Physics: Post your doubts here!

[white rose]

No, the support/equilibrium is not what we're looking for here.
Here we just want the dist at which the man's head will be level with the barrel's bottom.
That will only happen when the barrel is at 9m.

hhhmm..ok

 

[Syed Abdul Haseeb]

12) We know that loss in P.E. = gain in K.E.
For the man's head to be level with the bottom of the barrel, the barrel must fall halfway → 9 m.
So, loss of P.E. of the barrel = mgh = 120 * 10 * 9 → 10800 J
Now, gain in K.E. = ½mv² = 10800 J
So, velocity → ½ * 80 * v² = 10800 ------> v = 16 m/s

Since the gravity also acts on the man in the OPPOSITE direction,
the resultant velocity = 16 - 10 → 6 m/s
Ans = A

Hope u get it!

Why did u subtracted 10 from 16, can u explain please, 16 is speed and 10 is acceleration

 

[Sara syed]

pls explai winter 2006 question 27,winter 2005 question8,summer2007 question 23

 

[Minato112]

Thank you so much, I owe you one

Mention not

 

[Nibz]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
Question 18, 28 Nibz Sorry I desperately needs your help

Hi, sorry for the late response.

Q.18
Current is given as 32A
Voltage as 230 V

Calculate Power = I x V = 7360 W
Energy = Power x time
From flow rate, take time as 1 second.
Therefore, energy = 7360 x 1 = 7360J

For potential energy = m x g x h
m = 200
g = 9.81
h = 8m
P.E = 15696J

Efficiency = (7360/15696) x 100 = 47%

Q28. It's simple, really. Half (1/2) of one wavelength is 180 degrees. The part shown is 3/4 of a full wave. Full wave = 360
So 3/4th of the wave would be 3/4 x 360 = 270

 

[Nibz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
Question 11, 12,

Q11.

Same masses, same velocity. After the collision, they'd have have the same velocity, too.
So the kinetic energy for each sphere would be 1/2mv^2.
Total K.E = 2 (1/2 mv^2) = mv^2

Q12.

Force due to mass = m x g = 1200 N
Force due to person = m x g = 800 N
Net force for both = 1200 - 800 = 400 N
Mass of person + mass = 120 + 80 = 200 kg
Acceleration = force / mass = 400 / 200 = 2 ms-2
Distance person travels to meet mass = 9 meters
Velocity : v^2 - u^2 = 2as
u = 0
v = unknown
a = 2
s = 9

v^2 = 2(2 x 9) = 36
v = 6 m/s

 

[sma786]

Hey how are ya?

Mass? I think you have linked the wrong paper :\

E=V/d
E=4.8x10^5 NC^-1

The Value of Charge on an electron is .. 1.6 * 1o^-19 C .. from the data booklet..

E=F/Q
F=EQ
F=4.8x10^5 * 1.6*10^-19
F=7.68*10^-14 N

So answer is D.

i meant force lol..
thanks, i got it !

 

[sma786]

so did u get it?

Yeah i got it

 

[sma786]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf
Question 5, uncertainity?

 

[Nibz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
Question 5, uncertainity?

Speed = Distance / time = 16.0 ms-1

Uncertainty in speed = (uncertainty in distance + uncertainty in time) x speed
= (0.1/40 + 0.05/2.50)x 16 = 0.36, should be one d.p, so = 0.4


D answer.

 

[Martee100]

Can anyone help

 

[Nibz]

Can anyone help

Resistance along 3 ohms, and two 6 ohms resistors:
1/6 + 1/6 = 1/R
R = 3 ohms
3 ohms + 3 ohms = 6 ohms total.

The 6 ohms calculated and the other 6 ohm resistors are in parallel.

1/6 + 1/6 = 1/R
R = 3 ohms

C.

 

[Zarif009]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf

Someone please explain no 6 : ) , thanks

 

[sma786]

Speed = Distance / time = 16.0 ms-1

Uncertainty in speed = (uncertainty in distance + uncertainty in time) x speed
= (0.1/40 + 0.05/2.50)x 16 = 0.36, should be one d.p, so = 0.4


D answer.

why one d.p ?

 

[sma786]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf
question 12, answer?

 

[TaffsAsLevel]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
question 12, answer?

Atleast show your workings so I can spot your mistake

 

[stareaj]

Physic As oct 2010 / 12 mcq 22 and 34
May 2011/12 mcq 15,16,24
Pleaseee

 

[stareaj]

Physic As oct 2010 / 12 mcq 22 and 34
May 2011/12 mcq 15,16,24
Pleaseee

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf

Q28 plzz

yar its A, because, the distance between the charges does not change, so the work-done in moving the charge away or closer to each other is 0
remember.. work is only done when the distance 'between' the charges is changed

 

[daredevil]

yar its A, because, the distance between the charges does not change, so the work-done in moving the charge away or closer to each other is 0
remember.. work is only done when the distance 'between' the charges is changed

ooh i get it now... thankss