Physics: Post your doubts here!

[h4rriet]

Ty, I still don't understand 14, 19

You have to understand projectile motion to understand 14. The vertical component of the velocity at the highest point during a parabolic trajectory will be 0, and its horizontal component will stay the same throughout.
In 19, all you have to do is put in numbers into the formulas I've given you. You're given the pressure, which is F/A. They want you to find the density. Density=m/v. To get m, you'll need to divide F/g (g is a constant). To get V, you'll need to multiply A with the height, which is given in the question.

 

[h4rriet]

ahmed abdulla I is proportional to (A^2)(f^2). For wave Q, the frequency is halved, so the intensity is divided by four. Then its amplitude is doubled, so the intensity is quadrupled. 4 and 4 cancel out.

 

[Sohail A. Razzak]

...Can anybody help..?

 

[abruzzi]

Ah my bad.. you'd use

E=Stress/Strain
Strain*E=F/A
If strain increases then Area decreases.

Got it.. thanks

 

[zackle09]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf

can someone please do these questions 12 and 21
for 11..i dont understand why its 2sina shouldnt it just be sina?

 

[abrar]

View attachment 28612 ...Can anybody help..?

c 13.1 right? use v^2=u^2+2as, u=0 for the 1kg mass , a=9.81 and s= 0.5m find v.

 

[Syed Mohammad Ali]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf
can anybody tell da reason for Q 31 nd 36... puleajh!

 

[white rose]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf
can anyone explain ques. 7, 10, 16, 24, 27, & 33.....ur help is highly appreciated

 

[Warrior66]

Can some one please explain these questions :Q9, Q11, Q14
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf

Q9: i found this answer posted by someone else, so not taking any credits!
"initial momentum of the ball is 2mv before collision... if collision was perfectly elastic one, all the momentum would have been gained by the ball... and this would have made the change in the momentum equal to 4mv.. 2mv-(-2mv)... now it is given that collision is inelastic, it will lose some of its momentum to the wall.. so now change in momentum will be less than 4mv but cant be less than the initial momentum... so the answer is C.. i think you cannot verify the answer by the calculations..."
Q11:
The tension in the rope needs to balance out the weight of the student acting downwards, so since the total tension needs to be opposite and equal to the weight, Let tension in rope on left side be T1 and the tension on rope on right side T2 and :. Total tension = T1sino + T2sino
add them up... W= 2Tsino, rearrange to get T= W/ 2sino. Answer: B
Q14:
Weight of box on slope with inclination of 30 degrees, mgsino= 200*sin30 and this acts downwards along the slope and the frictional force also act downwards along the slope so...(200*sin30) + 150= 250
And since W=F*d, we have to find out the distance d along the slope, so hypotenuse...1.5/sin30
The forward force to maintain a steady speed is the frictional force plus the component of weight along slope
W= ((200*sin30) + 150)*(1.5/sin30 )
Answer: D
Hope this helps!

 

[bogus]

Increasing the variable resistor's resistance will make more PD fall on the variable resistor and less resistance on the fixed resistor! so Voltmeter reading will decrease..

But The ammeter reading will stay unchanged.. since the voltage in parallel circuits is same.. so according to the formula V=IR .. if R remains same I will remain same too.

so Answer should be C.

how is r remaining the same ... its increasing so effective resistance of parallel combination is decreasing i dont get it . I know that the voltage willl remain the same but resistance is changing

 

[bogus]

The reading on the voltmeter decreases because more voltage goes to the variable resistor. The reading on the ammeter isn't changed because the resistance of the resistor it is connected in series with is the same.

doesnt the overall resistance of the parallel circuit reduce causing more current to flow from the battery ... i'm lost

 

[Sohail A. Razzak]

c 13.1 right? use v^2=u^2+2as, u=0 for the 1kg mass , a=9.81 and s= 0.5m find v.

Thank u soo much.....could u please explain these questions 2...
M j 2012 paper 11 question 12,18,23
M j 2012 paper 12 question 6,13,23,24,26,28,29,30.
May ALLAH bless u.

 

[h4rriet]

doesnt the overall resistance of the parallel circuit reduce causing more current to flow from the battery ... i'm lost

The overall resistance of the circuit increases, so the current in the circuit decreases. But since the current through the first loop also decreases, it remains the same through the second loop.

 

[bogus]

The overall resistance of the circuit increases, so the current in the circuit decreases. But since the current through the first loop also decreases, it remains the same through the second loop.

put in values and you'll realise current increases ... its parallel combination remember!

 

[h4rriet]

put in values and you'll realise current increases ... its parallel combination remember!

I did put in values. Say the variable resistor is 6 ohms and the other two resistors are each 2. 6+2=8. 1/8+1/2=1.6, Increase the resistance of the variable resistor to 8, for instance. 8+2=10. 1/10+1/2=1.7. Current through first loop assuming the EMF=12V=12/8=1.5 and through the second=12/2=6. Then afterwards the current through the first loop = 12/10=1.2 and the second=12/2=6.

 

[h4rriet]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf
can anyone explain ques. 7, 10, 16, 24, 27, & 33.....ur help is highly appreciated

7. Diagram 2 shows that there's a zero error. Minus this from the reading on Diagram 1.
10. Final momentum must = initial momentum. Make sure you consider the directions for the velocities.
16. Since it's moving at constant velocity, the force upwards = the force downwards, which is the component of the weight parallel to the slope.
24. When is diffraction less?
27. W=Vq which is the same for all 4 point charges.
33. Put in numbers.

 

[TaffsAsLevel]

7. Diagram 2 shows that there's a zero error. Minus this from the reading on Diagram 1.
10. Final momentum must = initial momentum. Make sure you consider the directions for the velocities.
16. Since it's moving at constant velocity, the force upwards = the force downwards, which is the component of the weight parallel to the slope.
24. When is diffraction less?
27. W=Vq which is the same for all 4 point charges.
33. Put in numbers.

Hi do you mind giving me reason on november 2011 paper 12 q.35 why is it A? I don't get it isn't it supposed to increase linearly?

 

[white rose]

7. Diagram 2 shows that there's a zero error. Minus this from the reading on Diagram 1.
10. Final momentum must = initial momentum. Make sure you consider the directions for the velocities.
16. Since it's moving at constant velocity, the force upwards = the force downwards, which is the component of the weight parallel to the slope.
24. When is diffraction less?
27. W=Vq which is the same for all 4 point charges.
33. Put in numbers.

thnxx.....but if u brief me how to solve it then ill understand btr...ill b very grateful...

 

[HorsePower]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_12.pdf
cud any one help me out in Q 20, 30, 38, 39

 

[A*(a*)]

Amyone to solve these challenging, mind-blowing question?