Physics MCQs thread.

[abrraza]

A.C.W.M - C.W.M = resultant (ignore sign)
(5x2)+(10x2) - (15x3)= -15

step1 note 5N and 10N in A.C.W and 15N in C.W
step2 for every force take perpendicular distance

 

[yubakkk]

my last question to u all...

 

[Xthegreat]

my last question to u all...

the ans is B

but you never give your answers for the previous questions!

 

[yubakkk]

ok fromnext time i will give
will u xplain this 2 me????

 

[shyqueen]

i have the answers for june 01.

cud u plzzz post em.

 

[airborne1944]

oh, its that question from O/N 2010 right? i don't get it either.[/quote]
actually rate of energy is Power ... E/t=P ... and Power is equal to Force x Velocity ... this means uptill now rate of energy provided=Fv .. here force is resultant weight (m1g-m2g) takimg g common (m1-m2)g x Velocity .. so answer is D ...[/quote]
why m1-m2? isnt it m2-m1?[/quote]
Please clarify

 

[rabib2013]

i have the answers for june 01.

dude can u give them, or the link to the answers.. if u dont have any link then just tell me the answers of 2, 11 and 23

 

[Xthegreat]

ok fromnext time i will give
will u xplain this 2 me????

the graph are all decreasing because the question says POTENTIAL
the POTENTIAL DIFFERENCE increases, but the POTENTIAL decreases.
potential difference is between two points, say between P and Q
take potential difference at P = V and at Q = 0 ( means the potential decreases from P to Q )
as we move closer to P, the potential difference between P and the certain point increases
but the potential is actually decreasing.
* i dunno whether you get it or not, but i tried my best*

R = pl/A
so with increasing p, the R increases from left to right
so for the same material, as the length increases, the resistance increases.
resistance increases, potential at that certain point decreases further, making it's potential difference between that point and P greater.

 

[rabib2013]

yo zishi, how abt this one?

 

[rabib2013]

alright anyone with the solution to the problem i just posted? :S

 

[hassam]

 

[hassam]

well cn this be an explantion...for this one just wanna knw abt ma concept....energy supplied by moter is used to increas p.e of m1 and decrease p.e of m2.......so total energy suppplied m1gh-m2gh...for rate...we cn replace h wid v....so migv-m2gv

 

[Zizu1992]

http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf
Please explain question fourteen.

 

[examfearz]

alright anyone with the solution to the problem i just posted? :S

Is the answer D?

 

[rabib2013]

alright anyone with the solution to the problem i just posted? :S

Is the answer D?

yeah, can u explain it to me?

 

[examfearz]

I = A^2.....so for Amp 8.0 the Intensity shud be 8^2 = 64....in the equation given this I is for I = 1\r^2
now for distance 2r..the eq gives us I = 1/(2r)^2 whch is equal to 1\4r^2 which means that the intensity decreases by 4 times
so for 2r I becomes 64\4 = 16...and Amplitude= root of Intensity so it becomes 4.... hope I helped....

 

[Zizu1992]

Please somebody explain my question too!!

 

[examfearz]

well cn this be an explantion...for this one just wanna knw abt ma concept....energy supplied by moter is used to increas p.e of m1 and decrease p.e of m2.......so total energy suppplied m1gh-m2gh...for rate...we cn replace h wid v....so migv-m2gv

I thnk u r rite....though i hav a confusion...y cant we use the K.E equation in this??

 

[examfearz]

Please somebody explain my question too!![/qu
suppose initial velocity is v...so E=1/2 mv^2
At the highest point only the horizontal comp acts which is vcos45...so the K.e at the top will be 1/2(vcos45)^2=1/2v^2(0.5)
As 1/2 mv^2 = E so it becomes 0.5E which is A

 

[rabib2013]

I = A^2.....so for Amp 8.0 the Intensity shud be 8^2 = 64....in the equation given this I is for I = 1\r^2
now for distance 2r..the eq gives us I = 1/(2r)^2 whch is equal to 1\4r^2 which means that the intensity decreases by 4 times
so for 2r I becomes 64\4 = 16...and Amplitude= root of Intensity so it becomes 4.... hope I helped....

yo thanks man