Physics: Post your doubts here!

[Kamihus]

Link

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

Well I considered both of the distance.
i) For Tension on top belt (i.e is Q, as given in question "Wheel Q puts tension in the top portion of the belt") it has distance of 0.075 so I used that distance.
ii) Torque of P, we need distance of P here so I used 0.05 here.

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_12.pdf Q5 , 7 ,8 ,11 and39 please help

 

[Asad Moosvi]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_13.pdf

May someone explain question 25 to me quick? I have to do other questions too :/

 

[Kamihus]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_13.pdf

May someone explain question 25 to me quick? I have to do other questions too :/

Extension is proportional to length but here it is compression so length will decrease hence inversely proportional.

 

[tania]

q13 please

 

[Asad Moosvi]

Extension is proportional to length but here it is compression so length will decrease hence inversely proportional.

So normally, more extension means more length, but here, more compression means less length?

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf Q5 , 7 ,8 ,11 and39 please help

5)
IDK

7)
At R jumper will be at rest, coz no velocity
At Q accn is zero as Q is the maximum height gained by jumper.

8)
Short answer, because the time taken with each speed is different.

Let's do the math behind this - suppose the distance traveled with both speeds is denoted by "s".
Suppose the time taken at 600 kmph = t(1).

Then, since Speed = Distance/Time, we can write

600 = s/t(1)
So that
t(1) = s/600

Similarly, let's do the same for the other speed. The distance traveled is still "s", the speed = 400 kmph, and the time taken to traverse this distance = t(2). So:

400 = s/t(2)
So that
t(2) = s/400

The average speed is the total distance traveled dividing by the time taken (the average velocity is different - it is equal to total displacement divided by time taken).
= (Total Distance traveled)/(Time taken to cover that distance).

Since the plane travels a distance "s" in one direction and returns the same distance "s", the total distance traveled = s+s = 2s.

The time taken = t(1) + t(2) = s/400 + s/600 = 3s/1200 + 2s/1200 = 5s/1200 = s/240

Therefore, the average speed = 2s/(s/240) = 480 kmph = C.

11)
Mass of α = 6.6 x 10⁻²⁷
Speed of α = 1.5 x 10⁷
No. of α per sec = 50000 in area of 0.0001 m²
Use P = f / A
P = ma / A
mass of 50000 = 50000 * 6.6 x 10⁻²⁷
acceleration = V / t Where V = 1.5 x 10⁷ and t = 1
P = [ (50000 * 6.6 x 10⁻²⁷) x 1.5 x 10⁷ ] / 0.0001
P = 4.95 x 10⁻¹¹ ≈ 5 x 10⁻¹¹

39)
using a double thickness foil
Theory based answer, come on

1. After voltage passes through some resistance it will decrease
2. The higher the resistance the greater the decrease in voltage

-->So it will decrease less after passing through 2 ohm than when passing through 4 ohm
Hence it is D

 

[Thought blocker]

q13 please

Mass in Kg...
so
m1v1 = (m1+m2)v2
2.5 = 0.11v
v = 22.72 ≈ 23 m/s

 

[not.maria]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_12.pdf
q 11 28 and 39 please

 

[tania]

Mass in Kg...
so
m1v1 = (m1+m2)v2
2.5 = 0.11v
v = 22.72 ≈ 23 m/s

JAZALKAllAH KHAIR. Q15 bhi bta dien please

 

[Thought blocker]

JAZALKAllAH KHAIR. Q15 bhi bta dien please

Take moments from da hinge, let F b da reaction force of da spring.
15)
F × 2 = (40 × 9.8) × 5
2F = 1960 N
F = 980 N
------------------------------------
k = 10 kN/m = 10 000 N/m
k = F/x
x = F/k
x = 980/(10 000)
x = 0.098 m now convert it to cm i.e is 98 cm.---> C

 

[ZaqZainab]

How 2 amplitudes and then 1 ? ?
Is it given in book ._- ?

Suppose light goes from the source L, it has an amplitude A (suppose). Then, some waves pass through the upper slit and some waves pass through the lower slit (on the diagram. In real life, this is a top-down view, so the slits are right next to each other horizontally).

The waves that pass through the upper slit (on the diagram) have amplitude A (diffraction doesn't change amplitude), and the waves that pass through the lower slit (on the diagram) also have amplitude A.

Therefore, the waves that pass through either slit have an amplitude of A. At the center of the screen S, one set of waves from each slit (one set from upper slit, one set from lower slit) overlap to form a central maxima. Since each of those sets has wavelength A, the superimposing constructive interference means that the waves at the center have amplitude A + A = 2A (since the waves at the center are in phase and so interfere constructively, their amplitudes add up).

So, the Intensity here is I. Since I = kA², we can write for this situation

I = k * (2A)²
I = k * (4A²)
I = 4kA²

So that k = I/(4A²).

This constant k remains the same for any particular wave, and since the wave does not change when one slit is blocked, this can be applied in the final condition as well.

In the final condition, only waves coming out of one slit reach the center. This set of waves has an amplitude of A, and since they do not interfere with themselves (and since there are no waves coming from the other slit to interfere with them), the final amplitude in this final situation is equal to A.

We again write, using I = kA² for this situation,

I(final) = k * (A²)

Since k = I/(4A²), we put this value into the equation above to get

I(final) = I/(4A²) * (A²)
I(final) = I/4 = D.

Hope this helped!
Good Luck for all your exams!

I am not good at explaining

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_12.pdf
q 11 28 and 39 please

11)
The angle that T1 makes with R is greater than the angle that T2 makes with R.
R has to be balanced by the horizontal componts of T1 and T2
Let the angle be θ
R= T1cosθ + T1cosθ and R= T2cosθ + T2cosθ
We can see that for cos,, the greater the angle.. less the value of cosθ
so in order to cancel out R the value of T1 must be greater as cos θ is less in diagram 1. Therefore T1>T2

28)
F= Eq
F = 3 x 10^7 x 1.6 x 10 ^(-19)
F = A

the q is always the charge of the charge that is moving, not the charge of the electric field

39)
It can not be A cause there is suppose to be a deflection not B cause they weight the same not D cause alpha particles are postive and so is the gold necule they repel and not attract so it is JUST C

 

[Thought blocker]

I am not good at explaining

No you are just

 

[crazytaylorfanXD]

w12_12
5)
Simple concept, use your brain

19)
Use Pressure = force / area and find force ( dont forget to keep basic units of area )
Then w.d = f x d = A ( distance in meter)

27)

30)
Apply d*sin thetha = n*lamida
d = 1/n where n is no. of line/m. Metre not millimetre.
Grating's got 300lines/mm. So this means it has 300*1000 lines/m
(Coz 1m = 1000mm)
d = 1/n ie 1/300,000
U wanna see all the maximas possible, so to get max. no. of maximas on one half of the screen, u put thetha = 90
lamida is given. Convert it into m from nm.

U get n = 4
This means 4 maximas on one half of the screen.
On whole (both halves) of screen, maximas are = 4*2 = 8

BUT, u also have to consider the central maxima.
So total maximas are 8+1 = 9

36)
This is also a concept, we dont have fix values of currents, it varies, so we took a mean of power :¬

Find power at -1 and then at 2
For -1 :
P = 100W
For 2 :
P = 400W

Now as current is varying we have to take mean value of power : i.e 400 +100 = 500 / 2 = 250W

38)
remember when two resistor are in series, more voltage drop is at resistor with more value, so their a flashing light then R = 5 mega ohm when dark, then 5M is very large than 1 K of other resistor, then almost all voltage drop at 5M, hope you get this, it is a very simple concept.

Thanks A lot :****
i realized during ur explanation that these are just simple concepts that i couldnt grasp at the time i was solving the paper.
i was doing the paper at about 4 in the morning so my mind wasnt all there xD
Thanks again ur the best :3

 

[tania]

Take moments from da hinge, let F b da reaction force of da spring.
15)
F × 2 = (40 × 9.8) × 5
2F = 1960 N
F = 980 N
------------------------------------
k = 10 kN/m = 10 000 N/m
k = F/x
x = F/k
x = 980/(10 000)
x = 0.098 m now convert it to cm i.e is 98 cm.---> C

But they said extra compression. samajh nae aya

 

[Rockstar RK]

A quantity x is to be determined from the equation
x = P – Q.
P is measured as 1.27 ± 0.02 m.
Q is measured as 0.83 ± 0.01 m.
What is the percentage uncertainty in x to one significant figure?
A 0.4 % B 2 % C 3 % D 7 %

 

[ZaqZainab]

A quantity x is to be determined from the equation
x = P – Q.
P is measured as 1.27 ± 0.02 m.
Q is measured as 0.83 ± 0.01 m.
What is the percentage uncertainty in x to one significant figure?
A 0.4 % B 2 % C 3 % D 7 %

Subtract p from q and add their uncertainity!! 1.27-.83=.44
.01+.02=.03
x=.44±.03
now
0.03/.44 *100 = 7%??

 

[Thought blocker]

But they said extra compression. samajh nae aya

That is what we found