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Suchal Riaz I'm waiting for the diagram of http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_23.pdf
Q1 e i and ii
Q1 e i and ii
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So we can say that when a solid turns into a liquid .. The internal energy increases cuz there is an increase in the temp as well as the molecular separation. Right?well .. internal energy is the sum of the random KE and PE of all particles of a substance
KE depends on the temperature and PE depends on molecular sepration
So when temp increase ... KE increase so Internal energy increase and vice versa and same goes to the PE ... more te molecular sepration more PE more internal energy
P.E=mghcan anyone please explain Q4 c ii) 2. from
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_22.pdf
Yep ..So we can say that when a solid turns into a liquid .. The internal energy increases cuz there is an increase in the temp as well as the molecular separation. Right?
which year is this from?IS THIS OK?
i would say yesIS THIS OK?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_23.pdf
Q1 e (i),(ii) should we convert the speed into forces ? please someone explain ! .. thanks
Suchal Riaz
it's simple. you already know the weight of the load just simply multiply it with the extension(17.8-16.3)
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now use the cosine rule. angle is 90+45. scale drawing will not give accurate result because the velocity of wind gives 1.8cm only if i took 1 cm = 20 m/s
change in gravitational potential energy = 3.8*1.5*10^-2(because extension is in centimetre)but I am not getting the answer
But the question didn't ask for drawing such diagrams ! I did this way and got the answer ! but I'm not sure did this answer 1 e (i) or not ! please someone correct me If I'm wrong !![]()
now use the cosine rule. angle is 90+45. scale drawing will not give accurate result because the velocity of wind gives 1.8cm only if i took 1 cm = 20 m/s
What about the second part (2.)change in gravitational potential energy = 3.8*1.5
can u plz explainSnow Angel
question number 4 (b) (iv)
current = 0.24 A
(a) p.d = IR = 0.24*5.5=1.32V
(b) terminal p.d = emf - lost volts = 4.4 V - (0.24 A * 2.3 ohm) = 0.552 V
(c) terminal p.d = emf - lost volts = 2.1 V - (1.8 ohm * 0.24 A) = 1.67 V
acceleration = gsin15http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_23.pdf
Q 1 e both i and ii
and Question 2 a ii
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