Physics: Post your doubts here!

[DeadbeatCIE]

The Electric Field is upwards hence the Potential of the upper plate should be lower than the lower plate.
EFS= V/d
60000= - x - 80/4x10^-3
x= -320 (A)

I'm getting +320 instead of -320
x-80/0.004=60000 then x is 320
is it -320+-80 or -320--80?
bcos x-80=240

 

[amina1300]

View attachment 62392
2013 ... may...13

E = (F/A) /( extension /length)
--> Stress : F/A
F = W = mg ( g is constant)
m = density * volume (density is constant as same material )
V = l * l* l ( The Load is a cube )
[The model is one tenth full-size in all linear dimensions.]
V = 1/10*1/10*1/10
V = 1/1000
When V decreases mass decreases and hence Force decreases
F = 1/1000
Area = A = l*l
= 1/10 * 1/10
= 1/100
So Stress = F/A = (1/1000)/(1/100) ---->1/10

----> Strain : e/l
Suppose extension = 1
length = 1/10
Strain = 10

E = stress/ strain = 0.1 / (10)
=> 1/100

Ration of original to model = 1/(1/100)
= 10^2 ANSWER = C

 

[amina1300]

op

I'm getting +320 instead of -320
x-80/0.004=60000 then x is 320
is it -320+-80 or -320--80?
bcos x-80=240

Opps I wrote it the opposite way around -80--x / 4x10^-3 = 60 ,000
We always subtract the smaller potential from the bigger -80 > -320 .
My apologies.

 

[amina1300]

View attachment 62390
O/N 2015 (12)

p.d across 80 ohm resistor ;
V= IR
V = 1.2 * 80 = 96
Lost Volts = 120 - 96 = 24 V
Internal resistance = r = ?
24 = ( 1.2+0.4) r
r = 15 ohm.

 

[amina1300]

View attachment 62389
O/N 2015 (13)
help out

This is the hardest question!!! I spent a 5 hours trying to solve this!



First we calculate the Total resistance :
R2 = (R3 + R1) = [1+ 1/2]^-1 = 2/3
R5 + (R4 + 2/3 ) =[ 1 + 1 / (2/3 + 1)]^-1 = 0.625
0.625 + R6 = 1.675 ohm
Total R = 1.675 ohm.

Now we will use V= IR to calculate potential and Current across the resistors and loops.

Current R1 + R3 = 1 A
P.D R1 + R3 = 2 V (( V = 1*(1+10)) = 2V)

(R1 R3) are in parallel with R2 so P.D R2 = 2V

R2 Current = 2/1 = 2 A

Kirchoffs 1st law ; Current In = Current out

Looking at first junction from the left we have 2 A + 1A = 3A ( 2 A--> R2 and 1A--->R3+R1)

3A is leaving hence Current in R 4 = 3A
P.D across R4 = V= IR
V = 3*1 = 3V

Now lwt R1+R2+R3 = Resistor X

Resistor X is in series with R4( P.D = 3V)

Total potential = 2V + 3V = 5V across the whole loop.

So 5V is the potential across R5 as it is in parallel with (R4 + X)
Current across R5 = 5/1 = 5A

Second junction from left :
I in = I out
5 A + 3 A = 8 A

8 A iss the current across the resistor R 6
P.d across R6 = 8/1 = 8 V

E.M.F of the supply = Total potantial across all the loops = 3 V + 2 V + 8 = 13 V

OR

V = I R
V = total R * 8
= 1.635 * 8
= 13 V


AND YOUR ANSWER IS D!! !!! WOW I CAN'T BELIEVE I DID IT!!

You can do it by the Emf thing too for that you dont need to calculate the total resistance but I guess its harder to understand that.

I'd recommend you note all of this down on a note book view it step by step and solve it again for yourself because damn it was hard.

 

[hassan77]

Physics Feb / March 2017 var 12
Q9 Ans is D

Please help.......

 

[selrey]

How do i do this question?
The density of air on the Earth decreases almost linearly with height from 1.22 kg m –3 at sea level
to 0.74 kg m –3 at an altitude of 5000 m.
Atmospheric pressure at the Earth’s surface on a particular day is 100 000 Pa. The value of g
between the Earth’s surface and an altitude of 5000 m can be considered to have a constant
value of 9.7 m s –2 .
What will be the atmospheric pressure at an altitude of 5000 m?
OPTIONS:
36000 Pa
48000 Pa
54000 Pa
59000 Pa

 

[amina1300]

How do i do this question?
The density of air on the Earth decreases almost linearly with height from 1.22 kg m –3 at sea level
to 0.74 kg m –3 at an altitude of 5000 m.
Atmospheric pressure at the Earth’s surface on a particular day is 100 000 Pa. The value of g
between the Earth’s surface and an altitude of 5000 m can be considered to have a constant
value of 9.7 m s –2 .
What will be the atmospheric pressure at an altitude of 5000 m?
OPTIONS:
36000 Pa
48000 Pa
54000 Pa
59000 Pa

View attachment 62388
M/J 2016 (13)
Explain please

Im trying but all I get is 60655...Pa <_> SOMEONE SOLVE THIS QUESTION!!

 

[amina1300]

Physics Feb / March 2017 var 12
Q9 Ans is D

Please help.......

F = ma
F = m * v/t
= density * Volume * v/t
= p * A* l*v/t
= p A v * l/t
= pAv^2

 

[amina1300]

selrey Which formula are you using? Post the methods you tried.

 

[amina1300]

View attachment 62386

O/N 2016 (13)

SOMEONE EXPLAIN THIS QUESTION!! ANSWER = D

 

[amina1300]

View attachment 62387
O/N 2016 (13)
with vivid working please !

Ammeter reading = 0
P.D across junction = 0
P.D(R4) = P.D(R2 )
R4/(R3 + R4) * V = R2 /( R2 + R1 ) *V
(R2 + R1)* R4 = R2 *(R3 + R4)
R2R4 +R4R1 = R2R3 + R4R2
R4R1 = R2R3

AND = D

 

[amina1300]

View attachment 62384
M/J 2016 (13)

 

[selrey]

selrey Which formula are you using? Post the methods you tried.

I got 60655 as well through cross multiplication idk .-. I couldnt think of any proper method

 

[amina1300]

http://papers.gceguide.com/A Levels/Physics (9702)/9702_w16_qp_13.pdf
Q16,13 ,29,24,25

 

[amina1300]

http://papers.gceguide.com/A Levels/Physics (9702)/9702_w16_qp_12.pdf
Q4 , 10 26 ,29

 

[selrey]

http://papers.gceguide.com/A Levels/Physics (9702)/9702_w16_qp_13.pdf
Q16,13 ,29,24,25

I Have the same qs!! Can you explain mcq no. 20?

 

[Shimmery woods]

que 2 Anyonee

 

[Shimmery woods]

que 25!!!!!

 

[selrey]

que 2 Anyonee

Is the answer D?