Physics: Post your doubts here!

[A.ELWY 7]

like I said I ---> A^2
2I ----> 2 A^2
So A = sqauare root of (2A^2) = root 2 A

Q3
Use SI units
Q13
torque = force x PERPENDICULAR distance
There are 2 ways:-
1- find the vertical component of the force
2- find the perpendicular distance by trigonometry
Q33
V1 = 5/10 x 2 = 1V
V2 = 3/5 x 2 = 1.2V
V1-V2 = -0.2V!

well thanx...but for quest 3 i no that we must use THE SI units...but i also couldn'y figure it out, can u explain it pls???

 

[elbeyon]

P reduction=V1^2-V2^2 /V1 multiplied by 100.--------->P is proportional to V^2

I guess it would be:
P reduction=V1^2-V2^2 /(V1)^2 multiplied by 100. [As P ∝ V^2].

 

[SkyPilotage]

well thanx...but for quest 3 i no that we must use THE SI units...but i also couldn'y figure it out, can u explain it pls???

g x h = ms^-2 x m = m^2 x s^-2
Square root = ms^-1 = velocity SI units --> A

 

[hassam]

pls tell mo how u solve it

 

[A.ELWY 7]

g x h = ms^-2 x m = m^2 x s^-2
Square root = ms^-1 = velocity SI units --> A

thanx very much

 

[SkyPilotage]

pls tell mo how u solve it

An ultrasound is emitted to reach the fetus front edge of the head and is reflected back to the source and shown as the first pulse on the graph. The distance travelled is from the source to the front edge back to the source.
Another ultrasound is emitted to reach the fetus rear edge of the head and is reflected back to the source and shown as the second pulse on the graph. The distance travelled is from the source to the rear edge fo the face and back to the source. It covers the distance travelled by the first ultrasound and travels twice over the fetus's head.
There is a 5.5 x 20 microseconds difference between the reflection of the ultrasounds . i.e size of the fetus head x 2
The distance is = speed x time = 1.5 x10^3 x 5.5 x 20 micro seconds
Therefore the fetus head size is the distance diveded by 2!
This is my interpretation of the question! Hope that helped

 

[hassam]

thank you brother for clearing my cnfusion

 

[JD REBORN]

Q9 with proper explanation cuz i dont know what it is asking.
For Q26 why cant D be correct cuz at antinodes displacement is maximum ....???

 

[SkyPilotage]

Q9 with proper explanation cuz i dont know what it is asking.
For Q26 why cant D be correct cuz at antinodes displacement is maximum ....???

Q9
Its movin up and down. Lowest point of motion is at the bottom where the speed is zero!
Point D from the graph is the lowest point since the speed is increasing and is getting ready to move UP!
Q26
Yes , but the question is asking for vibrations not necessarily maximum vibrations!
At Nodes ! There are no vibrations.
At all the other points, there is! D stats that there is between antinodes. Between 2 antinodes there is a node where there is no vibratioN! so its wrong!

 

[anonymous123]

A ball is kicked towards goal posts from a position 20m from and directly in front of the posts. The ball takes 0.60 s from the time it is kicked to pass over the cross-bar, 2.5 m above the ground. The ball is at its maximum height as it passes over the cross bar. You may ignore the air resistance

a) Calculate the horizontal component of the ball's velocity

 

[darknessinme]

A ball is kicked towards goal posts from a position 20m from and directly in front of the posts. The ball takes 0.60 s from the time it is kicked to pass over the cross-bar, 2.5 m above the ground. The ball is at its maximum height as it passes over the cross bar. You may ignore the air resistance

a) Calculate the horizontal component of the ball's velocity

20/0.6=33.3 m/s

 

[Mustehssun Iqbal]

Assalamu alaikum,
How to solve Q.3 part b in June 2003, Paper 2??
http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s03_qp_2.pdf
Here's its marking scheme: http://www.xtremepapers.com/papers/...el/Physics (9702)/9702_s03_ms_1+2+3+4+5+6.pdf
Thanks!

 

[anonymous123]

20/0.6=33.3 m/s

i get the same ans but in ms its 2.5/0.6 = 4.17

 

[darknessinme]

i get the same ans but in ms its 2.5/0.6 = 4.17

The ms is wrong. They should've used 20m instead of 2.5m.

 

[farrukh]

pls solve this

 

[Zishi]

pls solve this

Assuming that you're facing a problem in part (a), the lowest wavelength photon would be emitted when an electron makes a transition from the ground state to "zero" level.

 

[farrukh]

Assuming that you're facing a problem in part (a), the lowest wavelength photon would be emitted when an electron makes a transition from the ground state to "zero" level.

i am confused that why are we using the zero level? why not the transition from -1.56 eV to -10.43eV

 

[Zishi]

i am confused that why are we using the zero level? why not the transition from -1.56 eV to -10.43eV

Because that is the Maximum Energy change we can get. According to E=hf max energy means maximum frequency, so minimum wavelength. Got it?

 

[farrukh]

thanks

 

[hmlahori]

http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w07_qp_1.pdf

Can anyone please explain Q 11 of this paper? Does the answer have anything to do with centripetal force?