Physics: Post your doubts here!

[danial 234]

can u also help me with JUN 11 P12 Qs ( 4 , 6 , 10 , 13( why isnt it A) , 14 (why not C))

4) using metre rule with mm divisions , using light gates, using balance which measures to nearest gram .. will increase the accuracy of reading... hence less uncertainty
5) horizontal distance = vt
vertical = 1/2gt^2 (using 2nd eq of motion)
horizontal/vertical= 2v/gt
10) acceleration and force will be in same direction
newtons second law states that" .... resultant force produces an acceleration in the direction of force..."

 

[chymera]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf



question 36

 

[chymera]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_1.pdf

Q 9

can anyone explain what is the difference between B and D and why the answer is not C .. inshort explain all !! Thanks in advance

when velocity is zero, particle is either at its highest piont or at its lowest point. at B the particle is at highest point and at D particle is at lowest point

 

[danial 234]

can u also help me with JUN 11 P12 Qs ( 4 , 6 , 10 , 13( why isnt it A) , 14 (why not C))

13) if we choose A
k.e wont be conserved
initial k.e = 1/2mv^2
final= 1/2m(0.5v^2) + 1/2m(0.5v^2)
= 1/4mv^2
14) the only force acting is gravity.. no air resistance because it is in vacuum .. so force will be in XV only

 

[crazytaylorfanXD]

4) using metre rule with mm divisions , using light gates, using balance which measures to nearest gram .. will increase the accuracy of reading... hence less uncertainty
5) horizontal distance = vt
vertical = 1/2gt^2 (using 2nd eq of motion)
horizontal/vertical= 2v/gt
10) acceleration and force will be in same direction
newtons second law states that" .... resultant force produces an acceleration in the direction of force..."

in question 6 , isnt the equation of 2nd motion s=ut+0.5at^2
how exactyl did u get the vertical??
also thanks anyways

 

[crazytaylorfanXD]

13) if we choose A
k.e wont be conserved
initial k.e = 1/2mv^2
final= 1/2m(0.5v^2) + 1/2m(0.5v^2)
= 1/4mv^2
14) the only force acting is gravity.. no air resistance because it is in vacuum .. so force will be in XV only

i didnt get question (14)
can u explain it again??

 

[danial 234]

in question 6 , isnt the equation of 2nd motion s=ut+0.5at^2
how exactyl did u get the vertical??
also thanks anyways

the eq is this s=ut+0.5at^2
but as the initial vertical velocity (u) was zero so the eq became 0.5at^2
no problem

 

[danial 234]

i didnt get question (14)
can u explain it again??

the question asked abt the direction of forces acting on stone.. stone is in vacuum so there is no air resistance .. so no force will be acting on it in horizontal direction
only force acting on the stone is gravity which is vertical..
so the ans will be vertical (XV)

 

[Snowysangel]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_12.pdf 9

 

[danial 234]

5)
We know S = d / t = 40 / 2.5 = 16 m/s
Uncertainty in speed = (uncertainty in distance + uncertainty in time) x speed
--------"--------------- = ( (0.1/40 ) + (0.05/2.5) ) x 16 = 0.36 ≈ 0.4 (1dp) = C

9)
For projectile motion, the vertical acceleration is the same throughout and is equal to g. There is no component of acceleration in the horizontal direction, so horizontal velocity is unchanged. But vertical speed changes due to which the resultant velocity changes.
15)
L is moved further to the right, so there is a bigger clockwise moment, meaning there need to be a bigger anticlockwise moment to counter it.
note the system is in equilibrium so there must be no net force and no net torque
immediately eliminate A and B because there is a horizontal H force that is unbalanced
now the logic comes for C, D.
for D, the reaction is shifted closer to the right (you can consider centre of mass shifted closer to right), so clockwise moment decreases, anticlockwise moment increases. Thus balances the system. so D is correct

24)
http://www.s-cool.co.uk/a-level/phy...finitions-of-stress-strain-and-youngs-modulus
We know that E is inversely proportional to the area, and directly proportional to the length.
Now, when we increase the length to 3L, the extension also increases 3 times → 3x.
But becoz we increase the diameter to 2D, we have increased the area by 4 times. A = π(d/2)^2 → A is now 4A.
And since extension is INVERSELY proportional to area, the extension now falls → x/4

So, the new wire's extension = 3x/4
Ans = B

30)
Lambda = 574nm just calculator's mistake

31)
What you don't get in this ?

33)
Q = It
Q = 0.01 * 1800 = 18C

34)

one question!!!!!
how do you type so much?
#tiring

 

[chymera]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_ms_11.pdf


question 36 pleaassseee

 

[Ahmed H. Al-Neel]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s03_qp_1.pdf
questions 10 and 40 please!

 

[danial 234]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf 9

Force = density * area * velocity^2
=1.2*12*(33^2)

 

[Thought blocker]

hey can you help me? asap thankssssssomuch

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf #7,9,20,23,29,32,33,36

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_13.pdf #15,16,19,29,31,34

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf Q11,12,18,19,21,31,32,35 PLEASEEEPLEASSEE

s13_13
15)
Use formula :
0.5 * F * extension
0.5 * F * ((40) - (40*5)) ≈ 1.3 J

16)
IDK

19)
Solved

29)
F = qV / D
F = [(1.6 x 10^-19) x (80)] / (5*10^-2) ≈ 2.6 x 10^-16

31)
I = V / R
I = ( 2 - 1 ) / ( 9+2+1 )
I = 0.083A

34)
The wires are connected in parallel - that's the first thing you need to notice, and the best way of noticing this is looking at the arrangement - none of the wires are connected directly to each other, but each one of them has the same starting point and the same ending point. This usually means they have the same potential difference applied across their ends, which means they are connected in parallel.

So each cable has a resistance as a function of length, and for 1.0 km, we need to calculate the overall resistance.
The resistance formula for multiple resistances connected in parallel to each other is

1/R(equivalent) = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅ + 1/R₆.........

So we have 1 resistance of 100Ω (the steel core) and 6 resistances of 10Ω (the copper wires). Let's put this into the equation:

1/R(equivalent) = 1/100 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10
= 1/100 + 6/10
= 1/100 + 60/100
= 61/100
= 0.61
So since 1/R(equivalent) = 0.61, R(equivalent) = 1/0.61 = 1.63Ω = 1.6Ω = B.

s13_11
11)
D,
as From the law of conservation OF momentum we will get the final speed when objects combine, that is 2 ms^-1,
Now,
final kinetic of the system is (.5)(1+2)(2)^2= 6 J
Initial K.E= .5(2)(3^2)= 9J
By subtracting we will get change that is 3 J

12)

2000 - R = 750 * 2
R = 500 N = 0.50 kN

18)
Let's take one second. A column of air 10 meters long and with an area of 2000 m² or a volume of 20000 m³ moves past the blades. That has a mass of
20000 m³ x 1.3 kg/m³ or 26e4 kg

KE of that air is ½mV² = ½(26e4)(10)² = 1.3e7 J
50% of that is 6.5e6 J
since this is every second, the power is 6.5e6 J/s or 6.5 MW

19)
sagar helped
Take the load on the right-hand side string. Since that load is in equilibrium, the tension force on it has to be equal to it's weight, which we are given is 100 Newtons.

Now take the scale on the right. The scale will tell us how hard the string is pulling on it, which is the definition of the tension in the string. So the tension in the left-hand string is given to us by the reading on the scale. Since this is 20 Newtons, the tension in the left-hand side string is also 20 Newtons.

Whenever the disc rotates a little, it loses energy because the tension in the string does a little work on the disc - take the right-hand string. The disc rotates in an anti-clockwise direction, such that the point of contact with the right-hand string moves upwards (Imagine it - the disc rotates, so the point on the right moves a little bit up). Since the force of tension acts downwards, the work done by the force is negative (because force and displacement are in the opposite direction, F.s becomes negative).

The work done by the force in 1 second is the (Magnitude of Force) * (distance traveled). The distance traveled by the point where the string contacts the disc is equal to 50 revolutions (every second, it goes around 50 times). This distance = 50 * 2πr.
We are told that circumference = 0.30 meters. Since circumference = 2πr, we can say that 2πr = 0.3 meters.
So distance = 50 * 0.3 = 15 meters.

Therefore, the work done by this 100 Newton force = - 100 * 15 = - 1500 Joules per Second. = -1500 Watts.

On the other side, the tension in the left-hand side string does positive work (because it acts downwards, and the point in contact with the string also moves downwards with the rotation). The distance traveled is the same, 15 meters, but the work done is positive because both the distance and force are in the same direction (downwards).

This work = + 20 Newtons * 15 meters = 300 Joules per Second = 300 Watts.

Therefore, the net power by external force = -1500 + 300 = -1200 Watts.
Since the disc keeps spinning at a constant rate and is not slowing down, the motor has to provide this much power per second to counter the effects of the external force. Therefore, the power of the motor is 1200 Watts = 1.2 kW = B.

21)
Pressure=height*density*gravitational pull
pressure is already given which is due to BOTH the liquids
if the height for oil is x then for water it is 2000-x
17.5*10^6=(pressure of oil)+(pressure of water)
17.5*10^6=(830*x*9.81)+(1000*(2000-x)*9.81)
17.5*10^6=8142.3x+19620000-9810x
-2120000/-1667.6=x

31)
What you dont get ?
Force will be same thorough out the distance... Why will it vary ?

32)
Find lost volts..
V = E - Ir
I = E / (R +r)
I = 0.8A
V = 12 - (0.8 * 2)
V = 10.4V
Now P = VI
P = 10.4 * 0.8 = 8.3W

35)
v1 = 600 / (600+3000/7) * 3 = 1.75V
v2 = 3000/7 / ( 3000/7 +600) * 3 = 1.25V
v = v1 - v2 = 0.5V

 

[ashcull14]

ok in this how do we do that counting the boxes method....or the area one plx explain

 

[crazytaylorfanXD]

the question asked abt the direction of forces acting on stone.. stone is in vacuum so there is no air resistance .. so no force will be acting on it in horizontal direction
only force acting on the stone is gravity which is vertical..
so the ans will be vertical (XV)

thanks

 

[Thought blocker]

View attachment 45243 ok in this how do we do that counting the boxes method....or the area one plx explain

B is the answer ?
If yes, then use the formula :
0.5 * F * extension
0.5 * 17 * (30 * 10 ^-2) = 2.6

 

[danial 234]

thanks

np

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_ms_11.pdf


question 36 pleaassseee

Here R = V/I
so, as R increase, V increases and I decreases...
So here Variable resistor is connected in series with other resistor, so its resistance would decrease hence V decreases
And Ammeter readings will be unchanged as it is in parallel...

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
questions 10 and 40 please!

10)
Force = mass times acceleration
total mass = 4m so,
F=4ma
a=F/4m
F(Y)=3m x F/4m = 3F/4

40)
Ok..This is Tricky But involves Math Skills..!!
Look Proton Has Charge of (+)1.! We need to end Up at One Using Values of Up quark and Down Quark.!

If U Take 2 UpQuarks = (+)(2/3) * 2 = (+)(4/3)
And One Down Quark = (-)(1/3) * 1 = (-)(1/3)
When You add them Together..
(+)(4/3) + (-)(1/3)= (+)1