Physics: Post your doubts here!

[smzimran]

http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9702 - Physics/&file=9702_s07_qp_1.pdf

help needed in Q12 ? Thank you

According to the law of conservationof momentum,
m1v1 + m2v2 = (m1+m2)V
2*4 + 4*1 = (2+4)V
V= 2 m/s
Now,
Total energy after collision = 1/2 (m1+m2) (V^2)
= 1/2 (6) (2^2)
= 1/2 (6) (4)
= 12 J
So, B is the answer.

 

[leadingguy]

Re: Physics questions? please?

Question 10 Please.

Thanks!

u can easily learn the formulas and definations being asked in this question from the application support booklet on page 22! its in detail and easy chek it.

 

[Silent Hunter]

http://www.xtremepapers.com/CIE/ind...Level/9702 - Physics/&file=9702_s10_qp_11.pdf

Q13 AND 11 THANK YOU

 

[maybeitwasnoone]

http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9702 - Physics/&file=9702_s10_qp_11.pdf

Q13 AND 11 THANK YOU

11 is C. Using F=ma, where F is the resultant force So, F would be 10-4 and m would be 20/10=10, substitute these values to find a.

 

[Silent Hunter]

 

[Silent Hunter]

11 is C. Using F=ma, where F is the resultant force So, F would be 10-4 and m would be 20/10=10, substitute these values to find a.

wont 20/10 = 2 ?

 

[maybeitwasnoone]

wont 20/10 = 2 ?

My bad, sorry. You're right. You get the right answer with 6/2=3 don't you?

 

[maybeitwasnoone]

View attachment 4862

Is the answer A?

 

[Silent Hunter]

Is the answer A?

answer is D

 

[1357913579]

http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9702 - Physics/&file=9702_s10_qp_11.pdf

Q13 AND 11 THANK YOU

first of all regarding no. 11 f=ma the weight of the substance is 20kg an its mass is 2 as g=10 in this case
two forces in opposite direction so 10-4=6
6=ma
6=2a
a=3
regarding 13
first thing what could be the minimum value of cos its o when all the forcse all prependicular to each other so by resolution when all forcse will get 0 as cos will get 0 so A is the answer
if you still have a confusion do tell me

 

[OakMoon!]

View attachment 4862

Total work done= Work done against friction+Work done to increase the Potential Energy
=150*1.5/sin30+200*1.5
=450+300
=750J (D)

 

[MEGUSTA_xD]

another question when we make a solution of two miscible liquids the boling point of the solution is effected...now ignore that whether it increases or decreases but the whole sol. has a uniform boiling point...if it is so then how we will perform the process of fractional distillation????

Chemistry question, i believe this is a application question, it's about successive extraction, pg 82. In chem app booklet.
They might have same bp, but not solubility in certain solvent.

 

[leadingguy]

a beam of monochromatic lite of wavelengh 630nm trnsports energy at the rate of 0.25mW. calculate the no. of photons passing a given cross section of beam each second.(planck cnstant=6.6*10-34,speed of light=3.0*108)

ans is [8.0*1014] ..................*10 stands for exp

 

[leadingguy]

View attachment 4862

did u find ur ans?? is it really option D???? let me know abt it??

 

[Duanoor]

Chemistry question, i believe this is a application question, it's about successive extraction, pg 82. In chem app booklet.
They might have same bp, but not solubility in certain solvent.

i did o lvls and thn doing fsc and dnt have tht book...cn u pls upload tht page...​

 

[maybeitwasnoone]

a beam of monochromatic lite of wavelengh 630nm trnsports energy at the rate of 0.25mW. calculate the no. of photons passing a given cross section of beam each second.(planck cnstant=6.6*10-34,speed of light=3.0*108)

ans is [8.0*1014] ..................*10 stands for exp

Use E=(hc)/lambda. where E=(6.63*10^-34*3*10^8)/(630*10^-9) E=3.157*10^-19 J is the energy of one photon.

You have the beam transporting energy at the rate of 0.25mW=0.25mJ So Divide 0.25*10^-3 by 3.157*10^-10 to obtain the number of photons. You get 7.92*10^14 photons being transported.

 

[Silent Hunter]

Total work done= Work done against friction+Work done to increase the Potential Energy
=150*1.5/sin30+200*1.5
=450+300
=750J (D)

Can you please explain the highlighted part? Thank You

 

[Silent Hunter]

did u find ur ans?? is it really option D???? let me know abt it??

yes....option D.

 

[MEGUSTA_xD]

i did o lvls and thn doing fsc and dnt have tht book...cn u pls upload tht page...​

http://www.xtremepapers.com/community/attachments/9701_chemistry_applications_booklet-pdf.3845/ here you go

 

[leosco1995]

November 2010, P11, Q24.

How is the answer A?