-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
discussion of physics ppr 2 AS
- Thread starter SAN
- Start date
- Messages
- 54
- Reaction score
- 0
- Points
- 0
Raptor said:
few alpha particles deviated by an angle of 10 degree
and the second answer is extremely few (or fewer) alpha particles deviated by an angle of more than 90 degree
- Messages
- 54
- Reaction score
- 0
- Points
- 0
i guess between 45-47 for A ( they never go up above 47 in past papers)
and 42-45 for B cuz A and B is not very much difference in marks but grade.
and 42-45 for B cuz A and B is not very much difference in marks but grade.
yes,even i got 3.33 as a ratio!!!
messed up alot in electricity question....lyk first time i solved the question,i thot it was wrong,then i cut it out and show the correct ans,dunno..it was quite messy and still nt sure ab the answers...
you guys,u know what is the percentile going to be this time???to get an A?any idea?
and fr the shading of the graph,frm the previous discusssion,i've seen they shaded the area between the two curves as the Es!..HOWCUM?
It has said that this energy was the energy needed for the permenant deformation,so isn't it going to be the area under the whole graph,lyk shade included both 1st and the second graph,everything!...as the area between two curves is the energy DISSIPATED,not fr the permenant deformation....guess!!!
but im STILL not sure!
suggestions plz??????
im really worried...the paper overall was bad!
messed up alot in electricity question....lyk first time i solved the question,i thot it was wrong,then i cut it out and show the correct ans,dunno..it was quite messy and still nt sure ab the answers...
you guys,u know what is the percentile going to be this time???to get an A?any idea?
and fr the shading of the graph,frm the previous discusssion,i've seen they shaded the area between the two curves as the Es!..HOWCUM?
It has said that this energy was the energy needed for the permenant deformation,so isn't it going to be the area under the whole graph,lyk shade included both 1st and the second graph,everything!...as the area between two curves is the energy DISSIPATED,not fr the permenant deformation....guess!!!
but im STILL not sure!
suggestions plz??????
im really worried...the paper overall was bad!
- Messages
- 54
- Reaction score
- 0
- Points
- 0
Edifier said:
intensity is proportional to ampliotude squred and frequency squred
but frequency is constant so only take amplitude^2
at source the amplitude was 2mm and 6.0cm apart was 1.2mm
therefore (2^2)/(1.2^2) = 2.78 not 3.33
idk how did u get 3.33 i guess u did (2^2)/1.2 = 3.33 ?
no no,i acatually ddnt get the second amplitude as 1.2mm...as far as i rmr was 1.1mm(think the difference is due to the error in the graph reading?)
so (2^2)/(1.1^2)=3.3!!!!
hey am i going to get marks on this still,even though the way of working out the intensity is absolutely ryt!
so (2^2)/(1.1^2)=3.3!!!!
hey am i going to get marks on this still,even though the way of working out the intensity is absolutely ryt!
question1.ii scalar vectr scalr. b) 25sin35
question3:reaction force of the rod and the action force provided by the hand...(sum of my frnds wrote the friction btwn the rod nd paper thingy too)
it will come to rest cuz of no resultant torque nd will come to rest in along the rod passin through G
question2rojectile: 1st part=1.71m by using s=vcos theta t......2nd aprt by using 2as=v^2-usin theta^2
question4:i dnt remember which question was it lol....
question5:electricity.....r total for the 1st part i got 857.3 sumthn.....voltage=3.2voltage......thn used potential divider wala formula to calcutale the new r total.....nd calcultaed the temperature by the ratio
ques 6:waves....as power/energy is directly propertional to the intensity of a wave nd intensity is dirctly proportional to amp square therefore as the amp was decreasin along the wave so will the power/energy
ratio=1/9 every1 in here got the same answer
question7:nucleusis small: small number of particles got deviated
nucleus is charged:deviations were greater thn 90 degrees
y isnt B particle used:less mass so more deviation......negatively charged so will be highle attracted towards the nucleus
question3:reaction force of the rod and the action force provided by the hand...(sum of my frnds wrote the friction btwn the rod nd paper thingy too)
it will come to rest cuz of no resultant torque nd will come to rest in along the rod passin through G
question2
rojectile: 1st part=1.71m by using s=vcos theta t......2nd aprt by using 2as=v^2-usin theta^2question4:i dnt remember which question was it lol....
question5:electricity.....r total for the 1st part i got 857.3 sumthn.....voltage=3.2voltage......thn used potential divider wala formula to calcutale the new r total.....nd calcultaed the temperature by the ratio
ques 6:waves....as power/energy is directly propertional to the intensity of a wave nd intensity is dirctly proportional to amp square therefore as the amp was decreasin along the wave so will the power/energy
ratio=1/9 every1 in here got the same answer
question7:nucleusis small: small number of particles got deviated
nucleus is charged:deviations were greater thn 90 degrees
y isnt B particle used:less mass so more deviation......negatively charged so will be highle attracted towards the nucleus
Hey a little confusion for the projectile question look t my work:
s = ut + 1/2 a t ^2
= (12.4)sin34 (0.7) + (1/2)(-9.81)(0.7)^2 which is equal to 7.1 m so how come some got 1.3 m or i am not correct.
The gt will be 37-40 because the level of physics has indeed lowered in the last 2 years so just chill...!!!!!!!!
s = ut + 1/2 a t ^2
= (12.4)sin34 (0.7) + (1/2)(-9.81)(0.7)^2 which is equal to 7.1 m so how come some got 1.3 m or i am not correct.
The gt will be 37-40 because the level of physics has indeed lowered in the last 2 years so just chill...!!!!!!!!