

Note: The angle is 40 degrees, not 50 degrees because you must measure it relative to the normal.

 

[B333]

What did you guys get for the wavefronts question? I wrote CD then cancelled it at the last minute and wrote CE
Also, I know this will sound dumb but what was the power of the lamp?

 

[BumbleBEA]

What did you guys get for the wavefronts question? I wrote CD then cancelled it at the last minute and wrote CE
Also, I know this will sound dumb but what was the power of the lamp?

The answer was CD, if memory serves me right, because waves in shallow water are faster than in deep water.

From a slow to a fast medium, it refracts AWAY from the normal.

From a fast to a slow medium, it refracts TOWARDS the normal.

Note: I don't remember the numbers, but...

Power=VI
Power=Energy/Time
Power=RI^2

 

[kboss]

The answer was CD, if memory serves me right, because waves in shallow water are faster than in deep water.

From a slow to a fast medium, it refracts AWAY from the normal.

From a fast to a slow medium, it refracts TOWARDS the normal.

Note: I don't remember the numbers, but...

Power=VI
Power=Energy/Time
Power=RI^2

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