Physics: Post your doubts here!

[étudiante]

object moving in a circular path experience a so called centripetal force.... they why objects leave the circular path making a tangent to its position of release...??? rather it should fall toward the centre...????

da object moves in a straight line BECAUSE the centripetal force has been removed...
Say if u f a string wid a ball on da end n ur spinning it..... ur hand provides da centripetal force..... as long as u continue to provide da force, the ball moves in a circle.....
BUT da minuite u leave the string (no centripetal force) it flies of at a tangent.....

 

[ABDSyed]

When to use PV=nrt and PV = NKt
Whats The differnce

 

[étudiante]

When to use PV=nrt and PV = NKt
Whats The differnce

same confusion

 

[smzimran]

same confusion

When to use PV=nrt and PV = NKt
Whats The differnce

You can use any one...

pV = nRT is used more because no of moles are usually given instead of no. of molecules

 

[Mustehssun Iqbal]

Assalamu alaikum,
Q.26) A point source radiates energy uniformly in all directions. At a distance of 3.0 m from the source, the amplitude of vibration of air molecules 1 * 10^-7 m. Assuming that no sound energy is absorbed, calculate the aplitude of vibration 5.0 m from the source.
Ans to Q.26 6 * 10^-5

 

[Karimgenena]

Guys I was wondering if someone could help me with a small question!
nov 08 paper4 question 1 a iii ... I dont get the answer on the markscheme S:

 

[hassam]

see brother.,....there will be weight toward centre of earth and normal prependicular to surface in direction opposite to weight........net force actinng on object is centripetal force........so mg-N=mv^2/r
and N=mg+mv^2/r

 

[no.mercy]

Can someone give me the worked solutions of these questions pls

 

[NokiaN95638]

Guyz can ne1 help me with these Q
http://www.xtremepapers.com/CIE/International A And AS Level/9702 - Physics/9702_s10_qp_11.pdf Q no 27, 29

 

[XPFMember]

Assalamoalaikum wr wb!!

got some doubts in p4...plz help..

June 2002

· Q:4 b & c​

· Q:6a​

​

Nov 2002

· Q:3 b(ii)​

​

June 2003

· Q:6 c (i) 2​

· A serious error was evident in a significant number of scripts. Candidates stated, quite correctly, that E = -_V/_x but then went on to state that, based on this equation, the field due to a spherical charge is given by E = V/r. (J’03) *Remember to use V= kQ/r and NOT V = E x r!​

· Q:5 c​

Nov 2003

· Q:5 b (ii)​

· Q:4 c​

· Q:2 b (ii) –ve sign?​

​

June 2004

· Q:4 c​

 

[unique840]

Assalamoalaikum wr wb!!

got some doubts in p4...plz help..

June 2002

· Q:4 b & c​

· Q:6a​

​

Nov 2002

· Q:3 b(ii)​

​

June 2003

· Q:6 c (i) 2​

· A serious error was evident in a significant number of scripts. Candidates stated, quite correctly, that E = -_V/_x but then went on to state that, based on this equation, the field due to a spherical charge is given by E = V/r. (J’03) *Remember to use V= kQ/r and NOT V = E x r!​

· Q:5 c​

Nov 2003

· Q:5 b (ii)​

· Q:4 c​

· Q:2 b (ii) –ve sign?​

​

June 2004

· Q:4 c​

june 2002 q4b) at max displacement the p.e is maximum. and at mean position it is zero. so the graph will be at max at max displacement. but it will be all above xaxis because energy can not be negative.
c) t = 2pie * sqr root of mass / sqr root of 'k'
when mass decreases time period also decrease.
sorry i dont knw abt amplitude

 

[unique840]

Assalamoalaikum wr wb!!

got some doubts in p4...plz help..

June 2002

· Q:4 b & c​

· Q:6a​

​

Nov 2002

· Q:3 b(ii)​

​

June 2003

· Q:6 c (i) 2​

· A serious error was evident in a significant number of scripts. Candidates stated, quite correctly, that E = -_V/_x but then went on to state that, based on this equation, the field due to a spherical charge is given by E = V/r. (J’03) *Remember to use V= kQ/r and NOT V = E x r!​

· Q:5 c​

Nov 2003

· Q:5 b (ii)​

· Q:4 c​

· Q:2 b (ii) –ve sign?​

​

June 2004

· Q:4 c​

nov 2002) when 1/2 f0 is impressed then the time period will be 2T. so every impulse is after 2 oscillations. means alternate oscillations are energized. so there is continuous increase of amplitude at every impulse.

 

[unique840]

Assalamoalaikum wr wb!!

got some doubts in p4...plz help..

June 2002

· Q:4 b & c​

· Q:6a​

​

Nov 2002

· Q:3 b(ii)​

​

June 2003

· Q:6 c (i) 2​

· A serious error was evident in a significant number of scripts. Candidates stated, quite correctly, that E = -_V/_x but then went on to state that, based on this equation, the field due to a spherical charge is given by E = V/r. (J’03) *Remember to use V= kQ/r and NOT V = E x r!​

· Q:5 c​

Nov 2003

· Q:5 b (ii)​

· Q:4 c​

· Q:2 b (ii) –ve sign?​

​

June 2004

· Q:4 c​

june 2003 q6ci) number of atoms = number of moles * 6.02*10^23
4.67*10^15 = (mass/mr) * 6.02*10^23
4.67*10^15 = (mass/90) * 6.02*10^23
mass = mass will be in grams so change it in kg

 

[unique840]

Assalamoalaikum wr wb!!

got some doubts in p4...plz help..

June 2002

· Q:4 b & c​

· Q:6a​

​

Nov 2002

· Q:3 b(ii)​

​

June 2003

· Q:6 c (i) 2​

· A serious error was evident in a significant number of scripts. Candidates stated, quite correctly, that E = -_V/_x but then went on to state that, based on this equation, the field due to a spherical charge is given by E = V/r. (J’03) *Remember to use V= kQ/r and NOT V = E x r!​

· Q:5 c​

Nov 2003

· Q:5 b (ii)​

· Q:4 c​

· Q:2 b (ii) –ve sign?​

​

June 2004

· Q:4 c​

june 2003 q5c) r = mv/Bq
v and B is given. charge of alpha is 2 times that of proton. and mass of alpha is 4 times that of proton.
and for beta, the charge and mass is of electron. use the formula to find the radius

 

[hmlahori]

need help with paper 1 questions. m/j 05 q14, o/n 05 q5,7,16, o/n 09 q9,14,28, m/j 10 q22,26,27, o/n 10 q18,24,37, m/j 11 q15,34

 

[unique840]

Assalamoalaikum wr wb!!

got some doubts in p4...plz help..

June 2002

· Q:4 b & c​

· Q:6a​

​

Nov 2002

· Q:3 b(ii)​

​

June 2003

· Q:6 c (i) 2​

· A serious error was evident in a significant number of scripts. Candidates stated, quite correctly, that E = -_V/_x but then went on to state that, based on this equation, the field due to a spherical charge is given by E = V/r. (J’03) *Remember to use V= kQ/r and NOT V = E x r!​

· Q:5 c​

Nov 2003

· Q:5 b (ii)​

· Q:4 c​

· Q:2 b (ii) –ve sign?​

​

June 2004

· Q:4 c​

nov 2003 q5b)ii 1) at first the plane was normal to the magnetic field producing maximum flux. now it is parallel producing zero flux so change will be max - 0. that is the flux u calculated in parti i.e 9.18*10^-5
2) emf is change in flux/time
9.18*10^-5 / 0.2

4c) capacitor linearise the voltage signal. the signal given in the ques is already linearised once. if it is linearised again with the second capacitor the signal's ripple is further reduced but the time period and freq remains the same
2bii) force = -kx
ma = -kx
a = -kx/m
-ve sign is because it is the restoring force and restoring force is always negative

 

[unique840]

need help with paper 1 questions. m/j 05 q14, o/n 05 q5,7,16, m/j 06 q2,12,14, o/n 08 q27, o/n 09 q9,14,28, m/j 10 q22,26,27, o/n 10 q18,24,37, m/j 11 q15,34

june 06) q2: we have to remember the reasonable estimates. option D is correct cox the wavelength of visible light is in nanometres.
12) momentum before and after remains same. before collision it was:
3mv - 2mv
= mv
before = after
mv = 4mv'
where v' is the new speed
m cancels out
v' = v/4
14) turning effect = F.d costheta
for a larger turning effect, F and d should be larger and "cos theta" must be greater
cos theta increases as angle decreases.
so option B is correct

 

[unique840]

need help with paper 1 questions. m/j 05 q14, o/n 05 q5,7,16, m/j 06 q2,12,14, o/n 08 q27, o/n 09 q9,14,28, m/j 10 q22,26,27, o/n 10 q18,24,37, m/j 11 q15,34

of may june 10 and 11, it is p11, 12 or 13?

 

[hmlahori]

of may june 10 and 11, it is p11, 12 or 13?

It is Paper 11

 

[unique840]

need help with paper 1 questions. m/j 05 q14, o/n 05 q5,7,16, m/j 06 q2,12,14, o/n 08 q27, o/n 09 q9,14,28, m/j 10 q22,26,27, o/n 10 q18,24,37, m/j 11 q15,34

nov 08 q27) it is a stationary wave. so there is half wave between 2 maximum. half wave is 15mm so full wave will be 30mm. v = f*lambda
3*10^8 = f * 30*10^-3
f = 1*10^10
option C