Physics: Post your doubts here!

[unseen95]

please help me with question 4 (d)(i) and (ii). http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_2.pdf

 

[Tkp]

can anyone help me with these two questions
with working
thnk u in advance

fr question no. 35 the ans is d as the current 5 is divided in 3 parts so the ammeter reading would be 1.66 approx to 1.7(hope its right)

 

[Ariel Robert]

What is the meaning of SHM? and what is Angular Frequency?

 

[taimoor.08]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf Q5

 

[anmolareeba]

Plz if anybody could help me with October 2011 paper 22 Q.5 c part 2 and oct 2009 paper 22 Q.3 part b

 

[Ariel Robert]

Does Omega only stand for Angular Speed?

 

[anmolareeba]

Plz ans my questions

 

[Muhammad Bin Anis]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf Q5

the ques says
3000 revolution per minute

converting it to seconds
3000/60=50 revolutions per second

this means that the disc passes through the coil 50 times a second.
each time it passes, a signal is generated and so a voltage peak is seen on the C.R.O. consider this is a signal sent on the wave .
So 50 signal waves in a second ... and reciprocate this to find the time each wave stays on the screen... (1/50 = 0.02 seconds that is 20 ms)
so if you select the time base as 10ms per cm, you will get half of the wave in that time period. The total length is 10 cm so you can get 5 complete separate waves.
The answer is B
hope it helps

 

[Ariel Robert]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf Q5

If anyone need help, please reply on me...

 

[anmolareeba]

Plz if anybody could help me wid some questions

 

[gary221]

Plz if anybody could help me wid some questions

wht ques??

 

[anmolareeba]

I hav some questions they r written on the top of this page

 

[applepie1996]

could someone do this

 

[anmolareeba]

plz if anybody could help me wid oct 2008 Q7 b part 2 and may 2008 paper2 Q6 part b

 

[attitudehed]

Hi can someone help me with these two questions from MJ/07/p1?
Q33, I want to know why the answer is NEGATIVE of 0.2
and Q36, why will the voltage be 2.0, in detail if possible please explain?
thankyou.
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_1.pdf

 

[Nagaanusan]

[efficiency=useful energy/ total energy used x100]
P.E=mgh= 50x10x0.35= 175 J since she does it 30 times every minute and does for 4 mins total...then 175x30x4= 21000J <----remember this is the useful energy.
the total energy is the useful energy plus the energy wasted as heat energy which is 120000J so total energy is (21000J + 120000J = 141 000J)
:. 21000/141000 x 100 = 14.89% ------> (2 significant figures) 15%

thank u so much

 

[salvatore]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_2.pdf
Could someone please explain the concepts in qn no. 7(b)(ii & iii).. I totally don't get it!
Thanks

 

[minie23]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_2.pdf
Could someone please explain the concepts in qn no. 7(b)(ii & iii).. I totally don't get it!
Thanks

​

7(b)(i)
R = qL/A , where q = resistivity
V = IR
Substituting R in V = IR
we get, V = IqL/A => cancel out the constants (that is, I, q & A)
that leaves us with V proportional to L
(ii)
1. p.d between contact M and point Q = 2.70 V
2. Since V is proportional to L,

we have V1/V2 = L1/L2
4.50/2.70 = 1.00/L2

L2 = 2.70/4.50 = 0.600m = 60.0 cm.

(iii) When thermistor is heated, the resistance of the thermistor decreases. V across thermistor is reduced. Hence Vqm is smaller. Thus, the length of the wire between Q & M is smaller.

Hope this helps Peace and Love

 

[salvatore]

​

7(b)(i)
R = qL/A , where q = resistivity
V = IR
Substituting R in V = IR
we get, V = IqL/A => cancel out the constants (that is, I, q & A)
that leaves us with V proportional to L
(ii)
1. p.d between contact M and point Q = 2.70 V
2. Since V is proportional to L,

we have V1/V2 = L1/L2
4.50/2.70 = 1.00/L2

L2 = 2.70/4.50 = 0.600m = 60.0 cm.

(iii) When thermistor is heated, the resistance of the thermistor decreases. V across thermistor is reduced. Hence Vqm is smaller. Thus, the length of the wire between Q & M is smaller.

Hope this helps Peace and Love

Thanks a lot for your help..

I don't understand part (ii).. why is the p.d 2.7?

 

[Malik777]