Physics: Post your doubts here!

[Shimmery woods]

que 8

 

[Shimmery woods]

que 3 ....last doubt of this paper

 

[Mstudent]

25 and 26 please

For q. 26:

equation for doppler effect:

[Speed of sound X Frequency of source]/[Speed of sound + or - speed of source]

so just put in the values and subtract when speed of sound - speed of source (approaching) from speed of sound + speed of source (Receding) ANS= C

for q. 25

for the length of wavelength: I used d = S X T

so wavelenght = 330 X 500 X 10^-6 = 0.165 m

so a translation of 0.165m means shift the graph by 1 lambda or 2 pi or 360 degrees

so basically it will be the same graph ANS= D

hope that helps

 

[janedoe]

I=A^2
Intensity for wave 1 is given which is equal to amplitude (2*10^-2)^2
Amplitude for wave 2 is (3*10^-2)^2. Find the intensity of wave 2 by cross multiplying and the answer you'll get is C.

The y-axis is labelled pressure instead of displacement; so how can we read off amplitude directly?

 

[janedoe]

Under what conditions do you suggest that the time taken is equal in both waves?

You're right. So my method is invalid. Can you please explain the correct method to obtain the answer?

 

[Thelastmoment]

You're right. So my method is invalid. Can you please explain the correct method to obtain the answer?

furuta said it so there's no need repeating the same thing

 

[nigeller]

dude i love you

 

[amina1300]

Do u want me to show you how I got the answers?
of course yes

Umm aren't u going to show working?

Q13 F1d1 + F2d2 =Ftdt
16000*0.6 +9000*2.1 =(16000+9000)*
28500÷2500=s
d=1.14m

Q10.
velocity upwards = +ve Total momentum before = 0 After = 0
100*-8 + 50 *v1sin60 + 50*v2sin60 = 0
800 = v1 sin 60 + vs sin60
(800/50)/sin60 = v1 + v2

v1 + v2 = 18.4752 (v1 and v2 have equal magnitudes)
18.4752/2 = 9.2

ans = B

Q28
Distance between the 2 sucessive wavefronts = lamda ( For both curved or straight waves)
v = f *lamda
f = 1/T c = v
c/lamda =1/T
lamda/c = T

Time when wavefront GH reaches point P;

3 lamda;
so,T = 3* lamda/c

ans = C

Hope this helps

 

[amina1300]

que 19 please

Loss in Potential Energy + work done by frictional force = Gain in Kinetic Energy
mgh + 400*250 = 0.5 *500*v^2
(500 *9.81 *100000)/250 = v^2
sqrt (ans) = 13.87

Ans = A

 

[amina1300]

O

que 3 ....last doubt of this paper

Only Ans A is correct as if wind blows, the aeroplane will move in the same direction as wind ( accelerate in that direction) and the velocity in wind will have a lower magnitude that without wind effect.

 

[amina1300]

que 8

Areas under the graph (5-0.8)*20/2 + 0.8*20

 

[Holmes]

Q13 F1d1 + F2d2 =Ftdt
16000*0.6 +9000*2.1 =(16000+9000)*
28500÷2500=s
d=1.14m

Q10.
velocity upwards = +ve Total momentum before = 0 After = 0
100*-8 + 50 *v1sin60 + 50*v2sin60 = 0
800 = v1 sin 60 + vs sin60
(800/50)/sin60 = v1 + v2

v1 + v2 = 18.4752 (v1 and v2 have equal magnitudes)
18.4752/2 = 9.2

ans = B

Q28
Distance between the 2 sucessive wavefronts = lamda ( For both curved or straight waves)
v = f *lamda
f = 1/T c = v
c/lamda =1/T
lamda/c = T

Time when wavefront GH reaches point P;

3 lamda;
so,T = 3* lamda/c

ans = C

Hope this helps

Thanks! amina1300

 

[dumbledore.]

Answer:

I guess,
K . E= 1/2 x m x v2(squared)
= 1/2 x 500 x 2 (2)
=1000 J

then,
G.p.e = mgh
re-arrange to
m= G.p.e/( g x h)
=1000J / (9.81 x 15 )
= 6.8kg


6.8kg for 100%
m1 kg for 10%
cross multiply to get 0.68kg so answer is B.

Thankyouuu

 

[Shimmery woods]

please help me with que no 35

 

[Mahad Tarar]

May/June 2016 qp 12 Q.11

The diagram shows a man standing on a platform that is attached to a flexible pipe. Water is
pumped through the pipe so that the man and platform remain at a constant height

The resultant vertical force on the platform is zero. The combined mass of the man and platform
is 96 kg. The mass of water that is discharged vertically downwards from the platform each
second is 40 kg.
What is the speed of the water leaving the platform?
see Image below!!
Can Someone Please Answer this Question with explanation .

 

[Mahad Tarar]

please help me with que no 35

Cannot be option B and D because p.d b/w S and R can only be 10 if p.d b/w Q and S is zero because then ratio will be same. (WheatStone Bridge theory)
Also p.d b/w Q and S is 2V (5V -3V) ..Therefore A should be correct answer..

 

[Shimmery woods]

May/June 2016 qp 12 Q.11

The diagram shows a man standing on a platform that is attached to a flexible pipe. Water is
pumped through the pipe so that the man and platform remain at a constant height

The resultant vertical force on the platform is zero. The combined mass of the man and platform
is 96 kg. The mass of water that is discharged vertically downwards from the platform each
second is 40 kg.
What is the speed of the water leaving the platform?
see Image below!!
Can Someone Please Answer this Question with explanation .

Use the formula F = m/t * (v-u)
applied force = rate of change of mass * change in velocity
96×9.81 = 40 × (V -0)
rearrange it and you will find the answer

 

[furuta]

hello~ can someone please explain/solve questions: 6,9,10,18 & 37 to me, thank you

http://papers.gceguide.com/A Levels/Physics (9702)/9702_s15_qp_12.pdf

 

[Shimmery woods]

help me with que 15 please

 

[Shimmery woods]

que 30 ...i could get to the part mgd/V but why do we multpily it with e ?
F = E×q
m×g = E ×q
m×g = V÷d × q
mgd/V = q ...how do i get the value of n ?