Physics: Post your doubts here!

[Saly Elkady]

Is there a site for Feb/March 2016 papers ?

 

[holoholo]

Part 2 please ?

 

[The Sarcastic Retard]

View attachment 59907

Part 2 please ?

1 mol -----------> avogedro number N
? <---------------- 4.67 x 10^15 N

m = n * Mr
Mr = 0.09kg
n = 7.76 x 10^-9
m = 6.98 x 10^-10 kg.

 

[holoholo]

1 mol -----------> avogedro number N
? <---------------- 4.67 x 10^15 N

m = n * Mr
Mr = 0.09kg
n = 7.76 x 10^-9
m = 6.98 x 10^-10 kg.

how did you get 0.09 ?

 

[The Sarcastic Retard]

how did you get 0.09 ?

The atomic mass of an element, measured in amu, is the same as the mass in grams of one mole of an element.
To convert gram to kilogram we should divide the mass by thousand so 90/1000 = 0.09 kg.

 

[funky brat]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s15_qp_11.pdf
Can anyone please explain 6, 13, 19, 29 and 36?

 

[Rizwan Javed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s15_qp_11.pdf
Can anyone please explain 6, 13, 19, 29 and 36?

6. In this question you're asked to calculate the percentage uncertainty.
First find the fractional uncertainties in k and x.
The fractional uncertainty in k is 2/100
The fractional uncertainty in x is 0.002/0.050
Now you can see that x is a squared term, so you'll multiply the fractional uncertainty of x with its power (that is 2 here): 2*0.002/0.050

Add up the fractional uncertainties and multiply with 100 to find the percentage uncertainy like this:

( 2/100 + 2*0.002/0.050 ) * 100 = 10%

So the answer is B.

 

[Rizwan Javed]

13. The force accelerating the box is actually the weight of the 2.0 kg mass. So first find this force (i.e. the weight of the 2.o kg mass).
It is 2 * 9.81 = 19.62 N
There's also a frictional force, so subtract that from the force calculated above.
19.62 - 6.0 = 13.62N

^ this force that you have calculated is accelerating two objects, the 8.0 kg box and the 2kg mass.

So use the equation F=ma (taking m = 8+2 = 10kg)

a = 13.62/10 = 1.362 ~ 1.4m/s^2

So the answer is A

 

[Rizwan Javed]

19. In this question, the sand is falling at a rate m kg/s. That means in 1 second, m kg undergoes a change in momentum which is mv. Force is the rate of change of momentum. So find the force, which is: mv/s.

Power is found by P = FV.
Substitute the value of F here. you get:
P = (mv/s)(v)
Since time (s) is 1 second, so the addition power will become
P = mv^2

So the answer is D.

 

[Rizwan Javed]

36.
Option A: The resistance of variable resistor is increasing, so the p.d. across it will also increase as R is directly proportional to V. So this statement is wrong.

Option B: It is a similar case. The resistance of variable resistor R is increasing, so p.d. will increase across R not r. So this statement is also wrong.

Option C & D are confusing here.

The power dissipated in R is (I^2)(R) and so it must be zero when R is zero, and must also be zero if R is infinite (as the current I will then be zero). The power must increase and then decrease as R increases from zero. As R increases, the current I decreases, the potential difference across R increases, the potential difference across r decreases and the power dissipated in r (given by I^2 * r) decreases.

So the correct option here is D.

 

[ashcull14]

need help to represent light gates or data logger /ticker tape to find the time of a falling body in nov 2003 paper 5 experiment
can anyone draw the diagram?

 

[funky brat]

13. The force accelerating the box is actually the weight of the 2.0 kg mass. So first find this force (i.e. the weight of the 2.o kg mass).
It is 2 * 9.81 = 19.62 N
There's also a frictional force, so subtract that from the force calculated above.
19.62 - 6.0 = 13.62N

^ this force that you have calculated is accelerating two objects, the 8.0 kg box and the 2kg mass.

So use the equation F=ma (taking m = 8+2 = 10kg)

a = 13.62/10 = 1.362 ~ 1.4m/s^2

So the answer is A

I get this but isn't the weight of box itself responsible for the acceleration too? Why isn't it?

 

[funky brat]

36.
Option A: The resistance of variable resistor is increasing, so the p.d. across it will also increase as R is directly proportional to V. So this statement is wrong.

Option B: It is a similar case. The resistance of variable resistor R is increasing, so p.d. will increase across R not r. So this statement is also wrong.

Option C & D are confusing here.

The power dissipated in R is (I^2)(R) and so it must be zero when R is zero, and must also be zero if R is infinite (as the current I will then be zero). The power must increase and then decrease as R increases from zero. As R increases, the current I decreases, the potential difference across R increases, the potential difference across r decreases and the power dissipated in r (given by I^2 * r) decreases.

So the correct option here is D.

And thanks a lot. It really did help.

 

[Rizwan Javed]

I get this but isn't the weight of box itself responsible for the acceleration too? Why isn't it?

No, the weight of the box (8kg) will not be responsible for acceleration, since the force that box exerts on the ground is balanced by the contact force. So it will not undergo any acceleration downwards. The only force that is acting upon the box is by that mass.

 

[funky brat]

No, the weight of the box (8kg) will not be responsible for acceleration, since the force that box exerts on the ground is balanced by the contact force. So it will not undergo any acceleration downwards. The only force that is acting upon the box is by that mass.

Alright. Thanks.

 

[ashcull14]

need help to represent light gates or data logger /ticker tape to find the time of a falling body in nov 2003 paper 5 experiment
can anyone draw the diagram?View attachment 59925

ANYONE PLEASE ANSWER THIS ?????????

 

[funky brat]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s15_qp_13.pdf
For question 13 why isn't this the right way to solve it?
5*200= 95*x
x= 10.53
then using the formula v^2= u^2=2as where v=0 This gives 5.6 as an answer and adding the 4 m gives 9.6. which is D. :/

 

[The Sarcastic Retard]

ANYONE PLEASE ANSWER THIS ?????????

SAME DOUBT!

 

[Rizwan Javed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s15_qp_13.pdf
For question 13 why isn't this the right way to solve it?
5*200= 95*x
x= 10.53
then using the formula v^2= u^2=2as where v=0 This gives 5.6 as an answer and adding the 4 m gives 9.6. which is D. :/

Your method is correct. But you are making a mistake. The mass after the collision is 100g instead of 95g because the bullets embeds itself in that block.

 

[funky brat]

Your method is correct. But you are making a mistake. The mass after the collision is 100g instead of 95g because the bullets embeds itself in that block.

Oh yeah.