Physics: Post your doubts here!

[Asad Moosvi]

Do you know the electric firld is caontant between 2 parrale plates if you see a pic of the electric lines you will see that it looks like those lines near the plate in your question but as you go future they change so its suppose to be really close to the plate.
electric field decrease with distance from the positive test charge

So is it constant near the plate?

And it decreases as we move closer to the plate? Both points would be in the same location then.

 

[Asad Moosvi]

Do you know the electric firld is caontant between 2 parrale plates if you see a pic of the electric lines you will see that it looks like those lines near the plate in your question but as you go future they change so its suppose to be really close to the plate.
electric field decrease with distance from the positive test charge

I just want to clarify this. Both C and D would be near the plate, am I correct? Please reply.

 

[ZaqZainab]

So is it constant near the plate?

And it decreases as we move closer to the plate? Both points would be in the same location then.

they aren't points they are regions and not at the same location
region C will be really close while D is the region just after C all the part near the negative charged sphere

 

[Asad Moosvi]

they aren't points they are regions and not at the same location
region C will be really close while D is the region just after C all the part near the negative charged sphere

Got it, thank you.

 

[Muskan Achhpilia]

You have to find the components for the 10 N force that is going upwards and to the left.

Thanks

 

[Muskan Achhpilia]

Hi Muskan,

Do you understand Q5 part b(ii) from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_21.pdf
?

If you do, may you please explain it to me?

Actually, in my school we have not started that chapter, so I may not be the best person to explain...

 

[Muskan Achhpilia]

I have a few doubts can someone please help me out,thanks in advance,

The first one is a easy question for many although I am sort of confused a little...


I thought the answer of this should be C but it is B!




Now in this question they have mentioned take vertically upwards as positive(contrary to -9.81m/s^-2) so shouldn't the answer be D?In the marking scheme however, it is mentioned B

 

[Asad Moosvi]

I have a few doubts can someone please help me out,thanks in advance,

The first one is a easy question for many although I am sort of confused a little... View attachment 37939

View attachment 37940
I thought the answer of this should be C but it is B!

View attachment 37941

View attachment 37942
Now in this question they have mentioned take vertically upwards as positive(contrary to -9.81m/s^-2) so shouldn't the answer be D?In the marking scheme however, it is mentioned B

For #11, the tires exert a force backwards on the road, therefore the road must be exerting a force in the opposite direction, that is forward. There is also a normal force exerted by the road on the tires. So now you know that the road is exerting a force up (the normal force) and forward to the right. The resultant of these forces must be B.

 

[ZaqZainab]

I have a few doubts can someone please help me out,thanks in advance,

The first one is a easy question for many although I am sort of confused a little... View attachment 37939

View attachment 37940
I thought the answer of this should be C but it is B!

View attachment 37941

View attachment 37942
Now in this question they have mentioned take vertically upwards as positive(contrary to -9.81m/s^-2) so shouldn't the answer be D?In the marking scheme however, it is mentioned B

For the first one you use the equation v^2 = u^2 + 2as
we have V=20
U=10
S=100 have to find a
so a=(v^2-u^2)/(2s)
a=1.5
for the next one
for every action there is an opposite and equal reaction you sitting on a chair you are applying a force directly downwards while the chair is applying the force directly up wards but in this cause the car is moving so it isn't directly upwards and so a little to the side
the next one
you use the equation s=0.5at^2 to get time
we have s as 1.25 and a is =9.81
we will t=0.5048
then we use the equation speed =distance/time
distance is 10 as we see so the answer is 20 from 10/0.5
and the last one
the laceration of free fall which is g is always vertically downwards so as they said take up wards + then downwards - whatever the case the g on earth will always be downwards like if it was + you be flying and if 0 you won't see an apple fall on the ground from the tree you get it right
I guess it helps

 

[ZaqZainab]

Also these are my doubts please help,
View attachment 37917
Answer is C

View attachment 37918
Answer is D
View attachment 37919
Answer is A

Please explain as much as possible,

Thanks a tonne!

the velocity in straight line is speed
Speed=distance/time
and so distance=speed*time
speed and time increases the distance increases (first part of the graph) (here A,C and D are correct so C is out)
speed stay constant time increases so distance increases but not like that much not like it did in the first part but if you look at A and D the steepness is more which is wrong but C the steepness stays constant
last part the velocity decreases but yet the time is increasing so distance covered in time will be less and eventually as velocity become zero the distance will stop increasing so that doesn't mean the car will go back and stand at the same position as it started that's another reason why A is wrong leaving us with C being the answer
next one
the total mass= 4m
F=MA a is constant as smooth horizontal
a little force goes to X and a little to Y but total is F
how do we divide we divide according to there m
F/4 goes to X and 3F/4 to Y simple
LAST ONE
momentum before=momentum after
momentum= velocity*mass
(8*2)+(4*2)= (v)*(4+2)
you will get v=4

 

[Asad Moosvi]

How do the particles in a standing wave vibrate in phase within half a wavelength? Can someone explain this to me?

 

[Thought blocker]

the velocity in straight line is speed
Speed=distance/time
and so distance=speed*time
speed and time increases the distance increases (first part of the graph) (here A,C and D are correct so C is out)
speed stay constant time increases so distance increases but not like that much not like it did in the first part but if you look at A and D the steepness is more which is wrong but C the steepness stays constant
last part the velocity decreases but yet the time is increasing so distance covered in time will be less and eventually as velocity become zero the distance will stop increasing so that doesn't mean the car will go back and stand at the same position as it started that's another reason why A is wrong leaving us with C being the answer
next one
the total mass= 4m
F=MA a is constant as smooth horizontal
a little force goes to X and a little to Y but total is F
how do we divide we divide according to there m
F/4 goes to X and 3F/4 to Y simple
LAST ONE
momentum before=momentum after
momentum= velocity*mass
(8*2)+(4*2)= (v)*(4+2)
you will get v=4

I was about to answer this, ty

 

[Thought blocker]

How do the particles in a standing wave vibrate in phase within half a wavelength? Can someone explain this to me?

Stationary waves are different from progressive waves in a number of respects:
There is no energy transfer along a stationary wave.
Within a half wavelength (one loop of a standing wave) all the particles vibrate in phase and they are all exactly out of phase (180° phase difference) with all the particles in the adjacent loop.


The amplitude of vibration varies with position within the loop.
There are nodes where the displacement is always zero and antinodes that vibrate with the same maximum amplitude.
The wavelength of a stationary wave is twice the distance between two adjacent nodes or antinodes.
Neighbouring nodes or antinodes are separated by lamda/2.

 

[Muskan Achhpilia]

For #11, the tires exert a force backwards on the road, therefore the road must be exerting a force in the opposite direction, that is forward. There is also a normal force exerted by the road on the tires. So now you know that the road is exerting a force up (the normal force) and forward to the right. The resultant of these forces must be B.

Thank you!

 

[Muskan Achhpilia]

For the first one you use the equation v^2 = u^2 + 2as
we have V=20
U=10
S=100 have to find a
so a=(v^2-u^2)/(2s)
a=1.5
for the next one
for every action there is an opposite and equal reaction you sitting on a chair you are applying a force directly downwards while the chair is applying the force directly up wards but in this cause the car is moving so it isn't directly upwards and so a little to the side
the next one
you use the equation s=0.5at^2 to get time
we have s as 1.25 and a is =9.81
we will t=0.5048
then we use the equation speed =distance/time
distance is 10 as we see so the answer is 20 from 10/0.5
and the last one
the laceration of free fall which is g is always vertically downwards so as they said take up wards + then downwards - whatever the case the g on earth will always be downwards like if it was + you be flying and if 0 you won't see an apple fall on the ground from the tree you get it right
I guess it helps

Thanks a lot, but I was wondering that in the first question had mentioned 2 marker posts so shouldn't s=(100*2)m?

 

[Muskan Achhpilia]

the velocity in straight line is speed
Speed=distance/time
and so distance=speed*time
speed and time increases the distance increases (first part of the graph) (here A,C and D are correct so C is out)
speed stay constant time increases so distance increases but not like that much not like it did in the first part but if you look at A and D the steepness is more which is wrong but C the steepness stays constant
last part the velocity decreases but yet the time is increasing so distance covered in time will be less and eventually as velocity become zero the distance will stop increasing so that doesn't mean the car will go back and stand at the same position as it started that's another reason why A is wrong leaving us with C being the answer
next one
the total mass= 4m
F=MA a is constant as smooth horizontal
a little force goes to X and a little to Y but total is F
how do we divide we divide according to there m
F/4 goes to X and 3F/4 to Y simple
LAST ONE
momentum before=momentum after
momentum= velocity*mass
(8*2)+(4*2)= (v)*(4+2)
you will get v=4

Thanks a lot!!!

 

[meerul264]

Need help with some doubts

(9702/02/O/N/07)
Q 4(c) and 4(d)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf

 

[Thought blocker]

Need help with some doubts

(9702/02/O/N/07)
Q 4(c) and 4(d)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_2.pdf

c)
Stress=f/a
A=f/stress
Stress=9.5*10^8(You got in b), f=1.9*10^3
For max you need min area..
So, min area= [(1.9*10^3)/(9.5*10^8)]=2*10^-6.
Max area = Area given - min area = [(3.2*10^-6)-(2*10^-6)]=1.2*10^-6

d)
The thicker a rod is, the more stress there will be between one side and another when the rod is bent the same amount.

If you bend a rod, say, 15 degrees, you will tend to compress one side and stretch the other. The compressed side is being forced to be a bit shorter than it wants to be, and the stretched side a bit longer.

Now, double the thickness of the rod. The amount of compression on the compressed side is now double, and the amount of stretching on the other side is also doubled.

IF I AM WRONG, CORRECT ME ZaqZainab

 

[The Godfather]

Let me give you notes to prepare from.. If you need more, tell me..
I have attached questions too..
I got this awesome notes from this site It helped me alot, hope it help you too Tc...

Tnq again ♣♠ Magnanimous ♣♠ Ask him for physics notes as He gave me alot materials i gave u all that i had so contact him now

 

[♣♠ Magnanimous ♣♠]

Tnq again ♣♠ Magnanimous ♣♠ Ask him for physics notes as He gave me alot materials i gave u all that i had so contact him now

yes who need help here.