Physics, Chemistry and Biology: Post your doubts here!

[Iris Dylan Lane]

Thanx Iris ...that was helpful for me too... ..so will I get the mark if I do that way or not?

Welcome..... This is the Allocation:

(30.9 – 14.7 = )16.2N OR evidence of calculation of resultant C1
use of a = F/m C1
5.24m/s2 A1 [3]

 

[Abdurrahmanmuhammed]

oct/nov 13 paper 32 question 2 someone help me out quick please

 

[shingiechingz@8]

Guys. I really need help on question 2(C) in paper 0625/m/j/31/11
Pretty PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!

 

[Mustafa84922]

Guys. I really need help on question 2(C) in paper 0625/m/j/31/11
Pretty PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!

we can calculate the work done or useful energy while climbing 20 rungs in a similar way to 2(B) which is 4800J how did i get this? here you go
work done is equal to force*area here the force is 800N while the area is the 0.3m(30cm) 20 times which goes like: 800*20*0.3 = 4800
now in the question they said 15% efficient, this means that 4800 is 15% of the total energy as it is useful.......
so 4800/x=15/100
so x=4800*100/15

 

[Mustafa84922]

oct/nov 13 paper 32 question 2 someone help me out quick please

2(A) first of all we will calculate the speed of the engine when it breaks the beams
12mm=0.012m
speed=distance/time
speed=0.012/0.08 (we will divide the distance between the beams with 0.08, because 0.08 is the time that the engine takes to cut the second beam after the first one)
speed=0.15
now we know that the acceleration of the engine when it moved was 0.03m/s^2
the students calculates the time taken to cut the beam from when the engine started moving, the time it took the engine to reach a speed of 0.15 is the time we need to calculate as the speed of engine when it cut the beams was 0.15
therefore, 0.15/0.03 = 5s
wait for the other parts pls

 

[Mustafa84922]

oct/nov 13 paper 32 question 2 someone help me out quick please

part (b) states that we need to find the mass of the engine.....
well we are already given the force needed to hold the engine in equilibrium; 0.06N this means that both directions have the same amount of force acting
therefore, we can use F=ma to calculate the mass as we are also given the acceleration in part(a)
so m=0.06/0.03=2kg is the mass of the engine along with the tank

 

[sj0007]

How do you have the ms of that paper ???????

Magic!!!!
I went on the advanced search of adobe, typed in the ques and well I got the year and so the ms!

 

[Mustafa84922]

oct/nov 13 paper 32 question 2 someone help me out quick please

part (c)
The acceleration of the engine will be greater according to the equation F=ma as the mass of the engine will decrease because of the empty water tank, however the force it still the same

 

[ThePacifics]

Please i need help in May June 2013 Paper 33 Question 3 b) ... how to draw it? help me .

 

[Shandana khan]

part (c)
The acceleration of the engine will be greater according to the equation F=ma as the mass of the engine will decrease because of the empty water tank, however the force it still the same

Thaaaaanx for helping ...I actually had this doubt since 1 week...but had no one to ask...since I am a private candidate...

 

[Mustafa84922]

Thaaaaanx for helping ...I actually had this doubt since 1 week...but had no one to ask...since I am a private candidate...

no problem......im glad i was of use to someone.........

 

[shingiechingz@8]

Guys results are coming @ 11 Aug

 

[AnonymousX9]

October/November 2010 Paper 32 Question 2(b). Please help me guys I'm really stressed out.

 

[DANISH182]

some tips for physics p3??? to get an A*

 

[shingiechingz@8]

some tips for physics p3??? to get an A*

yes please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

 

[Mustafa84922]

yes please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

some tips for physics p3??? to get an A*

i guess.......just solve past papers and study the notes you have made in your class
it is too late to ask everyone for concentrating on their teachers lecture as the small stuff really matter but still its easy to get a*
DO NOT LEAVE ANY QUESTION EVEN IF YOU ARE NOT SURE MAKE A GUESS AT LEAST WRITE THE ANSWER YOU MIGHT GET 1 MARK
one more thing before entering the examination hall go through the formula book; here you go
and dont worry too much pls
i almost forgot try to read as much marking schemes and examiner reports as you can it helps alot trust me
what i usually do is, what im doing right now xD i just help others and its kinda of the best way to get good at something

 

[Mustafa84922]

Guys results are coming @ 11 Aug

thanks for the info

 

[Mustafa84922]

Are there any important definitions we are supposed to know??
There are soo many so! ?

well dont learn them by heart a few important ones are attached so dumb of me to say few but u only have to read them 2-3 times thats all............

 

[Mustafa84922]

Please i need help in May June 2013 Paper 33 Question 3 b) ... how to draw it? help me .

Here you go!
it took me some time to print it and then solve and then again transfer them to pc through my mobile so srry for the late reply

 

[Mustafa84922]

October/November 2010 Paper 32 Question 2(b). Please help me guys I'm really stressed out.

PART B(i)
Maximum Kinetic energy is equal to maximum gravitational potential energy, this is because when the max. g.p.e will be lost all of it will turn into kinetic energy as there is no air resistance
therefore, we will take the lowest height (0.1) in the formula mgh
max k.e = max g.p.e = mgh = 0.15*10*0.1 = 0.15

PART B(ii)
maximum velocity
k.e= 0.5*0.15*v^2
we know that maximum k.e is 0.15
so, 0.15=0.5*0.15*v^2
v^2=0.15/0.075
v^2=2
v=1.4 m/s

PART B(iii)
the maximum height will be 0.3, because there is no air resistance

PART B(iv)
its easy just
extend the 0.3m line mark from the ground from the ground and then with a ruler measure the length from the support to the peg and from the peg to the bob
(add these two lengths)
towards the left of the centre/vertical line make a straight line from the support to the horizontal line using eye make sure this line is equal to measurment we did here (add these two lengths)