Physics: Post your doubts here!

[The Sarcastic Retard]

Hey all. I had this doubt. I want to give my Physics CIE AS level exams in 2016 november session. Can I give A levels in 2017 May June session?

You can give your A level exams in 13 months period after your As level examinations. So yes.

 

[nehaoscar]

Ok so the answer is from right to left ... but how??
LHS - current is anti-clockwise so north pole
RHS thus will be south pole
So shouldn't it be left to right??

 

[Catherine_1]

View attachment 59473
Ok so the answer is from right to left ... but how??
LHS - current is anti-clockwise so north pole
RHS thus will be south pole
So shouldn't it be left to right??

See magnetic field lines would go out of north pole and enter south pole so bascially, on the left hand side the arrow would be like <--- so that is right to left and on the south pole it would be --<-- so yeah.

 

[Catherine_1]

Can anyone please help and tell me why there is deviation in the value of centripetal force and gravitational force cause wasn't it that the gravitational force only provides the centripetal force?

I remember someone had posted this, I can't remember tho? Pleaseee help.
Ms

 

[nehaoscar]

See magnetic field lines would go out of north pole and enter south pole so bascially, on the left hand side the arrow would be like <--- so that is right to left and on the south pole it would be --<-- so yeah.

Ohhh yess xP I was thinking of it like an actual magnet so like the field lines would go through the wire lol thankss

 

[Catherine_1]

Ohhh yess xP I was thinking of it like an actual magnet so like the field lines would go through the wire lol thankss

Haha.Welcome. Btw check in case you get my doubt too?

 

[nehaoscar]

View attachment 59474
View attachment 59475

Can anyone please help and tell me why there is deviation in the value of centripetal force and gravitational force cause wasn't it that the gravitational force only provides the centripetal force?

I remember someone had posted this, I can't remember tho? Pleaseee help.
Ms
View attachment 59476

I'm not sure if this is correct but maybe it's because the mass is on the equator i.e placed on the surface of the earth so centripetal force will be very small
cause cf = gf for a satellite above the earth's surface so in this case the mass is not above but on the surface if that seems logical?

 

[Catherine_1]

I'm not sure if this is correct but maybe it's because the mass is on the equator i.e placed on the surface of the earth so centripetal force will be very small
cause cf = gf for a satellite above the earth's surface so in this case the mass is not above but on the surface if that seems logical?

Yeah it kinda does since mv^2 is small and R is huge it is obvious it won't be equal. Thank youu.



In this the last part when the output is negative shouldn't diode B be conducting rather than G.:/

Ms


Thanks

 

[nehaoscar]

Yeah it kinda does since mv^2 is small and R is huge it is obvious it won't be equal. Thank youu.

View attachment 59477
View attachment 59478
In this the last part when the output is negative shouldn't diode B be conducting rather than G.:/

Ms
View attachment 59479

Thanks

Is this from the specimen paper??
And yes it is B

 

[Catherine_1]

Is this from the specimen paper??
And yes it is B

Yeah but the subsequent answer also mentions the same thing. :/ You're sure, no?



How does one ideally find out the mean pd from the graph tho?
I am not too sure.

 

[nehaoscar]

Yeah but the subsequent answer also mentions the same thing. :/ You're sure, no?

View attachment 59480
View attachment 59481
How does one ideally find out the mean pd from the graph tho?
I am not too sure.
View attachment 59482

yep... which website did you get the ms from cause the ms i have from the cie website says :

 

[nehaoscar]

Yeah but the subsequent answer also mentions the same thing. :/ You're sure, no?

View attachment 59480
View attachment 59481
How does one ideally find out the mean pd from the graph tho?
I am not too sure.
View attachment 59482

mean pd = 0.5(minpd + maxpd)
so basically just the average

 

[OmarKhan99]

Ah yeah, it's this question.

Once the string is cut, the log doesn't immediately fall down. (I believe you understood that.) In the log's motion upward after the string is cut, there is friction acting on the same side as the component of weight - along the slope. So the addition of both forces divided by the mass of the log would give you acceleration.
On the other hand, once the log reaches a zero velocity, that is the maximum height it can go up to. Then, it starts falling down. In this part of its motion, the component of weight along the slope is opposed by the friction. So, acceleration would be the difference in both forces over the mass of the log.
Hence, the two accelerations are different and you get two different v-t gradients.

Hope you got it.

Thanks once again. I know that acceleration of the log while traveling up an down the slope will be different. What I to meant to ask was, why are the accelerations different when traveling down the slope? In other words, why are the gradients of the red and yellow line different. Shouldn't it be a straight line?

 

[Catherine_1]

mean pd = 0.5(minpd + maxpd)
so basically just the average

Omg thank you so much.

yep... which website did you get the ms from cause the ms i have from the cie website says :
View attachment 59483

Wait wut. Lol I got from papersgceguide site .
Lol thanks, I will check it out once again now.

 

[nehaoscar]

Omg thank you so much.

Wait wut. Lol I got from papersgceguide site .
Lol thanks, I will check it out once again now.

maybe it was some old copy which cie later corrected and republished i guess

 

[Catherine_1]

maybe it was some old copy which cie later corrected and republished i guess

Lol I guess

 

[Catherine_1]

How would the graph be?

 

[Catherine_1]

Would the answer to this question be correct in this way? Tho neater and clearer.

 

[Wkhan860]

View attachment 59485
View attachment 59486

How would the graph be?

Normal 'curving' graph sloping upwards initially and then becoming constant.
As it says that it should be 1.5 times the value it was at t(1/2), then if the curve was at n=1 (first line of y-axis considered 1) at t(1/2) then at 2t(1/2) it should be at n=1.5 (middle of 1st nd 2nd line of y axis).

 

[Catherine_1]

Normal 'curving' graph sloping upwards initially and then becoming constant.
As it says that it should be 1.5 times the value it was at t(1/2), then if the curve was at n=1 (first line of y-axis considered 1) at t(1/2) then at 2t(1/2) it should be at n=1.5 (middle of 1st nd 2nd line of y axis).

Thank you so much. But actually I didn't understand why would it be 1.5 times, if you could elaborate please?