Author Topic: ALL CIE PHYSICS DOUBTS HERE !!!  (Read 171912 times)

Offline JACKRABBIT

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #735 on: June 05, 2011, 11:36:08 am »
Quote
11. Gimme some time
19. KE (original) = 0.5mv^2
v^2 = 2as = 3920
therefore KE = 1960m

40% loss = 40/100 * 1960m loss = 784m loss
Thus, remaining = 1176m

1176m = 0.5mv^2
v = sqrt(2*1176) ~ 49

36) Total resistance = 1/(1/15 + 1/15) = 7.5ohms
Resistance between X and Y = 1/(1/10 + 1/5) = 10/3
VOltage = (10/3)/7.5 V = 2/3 V


VOltage = (10/3)/7.5 V = 2/3 V

Which formula did yu used

Awesome..thx for number 19...but for 36 I kinda need a bit more detail on that please...and yeah which formula did you use for the last part of qn 36

Offline EMO123

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #736 on: June 05, 2011, 11:38:54 am »
11. Gimme some time
19. KE (original) = 0.5mv^2
v^2 = 2as = 3920
therefore KE = 1960m

40% loss = 40/100 * 1960m loss = 784m loss
Thus, remaining = 1176m

1176m = 0.5mv^2
v = sqrt(2*1176) ~ 49

36) Total resistance = 1/(1/15 + 1/15) = 7.5ohms
Resistance between X and Y = 1/(1/10 + 1/5) = 10/3
VOltage = (10/3)/7.5 V = 2/3 V
for Kinetic Energy = 1/2mv2
then v2=2as
for resistance
if in parallel then = 1/R1+1/R2=1/R
if in series then = R1+R2=R
Voltage as such = IR
but i dint get it here HIghburry

Offline Ghost Of Highbury

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #737 on: June 05, 2011, 03:25:05 pm »

VOltage = (10/3)/7.5 V = 2/3 V


Which formula did yu used
Potential divider formula.

V(out) = R1/R1+R2 *V(in)
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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #738 on: June 05, 2011, 03:43:14 pm »
I in my book the formula they say is

V0ut=V*R^1/(R^1 + R^2)

Offline Malak

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #739 on: June 05, 2011, 04:22:32 pm »
I in my book the formula they say is

V0ut=V*R^1/(R^1 + R^2)
Its the same, Multiply by V first or in the end, wont make a difference :)
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Offline Ghost Of Highbury

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #740 on: June 05, 2011, 04:45:11 pm »
Awesome..thx for number 19...but for 36 I kinda need a bit more detail on that please...and yeah which formula did you use for the last part of qn 36


I'm sry, i can't figure out Q11. Would be great if someone could help out. FIDATO? Ari? Astar?
« Last Edit: June 05, 2011, 04:51:34 pm by Ghost Of Highbury »
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Offline EMO123

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #741 on: June 05, 2011, 05:15:52 pm »
I'm sry, i can't figure out Q11. Would be great if someone could help out. FIDATO? Ari? Astar?
ask this to
master_key

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #742 on: June 05, 2011, 05:22:02 pm »
I'm sry, i can't figure out Q11. Would be great if someone could help out. FIDATO? Ari? Astar?

Take east as the positive direction and west as the negative direction.

Use the idea :

Vi of  Body A - Vi of  Body B = Vf of  Body B - Vf of  Body A

Where Vi = initial VELOCITY and Vf = final velocity.


Offline Ghost Of Highbury

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #743 on: June 05, 2011, 05:26:42 pm »
Argh, and I was trying to form simultaneous equations..thanks man. Appreciated! :D
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Offline EMO123

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #744 on: June 05, 2011, 05:27:05 pm »
Take east as the positive direction and west as the negative direction.

Use the idea :

Vi of  Body A - Vi of  Body B = Vf of  Body B - Vf of  Body A

Where Vi = initial VELOCITY and Vf = final velocity.


thanxx
ari

Offline bulono

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #745 on: June 06, 2011, 01:13:48 am »
can somebody explain the answers for Q9,13,15 & 28 O/N  2009/11
Thx :)

Offline Ghost Of Highbury

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #746 on: June 06, 2011, 06:33:13 am »
can somebody explain the answers for Q9,13,15 & 28 O/N  2009/11
Thx :)

9. otal Ke b4 collision => 0.5*2m*u^2 + 0.5*m*(-u)^2 = mu^2 + mu^2/2 = 3/2 mu^2
Total momentum b4 => 2mu - mu = mu

A. KE after collision -> 0.5*2m*(u/3)^2 + 0.5*m*(5u/3)^2 = mu^2 /9 + 25mu^2/18  = 3/2 mu^2
    Momentum = 2m*(-u/3) + m*(5u/3) = 5mu/3 - 2mu/3 = 3mu/3 = mu

13.Im not sure about variant 1. The torque is obviously 4.5Nm
{Tension = Torque/diameter = 3/0.1 = 30
  Torque = tension*diameter = 30*0.15=4.5}

Although, i'm not sure why the tension is 60. Maybe, because the lower belt has no tension, and for a torque two forces acting in opposite directions
are required, I guess, the tension in the lower belt is transferred to the upper belt which adds up to 60 (30+30).

15. First find out the velocity of the 1kg trolley
(2*-2)+(1*x) = 0
-4+x = 0
x = 4m/s
KE = 0.5mv^2 = 0.5*2*(-2)^2 + 0.5*1*(4)^2 = 12
D

28. Charge to mass
We know, that Force acting on it = Weight
E = F/q
F = Eq
Eq = mg
m = Eq/g

Charge to mass = q/(Eq/g) = g/E

Polarity should be negative because the positive plate will pull it upwards and its weight will pull it downwards. Hence they balance o ut
« Last Edit: June 06, 2011, 06:37:31 am by Ghost Of Highbury »
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Offline JACKRABBIT

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #747 on: June 06, 2011, 06:43:27 am »
Quote
VOltage = (10/3)/7.5 V = 2/3 V

isnt 10/3/7.5 equal to 4/9?? and its root is 2/3...so...i get that its the potential divider formula..and its resistance  you want to find the Vout of divided by total resistance into the Vin..which is 2 in this case right?? so is it 10/3 * 2 /7.5 ...which is 8/9

Offline Ghost Of Highbury

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #748 on: June 06, 2011, 07:06:24 am »
isnt 10/3/7.5 equal to 4/9?? and its root is 2/3...so...i get that its the potential divider formula..and its resistance  you want to find the Vout of divided by total resistance into the Vin..which is 2 in this case right?? so is it 10/3 * 2 /7.5 ...which is 8/9

you're right, my bad, should be (10/3)/7.5 * 2 = 8/9
Hmm, y is the answer 2/3 then? Gimme some time..
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Offline Ghost Of Highbury

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #749 on: June 06, 2011, 08:40:25 am »
k, im not sure about this method but i got the answer somewhow

the p.d across the 5 ohm resistor in the 1st part of the circuit = 5/15 * 2V = 2/3

THe p.d across the 2 5 ohms resistors in the second part of the circuit = 10/15 * 2 = 4/3

Hence the total potential "DIFFERENCE" = 4/3 - 2/3 = 2/3 V

------

If u apply this formula to find the p.d across, say the all three resistors in the 1st branch
The voltage across the point at the bottom and the negative terminal = 0V
the voltage across the point at the top and the negative terminal = 2V
Difference = 2-0 = 2V -> which is the P.d across the 3 resistors in the first branch..
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