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Physics: Post your doubts here!

Tkp

Tkp

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Salman Khalid said:
can anyone help me with these two questions
with working :)
thnk u in advance
Click to expand...
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)
 
Muhammad Bin Anis

Muhammad Bin Anis

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taimoor.08 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf Q5
Click to expand...
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
 
Nagaanusan

Nagaanusan

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Warrior66 said:
[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%
Click to expand...
thank u so much
 
minie23

minie23

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salvatore said:
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
Click to expand...
​
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

salvatore

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minie23 said:
​
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
Click to expand...
Thanks a lot for your help..

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