-
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
dollykhan said:
The height of the ball is 1.38....u cant use v^2=u^2+2as.....Because it didnt say that at that height the ball was at its max height or final vertical velocity is zero. U have to use T=0.17 s. and this equation s=ut+1/2at^2...Horizontal distance was 1.7 m and height was 1.38 m.
On the other end it didnot said energy needed to deform the particle. It said energy expended (used) in deforming the particle.
Ther are under the lower graph would not be Es . It would be energy recovered when load is removed. My answer to that one is 7mJ.....
11 degree was the temperature of thermistor...
B-particle cannot be used because of 3 reason :
.some B-particle would be repelled by orbiting electron around nucleus. This will result to wrong justification that nucleus is negatively charged.
.Some would fully be attracted to nucleus causing no scatter in scattering experiment/
.B-particle travel at 99% of speed of light..By no chance it would get deviated...
the projectile was shown...and as the ball had crossed the wall so we can apply 2as=v2 -u2 cuz the half projection shown gives us the max height
and what was the new r total?
mine was 1280 ohms.....did it by pot divider formula....
nd yap ull get the ecf bcuz the whole of the question was dependent on part 1
and what was the new r total?
mine was 1280 ohms.....did it by pot divider formula....
nd yap ull get the ecf bcuz the whole of the question was dependent on part 1
Thats what i am saying. It said it took the ball o.17s to just cross the wall. U should check at what time it would reach the max height. It could also go above that height of the ball. I am sure i have done right(can have some calculation mistake)....Because there is a huge difference b/w just crossing it and reaching the max height...
- Messages
- 45
- Reaction score
- 1
- Points
- 0
The threshold for an A grade should be around 40-44....the exam's standard was quite average though it did contain some hell tricky questions....p.s....wasnt the answer to that slope wala question something like sin55 x 35?......nd how much marks would i lose if ive written (9.81) instead of (-9.81)????
- Messages
- 45
- Reaction score
- 1
- Points
- 0
huh?...how is it possible...if u use (-9.81)....this expression will get subtracted from the preceding one....nd so will produce a different answer.....nd hw abt sin55 * 35???.....
hmmm..i got it wrong..1.1 shud be the answer...My working is right...i also used a=+9.8.....
B/w the answer was 25sin(35)....Tension ki line ko backward extend karna tha...Move the 25 newton force and make a closed triangle...With simple geometry the answer would work out to be 14. something....
B/w the answer was 25sin(35)....Tension ki line ko backward extend karna tha...Move the 25 newton force and make a closed triangle...With simple geometry the answer would work out to be 14. something....