Physics: Post your doubts here!

[Scafalon40]

your absolute error should be correct to 1s.f. and then the value should be correct to the same no. od d.p. as ur resulting absolute error is.
like if u hav calculated a length to be 20.324 +/- 0.346cm

then ur correct absolute error should be +/- 0.3
and thus the resulting value should be 20.3 as the absolute error comes to 1 d.p.


for another case you have the speed to be 326.55 +/- 1 m/s
then ur answer should be:

327 m/s
because the absolute errror is correct to 0 s.f.

even if the value does not have a number after the decimal but the absolute error is correct to 1 d.p. or so u write abc.0 insead of just writing abc

hope u get it
1357913579 u might wanna check this out ... maybe it will help ^_^

Hey thanks dude! I'd totally forgotten about this stuff, even though I've already taken my AS papers. Thanks for the refresher! I owe you one

 

[Ahmedraza73]

HI
does anyone know how 2 do this?
9702_s12_qp_12
question 14
thnx in advance!

The Answer will be B that i think so,PLEASE do tell me if i am right?
A: In A weight is acting downward and the normal reaction force at the end of ladder is upward and the force P is the normal reaction force by the wall in straight direction.Which means the ladder is not sliding and on the equilibrium postion
B: In B weight is acting downward and the normal reaction force at the end of ladder is upward and the force P force P isacting slightly upward because the ladder is sliding .
C: In C weight is acting downward again and the normal reaction force at the end of ladder is now not acting against the direction of weight it is slighlt inward.
D : In D weight is acting downward again and the normal reaction force at the end of ladder is again not acting against the direction of weight it is slightly Outward
Thus the Answer will be B

 

[Hassan Ali Abid]

The Answer will be B that i think so,PLEASE do tell me if i am right?
A: In A weight is acting downward and the normal reaction force at the end of ladder is upward and the force P is the normal reaction force by the wall in straight direction.Which means the ladder is not sliding and on the equilibrium postion
B: In B weight is acting downward and the normal reaction force at the end of ladder is upward and the force P force P isacting slightly upward because the ladder is sliding .
C: In C weight is acting downward again and the normal reaction force at the end of ladder is now not acting against the direction of weight it is slighlt inward.
D : In D weight is acting downward again and the normal reaction force at the end of ladder is again not acting against the direction of weight it is slightly Outward
Thus the Answer will be B

how to solve Q13 of same ppr ????

 

[daredevil]

how to solve q13 ..m/j12 (12) ?

calculate the weight acting from the 2 kg block because that is the force moving the 8 kg block too:
2*9.81 = 19.62 N

the resultant force on the block would be this force minus the frictional force:
19.62 - 6 = 13.62 N

thus after calculating the resultant force, now calculate the acceleration with the formula F=ma :
a = F/m
= 13.62/8 = 1.7 m/s^2

is that the ryt answer ?? ^_^
if u don't get it then ask away

 

[Reh_456]

The Answer will be B that i think so,PLEASE do tell me if i am right?
A: In A weight is acting downward and the normal reaction force at the end of ladder is upward and the force P is the normal reaction force by the wall in straight direction.Which means the ladder is not sliding and on the equilibrium postion
B: In B weight is acting downward and the normal reaction force at the end of ladder is upward and the force P force P isacting slightly upward because the ladder is sliding .
C: In C weight is acting downward again and the normal reaction force at the end of ladder is now not acting against the direction of weight it is slighlt inward.
D : In D weight is acting downward again and the normal reaction force at the end of ladder is again not acting against the direction of weight it is slightly Outward
Thus the Answer will be B

yep it is

 

[Hassan Ali Abid]

calculate the weight acting from the 2 kg block because that is the force moving the 8 kg block too:
2*9.81 = 19.62 N
the resultant force on the block would be this force minus the frictional force:
19.62 - 6 = 13.62 N
a = F/m= 13.62/8 = 1.7 m/s^2

bro i also calculated like this but the ans is 1.4 ...option a .....

 

[daredevil]

bro i also calculated like this but the ans is 1.4 ...option a .....

what?? howww!!??!! :O

 

[Hassan Ali Abid]

what?? howww!!??!! :O

idk....just check its ms ......the ans is A

 

[daredevil]

idk....just check its ms ......the ans is A

yeah i just saw.... if anyone tells the answer plz tag me in it becuz i reaallllyy wanna know .....

 

[Hassan Ali Abid]

yeah i just saw.... if anyone tells the answer plz tag me in it becuz i reaallllyy wanna know .....

i wasted about 1 hr on this single question but was unable to calculate Ans A

 

[RadzMau]

Can someone pls help me with this question?
Question: A motorist travelling at 10 m/s can bring his car to rest in a distance of 10 m. If he had been travelling at 30 m/s, in what distance could he bring the car to rest using the same braking force?

 

[Ali_00921]

Distance between 2 successive heaps (minimum points) = λ/2

39cm covers 5 times of this distance, therefore:
5(λ/2) = 39 cm
λ = 15.6 cm = 15.6 x 10^-2 m

v = fλ
v = 2.14 x 10^3 x 15.6 x 10^-2 = 334 ms-1

Din't get the λ part..

 

[Dug]

how to solve q13 ..m/j12 (12) ?

T - 6 = 8a --- i
20 - T = 2a --- ii
We want to find acceleration so eliminate T:
8a + 6 = 20 - 2a
10a = 14
a = 1.4 ms-1

 

[Ahmedraza73]

bro i also calculated like this but the ans is 1.4 ...option a .....

Sorry dude i made a big mistake and not considered the Tension in my first answer: really apologize for tht:
here is the correct solution:
W=mg=2g(for the hanging part)
EQAUATION 1 : 2g-T=2a
Equa 2: T-6=8a
ADDING BOTH EQUATION WE GET:
2g-6=10a ---equation 3
puttin values in equation 3 we have:
2(9.81)-6/10=1.362 ROUNDED OFF TO 1.4m/s ^2

And thanks brood to helping me:

 

[Ahmedraza73]

yep it is

Did u understand?

 

[sweetjinnah]

plz i wanted explanations of the following mcqs:
o/n 2008 mcq 4 , 32 :
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf
m/j 2008 mcq 11 , 16 :
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s08_qp_1.pdf
o/n 2007 mcq 23 , 25 :
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_1.pdf
o/n 2009 mcq 12 , 13 , 25 :
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_12.pdf

 

[Mikaila]

W'12 P13
Q 25.) A lift is supported by two cables, each of length 10m and diameter o.5cm. The lift drops 1mm when a man of mass 80kg steps into the lift. What is the best estimate of the value of the Young Modulus of steel?
The answer is C which is 2x10^11...
Please can u explain how to arrive at the answer.

 

[salvatore]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Please help me solve qn no. 3 (b). It seems easy but I just can't think of the solution..
Thanks

 

[Tkp]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf
Please help me solve qn no. 3 (b). It seems easy but I just can't think of the solution..
Thanks

at first the initial height was 61.then when the pendulum swings the vertical height would be 61cos 18.so u need to to find the gain in potential energy.mg(h2-h2)

 

[salvatore]

at first the initial height was 61.then when the pendulum swings the vertical height would be 61cos 18.so u need to to find the gain in potential energy.mg(h2-h2)

Thank you for your reply..
This is where I'm stuck.. how did you get 61cos18??