Physics: Post your doubts here!

[hamzashariq]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_13.pdf
Q11.

 

[ShreeyaBeatz]

PLEASE HELP!!

A ball is thrown horizontally in still air from the top of a very tall building. The ball is affected by air
resistance.

What happens to the horizontal and to the vertical components of the ball’s velocity?

MARKSCHEME SAYS:
horizontal component of velocity decreases to zero AND
vertical component of velocity increases to a constant value

I SAY:

horizontal component of velocity remains constant AND
vertical component of velocity increases to a constant value

Am I wrong or markscheme wrong??

 

[Thought blocker]

Wait ._.

I quit

 

[mehria]

Since charge on proton is +ive e, the sum of both 'up quarks' and 'down quarks' should equal to 1.

hmm...

 

[Thought blocker]

hmm...

aya na shamajh!

 

[mehria]

aya na shamajh!

ye quarts hai kya?

 

[huh]

S13 QP 12: Q4

Can someone please explain the answer for this question.

 

[Thought blocker]

ye quarts hai kya?

A quark is one of two currently recognized groups of fundamental particles, which are subatomic, indivisible (at least as far as we know today) particles that represent the smallest known units of matter

 

[mehria]

A quark is one of two currently recognized groups of fundamental particles, which are subatomic, indivisible (at least as far as we know today) particles that represent the smallest known units of matter

hmm... thnx

 

[ShreeyaBeatz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_13.pdf
Q11.

Since it is asked for an elastic collision, remember: speed of approach= speed of separation
Again, look here :
This is a similar concept!!! A simple question! Now put the values of your velocities and see whether speed of approach= Speed of seperation.
If they both are equal it is elastic. I did this question yesterday. And remember to draw diagrams Forward direction is + and backward is -
Light and God bless you good luck !!
I hope this helps.

 

[Thought blocker]

PLEASE HELP!!

A ball is thrown horizontally in still air from the top of a very tall building. The ball is affected by air
resistance.

What happens to the horizontal and to the vertical components of the ball’s velocity?

MARKSCHEME SAYS:
horizontal component of velocity decreases to zero AND
vertical component of velocity increases to a constant value

I SAY:

horizontal component of velocity remains constant AND
vertical component of velocity increases to a constant value

Am I wrong or markscheme wrong??

The horizontal velocity will decrease to Zero and the veritcal velocity will increase until it reaches a constant value. This question involves the concept of terminal velocity. The vertical velocity will increase initially due to the object's weight by as it is countered by drag the velocity will reach a constant value when the object is in vertical equilibrium. Whereas, the horizonal velocity will decrease to zero because of the consistent air resistance.

 

[hamzashariq]

PLEASE HELP!!

A ball is thrown horizontally in still air from the top of a very tall building. The ball is affected by air
resistance.

What happens to the horizontal and to the vertical components of the ball’s velocity?

MARKSCHEME SAYS:
horizontal component of velocity decreases to zero AND
vertical component of velocity increases to a constant value

I SAY:

horizontal component of velocity remains constant AND
vertical component of velocity increases to a constant value

Am I wrong or markscheme wrong??

The horizontal component of the velocity is also affected by the air resistance so it eventually becomes zero.

 

[Thought blocker]

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

Q27,33

27)
look, for you to determine the polarity you just see,weight of body should act downwards
therefore electric field should act upwards
body should be attracted upwards(towards positive pole)means it is negatively charged
.now since body is kept at rest..electric force=weight
electric force = qE and weight = mg qE = mg if u make q divide by m (which is the charge to mass ratio) you are left with g divided by E
hence answer is B
33)
Use formula itself to answer this question
I = Q / t
So it means, the amount of charge flowing past a point in XY per second

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf Q7 ,Q8 , Q9 , Q11 and Q15 please

 

[sagar65265]

View attachment 44307
What does this mean? Please explain. sorry for the two posts. I couldnt remove it View attachment 44307 View attachment 44307

We have to look quite hard to get the answer - why does the current seem to prefer the train rather than the relay? Note that a relay should have a low resistance to ensure maximum current flowing through it (and this causes the electromagnetic effect that turns the switch on or off).

So if current prefers the train rather than the relay, then shouldn't the train have an even lower resistance than the relay?

This resistance is practically Zero, since the train is preferred over a relay coil.
Therefore, without the resistor, the current would just flow out of the positive terminal and back into the negative terminal, short-circuiting the circuit. So, to ensure this does not happen, a resistor is placed in series with the train connection (the tracks) which will dispense of most of the energy supplied by the battery whenever the train is on the track.

Hope this helped!
Good Luck for all your exams!

 

[redd]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf

Q 24

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf

Q 24

By knowing area under Force - extension graph is the energy stored in the material answer is B

 

[redd]

By knowing area under Force - extension graph is the energy stored in the material answer is B

But what about option A? Area X is the energy that heats up the band?

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf

Q7,10

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf

Q7,10

When the stone is thrown and in the air, the only force acting on it is gravity, acting vertically downwards. Therefore, the net force F on the stone is also downwards and has a magnitude of (mg) Newtons.

By Newton's Second Law, since the net force F is downwards, the acceleration of the stone is F/m downwards. This translates to a downward acceleration of magnitude (mg)/m = g ms^-2 = 9.81 ms^-2.

There is no force on the stone in the horizontal direction, so the horizontal acceleration is zero, and the horizontal velocity is constant. Gathering our facts, we see that:

i) the stone has a constant downwards acceleration (a constant vertical acceleration, since the net force of gravity acting on the stone is constant), and
ii) the horizontal velocity of the stone does not change because there is no horizontal force component acting on the stone.

The only option that satisfies both these requirements is A.

10)
jumana94 has already answered this (3 years back!) at this link:

https://www.xtremepapers.com/community/threads/physics-mcq-help.8257/#post-111624

Thanks, jumana94 !

Good Luck for all your exams!