- Messages
- 172
- Reaction score
- 71
- Points
- 38
You're welcome.thankuu
I'm leaving for the paper 32 physics, I hope I will not run out of time.
All the best.
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)
You're welcome.thankuu
kool , my paper 31 is 8 hours laterYou're welcome.
I'm leaving for the paper 32 physics, I hope I will not run out of time.
All the best.
http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Physics (0625)/0625_s07_qp_3.pdf
last question , (b) ? how to :\
who's done with the physics ??!! i hope its easy
wel am done eccept see oct 08 31 q1 d iiare you done with your physics? Was it hard?
oh no am doing after 8 hrsare you done with your physics? Was it hard?
any 1wel am done eccept see oct 08 31 q1 d ii
http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Physics (0625)/0625_w08_qp_03.pdf
wel am done eccept see oct 08 31 q1 d ii
http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Physics (0625)/0625_w08_qp_03.pdf
any 1
w0.5 m/s^2 is acceleration due to 0.4N weight.....so to get m we multiply by s and then by s again, then we divide by 2 to get the answer as its 0.5m/s^2 and not 1m/s^2
howIn the table 0.4 N has acc 0.50 m/s^2 after 0.25 m/s^2
but in the question its starting from rest so it will have 0.25 m/s^2
v=u+at?
u= initial speed, acceleration = a=0.5 t=1.2
v=0 + 0.5(1.2) =0.6m/s
ii) Distance
can use average speed (s= [v+u]/2 * t]
which gives distance = 0.3*1.2=0.36
or s=ut+1/2at^2
=0+(0.5)(0.5)(1.2)^2=0.25*1.44=0.36
Thank you! Inshallah we'll both get A*You seem to be really good in Physics!
Mashallah!
Molecules which have higher kinetic energy have a higher chance to break their bonds and escape (evaporation). At 85*C more energetic molecules exist and thus more water has evaporated/more molecules have broken their bonds which is why the level is lower.
Zain beat me!
So anyone know how to calibrate a C.R.O?
When they ask to calibrate a C.R.O. i assume it means find the Y-Gain/Time Base?
So that would mean you'd put in a source of known voltage on the Y-inputs (with X-Inputs being an A.C. source)
Measure the amplitude of resultant wave
and Y-Gain would be input voltage/amplitude (since we know the voltage input)
And then if we know the frequency of the input we can use that to find the time base
THAT'S assuming that that's what the question wants from us.
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now