Physics paper 31 , (a death call)

[IGCSE O/L student]

thank u!!

is it ok if i ask, hw old r u?

 

[svenas]

yes, dats ryt.
plus, it ws 4 2 marks, so no mch thinkin ws needed.

but I didn't do it correctly in the exam....even I wrote for first part 6v so i will lose all here......tell me if I'll lose around 20 marks can I get A* or atleast an A??

 

[svenas]

is it ok if i ask, hw old r u?

16 yrs old

 

[IGCSE O/L student]

but I didn't do it correctly in the exam....even I wrote for first part 6v so i will lose all here......tell me if I'll lose around 20 marks can I get A* or atleast an A??

listen, bro! da threshold is gonna b really low dis tym, i guess coz mny ppl said da ppr ws shitty.
so, dnt wry. if ny of ur ans r wrng b'coz a previous ans ws wrng, u'll get @ least part of da marks 4 showin da correct workin.
u may evn get da complete marks

 

[svenas]

listen, bro! da threshold is gonna b really low dis tym, i guess coz mny ppl said da ppr ws shitty.
so, dnt wry. if ny of ur ans r wrng b'coz a previous ans ws wrng, u'll get @ least part of da marks 4 showin da correct workin.
u may evn get da complete marks

thnks..u made me feel calm now....
my problem is when I go home and think of the solution I get it ryt but at the exam tym I don't...I thnk that cauz I was worried about the tym...this wat makes me feel sad

 

[ThatReallyReallyWeirdDude]

We know that the current at the 3 ohm resistor was 0.8A (yeah 3 ohm and 2 ohm position switched in my diagram)
Since the voltage is the same in parallel, the voltage at 2ohm = voltage at 3 ohm
So I1R1=I2R2 where I1 = 0.8A, R1= 3 ohms and R2 = 2 ohms
so 0.8*3=2*I2
I2 = 1.2A
The voltage is split between R and the effective 1.2 ohm resistor
We can find the voltage there easily, V = IR, 0.8*3 = 2.4 V or 2*1.2 = 2.4 V
So therefore R gets the rest of the voltage, 6-2.4=3.6V
the total current is 0.8A + 1.2A = 2A
Resistance = Voltage/Current
V of R is 3.6 and current at r = 2, so therefore resistance = 3.6/2 = 1.8 ohms

 

[IGCSE O/L student]

thnks..u made me feel calm now....
my problem is when I go home and think of the solution I get it ryt but at the exam tym I don't...I thnk that cauz I was worried about the tym...this wat makes me feel sad

dnt wry, bro!
cn i ask u sumthin?
4 da reason y da level falls, wat did u ryt?

 

[svenas]

dnt wry, bro!
cn i ask u sumthin?
4 da reason y da level falls, wat did u ryt?

I wrote it wrong...but the correct answer is due to expansion of glass so level drops at first...i thnk this is the ryt ans

 

[IGCSE O/L student]

We know that the current at the 3 ohm resistor was 0.8A (yeah 3 ohm and 2 ohm position switched in my diagram)
Since the voltage is the same in parallel, the voltage at 2ohm = voltage at 3 ohm
So I1R1=I2R2 where I1 = 0.8A, R1= 3 ohms and R2 = 2 ohms
so 0.8*3=2*I2
I2 = 1.2A
The voltage is split between R and the effective 1.2 ohm resistor
We can find the voltage there easily, V = IR, 0.8*3 = 2.4 V or 2*1.2 = 2.4 V
So therefore R gets the rest of the voltage, 6-2.4=3.6V
the total current is 0.8A + 1.2A = 2A
Resistance = Voltage/Current
V of R is 3.6 and current at r = 2, so therefore resistance = 3.6/2 = 1.8 ohms

that's da xact concept i used, n dats da ans i got.
u rock, bro!

 

[svenas]

We know that the current at the 3 ohm resistor was 0.8A (yeah 3 ohm and 2 ohm position switched in my diagram)
Since the voltage is the same in parallel, the voltage at 2ohm = voltage at 3 ohm
So I1R1=I2R2 where I1 = 0.8A, R1= 3 ohms and R2 = 2 ohms
so 0.8*3=2*I2
I2 = 1.2A
The voltage is split between R and the effective 1.2 ohm resistor
We can find the voltage there easily, V = IR, 0.8*3 = 2.4 V or 2*1.2 = 2.4 V
So therefore R gets the rest of the voltage, 6-2.4=3.6V
the total current is 0.8A + 1.2A = 2A
Resistance = Voltage/Current
V of R is 3.6 and current at r = 2, so therefore resistance = 3.6/2 = 1.8 ohms

u r 200% ryt

 

[ThatReallyReallyWeirdDude]

that's da xact concept i used, n dats da ans i got.
u rock, bro!

Thanks Let's stop dwelling on this and focus on our next exams.
I'm personally worried about Accounting.

 

[IGCSE O/L student]

I wrote it wrong...but the correct answer is due to expansion of glass so level drops at first...i thnk this is the ryt ans

here we go again!
dats wrng 2. i wrote:
da molecules @ da surface
gain a higher avg kinetic energy
n so leave da surface of da liquid.

 

[IGCSE O/L student]

Thanks Let's stop dwelling on this and focus on our next exams.
I'm personally worried about Accounting.

accounting?? y?
i luv accountin vry mch

 

[IGCSE O/L student]

u r 200% ryt

heheheheheheheh... LOL!

 

[RSDEV]

EVEN I GOT 9.333333333M/S IN THE MAX SPEEED QUESTION. If the point with the highest speed was asked then it was C with 7.78m/s. But they asked for the max speed which means one has to take the speed of the steepest line in the graph which was B-C. Also they wont ask you the question for 7.78 which just involvs taking the distance and time at Point C for 3 Marks. In conclusion, it is (350-70)/(45-15) which will give you the answer 9.3333333333m/s.

 

[IGCSE O/L student]

EVEN I GOT 9.333333333M/S IN THE MAX SPEEED QUESTION. If the point with the highest speed was asked then it was C with 7.78m/s. But they asked for the max speed which means one has to take the speed of the steepest line in the graph which was B-C. Also they wont ask you the question for 7.78 which just involvs taking the distance and time at Point C for 3 Marks. In conclusion, it is (350-70)/(45-15) which will give you the answer 9.3333333333m/s.

pheww
i thought dat ws wrng
thnx

 

[svenas]

pheww
i thought dat ws wrng
thnx

it's correct i got it too!!!
for the fleming rule u used right or left?

 

[Ahmedboombox]

it's correct i got it too!!!
for the fleming rule u used right or left?

I used right because the current was uknown in this case.

 

[IGCSE O/L student]

it's correct i got it too!!!
for the fleming rule u used right or left?

left

 

[RSDEV]

yeah lol, LDR was in A i guess ..


u found the accelration , v had to find the speed

It was the other way around ...LDR was down and the resistor was up.


In the last one ......LDR should be replaced by resistor and the resistor should be replaced by a thermistor.