Physics: Post your doubts here!

[Bloodlines]

guys plz answer physics w 07 qp1 question no. 23...

 

[Bloodlines]

reply plzzz!....

 

[Arshiful]

someone pls kindly help me out, um stuck in communicating information in application chapter... donno why but i cant understand what the booklet says... i am seriously having problem digging deeper into its concepts..... someone can u pls provide me good links where i can find easy (understandable) notes or video lecture or anything which will make everything clear to me.... pls help me .....thnx in advance

 

[omg]

someone pls kindly help me out, um stuck in communicating information in application chapter... donno why but i cant understand what the booklet says... i am seriously having problem digging deeper into its concepts..... someone can u pls provide me good links where i can find easy (understandable) notes or video lecture or anything which will make everything clear to me.... pls help me .....thnx in advance

i was about to post the same qs!!!!!!!!!!!!!!!!!!!!!!!!
im stuck there toooo!!! -.-

 

[Oliveme]

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

 

[leadingguy]

i was about to post the same qs!!!!!!!!!!!!!!!!!!!!!!!!
im stuck there toooo!!! -.-

someone pls kindly help me out, um stuck in communicating information in application chapter... donno why but i cant understand what the booklet says... i am seriously having problem digging deeper into its concepts..... someone can u pls provide me good links where i can find easy (understandable) notes or video lecture or anything which will make everything clear to me.... pls help me .....thnx in advance

from whch book are u studying?? do u guys have the black one ???? "internalional A/AS level physics??" chriss MEE, mike crundel, brian arnold are the authors. of the book. well apart have not studied this topic bt yes have gone through some of it till a.m and f.m itx quite typical but not that bad!

 

[unique840]

guys plz answer physics w 07 qp1 question no. 23...

speed is 8m/s which means
8m in 1 s so according to unitary method
50 m in x s 50 m cox one cycle is of 50 m
x = 6.25s
time period is 6.25
frequency is 1/6.25 = 0.16 hz
for max speed: 2*pie*a*f
for max speed we will take max amplitude.
so it will be 2*pie*2*0.16 = 2.01
max kinetic energy = 1/2 (mv^2)
1/2 (0.002*2.01^2)
= 4mJ

 

[omg]

from whch book are u studying?? do u guys have the black one ???? "internalional A/AS level physics??" chriss MEE, mike crundel, brian arnold are the authors. of the book. well apart have not studied this topic bt yes have gone through some of it till a.m and f.m itx quite typical but not that bad!

YEAH i have dat book bt idk y is it so confusing!!
maybe cuz idk the basics!
:/

 

[unique840]

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

14) rebounding height is half of initial means potential energy is half after rebounding.
p.e = k.e therefore k.e is also half after the rebound
final k.e = 1/2 of initial k.e
1/2*m*v^2 = 1/2 (1/2*m*u^2)
v^2 = 1/2* u^2
taking square root on both sides
v = u/sqr rt of 2
v/u = (u/sqr rt of 2) / u
u cancels out
ratio is 1/sqr rt of 2

 

[leadingguy]

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

question 15
look the box at rest having potential energr mgh = 9.81*3*2=58.9J.


now this energy will be converted to kinetic energy at the bottom of the slope.
means at bottom 1/2mv^2 = 58.9J now subtract the work done against friction while coming down! W=f.d



means W= 5*7 = 35J now subtract it from the original ans. u will get the net kinetic energy at the bottom of the slope which is 58.9-35=23.9J



now look 1/2 mv^2 =23.9 find v??

v= 4.88 round off u will get 4.9ms-1Ans

 

[unique840]

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

35) total resistance is
1/r = 1/1 + 1/2 + 1/5
r= 0.588 ohms
total currrent is 5A
v = IR
v = 5*0.588
= 2.94
voltage is parallel does not divide
so voltage in 2ohms resistor will be 2.94 volts
v=ir
2.94 = i * 2
i = 1.5

 

[leadingguy]

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

we know that current entering circuit is equal to leaving the circuit according to khirchoff's law

now I =v/r thsi is the equation for current
now current in resistor of 1 ohm = I1=v/1
2 ohm = I2=v/2
5 ohm = I3 v/5 ..... remember V being same as voltage never divides in paralel circuit!

now total curren I is 5 amperes we have the eq. I1+ I2=I3 = 5

substitute values of I1 , I2 ,I3 and find the unknown voltage! which will be V = 50/17. = 2.94volts

now fr finding current in resistor of 2 ohms! V=IR I=V/R 2.94/2 = 1.47 round of to get 1.5amperes Ans.



hope u get it well???? let me know abt it???

 

[leadingguy]

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

sorry buddy in a rush at the moment cant tell u the ans of 14 as it needs some careful thoughts!

 

[omg]

//http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w07_qp_4.pdf
q.10! how do v know dat its amplitude modulated?/?

 

[smzimran]

//http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_w07_qp_4.pdf
q.10! how do v know dat its amplitude modulated?/?

Check the frequency:
Firstly, the frequency is in kHz not MHz
Secondly, the bandwith is +-5 that is a characteristic of AM

 

[hassam]

Check the frequency:
Firstly, the frequency is in kHz not MHz
Secondly, the bandwith is +-5 that is a characteristic of AM

i quote from the book:"The frequency spectrum of a frequency-modulated
(FM) carrier wave is more complex. In particular, there
are often more than two sideband frequencies for each
signal frequency."

 

[Mustehssun Iqbal]

14) rebounding height is half of initial means potential energy is half after rebounding.
p.e = k.e therefore k.e is also half after the rebound
final k.e = 1/2 of initial k.e
1/2*m*v^2 = 1/2 (1/2*m*u^2)
v^2 = 1/2* u^2
taking square root on both sides
v = u/sqr rt of 2
v/u = (u/sqr rt of 2) / u
u cancels out
ratio is 1/sqr rt of 2

Asalam-o-alaikum
please help me with these questions. Physics paper 11 may 2011 questions 14, 15, 35.
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s11_qp_11.pdf

thank you very much.

Another solution to Q.14:
you take the motion of the sphere in two parts. In the first part;
v^2 = u^2 +2as
v^2 = (0)^2 + 2(10)s
The ball is released from rest and u =0;
v^2 = 20s
s = v^2/20
the final velocity of this motion is the initial velocity of the second motion.v is replaced by u.
s = u^2/20
In the second part of the motion;
v^2 = u^2+2as
v^2 = u^2+2(-10)(s/2) - the height of the sphere in the second part of the motion is half to that in the first part of the motion.
v^2 = u^2 -10s
v^2 = u^2-10(u^2/20)
v^2 = u^2-u^2/2
v^2 = u^2/2
square root of v^2 = square root of (u^2/2)
v = u/square root of 2
v/u = u/square root of 2/u
v/u = 1/square root of 2
although Unique240 's solution to this question is faster!

 

[omg]

i quote from the book:"The frequency spectrum of a frequency-modulated
(FM) carrier wave is more complex. In particular, there
are often more than two sideband frequencies for each
signal frequency."

so it means that the frequency of FM is in Mhz?? and wht abt the bandwidth??

 

[omg]

Check the frequency:
Firstly, the frequency is in kHz not MHz
Secondly, the bandwith is +-5 that is a characteristic of AM

so it means that the frequency of FM is in Mhz?? and wht abt the bandwidth??

 

[smzimran]

so it means that the frequency of FM is in Mhz?? and wht abt the bandwidth??

Yes it is in MHz, the FM channels u listen to are FM 101 MHz, FM 107 MHz etc...
the bandwidth i do not remember exactly; check it from the booklet...