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1 mol -----------> avogedro number N
how did you get 0.09 ?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.
The atomic mass of an element, measured in amu, is the same as the mass in grams of one mole of an element.how did you get 0.09 ?
6. In this question you're asked to calculate the percentage uncertainty.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?
I get this but isn't the weight of box itself responsible for the acceleration too? Why isn't it?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
And thanks a lot. It really did help.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.
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.I get this but isn't the weight of box itself responsible for the acceleration too? Why isn't it?
Alright. Thanks.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.
ANYONE PLEASE ANSWER THIS ?????????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
SAME DOUBT!ANYONE PLEASE ANSWER THIS ?????????
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.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. :/
Oh yeah.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.
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