Physics: Post your doubts here!

[Thought blocker]

lol you couldn't

I'am successful. I'am eating that thought, too acerbic, Yuck

 

[Suchal Riaz]

lets not make this sacred thread a convo.

 

[Thought blocker]

lets not make this sacred thread a convo.

Was going to type the same. Guruji

 

[mutilated_grass]

Hey guys

This question is from may june 2008 paper 4 Question 6 part a) and b) i).

I do not understand why the coil must be PARALLEL to the direction of the magnetic field. The magnetic flux is the greatest when the coil is normal to the magnetic field. Is this question somehow linking the concept of the rate of change of flux with the torque? If that is so, then when the coil is parallel to the magnetic field the rate of change of flux is maximum. Just like the torque!

For part b i) I don't get why side BC must be taken instead of side AB. Is it because side BC is the side which is actually rotating....? (does this make sense?)

 

[Snow Angel]

please can someone explain the 13 and 14 question from the paper below
papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

 

[Snow Angel]

please can explain the 26 q from this paper

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

 

[Sam Ivashkov]

the answer is C becuz here we have 2 resistors that will half the value of the current...
Suppose that the value of one resistor is 2 ohms n the voltage supply is 12V
so for first circuit current will be:- 12/2= 6A
while for the second circuit it will be:- 12/4=3A (NOTE: resistors are in series so they will add up to 4)
thats the reason y the current value won't remain the same..so it is incorrect

Okay. So for option A to be correct, it should also have a reading of 6A as per your example. Can you please explain how it will show a reading of 6A if the two resistors are of 2 ohms each and the power supply of 12V? Thanks!

 

[Sam Ivashkov]

Maybe you didn´t read the question properly? They are asking for which row is NOT correct. Maybe that´s why you dont understand why the answer is C? The answer is C because it´s wrong! - Thats what they are asking for!

um, no. my question was to explain WHY its wrong? I know what they're asking for. I need a logical explanation behind it.

 

[Snackbox86]

Hey again, More questions involving MCQs questions, Can i have help with explaining q 22 (Answer C), 32(answer C),37 (Answer D) Year 2012s p12, Thanks

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf

 

[Suchal Riaz]

Hey again, More questions involving MCQs questions, Can i have help with explaining q 22 (Answer C), 32(answer C),37 (Answer D) Year 2012s p12, Thanks

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf

q22 density=mass/volume so volume=mass/denxity = 3.5 × 10–25 kg/9.2 × 10 3 kg m-3 = 3.8 x 10 -29
seperation = cube root(3.8 x 10 -29=3.36 x 10 -10

q32
I=Q/t so Q=It. in option C, Q=0.005*200=1 C

 

[Suchal Riaz]

q37
current is passing through both lamps so they will glow dimly. dimly because the current will be less as now there are two compared to one light bulbs.

Hey again, More questions involving MCQs questions, Can i have help with explaining q 22 (Answer C), 32(answer C),37 (Answer D) Year 2012s p12, Thanks

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf

 

[Suchal Riaz]

please can someone explain the 13 and 14 question from the paper below
papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

Q13
distance of upper string from centre opf Q = 100/2 = 50mm=0.05m
distance of upper string from centre of p = 150/2=75mm=0.075m
tension in upper string = torque/distance = 3 Nm / 0.05 m = 60 N
torque on P = force*distance = 60 * 0.075 = 4.5
Q14
the horizonal velocity will remain the same. the vertical velocity will be zero at that time. so the kinetic energy will be due to horizontal velocity alone which is vcos45 = v √2/2
kinetic energy at max point = 1/2 * m * (v √2/2)² = [1/2 m v² ]* 1/2 (the thing in square brackets is equal to kinetic energy at the begining) so k(max)=0.5K(initial)

 

[sitooon]

Q1 ) A gas cylinder contains 4.00 × 104cm3 of hydrogen at a pressure of 2.50 × 107Pa and a
temperature of 290 K.
The cylinder is to be used to fill balloons. Each balloon, when filled, contains
7.24 × 103cm3 of hydrogen at a pressure of 1.85 × 105Pa and a temperature of 290 K.
Calculate, assuming that the hydrogen is ideal gas ,
(i) total amount of hydrogen in the cylinder ( i Found it already )

(ii) the number of balloons that can be filled from the cylinder

Q2)
At one particular temperature, the resistance of the thermistor is 2040 ± 20 Ω.
Determine this temperature, in kelvin,( using graph ) assume 15 degree celcius
Then how to find its error ?

 

[Suchal Riaz]

d=1/N = 1*10^-3/300=3.3*10^-6

d sinΘ = n λ
the maximum order will be at 90 degrees sin of 90 = 1 so max order = d/λ = 3.3*10^-6/450*10^-9 = 7.33
max order = 7
one maxima at centre and 7 above it and 7 below it. giving 15 maxima.

 

[Thought blocker]

Ricky Martin - Livin' La Vida Loca
Hear this while studying

 

[Suchal Riaz]

Ricky Martin - Livin' La Vida Loca
Hear this while studying

you posted in the wrong thread. this is physics doubts thread

 

[Thought blocker]

you posted in the wrong thread. this is physics doubts thread

 

[Hamza Khan]

Hey guys, can someone please provide me with notes of the AS level topic regarding electric field,electric charge, etc ?

 

[Al. Ahmad]

May june 2009 paper 21
Q5..
can anyone help?

 

[Snow Angel]

Q13
distance of upper string from centre opf Q = 100/2 = 50mm=0.05m
distance of upper string from centre of p = 150/2=75mm=0.075m
tension in upper string = torque/distance = 3 Nm / 0.05 m = 60 N
torque on P = force*distance = 60 * 0.075 = 4.5
Q14
the horizonal velocity will remain the same. the vertical velocity will be zero at that time. so the kinetic energy will be due to horizontal velocity alone which is vcos45 = v √2/2
kinetic energy at max point = 1/2 * m * (v √2/2)² = [1/2 m v² ]* 1/2 (the thing in square brackets is equal to kinetic energy at the begining) so k(max)=0.5K(initial)

thanks