Physics: Post your doubts here!

[daredevil]

Q3ii)since there is no accel. therefore log is said to be in equilibrium i.e. not external force acting related with newtons first law
Q3 iii) weight component downward + frictional force equal to tension
Q3 iv) since friction is involved there will be loss of energy due to friction

sorry i did not get ii) how do u work with just adding 2 forces in different directions and not even considering the angle? is that even possible??

 

[Tkp]

can anyone help q 23 .. for force/extension ...?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf
Tkp

c u need to find the spring constant
so the 1st 2 spring constant is in series
so 2k
and the other 1 is paralled
so 1/k=1/2+1/3
k=6/5
so w=kx
w/(6k/5)=x
w*5/6k=5w/6k

 

[asd]

sorry i did not get ii) how do u work with just adding 2 forces in different directions and not even considering the angle? is that even possible??

He actually meant the component down the slope, he just used the wrong word "downward"

 

[asd]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf
6b (i) and (ii)
If for the first part we're considering that there will be a short circuit and therefore the ammeter reading is going to be 0, why doesnt the same reason go for the second part? (cause the question says the wire has a negligible resistance)
littlecloud11 , since you seem to be a smart guy -_-
Oh, and gary221

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_22.pdf
6b (i) and (ii)
If for the first part we're considering that there will be a short circuit and therefore the ammeter reading is going to be 0, why doesnt the same reason go for the second part? (cause the question says the wire has a negligible resistance)
littlecloud11 , since you seem to be a smart guy -_-
Oh, and gary221

regardless of the resistance of the wire... we r measuring the current which passes thru the wire ---> V = IR (ohm's law)
So, the current passing thru the wire will hv 2 pass thru the ammeter itself which has resistance of ohms.

The reason the reading is 0 in the 1st one is bcoz p.d = 0 (as both the points(A n C) r at the same potential ie A ---> Since V = 0V , so V = IR
I = V/ R
= 0/5
I = o A

Hope u gt it!!

 

[asd]

regardless of the resistance of the wire... we r measuring the current which passes thru the wire ---> V = IR (ohm's law)
So, the current passing thru the wire will hv 2 pass thru the ammeter itself which has resistance of ohms.

The reason the reading is 0 in the 1st one is bcoz p.d = 0 (as both the points(A n C) r at the same potential ie A ---> Since V = 0V , so V = IR
I = V/ R
= 0/5
I = o A

Hope u gt it!!

Umm, why isnt the ammeter in series, firstly?
and shouldnt they be at the same potential even if C is connected to B, cause 0 resistance should give 0 p.d?

 

[asd]

or are we considering there has to be a p.d between A and B because they're a distance apart, even though its stated the wire has negligible resistance?

 

[Nidz-Ahmed]

Can some one help me with this ? http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s06_qp_2.pdf
Question 2 c (iii)
Thanks

 

[asd]

Can some one help me with this ? http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf
Question 2 c (iii)
Thanks

Length of AC=2/3 AB
Moment of W about A= 2/3(AB) * W (clockwise)
Moment of P about A= (AB) * Tsin(beta) (anti-clockwise)
2/3(AB)W = (AB)Tsin(beta)
2/3W = Tsin(beta)

 

[gary221]

Umm, why isnt the ammeter in series, firstly?
and shouldnt they be at the same potential even if C is connected to B, cause 0 resistance should give 0 p.d?

its in sries coz its a potentiometer... see txtbk!

 

[asd]

its in sries coz its a potentiometer... see txtbk!

LOL okay.

 

[gary221]

asd... no they aren't at the same potential..coz pd (= the energy given up bt unit charge as it moves from pt to pt) n potential B will defn be differnt from potential at A, hence thr is a pd btw the 2 two pts.. if the potential at both the points is same, current will not flow...

 

[asd]

asd... no they aren't at the same potential..coz pd (= the energy given up bt unit charge as it moves from pt to pt) n potential B will defn be differnt from potential at A, hence thr is a pd btw the 2 two pts.. if the potential at both the points is same, current will not flow...

That's exactly what Im saying man.
why would the charge give up any energy if there's no resistance in the wire?
And my textbook doesnt say its series :/
Im confused.

 

[gary221]

That's exactly what Im saying man.
why would the charge give up any energy if there's no resistance in the wire?
And my textbook doesnt say its series :/
Im confused.

nah..it says 'negligible resistance' --->although the resistance is present it is quite small n u shudnt consider it for ur calculations
wht coursbk do u hv??

 

[asd]

That was the only confusion. I Kinda knew that.
Thanks.
By David Sang.

 

[Tabi Sheikh]

sorry i did not get ii) how do u work with just adding 2 forces in different directions and not even considering the angle? is that even possible??

well its weight's component down the slope + frictional force
mathematically
W*sin(12)+frictional force =Tension

 

[Tabi Sheikh]

He actually meant the component down the slope, he just used the wrong word "downward"

well done bro

 

[gary221]

cyclone94..

 

[gary221]

That was the only confusion. I Kinda knew that.
Thanks.
By David Sang.

did u gt it now???

 

[asd]

I guess yeah
well, not the series part.